Upload
chemdummy
View
1.767
Download
4
Embed Size (px)
Citation preview
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-1
1 LITER OF GASOLINE CONTAINS 8000 CALORIES.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-2
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-3
A PERSON USES AN AVERAGE OF
2000 CALORIES IN A DAY.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-4
EXCESS CALORIESIS STORED BY THE BODY
AS FAT.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-5
SO WHAT?
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-6
WHAT DOES CALORIES MEAN ANYWAY?
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-7
NO. CALORIES IS NOT FAT.IT IS THE ENERGY
FROM THE FOOD WE EAT THAT IS MEASURED IN
CALORIES.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-8
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-9
SO HOW DO THEY KNOW HOW MUCH CALORIES THERE ARE
IN STUFF?
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-10
THEY KNOW THROUGHTHERMOCHEMISTRY
ANDCALORIMETRY
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-11
Thermochemistry: Energy Flow and Chemical Change
6.1 Forms of Energy and Their Interconversion
6.2 Enthalpy: Heats of Reaction and Chemical Change
6.3 Calorimetry: Laboratory Measurement of Heats of Reaction
6.4 Stoichiometry of Thermochemical Equations
6.5 Hess’s Law of Heat Summation
6.6 Standard Heats of Reaction (H0rxn)
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-12
Thermochemistry is a branch of thermodynamics that deals withthe heat involved with chemical and physical changes. (Duhh!!)
Thermodynamics is the study of heat and its transformations.
Fundamental premise
When energy is transferred from one object to another, it appears as work and/or as heat.
For our work we must define a system to study; everything elsethen becomes the surroundings.
The system is composed of particles with their own internal energies (E or U).Therefore the system has an internal energy. When a change occurs, theinternal energy changes.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-13
A boundary separates a system from the surrounding. It is arbitrary in character, real or imaginary.
Types of System 1. Open system – separated by an imaginary boundary that allows matter
and energy exchange between the system and surroundings 2. Closed system – separated by a real diathermal and non-permeable boundary that only allows energy exchange to and from the surroundings 3. Isolated system – a system enclosed by a real adiabatic, non- permeable boundary where no matter and energy exchange occur
State and Non-state PropertiesState properties – observed when a system is in an equilibrium state
e.g. P, T, V, U, H, G, SNon-state properties – path dependent properties that are measured or
observed during a change in statee.g. q, w,
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-14
Thermodynamics
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 , pressure, volume, temperature
6.7
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-15
Thermochemistry is the study of heat change in chemical reactions.
The system is the specific part of the universe that is of interest in the study.
open
mass & energyExchange:
closed
energy
isolated
nothing
SYSTEMSURROUNDINGS
6.2
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-16
E = Efinal - Einitial = Eproducts - Ereactants
Figure 6.2 Energy diagrams for the transfer of internal energy (E) between a system and its surroundings.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-17
1st Law of Thermodynamics
• This is a generalization of the Conservation of Energy Law and it also serves to define the change in internal energy of a system.
• Energy cannot be created nor destroyed. Energy can only be transferred and or transformed.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-18
1st Law of Thermodynamics (cont.)
Equation:
Q = W + ΔU
U = internal energy
Q = heat energy (=heat transfer)
W = work done• The equation states that heat supplied to a system goes
to increase the internal energy of the system and perform mechanical work on the surroundings.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-19
Internal Energy & Heat
Internal Energy:• Is the total molecular energy of a system
= U
Heat:• Is the movement of energy = transfer of energy = flow
of energy = Q
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-20
Isothermal Process
• Any process in which the temperature of a system remains constant.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-21
Isothermal Process (cont.)
1st Law as applied to an Isothermal Process:
Q = W
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-22
Isobaric Process
• Is any process in which pressure of the system remains constant.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-23
Isobaric Process (cont.)
1st Law as applied to an Isobaric Process:
W = P (Vf – Vi)
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-24
Adiabatic Process
• Is any process in which no heat enters or leaves the system.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-25
Adiabatic Process (cont.)
1st Law as applied to an Adiabatic Process:
ΔU = - W
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-26
Thermodynamics
6.7
E = q + w
E is the change in internal energy of a system
q is the heat exchange between the system and the surroundings
w is the work done on (or by) the system
w = -PV when a gas expands against a constant external pressure
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-27
Enthalpy and the First Law of Thermodynamics
6.7
E = q + w
At constant pressure, q = H and w = -PV
E = H - PV
H = E + PV
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-28
Figure 6.3 A system transferring energy as heat only.
Figure 6.4 A system losing energy as work only.
Energy, E
Zn(s) + 2H+(aq) + 2Cl-(aq)
H2(g) + Zn2+(aq) + 2Cl-(aq)
E<0work done onsurroundings
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-29
Table 6.1 The Sign Conventions* for q, w and E
q w+ = E
+
+
-
-
-
-
+
+
+
-
depends on sizes of q and w
depends on sizes of q and w
* For q: + means system gains heat; - means system loses heat.
* For w: + means word done on system; - means work done by system.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-30
Limitations of the First Law of Thermodynamics
E = q + w
Euniverse = Esystem + Esurroundings
Esystem = -Esurroundings
The total energy-mass of the universe is constant.
However, this does not tell us anything about the direction of change in the universe.
Esystem + Esurroundings = 0 = Euniverse
Euniverse = Esystem + Esurroundings
Units of Energy
Joule (J)
Calorie (cal)
British Thermal Unit
1 cal = 4.18J
1 J = 1 kg*m2/s2
1 Btu = 1055 J
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-31
Sample Problem 6.1 Determining the Change in Internal Energy of a System
PROBLEM: When gasoline burns in a car engine, the heat released causes the products CO2 and H2O to expand, which pushes the pistons outward. Excess heat is removed by the car’s cooling system. If the expanding gases do 451 J of work on the pistons and the system loses 325 J to the surroundings as heat, calculate the change in energy (E) in J, kJ, and kcal.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-32
Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.
H = H (products) – H (reactants)
H = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
H < 0Hproducts > Hreactants
H > 0 6.3
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-33
The Meaning of Enthalpy
w = - PV
H = E + PV
qp = E + PV = H
H ≈ E in
1. Reactions that do not involve gases.
2. Reactions in which the number of moles of gas does not change.
3. Reactions in which the number of moles of gas does change but q is >>> PV.
H = E + PV
where H is enthalpy
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-34
Figure 6.8
Enthalpy diagrams for exothermic and endothermic processes.
Ent
halp
y, H
Ent
halp
y, H
CH4 + 2O2
CO2 + 2H2O
Hinitial
HinitialHfinal
Hfinal
H2O(l)
H2O(g)
heat out heat inH < 0 H > 0
A Exothermic process B Endothermic process
CH4(g) + 2O2(g) CO2(g) + 2H2O(g) H2O(l) H2O(g)
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-35
Sample Problem 6.2 Drawing Enthalpy Diagrams and Determining the Sign of H
PROBLEM: In each of the following cases, determine the sign of H, state whether the reaction is exothermic or endothermic, and draw and enthalpy diagram.
SOLUTION:
PLAN: Determine whether heat is a reactant or a product. As a reactant, the products are at a higher energy and the reaction is endothermic. The opposite is true for an exothermic reaction
(a) H2(g) + 1/2O2(g) H2O(l) + 285.8kJ
(b) 40.7kJ + H2O(l) H2O(g)
(a) The reaction is exothermic.
H2(g) + 1/2O2(g) (reactants)
H2O(l) (products)
EXOTHERMIC
(products)H2O(g)
(reactants)H2O(l)
H = -285.8kJ H = +40.7kJENDOTHERMIC
(b) The reaction is endothermic.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-36
Some Important Types of Enthalpy Change
heat of combustion (Hcomb)
heat of formation (Hf)
heat of fusion (Hfus)
heat of vaporization (Hvap)
C4H10(l) + 13/2O2(g) 4CO2(g) + 5H2O(g)
K(s) + 1/2Br2(l) KBr(s)
NaCl(s) NaCl(l)
C6H6(l) C6H6(g)
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-37
Constant-Pressure Calorimetry
No heat enters or leaves!
quniv = qwater + qcal + qmetal
quniv = 0qmetal = - (qwater + qcal)
qwater = mct
qcal = Ccalt
6.4
Reaction at Constant PH = qmetal
A piece of heated metal is dropped in an insulated cup with 75.0 grams of water. The temperature of the water increased from 29degC to 32degC. How much heat was lost by the piece of metal?
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-38
The specific heat (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 = mc
Heat (q) absorbed or released:
q = mct
q = Ct
t = tfinal - tinitial
6.4
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-39
Table 6.2 Specific Heat Capacities of Some Elements, Compounds, and Materials
Specific Heat Capacity (J/g*K)
SubstanceSpecific Heat Capacity (J/g*K)
Substance
Compounds
water, H2O(l)
ethyl alcohol, C2H5OH(l)
ethylene glycol, (CH2OH)2(l)
carbon tetrachloride, CCl4(l)
4.184
2.46
2.42
0.864
Elements
aluminum, Al
graphite,C
iron, Fe
copper, Cu
gold, Au
0.900
0.711
0.450
0.387
0.129
wood
cement
glass
granite
steel
Materials
1.76
0.88
0.84
0.79
0.45
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-40
How much heat is given off when an 869 g iron bar cools from 940C to 50C?
s of Fe = 0.444 J/g • 0C
t = tfinal – tinitial = 50C – 940C = -890C
q = mst = 869 g x 0.444 J/g • 0C x –890C = -34,000 J
6.4
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-41
Sample Problem 6.3 Finding the Quantity of Heat from Specific Heat Capacity
PROBLEM: A layer of copper welded to the bottom of a skillet weighs 125 g. How much heat is needed to raise the temperature of the copper layer from 250C to 300.0C? The specific heat capacity (c) of Cu is 0.387 J/g*K.
Sample Problem 6.4 Determining the Heat of a Reaction
PROBLEM: You place 50.0 mL of 0.500 M NaOH in a coffee-cup calorimeter at 25.000C and carefully add 25.0 mL of 0.500 M HCl, also at 25.000C. After stirring, the final temperature is 27.210C. Calculate qsoln (in J) and Hrxn (in J/mol). (Assume the total volume is the sum of the individual volumes and that the final solution has the same density and specfic heat capacity as water: d = 1.00 g/mL and c = 4.184 J/g*K)
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-42
Sample Problem 6.5 Calculating the Heat of Combustion
PROBLEM: A manufacturer claims that its new dietetic dessert has “fewer than 10 Calories per serving.” To test the claim, a chemist at the Department of Consumer Affairs places one serving in a bomb calorimeter and burns it in O2(the heat capacity of the calorimeter = 8.15 kJ/K). The temperature increases 4.9370C. Is the manufacturer’s claim correct?
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-43
PROBLEM: A balloon is filled with 3 liters of helium (atomic mass=4.0) at 1 atm pressure. It initially had a temperature of 25°C but when left under the sun the temperature changed to 28°C and the balloon expanded. The specific heat of helium is 5.1932 J/gK.
a. What is the amount of heat absorbed?
b. What is the amount of work done?
c. What is the amount of change in internal energy?
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-44
Because there is no way to measure the absolute value of the enthalpy of a substance, must I measure the enthalpy change for every reaction of interest?
Establish an arbitrary scale with the standard enthalpy of formation (H0) as a reference point for all enthalpy expressions.f
Standard enthalpy of formation (H0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm.
f
The standard enthalpy of formation of any element in its most stable form is zero.
H0 (O2) = 0f
H0 (O3) = 142 kJ/molf
H0 (C, graphite) = 0f
H0 (C, diamond) = 1.90 kJ/molf6.5
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-45
Calculate the standard enthalpy of formation of CS2 (l) given that:C(graphite) + O2 (g) CO2 (g) H0 = -393.5 kJRxn/f
S(rhombic) + O2 (g) SO2 (g) H0 = -296.1 kJRxn/f
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) H0 = -1072 kJrxn
1. 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) H0 = -393.5 kJ
2S(rhombic) + 2O2 (g) 2SO2 (g) H0 = -296.1x2 kJrxn
CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) H0 = +1072 kJrxn+
C(graphite) + 2S(rhombic) CS2 (l)
H0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJrxn6.5
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-46
Table 6.5 Selected Standard Heats of Formation at 250C(298K)
Formula H0f(kJ/mol)
calciumCa(s)CaO(s)CaCO3(s)
carbonC(graphite)C(diamond)CO(g)CO2(g)CH4(g)CH3OH(l)HCN(g)CSs(l)
chlorineCl(g)
0-635.1
-1206.9
01.9
-110.5-393.5-74.9
-238.6135
87.9
121.0
hydrogen
nitrogen
oxygen
Formula H0f(kJ/mol)
H(g)H2(g)
N2(g)NH3(g)NO(g)
O2(g)O3(g)H2O(g)
H2O(l)
Cl2(g)
HCl(g)
0
0
0
-92.30
218
-45.990.3
143-241.8
-285.8
107.8
Formula H0f(kJ/mol)
silverAg(s)AgCl(s)
sodium
Na(s)Na(g)NaCl(s)
sulfurS8(rhombic)S8(monoclinic)SO2(g)
SO3(g)
0
0
0
-127.0
-411.1
2-296.8
-396.0
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-47
The standard enthalpy of reaction (H0 ) is the enthalpy of a reaction carried out at 1 atm.
rxn
aA + bB cC + dD
H0rxn dH0 (D)fcH0 (C)f= [ + ] - bH0 (B)faH0 (A)f[ + ]
H0rxn nH0 (products)f= mH0 (reactants)f-
6.5
Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.
(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-48 6.5
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-49
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 49.04 kJ/mol.
2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
H0rxn nH0 (products)f= mH0 (reactants)f-
H0rxn 6H0 (H2O)f12H0 (CO2)f= [ + ] - 2H0 (C6H6)f[ ]
H0rxn = [ 12x–393.5 + 6x–187.6 ] – [ 2x49.04 ] = -5946 kJ
-5946 kJ2 mol
= - 2973 kJ/mol C6H6
6.5
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-50
Sample Problem 6.7 Using Hess’s Law to Calculate an Unknown H
PROBLEM: Two gaseous pollutants that form in auto exhaust are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following equation:
CO(g) + NO(g) CO2(g) + 1/2N2(g) H = ?
Given the following information, calculate the unknown H:
Equation A: CO(g) + 1/2O2(g) CO2(g) HA = -283.0 kJ
Equation B: N2(g) + O2(g) 2NO(g) HB = 180.6 kJ
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-51
Sample Problem 6.9 Calculating the Heat of Reaction from Heats of Formation
PROBLEM: Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia:
Calculate H0rxn from H0
f values.
4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g)
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-52
Sample Problem 6.6 Using the Heat of Reaction (Hrxn) to Find Amounts
PROBLEM: The major source of aluminum in the world is bauxite (mostly aluminum oxide). Its thermal decomposition can be represented by
If aluminum is produced this way, how many grams of aluminum can form when 1.000x103 kJ of heat is transferred?
Al2O3(s) 2Al(s) + 3/2O2(g) Hrxn = 1676 kJ
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-53
Entropy
• Is a measure of a system’s organizational order.
• A system that has some recognizable order has a low entropy value; the more DISORDER the system, the HIGHER its entropy value.
• Symbol: ΔS
• Units: Joules/Kelvin
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-54
Entropy (cont.)
2nd Law as applied to any process:– The entropy of an isolated system will always tend
to increase.– It is very easy to destroy (=high entropy) than to
build (=low entropy)
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-55
Entropy (cont.)
Equation:Δ S ≥ 0S = entropyΔS = change in entropyFor an isothermal process where a change in entropy occurs, the entropy change is given by:
ΔS = ΔQ / T
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-56
Entropy (cont.)
Ex: Phase change: going from ice (solid) to water (liquid), the increase in entropy value,
ΔS = ΔQ / T
T = melting or boiling point (in Kelvin)
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-57
Entropy (cont.)
2nd Law as applied to Entropy:
ΔS > 0 every system tends to get more disordered, hence the entropy increases
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-58
Entropy - Calculations
• Is also based on probability concepts. According to Boltzman, the entropy of a system taking place is given by:
S = K ln W
K = Boltzman constant
W = Probability of occurrence of an event
ln = natural log
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-59
Entropy (cont.)
Sample space for throwing 3 coins simultaneously:
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-60
Entropy (cont.)
Conclusion:
S (= entropy) for all heads is far less compared to S for only two heads. So the probability of obtaining 2 heads far exceeds all heads.
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
6-61
Entropy (cont.)
Relevance:
Can be used to determine direction of any process or invention.