Upload
gene
View
32
Download
2
Embed Size (px)
DESCRIPTION
1 November 2011. Objective : You will be able to: describe enthalpy and differentiate between endothermic and exothermic reactions and calculate the enthalpy of a reaction. Homework : p. 263 #21, 22, 23, 24, 25: tomorrow. Thermochemistry. - PowerPoint PPT Presentation
Citation preview
1 November 2011
Objective You will be able to describe enthalpy and differentiate
between endothermic and exothermic reactions and calculate the enthalpy of a reaction
Homework p 263 21 22 23 24 25 tomorrow
Thermochemistry
Thermodynamics study of interconversion of heat and other kinds of energy First Law of Thermodynamics energy
can be converted from one form to another but can not be created or destroyed
We can not accurately measure total energy of a system
Instead we measure changes in energy ΔE
Exothermic process is any process that gives off heat ndash transfers thermal energy from the system to the surroundings
Endothermic process is any process in which heat has to be supplied to the system from the surroundings
2H2 (g) + O2 (g) 2H2O (l) + energyH2O (g) H2O (l) + energy
energy + 2HgO (s) 2Hg (l) + O2 (g)energy + H2O (s) H2O (l)
Specifically wersquore interested in heat flow that occurs under constant pressure
Enthalpy of Chemical Reactions
Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressureEnthalpy change (∆H) is the heat absorbed during a physical or chemical processDH = H (products) ndash H (reactants)
Hproducts lt HreactantsDH lt 0
Hproducts gt HreactantsDH gt 0 64
Enthalpies of Reaction
Expressed in kJ or kJmol endothermic ∆H is always positive
endothermic changes absorb heat exothermic ∆H is always negative
exothermic changes liberate heat Calculating ∆H is the same for a wide variety
of processes (see table)
Thermochemical Equations
H2O (s) H2O (l) DHfus = 601 kJ
Is DH negative or positive
System absorbs heat
Endothermic
DH gt 0
601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm
Thermochemical Equations
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)DHcomb = -8904 kJ
Is DH negative or positive
System gives off heat
Exothermic
DH lt 0
8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm
H2O (s) H2O (l)DH = 601 kJmol ΔH = 601 kJ
bull The stoichiometric coefficients always refer to the number of moles of a substance
Thermochemical Equations
bull If you reverse a reaction the sign of DH changesH2O (l) H2O
(s)DH = -601 kJ
bull If you multiply both sides of the equation by a factor n then DH must change by the same factor n2H2O (s) 2H2O (l)
DH = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be
specified in thermochemical equations
Note on ldquoimplied signsrdquo
In a sentence the sign of ∆H is often implied
ldquo69 kJ of heat was liberated from the systemrdquo = minus69 kJ
ldquo69 kJ of heat was absorbed by the systemrdquo = +69 kJ
Phase transitions
solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ
gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ
Example 1
Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol
calculate the heat evolved when 879 g of SO2 is converted to SO3
Example 2
Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air
P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol
Example 3
How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air
Problem
Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain
Why are phase changes from solid to liquid and from liquid to gas always endothermic
Is the process of sublimation endothermic or exothermic Explain
2 November 2011
Objective You will be able to describe calorimetry and calculate
heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off
when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
Agenda
I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions
lab notebook set upp
Calorimetry
Calorimetry measurement of heat flow Calorimeter an apparatus that
measures heat flow Heat capacity C heat required to raise
the temperature of an object by 1 K (units are JK)
Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)
Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Thermochemistry
Thermodynamics study of interconversion of heat and other kinds of energy First Law of Thermodynamics energy
can be converted from one form to another but can not be created or destroyed
We can not accurately measure total energy of a system
Instead we measure changes in energy ΔE
Exothermic process is any process that gives off heat ndash transfers thermal energy from the system to the surroundings
Endothermic process is any process in which heat has to be supplied to the system from the surroundings
2H2 (g) + O2 (g) 2H2O (l) + energyH2O (g) H2O (l) + energy
energy + 2HgO (s) 2Hg (l) + O2 (g)energy + H2O (s) H2O (l)
Specifically wersquore interested in heat flow that occurs under constant pressure
Enthalpy of Chemical Reactions
Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressureEnthalpy change (∆H) is the heat absorbed during a physical or chemical processDH = H (products) ndash H (reactants)
Hproducts lt HreactantsDH lt 0
Hproducts gt HreactantsDH gt 0 64
Enthalpies of Reaction
Expressed in kJ or kJmol endothermic ∆H is always positive
endothermic changes absorb heat exothermic ∆H is always negative
exothermic changes liberate heat Calculating ∆H is the same for a wide variety
of processes (see table)
Thermochemical Equations
H2O (s) H2O (l) DHfus = 601 kJ
Is DH negative or positive
System absorbs heat
Endothermic
DH gt 0
601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm
Thermochemical Equations
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)DHcomb = -8904 kJ
Is DH negative or positive
System gives off heat
Exothermic
DH lt 0
8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm
H2O (s) H2O (l)DH = 601 kJmol ΔH = 601 kJ
bull The stoichiometric coefficients always refer to the number of moles of a substance
Thermochemical Equations
bull If you reverse a reaction the sign of DH changesH2O (l) H2O
(s)DH = -601 kJ
bull If you multiply both sides of the equation by a factor n then DH must change by the same factor n2H2O (s) 2H2O (l)
DH = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be
specified in thermochemical equations
Note on ldquoimplied signsrdquo
In a sentence the sign of ∆H is often implied
ldquo69 kJ of heat was liberated from the systemrdquo = minus69 kJ
ldquo69 kJ of heat was absorbed by the systemrdquo = +69 kJ
Phase transitions
solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ
gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ
Example 1
Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol
calculate the heat evolved when 879 g of SO2 is converted to SO3
Example 2
Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air
P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol
Example 3
How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air
Problem
Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain
Why are phase changes from solid to liquid and from liquid to gas always endothermic
Is the process of sublimation endothermic or exothermic Explain
2 November 2011
Objective You will be able to describe calorimetry and calculate
heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off
when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
Agenda
I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions
lab notebook set upp
Calorimetry
Calorimetry measurement of heat flow Calorimeter an apparatus that
measures heat flow Heat capacity C heat required to raise
the temperature of an object by 1 K (units are JK)
Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)
Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Thermodynamics study of interconversion of heat and other kinds of energy First Law of Thermodynamics energy
can be converted from one form to another but can not be created or destroyed
We can not accurately measure total energy of a system
Instead we measure changes in energy ΔE
Exothermic process is any process that gives off heat ndash transfers thermal energy from the system to the surroundings
Endothermic process is any process in which heat has to be supplied to the system from the surroundings
2H2 (g) + O2 (g) 2H2O (l) + energyH2O (g) H2O (l) + energy
energy + 2HgO (s) 2Hg (l) + O2 (g)energy + H2O (s) H2O (l)
Specifically wersquore interested in heat flow that occurs under constant pressure
Enthalpy of Chemical Reactions
Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressureEnthalpy change (∆H) is the heat absorbed during a physical or chemical processDH = H (products) ndash H (reactants)
Hproducts lt HreactantsDH lt 0
Hproducts gt HreactantsDH gt 0 64
Enthalpies of Reaction
Expressed in kJ or kJmol endothermic ∆H is always positive
endothermic changes absorb heat exothermic ∆H is always negative
exothermic changes liberate heat Calculating ∆H is the same for a wide variety
of processes (see table)
Thermochemical Equations
H2O (s) H2O (l) DHfus = 601 kJ
Is DH negative or positive
System absorbs heat
Endothermic
DH gt 0
601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm
Thermochemical Equations
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)DHcomb = -8904 kJ
Is DH negative or positive
System gives off heat
Exothermic
DH lt 0
8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm
H2O (s) H2O (l)DH = 601 kJmol ΔH = 601 kJ
bull The stoichiometric coefficients always refer to the number of moles of a substance
Thermochemical Equations
bull If you reverse a reaction the sign of DH changesH2O (l) H2O
(s)DH = -601 kJ
bull If you multiply both sides of the equation by a factor n then DH must change by the same factor n2H2O (s) 2H2O (l)
DH = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be
specified in thermochemical equations
Note on ldquoimplied signsrdquo
In a sentence the sign of ∆H is often implied
ldquo69 kJ of heat was liberated from the systemrdquo = minus69 kJ
ldquo69 kJ of heat was absorbed by the systemrdquo = +69 kJ
Phase transitions
solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ
gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ
Example 1
Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol
calculate the heat evolved when 879 g of SO2 is converted to SO3
Example 2
Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air
P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol
Example 3
How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air
Problem
Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain
Why are phase changes from solid to liquid and from liquid to gas always endothermic
Is the process of sublimation endothermic or exothermic Explain
2 November 2011
Objective You will be able to describe calorimetry and calculate
heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off
when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
Agenda
I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions
lab notebook set upp
Calorimetry
Calorimetry measurement of heat flow Calorimeter an apparatus that
measures heat flow Heat capacity C heat required to raise
the temperature of an object by 1 K (units are JK)
Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)
Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Exothermic process is any process that gives off heat ndash transfers thermal energy from the system to the surroundings
Endothermic process is any process in which heat has to be supplied to the system from the surroundings
2H2 (g) + O2 (g) 2H2O (l) + energyH2O (g) H2O (l) + energy
energy + 2HgO (s) 2Hg (l) + O2 (g)energy + H2O (s) H2O (l)
Specifically wersquore interested in heat flow that occurs under constant pressure
Enthalpy of Chemical Reactions
Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressureEnthalpy change (∆H) is the heat absorbed during a physical or chemical processDH = H (products) ndash H (reactants)
Hproducts lt HreactantsDH lt 0
Hproducts gt HreactantsDH gt 0 64
Enthalpies of Reaction
Expressed in kJ or kJmol endothermic ∆H is always positive
endothermic changes absorb heat exothermic ∆H is always negative
exothermic changes liberate heat Calculating ∆H is the same for a wide variety
of processes (see table)
Thermochemical Equations
H2O (s) H2O (l) DHfus = 601 kJ
Is DH negative or positive
System absorbs heat
Endothermic
DH gt 0
601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm
Thermochemical Equations
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)DHcomb = -8904 kJ
Is DH negative or positive
System gives off heat
Exothermic
DH lt 0
8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm
H2O (s) H2O (l)DH = 601 kJmol ΔH = 601 kJ
bull The stoichiometric coefficients always refer to the number of moles of a substance
Thermochemical Equations
bull If you reverse a reaction the sign of DH changesH2O (l) H2O
(s)DH = -601 kJ
bull If you multiply both sides of the equation by a factor n then DH must change by the same factor n2H2O (s) 2H2O (l)
DH = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be
specified in thermochemical equations
Note on ldquoimplied signsrdquo
In a sentence the sign of ∆H is often implied
ldquo69 kJ of heat was liberated from the systemrdquo = minus69 kJ
ldquo69 kJ of heat was absorbed by the systemrdquo = +69 kJ
Phase transitions
solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ
gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ
Example 1
Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol
calculate the heat evolved when 879 g of SO2 is converted to SO3
Example 2
Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air
P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol
Example 3
How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air
Problem
Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain
Why are phase changes from solid to liquid and from liquid to gas always endothermic
Is the process of sublimation endothermic or exothermic Explain
2 November 2011
Objective You will be able to describe calorimetry and calculate
heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off
when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
Agenda
I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions
lab notebook set upp
Calorimetry
Calorimetry measurement of heat flow Calorimeter an apparatus that
measures heat flow Heat capacity C heat required to raise
the temperature of an object by 1 K (units are JK)
Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)
Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Specifically wersquore interested in heat flow that occurs under constant pressure
Enthalpy of Chemical Reactions
Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressureEnthalpy change (∆H) is the heat absorbed during a physical or chemical processDH = H (products) ndash H (reactants)
Hproducts lt HreactantsDH lt 0
Hproducts gt HreactantsDH gt 0 64
Enthalpies of Reaction
Expressed in kJ or kJmol endothermic ∆H is always positive
endothermic changes absorb heat exothermic ∆H is always negative
exothermic changes liberate heat Calculating ∆H is the same for a wide variety
of processes (see table)
Thermochemical Equations
H2O (s) H2O (l) DHfus = 601 kJ
Is DH negative or positive
System absorbs heat
Endothermic
DH gt 0
601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm
Thermochemical Equations
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)DHcomb = -8904 kJ
Is DH negative or positive
System gives off heat
Exothermic
DH lt 0
8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm
H2O (s) H2O (l)DH = 601 kJmol ΔH = 601 kJ
bull The stoichiometric coefficients always refer to the number of moles of a substance
Thermochemical Equations
bull If you reverse a reaction the sign of DH changesH2O (l) H2O
(s)DH = -601 kJ
bull If you multiply both sides of the equation by a factor n then DH must change by the same factor n2H2O (s) 2H2O (l)
DH = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be
specified in thermochemical equations
Note on ldquoimplied signsrdquo
In a sentence the sign of ∆H is often implied
ldquo69 kJ of heat was liberated from the systemrdquo = minus69 kJ
ldquo69 kJ of heat was absorbed by the systemrdquo = +69 kJ
Phase transitions
solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ
gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ
Example 1
Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol
calculate the heat evolved when 879 g of SO2 is converted to SO3
Example 2
Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air
P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol
Example 3
How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air
Problem
Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain
Why are phase changes from solid to liquid and from liquid to gas always endothermic
Is the process of sublimation endothermic or exothermic Explain
2 November 2011
Objective You will be able to describe calorimetry and calculate
heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off
when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
Agenda
I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions
lab notebook set upp
Calorimetry
Calorimetry measurement of heat flow Calorimeter an apparatus that
measures heat flow Heat capacity C heat required to raise
the temperature of an object by 1 K (units are JK)
Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)
Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressureEnthalpy change (∆H) is the heat absorbed during a physical or chemical processDH = H (products) ndash H (reactants)
Hproducts lt HreactantsDH lt 0
Hproducts gt HreactantsDH gt 0 64
Enthalpies of Reaction
Expressed in kJ or kJmol endothermic ∆H is always positive
endothermic changes absorb heat exothermic ∆H is always negative
exothermic changes liberate heat Calculating ∆H is the same for a wide variety
of processes (see table)
Thermochemical Equations
H2O (s) H2O (l) DHfus = 601 kJ
Is DH negative or positive
System absorbs heat
Endothermic
DH gt 0
601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm
Thermochemical Equations
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)DHcomb = -8904 kJ
Is DH negative or positive
System gives off heat
Exothermic
DH lt 0
8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm
H2O (s) H2O (l)DH = 601 kJmol ΔH = 601 kJ
bull The stoichiometric coefficients always refer to the number of moles of a substance
Thermochemical Equations
bull If you reverse a reaction the sign of DH changesH2O (l) H2O
(s)DH = -601 kJ
bull If you multiply both sides of the equation by a factor n then DH must change by the same factor n2H2O (s) 2H2O (l)
DH = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be
specified in thermochemical equations
Note on ldquoimplied signsrdquo
In a sentence the sign of ∆H is often implied
ldquo69 kJ of heat was liberated from the systemrdquo = minus69 kJ
ldquo69 kJ of heat was absorbed by the systemrdquo = +69 kJ
Phase transitions
solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ
gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ
Example 1
Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol
calculate the heat evolved when 879 g of SO2 is converted to SO3
Example 2
Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air
P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol
Example 3
How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air
Problem
Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain
Why are phase changes from solid to liquid and from liquid to gas always endothermic
Is the process of sublimation endothermic or exothermic Explain
2 November 2011
Objective You will be able to describe calorimetry and calculate
heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off
when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
Agenda
I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions
lab notebook set upp
Calorimetry
Calorimetry measurement of heat flow Calorimeter an apparatus that
measures heat flow Heat capacity C heat required to raise
the temperature of an object by 1 K (units are JK)
Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)
Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Enthalpies of Reaction
Expressed in kJ or kJmol endothermic ∆H is always positive
endothermic changes absorb heat exothermic ∆H is always negative
exothermic changes liberate heat Calculating ∆H is the same for a wide variety
of processes (see table)
Thermochemical Equations
H2O (s) H2O (l) DHfus = 601 kJ
Is DH negative or positive
System absorbs heat
Endothermic
DH gt 0
601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm
Thermochemical Equations
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)DHcomb = -8904 kJ
Is DH negative or positive
System gives off heat
Exothermic
DH lt 0
8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm
H2O (s) H2O (l)DH = 601 kJmol ΔH = 601 kJ
bull The stoichiometric coefficients always refer to the number of moles of a substance
Thermochemical Equations
bull If you reverse a reaction the sign of DH changesH2O (l) H2O
(s)DH = -601 kJ
bull If you multiply both sides of the equation by a factor n then DH must change by the same factor n2H2O (s) 2H2O (l)
DH = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be
specified in thermochemical equations
Note on ldquoimplied signsrdquo
In a sentence the sign of ∆H is often implied
ldquo69 kJ of heat was liberated from the systemrdquo = minus69 kJ
ldquo69 kJ of heat was absorbed by the systemrdquo = +69 kJ
Phase transitions
solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ
gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ
Example 1
Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol
calculate the heat evolved when 879 g of SO2 is converted to SO3
Example 2
Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air
P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol
Example 3
How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air
Problem
Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain
Why are phase changes from solid to liquid and from liquid to gas always endothermic
Is the process of sublimation endothermic or exothermic Explain
2 November 2011
Objective You will be able to describe calorimetry and calculate
heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off
when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
Agenda
I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions
lab notebook set upp
Calorimetry
Calorimetry measurement of heat flow Calorimeter an apparatus that
measures heat flow Heat capacity C heat required to raise
the temperature of an object by 1 K (units are JK)
Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)
Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Thermochemical Equations
H2O (s) H2O (l) DHfus = 601 kJ
Is DH negative or positive
System absorbs heat
Endothermic
DH gt 0
601 kJ are absorbed for every 1 mole of ice that melts at 00C and 1 atm
Thermochemical Equations
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)DHcomb = -8904 kJ
Is DH negative or positive
System gives off heat
Exothermic
DH lt 0
8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm
H2O (s) H2O (l)DH = 601 kJmol ΔH = 601 kJ
bull The stoichiometric coefficients always refer to the number of moles of a substance
Thermochemical Equations
bull If you reverse a reaction the sign of DH changesH2O (l) H2O
(s)DH = -601 kJ
bull If you multiply both sides of the equation by a factor n then DH must change by the same factor n2H2O (s) 2H2O (l)
DH = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be
specified in thermochemical equations
Note on ldquoimplied signsrdquo
In a sentence the sign of ∆H is often implied
ldquo69 kJ of heat was liberated from the systemrdquo = minus69 kJ
ldquo69 kJ of heat was absorbed by the systemrdquo = +69 kJ
Phase transitions
solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ
gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ
Example 1
Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol
calculate the heat evolved when 879 g of SO2 is converted to SO3
Example 2
Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air
P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol
Example 3
How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air
Problem
Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain
Why are phase changes from solid to liquid and from liquid to gas always endothermic
Is the process of sublimation endothermic or exothermic Explain
2 November 2011
Objective You will be able to describe calorimetry and calculate
heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off
when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
Agenda
I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions
lab notebook set upp
Calorimetry
Calorimetry measurement of heat flow Calorimeter an apparatus that
measures heat flow Heat capacity C heat required to raise
the temperature of an object by 1 K (units are JK)
Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)
Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Thermochemical Equations
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l)DHcomb = -8904 kJ
Is DH negative or positive
System gives off heat
Exothermic
DH lt 0
8904 kJ are released for every 1 mole of methane that is combusted at 250C and 1 atm
H2O (s) H2O (l)DH = 601 kJmol ΔH = 601 kJ
bull The stoichiometric coefficients always refer to the number of moles of a substance
Thermochemical Equations
bull If you reverse a reaction the sign of DH changesH2O (l) H2O
(s)DH = -601 kJ
bull If you multiply both sides of the equation by a factor n then DH must change by the same factor n2H2O (s) 2H2O (l)
DH = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be
specified in thermochemical equations
Note on ldquoimplied signsrdquo
In a sentence the sign of ∆H is often implied
ldquo69 kJ of heat was liberated from the systemrdquo = minus69 kJ
ldquo69 kJ of heat was absorbed by the systemrdquo = +69 kJ
Phase transitions
solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ
gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ
Example 1
Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol
calculate the heat evolved when 879 g of SO2 is converted to SO3
Example 2
Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air
P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol
Example 3
How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air
Problem
Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain
Why are phase changes from solid to liquid and from liquid to gas always endothermic
Is the process of sublimation endothermic or exothermic Explain
2 November 2011
Objective You will be able to describe calorimetry and calculate
heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off
when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
Agenda
I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions
lab notebook set upp
Calorimetry
Calorimetry measurement of heat flow Calorimeter an apparatus that
measures heat flow Heat capacity C heat required to raise
the temperature of an object by 1 K (units are JK)
Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)
Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
H2O (s) H2O (l)DH = 601 kJmol ΔH = 601 kJ
bull The stoichiometric coefficients always refer to the number of moles of a substance
Thermochemical Equations
bull If you reverse a reaction the sign of DH changesH2O (l) H2O
(s)DH = -601 kJ
bull If you multiply both sides of the equation by a factor n then DH must change by the same factor n2H2O (s) 2H2O (l)
DH = 2 mol x 601 kJmol = 120 kJThe physical states of all reactants and products must be
specified in thermochemical equations
Note on ldquoimplied signsrdquo
In a sentence the sign of ∆H is often implied
ldquo69 kJ of heat was liberated from the systemrdquo = minus69 kJ
ldquo69 kJ of heat was absorbed by the systemrdquo = +69 kJ
Phase transitions
solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ
gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ
Example 1
Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol
calculate the heat evolved when 879 g of SO2 is converted to SO3
Example 2
Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air
P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol
Example 3
How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air
Problem
Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain
Why are phase changes from solid to liquid and from liquid to gas always endothermic
Is the process of sublimation endothermic or exothermic Explain
2 November 2011
Objective You will be able to describe calorimetry and calculate
heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off
when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
Agenda
I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions
lab notebook set upp
Calorimetry
Calorimetry measurement of heat flow Calorimeter an apparatus that
measures heat flow Heat capacity C heat required to raise
the temperature of an object by 1 K (units are JK)
Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)
Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Note on ldquoimplied signsrdquo
In a sentence the sign of ∆H is often implied
ldquo69 kJ of heat was liberated from the systemrdquo = minus69 kJ
ldquo69 kJ of heat was absorbed by the systemrdquo = +69 kJ
Phase transitions
solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ
gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ
Example 1
Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol
calculate the heat evolved when 879 g of SO2 is converted to SO3
Example 2
Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air
P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol
Example 3
How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air
Problem
Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain
Why are phase changes from solid to liquid and from liquid to gas always endothermic
Is the process of sublimation endothermic or exothermic Explain
2 November 2011
Objective You will be able to describe calorimetry and calculate
heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off
when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
Agenda
I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions
lab notebook set upp
Calorimetry
Calorimetry measurement of heat flow Calorimeter an apparatus that
measures heat flow Heat capacity C heat required to raise
the temperature of an object by 1 K (units are JK)
Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)
Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Phase transitions
solid to liquid liquid to gas absorb heat endothermic positive ∆H H2O(l) rarr H2O(g) ∆Hvap= +44 kJ
gas to liquid liquid to solid liberate heat exothermic negative ∆H H2O(g) rarr H2O(l) ∆Hvap= minus44 kJ
Example 1
Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol
calculate the heat evolved when 879 g of SO2 is converted to SO3
Example 2
Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air
P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol
Example 3
How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air
Problem
Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain
Why are phase changes from solid to liquid and from liquid to gas always endothermic
Is the process of sublimation endothermic or exothermic Explain
2 November 2011
Objective You will be able to describe calorimetry and calculate
heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off
when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
Agenda
I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions
lab notebook set upp
Calorimetry
Calorimetry measurement of heat flow Calorimeter an apparatus that
measures heat flow Heat capacity C heat required to raise
the temperature of an object by 1 K (units are JK)
Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)
Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Example 1
Given the thermochemical equation2SO2(g) + O2(g) 2SO3(g) DH = -1982kJmol
calculate the heat evolved when 879 g of SO2 is converted to SO3
Example 2
Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air
P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol
Example 3
How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air
Problem
Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain
Why are phase changes from solid to liquid and from liquid to gas always endothermic
Is the process of sublimation endothermic or exothermic Explain
2 November 2011
Objective You will be able to describe calorimetry and calculate
heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off
when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
Agenda
I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions
lab notebook set upp
Calorimetry
Calorimetry measurement of heat flow Calorimeter an apparatus that
measures heat flow Heat capacity C heat required to raise
the temperature of an object by 1 K (units are JK)
Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)
Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Example 2
Calculate the heat evolved with 266 g of white phosphorus (P4) burns in air
P4(s) + 5O2(g) P4O10(s) DH= -3013kJmol
Example 3
How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air
Problem
Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain
Why are phase changes from solid to liquid and from liquid to gas always endothermic
Is the process of sublimation endothermic or exothermic Explain
2 November 2011
Objective You will be able to describe calorimetry and calculate
heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off
when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
Agenda
I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions
lab notebook set upp
Calorimetry
Calorimetry measurement of heat flow Calorimeter an apparatus that
measures heat flow Heat capacity C heat required to raise
the temperature of an object by 1 K (units are JK)
Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)
Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Example 3
How many kJ of heat are absorbed by the surroundings when 250 g of methane (CH4) burns in air
Problem
Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain
Why are phase changes from solid to liquid and from liquid to gas always endothermic
Is the process of sublimation endothermic or exothermic Explain
2 November 2011
Objective You will be able to describe calorimetry and calculate
heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off
when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
Agenda
I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions
lab notebook set upp
Calorimetry
Calorimetry measurement of heat flow Calorimeter an apparatus that
measures heat flow Heat capacity C heat required to raise
the temperature of an object by 1 K (units are JK)
Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)
Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Problem
Determine the amount of heat (in kJ) given off when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain
Why are phase changes from solid to liquid and from liquid to gas always endothermic
Is the process of sublimation endothermic or exothermic Explain
2 November 2011
Objective You will be able to describe calorimetry and calculate
heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off
when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
Agenda
I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions
lab notebook set upp
Calorimetry
Calorimetry measurement of heat flow Calorimeter an apparatus that
measures heat flow Heat capacity C heat required to raise
the temperature of an object by 1 K (units are JK)
Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)
Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
What scientific law requires that the magnitude of the heat change for forward and reverse processes be the same with opposite signs Explain
Why are phase changes from solid to liquid and from liquid to gas always endothermic
Is the process of sublimation endothermic or exothermic Explain
2 November 2011
Objective You will be able to describe calorimetry and calculate
heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off
when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
Agenda
I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions
lab notebook set upp
Calorimetry
Calorimetry measurement of heat flow Calorimeter an apparatus that
measures heat flow Heat capacity C heat required to raise
the temperature of an object by 1 K (units are JK)
Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)
Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Why are phase changes from solid to liquid and from liquid to gas always endothermic
Is the process of sublimation endothermic or exothermic Explain
2 November 2011
Objective You will be able to describe calorimetry and calculate
heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off
when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
Agenda
I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions
lab notebook set upp
Calorimetry
Calorimetry measurement of heat flow Calorimeter an apparatus that
measures heat flow Heat capacity C heat required to raise
the temperature of an object by 1 K (units are JK)
Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)
Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
2 November 2011
Objective You will be able to describe calorimetry and calculate
heat change and specific heat Homework Quiz Week of Oct 31Determine the amount of heat (in kJ) given off
when 126x104 g of NO2 are produced according to the equation
2NO(g) + O2(g) rarr 2NO2(g) ∆H = minus1146 kJmol
Agenda
I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions
lab notebook set upp
Calorimetry
Calorimetry measurement of heat flow Calorimeter an apparatus that
measures heat flow Heat capacity C heat required to raise
the temperature of an object by 1 K (units are JK)
Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)
Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Agenda
I Homework QuizII Go over homeworkIII Calorimetry calculations examplesIV Practice ProblemsV Pre-lab questions and lab set upHomework Finish pre-lab questions
lab notebook set upp
Calorimetry
Calorimetry measurement of heat flow Calorimeter an apparatus that
measures heat flow Heat capacity C heat required to raise
the temperature of an object by 1 K (units are JK)
Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)
Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Calorimetry
Calorimetry measurement of heat flow Calorimeter an apparatus that
measures heat flow Heat capacity C heat required to raise
the temperature of an object by 1 K (units are JK)
Molar heat capacity Cmolar amount of heat absorbed by one mole of a substance when it experiences a one degree temperature change (units are Jmol K)
Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Specific heat capacity the heat capacity of one gram of a substance (units Jg K) water 1 calg K = 4184 J gK
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Calorimetry
q=CΔt q = heat absorbed in Joules C = specific heat capacity of a substance
the amount of heat required to raise the temperature of a given quantity of a substance by 1 degree (=massspecific heat)
q=msΔtEx 1 A 466 g sample of water is heated from
850oC to 7460oC Calculate the amount of heat absorbed (in kJ) by the water
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Ex 2
1435 grams of naphthalene (C10H8) was burned in a constant-volume bomb calorimeter The temperature of the water rose from 2028oC to 2595oC If the heat capacity of the bomb plus water was 1017 kJoC calculate the heat of combustion of naphthalene on a molar basis (find the molar heat of combustion)
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Ex 3
A lead pellet having a mass of 2647 g at 8998oC was placed in a constant pressure calorimeter of negligible heat capacity containing 1000 mL of water The water temperature rose from 2250oC to 2317oC What is the specific heat of the lead pellet
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Ex 4
What is the molar heat of combustion of liquid ethanol if the combustion of 903 grams of ethanol causes a calorimeter to increase in temperature by 319 K
heat capacity of the calorimeter is 758 kJK
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Pre-Lab
Molar heat of crystallization aka Latent heat of fusion
Set up your lab notebook and answer the pre-lab questions (refer to your textbook)
Make a procedure summary and set up a data table
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Homework
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
3 November 2011
Objective You will be able to calculate the molar heat of crystallization
for a chemical handwarmer Homework Quiz A 100 gram piece of
copper (specific heat = 0385 Jg oC) has been heated to 100oC It is then added to a sample of water at 220oC in a calorimeter If the final temperature of the water is 280oC calculate the mass of the water in the calorimeter
specific heat of water = 4184 J g oC
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Lab
1 With your partner carefully follow the directions
Hand warmer pouch and metal disk 342 grams
2 Collect all data results and answers to questions in your lab notebook
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
7 November 2011
Objective You will be able to calculate the standard enthalpy of
formation and reaction for compounds
Homework Quiz (Week of Nov 7)An iron bar of mass 869 g cools from
94oC to 5oC Calculate the heat released (in kJ) by the metal
specific heat of iron=0444 JgoC
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Agenda
I Homework QuizII Return TestsIII Standard Enthalpy of
FormationReaction notesproblems Homework p 264 46 49 51 53 56
57 59 62 64 Tuesday Test on Thermochem Thurs Lab notebook due tomorrow (checklist) Test Corrections Mon
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
8 November 2011
Objective You will be able to review thermochemistry
Homework Quiz (Week of Nov 7)Calculate the heat of combustion for
the following reaction from the standard enthalpies of formation in Appendix 3
2H2S(g) + 3O2(g) rarr 2H2O(l) + 2SO2(g)
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Agenda
I Homework quizII Collect lab notebooksIII Homework answers and more
enthalpy problemsIV Problem set work time Homework Problem Set ThursTest on Thermochem ThursTest Corrections Mon
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
9 November 2022
Objective You will be able to practice thermochemistry for a
test Homework QuizCalculate the standard enthalpy of
formation of CS2 (l) given thatC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Agenda
Homework Quiz Problem Set work timeHomework Problem set due
tomorrowThermochemistry test tomorrow
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Test Corrections
Required 2nd quarter quiz grade Show work including for multiple
choice questions On a separate sheet of paper or
using a different color pen Due Mon
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Standard Enthalpy of Formation and Reaction
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Formation Reaction
a reaction that produces one mole of a substance from its constituent elements in their most stable thermodynamic state
frac12H2(g) + frac12I2(s) rarr HI(g) ∆H = +2594 KJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Standard heat of formation (DH0) is the heat absorbed when one mole of a compound is formed from its elements at a pressure of 1 atm at 25oC
f
The standard heat of formation of any element in its most stable form is zeroDH0 (O2) = 0f
DH0 (O3) = 142 kJmolfDH0 (C graphite) = 0f
DH0 (C diamond) = 190 kJmolf
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Example
Write the thermochemical equation associated with the standard heat of formation of AlCl3(s) using the values in the table (Hint Be sure to make only one mole of product)
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Based on the definition of a formation reaction explain why the standard heat of formation of an element in its most stable thermodynamic state is zero Write a chemical equation to illustrate your answer
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Hessrsquos Law
Hessrsquos Law if a reaction is carried out in a series of steps DH of the overall reaction is equal to the sum of the DHrsquos for each individual step(Enthalpy is a state function It doesnrsquot matter how you get there only where you start and end)
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Example
Calculate DH for the following reaction 2S(s) + 3O2(g) rarr 2SO3(g)
from the enthalpies of these related reactions
S(s) + O2(g) rarr SO2(g) DH = minus2969 kJ
2SO2(g) + O2(g) rarr 2SO3(g) DH = minus1966 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
The standard heat of reaction (DH0 ) is the heat change of a reaction carried out at 1 atmrxn
aA + bB cC + dD
DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]
DH0rxn DH0 (products)f=S DH0 (reactants)fS-
Standard Heat of Reaction
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Example 1
Calculate the standard enthalpy change for the combustion of one mole of liquid ethanol
(Hint Be really careful with your math and the + and minus signs)
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Example 2
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation of benzene is 4904 kJmol2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
DH0rxn DH0 (products)f= S DH0 (reactants)fS-
DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]
DH0rxn=[ 12 times -3935 + 6 times -2858 ] ndash [ 2 times 4904 ] = -6535 kJ
-6535 kJ2 mol = - 3267 kJmol C6H6
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Practice Problem 1
Calculate the heat of combustion for the following reaction
C2H4(g) + 3O2(g) 2CO2(g) + 2H2O(l)The standard enthalpy of formation of
ethylene (C2H4) is 523 kJmolThen calculate per mole of ethylene
combusted
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Problem 2
The thermite reaction involves solid aluminum and solid iron (III) oxide reacting in a single displacement reaction to produce solid aluminum oxide and liquid iron
This reaction is highly exothermic and the liquid iron formed is used to weld metals Calculate the heat released in kJg of Al
∆Hof for Fe(l) is 1240 kJmol
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
What if the reaction is not a single step Practice Problem 3
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxnCS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn
1 Write the enthalpy of formation reaction for CS2
2 Add the given rxns so that the result is the desired rxn
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Calculate the standard enthalpy of formation of CS2 (l) given that
C(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJrxnS(rhombic) + O2 (g) SO2 (g) DH0 = -2961 kJrxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn1 Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)2 Add the given rxns so that the result is the desired rxn
rxnC(graphite) + O2 (g) CO2 (g) DH0 = -3935 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -2961x2 kJrxnCO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)DH0 = -3935 + (2x-2961) + 1072 = 863 kJrxn
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Practice Problem 4
Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements
2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935
kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l)
DH0= -25988 kJmol
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Homework
p 264 46 49 51 53 56 57 59 62 64 Tuesday
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
4 Nov 2010 Take out homework p 264 42 43 46 51 53 58 63 Objective SWBAT review thermochemistry
for a test on Monday Do now Calculate the standard enthalpy of
formation of acetylene (C2H2) from its elements2C(graphite) + H2(g) C2H2(g)
Here are the equations for each stepC(graphite) + O2(g) CO2(g) DH0=-3935 kJmolH2(g) + 12O2(g) H2O(l) DH0=-2858 kJmol2C2H2(g) + 5O2(g) 4CO2(g) + 2H2O(l) DH0= -25988 kJmol
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Agenda
I Do nowII HomeworkIII Chapter 6 Problem Set Work TimeHomework Finish problem set MonTest MonBonus Tues
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Chapter 6 Problem Set
Due Mon p 263 16 20 33 37 38 52 54b
57 64 70 73 80 Please show all your work
BONUS p 269 119 Tues (Up to +5 on your quarter grade)
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Read p 258-261 Heats of solution and dilution on your own
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
8 Nov 2010
Hand in Ch 6 Problem Set Objective SWBAT show what you know
about thermochemistry on a test Do Now Review the sign conventions
for q and w A reaction absorbs heat from its
surroundings and has work done on it by the surroundings Is it endothermic or exothermic What are the signs for q and w
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Agenda
I Collect ch 6 problem setII QuestionsIII Thermodynamics TestHomework Bonus problem due
tomorrow p 269 119Read p 801-814
p 8291-5
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
9 Nov 2010
Take out homework Bonus problem Objective SWBAT predict and
calculate entropy changes to systems
Do now Give one example of a spontaneous process Is it spontaneous in the opposite direction
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Agenda
I Do nowII Notes on spontaneous processes
and entropyIII Practice ProblemsHomework p 829 10-14 (SR)
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Thermodynamics Part 2
1st Law of Thermodynamics energy can be converted from one form to another but can not be created or destroyed How to measure these changes ∆H Change in enthalpy amount of
heat given off or absorbed by a system
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
2nd Law of Thermodynamics explains why chemical processes tend to favor one direction
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Spontaneous Processes
Chemists need to be able to predict whether or not a reaction will occur when reactants are brought together under a specific set of conditions (temperature pressure concentration etc)
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Spontaneous Physical and Chemical Processesbull A waterfall runs downhillbull A lump of sugar dissolves in a cup of coffeebull At 1 atm water freezes below 0 0C and ice melts above 0 0Cbull Heat flows from a hotter object to a colder objectbull A gas expands in an evacuated bulbbull Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
bullProcesses that occur spontaneously in one direction can not under the same conditions also take place spontaneously in the opposite direction
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
spontaneous
nonspontaneous
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Does a decrease in enthalpy mean a reaction proceeds spontaneously
H+ (aq) + OH- (aq) H2O (l) ∆H0 = -562 kJmol
H2O (s) H2O (l) ∆H0 = 601 kJmol
NH4NO3 (s) NH4+(aq) + NO3
- (aq) ∆H0 = 25 kJmolH2O
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) ∆H0 = -8904 kJmol
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Does a decrease in enthalpy mean a reaction proceeds spontaneously
No Being exothermic (ldquoexothermicityrdquo) favors
spontaneity but does not guarantee it An endothermic reaction can be spontaneous Not all exothermic reactions are
spontaneous So how do we predict if a reaction is
spontaneous under given conditions if ∆H doesnt help us
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Entropy (S) is a measure of the randomness or disorder of a system
order SdisorderS
If the change from initial to final results in an increase in randomness∆S gt 0
For any substance the solid state is more ordered than the liquid state and the liquid state is more ordered than gas stateSsolid lt Sliquid ltlt Sgas
H2O (s) H2O (l)∆S gt 0
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
74
Processes that lead to an increase in entropy (∆S gt 0)
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
∆S gt 0
Example Br2(l) Br2(g)
∆S gt 0
Example I2(s) I2(g)
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Entropy
State functions are properties that are determined by the state of the system regardless of how that condition was achieved
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths
energy enthalpy pressure volume temperature entropy
Review
Examples
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
77
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
How does the entropy of a system change for each of the following processes
(a) Condensing water vapor
(b) Forming sucrose crystals from a supersaturated solution
(c) Heating hydrogen gas from 600C to 800C
(d) Subliming dry ice
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
How does the entropy of a system change for each of the following processes
(a) Condensing water vaporRandomness decreases
Entropy decreases (∆S lt 0)(b) Forming sucrose crystals from a supersaturated
solutionRandomness decreases
Entropy decreases (∆S lt 0)(c) Heating hydrogen gas from 600C to 800C
Randomness increases
Entropy increases (∆S gt 0)(d) Subliming dry ice
Randomness increases
Entropy increases (∆S gt 0)
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Practice
Predict whether the entropy change is greater or less than zero for each of the following processesa freezing ethanolb evaporating a beaker of liquid
bromine at room temperaturec dissolving glucose in waterd cooling nitrogen gas from 80oC to
20oC
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
First Law of ThermodynamicsEnergy can be converted from one form to another but energy cannot be created or destroyed
Second Law of ThermodynamicsThe entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Entropy Changes in the System (∆Ssys)
aA + bB cC + dD
∆S0rxn dS0(D)cS0(C)= [ + ] - bS0(B)aS0(A)[ + ]
∆S0rxn nS0(products)= Σ mS0(reactants)Σ-
The standard entropy of reaction (∆S0 ) is the entropy change for a reaction carried out at 1 atm and 250C
rxn
What is the standard entropy change for the following reaction at 250C 2CO (g) + O2 (g) 2CO2 (g)S0(CO) = 1979 JKbullmol
S0(O2) = 2050 JKbullmolS0(CO2) = 2136 JKbullmol
DS0rxn= 2 x S0(CO2) ndash [2 x S0(CO) + S0 (O2)]
DS0rxn= 4272 ndash [3958 + 2050] = -1736 JKbullmol
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Practice
From the standard entropy values in Appendix 3 calculate the standard entropy changes for the following reactions at 25oCa CaCO3(s) CaO(s) + CO2(g)b N2(g) + 3H2(g) 2NH3(g)c H2(g) + Cl2(g) 2HCl(g)
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Entropy Changes in the System (∆Ssys)When gases are produced (or consumed)
bull If a reaction produces more gas molecules than it consumes ∆S0 gt 0
bull If the total number of gas molecules diminishes ∆S0 lt 0
bull If there is no net change in the total number of gas molecules then ∆S0 may be positive or negative BUT ∆S0 will be a small number
What is the sign of the entropy change for the following reaction 2Zn (s) + O2 (g) 2ZnO (s)The total number of gas molecules goes down ∆S is
negative
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Practice
Predict (not calculate) whether the entropy change of the system in each of the following reactions is positive or negativea 2H2(g) + O2(g) 2H2O(l)b NH4Cl(s) NH3(g) + HCl(g)c H2(g) + Br2(g) 2HBr(g)d I2(s) 2I(g)e 2Zn(s) + O2(g) 2ZnO(s)
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Homework
p 829 10-14 (SR)
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
10 Nov 2010
Objective SWBAT calculate Gibbs free energy and predict the spontaneity of a reaction
Do now Give one example of a change that increases entropy
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Agenda
I Do nowII Homework SRIII Gibbs Free Energy and Predicting
SpontaneityHW p 829 15 16 17 18 19 20 (TTL)
(Tues)Read lab and answer pre-lab questions in
lab notebook (Mon)AP Test Thermo questions (Tues)
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Entropy Changes in the Surroundings (∆Ssurr)
Exothermic Process∆Ssurr gt 0
Endothermic Process∆Ssurr lt 0
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
91
Third Law of ThermodynamicsThe entropy of a perfect crystalline substance is zero at the absolute zero of temperature
S = k ln W
W = 1
S = 0
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
∆Suniv = ∆Ssys + ∆Ssurr gt 0Spontaneous process
∆Suniv = ∆Ssys + ∆Ssurr = 0Equilibrium process
Gibbs Free Energy
For a constant-temperature process
∆G = ∆Hsys -T ∆SsysGibbs free energy (G)
∆G lt 0 The reaction is spontaneous in the forward direction∆G gt 0 The reaction is nonspontaneous as written The reaction is spontaneous in the reverse direction
∆G = 0 The reaction is at equilibrium
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
aA + bB cC + dD
∆G0rxn d ∆G0 (D)fc ∆G0 (C)f= [ + ] - b ∆G0 (B)fa ∆G0 (A)f[ + ]
∆G0rxn n ∆G0 (products)f= S m ∆G0 (reactants)fS-
The standard free-energy of reaction (∆G0 ) is the free-energy change for a reaction when it occurs under standard-state conditions
rxn
Standard free energy of formation (∆G0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states
f
∆G0 of any element in its stable form is zerof
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)DG0
rxn nDG0 (products)f= S mDG0 (reactants)fS-
What is the standard free-energy change for the following reaction at 25 0C
DG0rxn 6DG0 (H2O)f12DG0 (CO2)f= [ + ] - 2DG0 (C6H6)f[ ]
DG0rxn=[ 12xndash3944 + 6xndash2372 ] ndash [ 2x1245 ] = -6405 kJmol
Is the reaction spontaneous at 25 0C
DG0 = -6405 kJmollt 0
spontaneous
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Practice
Calculate the standard free-energy changes for the following reactions at 25oCa CH4(g) + 2O2(g) CO2(g) +
2H2O(l)b 2MgO(s) 2Mg(s) + O2(g)c H2(g) + Br2(l) 2HBr(g)
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
∆G = ∆H - T ∆S
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Recap Signs of Thermodynamic Values
Negative PositiveEnthalpy (ΔH) Exothermic Endothermic
Entropy (ΔS) Less disorder More disorder
Gibbs Free Energy (ΔG)
Spontaneous Not spontaneous
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Homework
p 829 15 16 17 18 19 20 (TTL) (Tues)
Read lab and answer pre-lab questions in lab notebook (Mon)
AP Test Thermo questions (Tues)
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
16 Nov 2010
Objective SWBAT calculate the temperature at which a reaction is spontaneous and review entropy and free energy problems
Do now If enthalpy is positive and entropy is positive at what temperatures will a reaction proceed spontaneously
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Agenda
I Do nowII Homework solutions (TTL)III AP Test question solutions -
collectedIV Thermochemistry review problemsHomework Lab notebook ThursCome after school to mass your filter
paper + M2CO3
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
CaCO3(s) CaO(s) + CO2(g)
CaO (quicklime) is used in steelmaking producing calcium metal paper making water treatment and pollution control
It is made by decomposing limestone (CaCO3) in a kiln at high temperature
But the reaction is reversible CaO(s) readily combines with CO2 to form CaCO3 So in the kiln CO2 is constantly removed to
shift equilibrium to favor the formation of CaO
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
At what temperature does this reaction favor the formation of CaO
Calculate ∆Ho and ∆So for the reaction at 25oC
Then plug in to ∆Go=∆Ho ndash T∆So to find ∆Go
What sign is ∆Go What magnitude Set ∆Go=0 to find the temperature at which
equilibrium occurs At what temperature can we expect this
reaction to proceed spontaneously Choose a temperature plug in and check
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Keep in mind that this does not mean that CaO is only formed above 835oC
Some CaO and CO2 will be formed but the pressure of CO2 will be less than 1 atm (its standard state)
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Practice Problems
1 Find the temperatures at which reactions will the following ∆H and ∆S will be spontaneous
a ∆H = -126 kJmol∆S = 84 JKmiddotmol
b ∆H = -117 kJmol∆S = -105 JKmiddotmol
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Solutions to AP test problems
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
17 Nov 2010
Objective SWBAT review enthalpy and entropy for a quiz
Do nowThe reaction between nitrogen and hydrogen to
form ammonia is represented belowN2(g) + 3 H2(g) rarr 2 NH3(g) ∆H˚ = ndash922 kJa Predict the sign of the standard entropy change
∆S˚ for the reaction Justify your answerb The value of ∆G˚ for the reaction represented in
part (a) is negative at low temperatures but positive at high temperatures Explain
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Do now solutions
(a) (ndash) the mixture of gases (high entropy) is converted into a pure gas (low entropy) and the 4 molecules of gas is reduced to 2
(b) ∆G˚ = ∆H˚ ndash T∆S˚ enthalpy favors spontaneity (∆H lt 0) negative entropy change does not favor spontaneity Entropy factor becomes more significant as temperature increases At high temperatures the T∆S factor becomes larger in magnitude than ∆H and the reaction is no longer spontaneous (∆G gt 0)
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Thermochemistry review
p 831 42 52 54 56 57 60 78 Quiz Thursday
Entropy (∆S) Free energy (∆G)
Study homework assignments from chapter 18 AP test questions these review problems
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
stop
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
During the course of a chemical reaction not all the reactants and products will be at their standard states (10 atm and 25oC)
How do you calculate free energy (and thus the spontaneity of the reaction)
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Gibbs Free Energy and Chemical Equilibrium∆G = ∆ G0 + RT lnQ
R is the gas constant (8314 JKbullmol)T is the absolute temperature (K)Q is the reaction quotient At Equilibrium
∆G = 0 Q = K0 = ∆G0 + RT lnK∆G0 = - RT lnK
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
What is Q
Reaction quotient
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
What is K
equilibrium constant
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
∆G0 = - RT lnK
186
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
The specific heat (s) [most books use lower case c] of a substance is the amount of heat (q) required to raise the temperature of one gram of the substance by one degree Celsius
The heat capacity (C) of a substance is the amount of heat (q) required to raise the temperature of a given quantity (m) of the substance by one degree Celsius
C = ms
Heat (q) absorbed or released
q = msDt
q = CDt
Dt = tfinal - tinitial
65
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
The boiling point is the temperature at which the (equilibrium) vapor pressure of a liquid is equal to the external pressure
The normal boiling point is the temperature at which a liquid boils when the external pressure is 1 atm
118
Phase Changes
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
The critical temperature (Tc) is the temperature above which the gas cannot be made to liquefy no matter how great the applied pressure
The critical pressure (Pc) is the minimum pressure that must be applied to bring about liquefaction at the critical temperature
118
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Wherersquos WaldoCan you findhellipThe Triple PointCritical pressureCritical temperatureWhere fusion occursWhere vaporization occursMelting point (at 1 atm)Boiling point(at 6 atm)
Carbon Dioxide
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Mel
ting
118Fr
eezin
g
H2O (s) H2O (l)
The melting point of a solid or the freezing point of a liquid is the temperature at which the solid and liquid phases coexist in equilibrium
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Subl
imat
ion
118
Depo
sitio
n
H2O (s) H2O (g)
Molar heat of sublimation (DHsub) is the energy required to sublime 1 mole of a solid
DHsub = DHfus + DHvap
( Hessrsquos Law)
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Molar heat of fusion (DHfus) is the energy required to melt 1 mole of a solid substance
118
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
118
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg C
Step 1 Heat the ice Q=mcΔTQ = 36 g x 206 Jg deg C x 8 deg C = 59328 J = 059 kJ
Step 2 Convert the solid to liquid ΔH fusion
Q = 20 mol x 601 kJmol = 12 kJ
Step 3 Heat the liquid Q=mcΔTQ = 36g x 4184 Jg deg C x 100 deg C = 15063 J = 15 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ
Sample Problem How much heat is required to change 36 g of
H2O from -8 deg C to 120 deg CStep 4 Convert the liquid to gas ΔH vaporization
Q = 20 mol x 4401 kJmol = 88 kJ
Step 5 Heat the gas Q=mcΔTQ = 36 g x 202 Jg deg C x 20 deg C = 14544 J = 15 kJ
Now add all the steps together
059 kJ + 12 kJ + 15 kJ + 88 kJ + 15 kJ = 118 kJ