1. Chapter 6 Lesson 2 Thermochemistry
Prentice Hall

2. The First Law of Thermodynamics: Energy is Conserved
UNIT 3
Chapter 5: Energy Changes
Section 5.1
The First Law of Thermodynamics: Energy is Conserved
The first law of thermodynamics states that:
energy can be converted from one form to another but cannot be created or destroyed.
Since any change in energy of the universe must be zero,
ΔEuniverse = ΔEsystem + ΔEsurroundings = 0
ΔEsystem = –ΔEsurroundings
if a system gains energy, that energy comes from the surroundings
if a system loses energy, that energy enters the surroundings

3. UNIT 3
Chapter 5: Energy Changes
Section 5.1
Enthalpy
One way chemists express thermochemical changes is by a variable called enthalpy, H. The change in enthalpy, ΔH, of a system can be measured.
It depends only on the initial and final states of the system, and is represented by
ΔH = ΔE + Δ(PV)
For reactions of solids and liquids in solutions, Δ(PV) = 0
If heat enters a system
ΔH is positive
the process is endothermic
If heat leaves a system
ΔH is negative
the process is exothermic

4. The Second Law of Thermodynamics
UNIT 3
Chapter 5: Energy Changes
Section 5.1
The Second Law of Thermodynamics
The second law of thermodynamics states that:
when two objects are in thermal contact, heat is transferred from the object at a higher temperature to the object at the lower temperature until both objects are the same temperature (in thermal equilibrium)
Image source: MHR, Chemistry 12 © ISBN ; page 283
When in thermal contact, energy from hot particles will transfer to cold particles until the energy is equally distributed and thermal equilibrium is reached.

5. Comparing Categories of Enthalpy Changes: Enthalpy of Solution
UNIT 3
Chapter 5: Energy Changes
Section 5.1
Comparing Categories of Enthalpy Changes: Enthalpy of Solution
Three processes occur when a substance dissolves, each with a ΔH value.
bonds between solute molecules or ions break
bonds between solvent molecules break
bonds between solvent molecules and solute molecules or ions form
Sum of the enthalpy changes: enthalpy of solution, ΔHsolution
The three fundamental types of processes in which enthalpy change is considered are: physical, chemical, and nuclear.
Recall: physical changes are changes to the condition of a substance that do not change the chemical properties of the substance.
Image source: MHR, Chemistry 12 © ISBN ; page 284
The orange arrow shows the overall ΔH.

6. Thermochemical Equations and Calorimetry
UNIT 3
Chapter 5: Energy Changes
Section 5.2
Thermochemical Equations and Calorimetry
Chemical reactions involve
initial breaking of chemical bonds (endothermic)
then formation of new bonds (exothermic)
ΔHr is the difference between the total energy required to break bonds and the total energy released when bonds form.
Image source: MHR, Chemistry 12 © ISBN ; page 292

7. UNIT 3
Chapter 5: Energy Changes
Section 5.3
Hess’s Law
The enthalpy change of nearly any reaction can be determined using collected data and Hess’s law.
The enthalpy change of any reaction can be determined if:
the enthalpy changes of a set of reactions “add up to” the overall reaction of interest
standard enthalpy change, ΔH°, values are used
Image source: MHR, Chemistry 12 © ISBN ; page

8. Combining Sets of Chemical Equations
UNIT 3
Chapter 5: Energy Changes
Section 5.3
Combining Sets of Chemical Equations
To find the enthalpy change for formation of SO3 from O2 and S8, you can use
Because only 1 mol of sulfur trioxide is in the final step:
Divide the first equation by 8 so 1 mol of sulfur dioxide is in the first step
Divide the second equation by 2
Image source: MHR, Chemistry 12 © ISBN ; page

9. Techniques for Manipulating Equations
UNIT 3
Chapter 5: Energy Changes
Section 5.3
Techniques for Manipulating Equations
Reverse an equation
the products become the reactants, and reactants become the products
the sign of the ΔH value must be changed
Multiply each coefficient
all coefficients in an equation are multiplied by the same integer or fraction
the value of ΔH must also be multiplied by the same number
Image source: MHR, Chemistry 12 © ISBN ; page

10. Standard Conditions
the standard state is the state of a material at a defined set of conditions
pure gas at exactly 1 atm pressure
pure solid or liquid in its most stable form at exactly 1 atm pressure and temperature of interest
usually 25°C
substance in a solution with concentration 1 M
the standard enthalpy change, DH°, is the enthalpy change when all reactants and products are in their standard states
the standard enthalpy of formation, DHf°, is the enthalpy change for the reaction forming 1 mole of a pure compound from its constituent elements
the elements must be in their standard states
the DHf° for a pure element in its standard state = 0 kJ/mol
by definition
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11. Standard Molar Enthalpies of Formation
UNIT 3
Chapter 5: Energy Changes
Section 5.3
Standard Molar Enthalpies of Formation
Data that are especially useful for calculating standard enthalpy changes:
standard molar enthalpy of formation, ΔH˚f
the change in enthalpy when 1 mol of a compound is synthesized from its elements in their most stable form at SATP conditions
enthalpies of formation for elements in their most stable state under SATP conditions are set at zero
since formation equations are for 1 mol of compound, many equations include fractions
standard molar enthalpies of formation are in Appendix B

12. Formation Reactions and Thermal Stability
UNIT 3
Chapter 5: Energy Changes
Section 5.3
Formation Reactions and Thermal Stability
The thermal stability of a substance is the ability of the substance to resist decomposition when heated.
decomposition is the reverse of formation
the opposite sign of an enthalpy change of formation for a compound is the enthalpy change for its decomposition
the greater the enthalpy change for the decomposition of a substance, the greater the thermal stability of the substance
Image source: MHR, Chemistry 12 © ISBN ; page

13. Using Enthalpies of Formation and Hess’s Law
UNIT 3
Chapter 5: Energy Changes
Section 5.3
Using Enthalpies of Formation and Hess’s Law
For example:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Image source: MHR, Chemistry 12 © ISBN ; page

14. UNIT 3
Chapter 5: Energy Changes
Section 5.3
Learning Check
Determine ∆H˚r for the following reaction using the enthalpies of formation that are provided.
C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
∆H˚f of C2H5OH(l): –277.6 kJ/mol
∆H˚f of CO2(g): –393.5 kJ/mol
∆H˚f of H2O(l): –285.8 kJ/mol
Answer on the next slide

15. UNIT 3
Chapter 5: Energy Changes
Section 5.3
Learning Check
∆H˚r = [(2 mol)(∆H˚f CO2(g)) + (3 mol)(∆H˚f H2O(l))] –
[(1 mol)(∆H˚f C2H5OH(l)) + (3 mol)(∆H˚fO2(g)]
∆H˚r = [(2 mol)(–393.5 kJ/mol) + (3 mol)(–285.8 kJ/mol)] –
[(1 mol)(–277.6 kJ/mol) + (3 mol)(0 kJ/mol)]
Image source: MHR, Chemistry 12 © ISBN ; page 98
∆H˚r = (– kJ) – (–277.6 kJ)
∆H˚r = – kJ

16. Relationships Involving DHrxn
when reaction is multiplied by a factor, DHrxn is multiplied by that factor
because DHrxn is extensive
C(s) + O2(g) → CO2(g) DH = kJ
2 C(s) + 2 O2(g) → 2 CO2(g) DH = 2( kJ) = kJ
if a reaction is reversed, then the sign of DH is reversed
CO2(g) → C(s) + O2(g) DH = kJ
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17. Sample – Hess’s Law [3 NO2(g)  3 NO(g) + 1.5 O2(g)] DH° = (+259.5 kJ)
Given the following information:
2 NO(g) + O2(g)  2 NO2(g) DH° = -173 kJ
2 N2(g) + 5 O2(g) + 2 H2O(l)  4 HNO3(aq) DH° = -255 kJ
N2(g) + O2(g)  2 NO(g) DH° = +181 kJ
Calculate the DH° for the reaction below:
3 NO2(g) + H2O(l)  2 HNO3(aq) + NO(g) DH° = ?
[3 NO2(g)  3 NO(g) O2(g)] DH° = ( kJ)
[1 N2(g) O2(g) + 1 H2O(l)  2 HNO3(aq)] DH° = (-128 kJ)
[2 NO(g)  N2(g) + O2(g)] DH° = -181 kJ
3 NO2(g) + H2O(l)  2 HNO3(aq) + NO(g) DH° = - 49 kJ
[2 NO2(g)  2 NO(g) + O2(g)] x 1.5 DH° = 1.5(+173 kJ)
[2 N2(g) + 5 O2(g) + 2 H2O(l)  4 HNO3(aq)] x 0.5 DH° = 0.5(-255 kJ)
[2 NO(g)  N2(g) + O2(g)] DH° = -181 kJ
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18. The Combustion of CH4
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19. Sample - Calculate the Enthalpy Change in the Reaction 2 C2H2(g) + 5 O2(g) ® 4 CO2(g) + 2 H2O(l)
1. Write formation reactions for each compound and
determine the DHf° for each
2 C(s, gr) + H2(g) ® C2H2(g) DHf° = kJ/mol
C(s, gr) + O2(g) ® CO2(g) DHf° = kJ/mol
H2(g) + ½ O2(g) ® H2O(l) DHf° = kJ/mol
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20. Sample - Calculate the Enthalpy Change in the Reaction 2 C2H2(g) + 5 O2(g) ® 4 CO2(g) + 2 H2O(l)
2. Arrange equations so they add up to desired reaction
2 C2H2(g) ® 4 C(s) + 2 H2(g) DH° = 2(-227.4) kJ
4 C(s) + 4 O2(g) ® 4CO2(g) DH° = 4(-393.5) kJ
2 H2(g) + O2(g) ® 2 H2O(l) DH° = 2(-285.8) kJ
2 C2H2(g) + 5 O2(g) ® 4 CO2(g) + 2 H2O(l) DH = kJ
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21. DH°reaction = S n DHf°(products) - S n DHf°(reactants)
Sample - Calculate the Enthalpy Change in the Reaction 2 C2H2(g) + 5 O2(g) ® 4 CO2(g) + 2 H2O(l)
DH°reaction = S n DHf°(products) - S n DHf°(reactants)
DHrxn = [(4•DHCO2 + 2•DHH2O) – (2•DHC2H2 + 5•DHO2)]
DHrxn = [(4•(-393.5) + 2•(-285.8)) – (2•(+227.4) + 5•(0))]
DHrxn = kJ
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22. Example – How many kg of octane must be combusted to supply 1
Example – How many kg of octane must be combusted to supply 1.0 x 1011 kJ of energy?
Given:
Find:
1.0 x 1011 kJ
mass octane, kg
Write the balanced equation per mole of octane
Concept Plan:
Relationships:
MMoctane = g/mol, 1 kg = 1000 g
DHf°’s
DHrxn° kJ
mol C8H18
g C8H18
kg C8H18
from
above
Solution:
C8H18(l) + 25/2 O2(g) → 8 CO2(g) + 9 H2O(g)
Material
DHf°, kJ/mol
C8H18(l)
-250.1
O2(g)
CO2(g)
-393.5
H2O(g)
-241.8
Look up the DHf°
for each material
in Appendix 4
Check:
the units and sign are correct
the large value is expected

23. The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm.
rxn
aA + bB cC + dD
DH0
rxn
dDH0 (D) f
cDH0 (C) = [ + ] -
bDH0 (B)
aDH0 (A)
DH0
rxn
nDH0 (products) f = S
mDH0 (reactants) -
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.)
6.5

24. 6.5

25. Calculate the standard enthalpy of formation of CS2 (l) given that:
C(graphite) + O2 (g) CO2 (g) DH0 = kJ
rxn
S(rhombic) + O2 (g) SO2 (g) DH0 = kJ
rxn
CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = kJ
rxn
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.
rxn
C(graphite) + O2 (g) CO2 (g) DH0 = kJ
2S(rhombic) + 2O2 (g) SO2 (g) DH0 = x2 kJ
rxn +
CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = kJ
rxn
C(graphite) + 2S(rhombic) CS2 (l)
DH0 = (2x-296.1) = 86.3 kJ
rxn
6.5

26. 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 kJ/mol.
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)
DH0
rxn
nDH0 (products) f = S
mDH0 (reactants) -
DH0
rxn
6DH0 (H2O) f
12DH0 (CO2) = [ + ] -
2DH0 (C6H6)
DH0
rxn =
[ 12x– x–187.6 ] – [ 2x49.04 ] = kJ
-5946 kJ
2 mol
= kJ/mol C6H6
6.5

27. DHsoln = Hsoln - Hcomponents
The enthalpy of solution (DHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.
DHsoln = Hsoln - Hcomponents
Which substance(s) could be used for melting ice?
Which substance(s) could be used for a cold pack?
6.6

28. The Solution Process for NaCl
DHsoln = Step 1 + Step 2 = 788 – 784 = 4 kJ/mol
6.6

29. Energy Use and the Environment
in the U.S., each person uses over 105 kWh of energy per year
most comes from the combustion of fossil fuels
combustible materials that originate from ancient life
C(s) + O2(g) → CO2(g) DH°rxn = kJ
CH4(g) +2 O2(g) → CO2(g) + 2 H2O(g) DH°rxn = kJ
C8H18(g) O2(g) → 8 CO2(g) + 9 H2O(g) DH°rxn = kJ
fossil fuels cannot be replenished
at current rates of consumption, oil and natural gas supplies will be depleted in 50 – 100 yrs.
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30. Energy Consumption the increase in energy consumption in the US
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31. The Effect of Combustion Products on Our Environment
because of additives and impurities in the fossil fuel, incomplete combustion and side reactions, harmful materials are added to the atmosphere when fossil fuels are burned for energy
therefore fossil fuel emissions contribute to air pollution, acid rain, and global warming
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32. Global Warming CO2 is a greenhouse gas
it allows light from the sun to reach the earth, but does not allow the heat (infrared light) reflected off the earth to escape into outer space
it acts like a blanket
CO2 levels in the atmosphere have been steadily increasing
current observations suggest that the average global air temperature has risen 0.6°C in the past 100 yrs.
atmospheric models suggest that the warming effect could worsen if CO2 levels are not curbed
some models predict that the result will be more severe storms, more floods and droughts, shifts in agricultural zones, rising sea levels, and changes in habitats
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33. CO2 Levels
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34. Renewable Energy our greatest unlimited supply of energy is the sun
new technologies are being developed to capture the energy of sunlight
parabolic troughs, solar power towers, and dish engines concentrate the sun’s light to generate electricity
solar energy used to decompose water into H2(g) and O2(g); the H2 can then be used by fuel cells to generate electricity
H2(g) + ½ O2(g) → H2O(l) DH°rxn = kJ
hydroelectric power
wind power
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