Upload
lindsay-hoosier
View
225
Download
3
Tags:
Embed Size (px)
Citation preview
Chapter 5
Principles of Chemical Reactivity
Basic PrinciplesThermodynamics: The science of heat
and workEnergy: the capacity to do work
-chemical, mechanical, thermal, electrical, radiant, sound, nuclear-affects matter by raising its temperature,
eventually causing a state change-All physical changes and chemical
changes involve energy
Potential Energy: energy that results from an object’s
position -gravitational, chemical, electrostatic
Kinetic Energy: energy of motion
Basic Principles
Law of Energy Conservation:Energy can neither be created nor
destroyed-a.k.a. The first law of thermodynamics
-The total energy of the universe is constant
Temperature vs. Heat:– Temperature is the measure of an object’s
heat energy– Heat ≠ temperature
The Measurement of Heat
Thermal Energydepends on temperature and the amount (mass or volume) of the object
-More thermal energy a substances has the greater the motion its atoms/molecules have-Total thermal energy of an object is the sum of the individual energies of all atoms/molecules/ions that make up that object
SI unit: Joule (J)1 calorie = 4.184 JEnglish unit = BTU
Converting Calories to Joules
Convert 60.1 cal to joules
System: object or collection of objects being studied– In lab, the system is the chemicals inside the
beaker
Surroundings: everything outside of the system that can exchange energy with the system– The surroundings are outside the beaker
Universe: system plus surroundings
Exothermic: heat transferred from the system to the surroundings
Endothermic: heat transferred from the surroundings to the system
Basic Principles
Specific Heat Capacity (C)
amount of heat required to raise the temperature of 1gram of a substance by 1 kelvin
SI Units: Specific heat capacity = J /g.K
Specific heat of water = 4.184J
g.K
Heat Transfer
Heat transfer equationused to calculate amounts of heat (q) in
a substance
T C mq J
g Jg.K
K
q1 + q2 + q3 … = 0 or qsystem + qsurroundings = 0
Heat Transfer
Calculate the amount of heat to raise the temperature of 200 g of water from 10.0 oC to 55.0 oC
Heat Transfer
Calculate the amount of heat energy (in joules) needed to raise the temperature of 7.40 g of water from 29.0°C to 46.0°C
Heat Transfer
Specific heat of gold is 0.13
Therefore the metal cannot be pure gold.
A 1.6 g sample of metal that appears to be gold requires 5.8 J to raise the temperature from 23°C to 41°C. Is the metal pure gold?
Jg.K
Changes of State
occurs when enough energy is put into a substance to over come molecular interactions
Solid-liquid: molecules in a solid when heated move about vigorously enough to break solid-solid molecular interactions to become a liquid
Liquid-gas: molecules in a liquid when heated move about more vigorously enough to break liquid-liquid molecular interactions to become a gas
Note: This happens in reverse by removing heat energy
Energy and Changes of StateHeat of fusion:
heat needed to convert a substance from a solid to a liquid (at its melting/freezing point)
333 J/g for water
Heat of vaporization: heat needed to convert a substance from a liquid to a gas (at its boiling/condensation point)
2256 J/g for water
Example: Calculate the amount of heat involved to convert 100.0 g of ice at -50.0°C to steam at 200.0°C.
The First Law of Thermodynamics
This law can be stated as, “The combined amount of energy in the universe is constant”
Also called-The Law of Conservation of Energy:– Energy is neither created nor destroyed in
chemical reactions and physical changes.
The First Law of Thermodynamics
There are two basic ideas for thermodynamic systems:
1. Chemical systems tend toward a state of minimum potential energy
Some examples of this include:-H2O flows downhill-Objects fall when dropped
Epotential = mg(h)
State FunctionThe value of a state function is independent
of pathway-An analogy to a state function is the energy required to climb a mountain taking two different paths:
E1 = energy at the bottom of the mountainE1 = mgh1
E2 = energy at the top of the mountainE2 = mgh2
E = E2-E1 = mgh2 – mgh1 = mg(h)
Examples of state functions: Temperature, Pressure, Volume, Energy, Entropy, and
enthalpy
Examples of non-state functions: Number of moles, heat, work
The First Law of Thermodynamics
2. Chemical systems tend toward a state of maximum disorder
Common examples of this are:- A mirror shatters when dropped and does
not reform- It is easy to scramble an egg and difficult to
unscramble it- Food dye when dropped into water disperses
The First Law of Thermodynamics
Thermodynamic questions:1. Will these substances react when they
are mixed under specified conditions?2. If they do react, what energy changes
and transfers are associated with their reaction?
3. If a reaction occurs, to what extent does it occur?
Changes in Internal Energy (E or U)
all of the energy contained within a substance– all forms of energy such as kinetic,
potential, gravitational, electromagnetic, etc.
First Law of Thermodynamics states that the change in internal energy, E, is determined by the heat flow (q) and the work (w)
E = q + wbook: U = q + w
Changes in Internal Energy (E)
E = Eproducts – Ereactants
E = q + w at constant pressure:
w = -P x V
q > 0 if heat is absorbed by the systemq < 0 if heat is absorbed by the surroundingsw > 0 if surroundings do work on the systemw < 0 if system does work on the surroundings
Changes in Internal Energy (E)
If 1200 joules of heat are added to a system, and the system does 800 joules of work on the surroundings, what is the:
1. energy change for the system, Esys?
2. energy change of the surroundings, Esurr?
Changes in Internal Energy (E)
E is negative when energy is released by a system
-Energy can be written as a product of the process
kJ 10 3.516- E
kJ 10 3.516 OH 6 CO 5 O 8 HC3
3)(22(g)2(g))(125
Changes in Internal Energy (E)
E is positive when energy is absorbed by a system undergoing a chemical or physical change
– Energy can be written as a reactant of the process
kJ 10 3.516 E
O 8 HC kJ 10 3.516 OH 6 CO 53
2(g))(1253
)(22(g)
Enthalpy Changes for Chemical Reactions
Exothermic reactions: release energy in the form of heat to the surroundings (H < 0)
-heat is transferred from a system to the surroundings
Endothermic reactions: gain energy in the form of heat from the surroundings (H > 0)
-heat is transferred from the surroundings to the system
For example, the combustion of propane:
Combustion of butane:
Enthalpy Change (H)Heat content of a substance at constant
pressure- Chemistry is commonly done in open beakers
on a desk top at atmospheric pressure- Therefore enthalpy change (H) is the change
in heat content: H = qp at constant pressure E = qv at constant volume
If E and H < 0:energy is transferred to the surroundings
If E and H > 0: energy is transferred to the system
- Enthalpy and energy differ by the amount of work
E = H + w and w = -PV
Enthalpy Changes for Chemical Reactions
Exothermic reactions generate specific amounts of heat– Because the potential energies of the products are
lower than the potential energies of the reactants
Endothermic reactions consume specific amounts of heat– Potential energies of the reactants are lower than the products
H for the reverse reactionis equal, but has the opposite sign to the forward reaction
Thermochemical Equations
balanced chemical reaction with the H value for the reaction
H < 0 designates an exothermic reaction: heat is a product, the container feels hot
H > 0 designates an endothermic reaction: heat is a reactant, the container feels cold
kJ 3523 - H OH 6 CO 5O 8 HC orxn)(22(g)2(g))12(5
Hess’s Law
If a reaction is the sum of two or more other reactions, H for the overall process is the sum of the H for the component reactions– Hess’s Law is true because H is a state function
If we know the following H values:
kJ 1648H OFe 2 O 3 Fe 4 3
kJ 454H FeO 2O Fe 2 2
kJ 560H OFe 2O FeO 4 1
o3(s)22(g)(s)
o(g)2(g)(s)
o(s)322(g)(s)
1.
2.
Target: ?
Hess’s LawWe can calculate the H for the reaction by
properly adding (or subtracting) the H for reactions 1 and 2– Notice that the target reaction has FeO and O2
as reactants and Fe2O3 as a product– Arrange reactions 1 and 2 so that they also have
FeO and O2 as reactants and Fe2O3 as a product• Each reaction can be doubled, tripled, or multiplied by
a half, etc.• H values are then doubled, tripled, etc.• If a reaction is reversed the H value is changed to the
opposite sign
H
2(2 FeO Fe O kJ
3 4 Fe O Fe O kJ
4 FeO O Fe O kJ
0
s s 2 g
s 2 g 2 3 s
s 2 g 2 3
2 2 2 2 544
3 2 1648
1 2 560
x [ ] ) ( )
1.
2.+
4 FeO 4 Fe 2 O2
+1088 kJ
1
Hess’s Law
Given the following equations and H values
calculate H for the reaction below:
H kJ
2 N O N O 164.1
[2] N + O NO 180.5
[3] N + 2 O NO 66.4
o
2 g 2 g 2 g
2 g 2 g g
2 g 2 g 2 g
[ ]1 2
2
2
You do it!
?H NO 3 NO + ON ogg2g2
Hess’s Law
-The + sign of the H value tells us that the reaction is endothermic.
-The reverse reaction is exothermic, i.e.
3 NO N O + NO H = -155.5 kJg 2 g 2 go
Standard Enthalpy of Formation
Thermochemical standard state conditionsThe thermochemical standard T = 298.15 KThe thermochemical standard P = 1.0000 atm
- Be careful not to confuse these values with STP
Thermochemical standard states of matterFor pure substances in their liquid or solid phase the
standard state is the pure liquid or solid- For aqueous solutions the standard state is 1.00 M
concentrationFor gases the standard state is the gas at 1.00 atm of
pressure- For gaseous mixtures the partial pressure must be
1.00 atm
Standard Molar Enthalpies of Formation, Hf
o
The enthalpy change for the formation of one mole of a substance formed directly from its constituent elements in their standard states– The symbol for standard molar enthalpy of formation is Hf
o (KJ/mol)
The standard molar enthalpy of formation for MgCl2 is:
Standard Molar Enthalpies of Formation (Hf
o)Standard molar enthalpies of formation have been
determined for many substances and are tabulated in Appendix L
Standard molar enthalpies of elements in their most stable forms at 298.15 K and 1.000 atm are zero.
Example: The standard molar enthalpy of formation for phosphoric acid is -1279 kJ/mol. Write the equation for the reaction for which Ho
f = -1279 kJ
Note: P in standard state is P4 (s)
Phosphoric acid in standard state is H3PO4(s)
Standard Molar Enthalpies of Formation
Calculate the enthalpy change for the reaction of one mole of H2(g) with one mole of F2(g) to form two moles of HF(g) at 25oC and one atmosphere.
Standard Molar Enthalpies of Formation
Calculate the enthalpy change for the reaction in which 15.0 g of aluminum reacts with oxygen to form Al2O3 at 25oC and one atmosphere.
You do it!You do it!
Hess’s LawVersion IIFor chemical reaction at standard conditions:
– the standard enthalpy change is the sum of the standard molar enthalpies of formation of the products minus the sum for the reactants
- each enthalpy of formation is multiplied by its coefficient in the balanced chemical equation
tscoefficien tricstoichiomen
Hn Hn H 0reactants f
0products f
0rxn
nn
Final - Initial
Hess’s Law
C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l)
Calculate the H°f for the following reaction fromthe data in appendix L:
Hess’s Law
Given the following information, calculate Hf
o for H2S(g)
2 H S + 3 O 2 SO + 2 H O H = -1124 kJ
H ? 0 - 296.8 - 285.8
(kJ / mol)
2 g 2 g 2 g 2 298o
fo
l
You do it!You do it!
CalorimetryAn experimental technique that measures the heat
transfer during a chemical or physical process
Constant pressure calorimetry:A styrofoam coffee-cup calorimeter is
used to measure the amount of heat produced (or absorbed) in a reaction
– This is one method to measure qP (called H) for reactions in solution
qreaction + qsolution = 0
Note: Assuming no heat transfer to the surroundings
CalorimetryIf an exothermic reaction is performed in a
calorimeter, the heat evolved by the reaction is determined from the temperature rise of the solution– This requires a two part calculation
When we add 25.00 mL of 0.500 M NaOH at 23.000oC to 25.00 mL of 0.600 M CH3COOH already in the calorimeter at the same temperature, the resulting temperature is observed to be 25.947oC. Determine heat of reaction and then calculate the change in enthalpy (as KJ/mol) for the production of NaCH3COO.
CH3COOH(aq) + NaOH(aq) NaCH3COO(aq) + H2O(l)
Calorimetry
Constant volume calorimetry:Or Bomb calorimetry measures measure the amount of
heat produced (or absorbed) in a chemical reaction
-this method is used for measuring qv (E)
qreaction + qbomb + qwater = 0