1. Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Copyright © 2010 Pearson Prentice Hall, Inc.
Copyright © 2010 Pearson Prentice Hall, Inc.

2. Energy and Its Conservation
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Energy and Its Conservation
Energy: The capacity to supply heat or do work.
Kinetic Energy (EK): The energy of motion.
Potential Energy (EP): Stored energy.
Potential energy comes in many forms. Drop something and it comes from a difference in height (gravity). Chemical bonds “store” energy.
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3. Energy and Its Conservation
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Energy and Its Conservation
Law of Conservation of Energy: Energy cannot be created or destroyed; it can only be converted from one form to another.
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4. Energy and Its Conservation
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Energy and Its Conservation
Total Energy = EK + EP
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5. Energy and Its Conservation
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Energy and Its Conservation
Thermal Energy: The kinetic energy of molecular motion, measured by finding the temperature of an object.
Heat: The amount of thermal energy transferred from one object to another as the result of a temperature difference between the two.
Heat transfers from hot to cold. Both of them are relative to each other.
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6. Energy and Its Conservation
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Energy and Its Conservation
First Law of Thermodynamics: Energy cannot be created or destroyed; it can only be converted from one form to another.
The first law is essentially the conservation law.
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7. Internal Energy and State Functions
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Internal Energy and State Functions
First Law of Thermodynamics: The total internal energy of an isolated system is constant.
∆E = Efinal - Einitial
Restating the first law.
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8. Internal Energy and State Functions
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Internal Energy and State Functions
First Law of Thermodynamics: The total internal energy of an isolated system is constant.
∆E = Efinal - Einitial
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9. Internal Energy and State Functions
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Internal Energy and State Functions
First Law of Thermodynamics: The total internal energy of an isolated system is constant.
∆E = Efinal - Einitial
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10. Internal Energy and State Functions
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Internal Energy and State Functions
CO2(g) + 2H2O(g) kJ energy
CH4(g) + 2O2(g)
∆E = Efinal - Einitial = -802 kJ
802 kJ is released when 1 mole of methane, CH4, reacts with 2 moles of oxygen to produce 1 mole of carbon dioxide and two moles of water.
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11. Internal Energy and State Functions
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Internal Energy and State Functions
State Function: A function or property whose value depends only on the present state, or condition, of the system, not on the path used to arrive at that state.
Internal energy, E, is a state function. You can also think of climbing a hill. Regardless of the path taken to climb the hill, the height difference is simply the height of the hill.
Copyright © 2010 Pearson Prentice Hall, Inc.

12. Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Expansion Work
Work = Force x Distance
w = F x d
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13. Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Expansion Work
3CO2(g) + 4H2O(g)
C3H8(g) + 5O2(g)
6 mol of gas
7 mol of gas
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14. Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Expansion Work
Expansion Work: Work done as the result of a volume change in the system. Also called pressure-volume or PV work.
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15. Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Energy and Enthalpy
∆E = q + w
q = heat transferred
w = work = -P∆V
q = ∆E + P∆V
Constant Volume (∆V = 0):
qV = ∆E
Constant Pressure:
qP = ∆E + P∆V
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16. Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Energy and Enthalpy
qP = ∆E + P∆V = ∆H
Enthalpy change or
Heat of reaction (at constant pressure)
Enthalpy is a state function whose value depends only on the current state of the system, not on the path taken to arrive at that state.
∆H = Hfinal - Hinitial
= Hproducts - Hreactants
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17. The Thermodynamic Standard State
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
The Thermodynamic Standard State
3CO2(g) + 4H2O(g)
C3H8(g) + 5O2(g)
∆H = kJ
3CO2(g) + 4H2O(l)
C3H8(g) + 5O2(g)
∆H = kJ
Thermodynamic Standard State: Most stable form of a substance at 1 atm pressure and at a specified temperature, usually 25 °C; 1 M concentration for all substances in solution.
The conversion from liquid to gas phases requires energy. Therefore, it’s important that the states are specified.
Putting the “°” superscripted next to the ∆H means standard state.
The standard pressure is actually 1 bar. The enthalpy difference is usually small, however.
3CO2(g) + 4H2O(g)
C3H8(g) + 5O2(g)
∆H° = kJ
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18. Enthalpies of Physical and Chemical Change
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Enthalpies of Physical and Chemical Change
Enthalpy of Fusion (∆Hfusion): The amount of heat necessary to melt a substance without changing its temperature.
Enthalpy of Vaporization (∆Hvap): The amount of heat required to vaporize a substance without changing its temperature.
Enthalpy of Sublimation (∆Hsubl): The amount of heat required to convert a substance from a solid to a gas without going through a liquid phase.
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19. Enthalpies of Physical and Chemical Change
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Enthalpies of Physical and Chemical Change
Copyright © 2010 Pearson Prentice Hall, Inc.

20. Enthalpies of Physical and Chemical Change
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Enthalpies of Physical and Chemical Change
2Fe(s) + Al2O3(s)
2Al(s) + Fe2O3(s)
∆H° = -852 kJ
Exothermic: Heat (enthalpy) flows from the system to the surroundings.
BaCl2(aq) + 2NH3(aq) + 10H2O(l)
Ba(OH)2 •8H2O(s) + 2NH4Cl(s)
The first reaction is the thermite reaction.
“Exo” = out
“Endo” = in
∆H° = kJ
Endothermic: Heat (enthalpy) flows from the surroundings to the system.
Copyright © 2010 Pearson Prentice Hall, Inc.

21. Enthalpies of Physical and Chemical Change
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Enthalpies of Physical and Chemical Change
2Fe(s) + Al2O3(s)
2Al(s) + Fe2O3(s)
∆H° = -852 kJ
Exothermic: Heat (enthalpy) flows from the system to the surroundings.
2Al(s) + Fe2O3(s)
2Fe(s) + Al2O3(s)
∆H° = +852 kJ
The enthalpy of reaction is for the reaction in the direction written.
Endothermic: Heat (enthalpy) flows from the surroundings to the system.
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22. Calorimetry and Heat Capacity
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Calorimetry and Heat Capacity
Measure the heat flow at constant pressure (∆H).
Calorimetry = study of heat flow.
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23. Calorimetry and Heat Capacity
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Calorimetry and Heat Capacity
Measure the heat flow at constant volume (∆E).
Bomb calorimeter.
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24. Calorimetry and Heat Capacity
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Calorimetry and Heat Capacity
Heat Capacity (C): The amount of heat required to raise the temperature of an object or substance a given amount.
q = C x ∆T
Molar Heat Capacity (Cm): The amount of heat required to raise the temperature of 1 mol of a substance by 1 °C.
∆T is the temperature change.
Heat capacity is an extensive property.
Specific heat is an intensive property.
q = (Cm) x (moles of substance) x ∆T
Specific Heat: The amount of heat required to raise the temperature of 1 g of a substance by 1 °C.
q = (specific heat) x (mass of substance) x ∆T
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25. Calorimetry and Heat Capacity
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Calorimetry and Heat Capacity
Note that the values for water are considerably higher than most other substances. Interesting discussion points could be how large bodies of water affect weather; the human body is approximately 60% water and the role it plays in homeostasis, etc.
Copyright © 2010 Pearson Prentice Hall, Inc.

26. Calorimetry and Heat Capacity
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Calorimetry and Heat Capacity
Assuming that a can of soda has the same specific heat as water, calculate the amount of heat (in kilojoules) transferred when one can (about 350 g) is cooled from 25 °C to 3 °C.
q = (specific heat) x (mass of substance) x ∆T
Specific heat = 4.18
g °C J
Mass = 350 g
Temperature change = 25 °C - 3 °C = 22 °C
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27. Calorimetry and Heat Capacity
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Calorimetry and Heat Capacity
Calculate the amount of heat transferred.
g °C
4.18 J
350 g
22 °C x x
Heat evolved =
= J
1000 J
1 kJ J
= 32 kJ x
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28. Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Hess’s Law
Hess’s Law: The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction.
Haber Process:
2NH3(g)
3H2(g) + N2(g)
∆H° = kJ
Multiple-Step Process
N2H4(g)
2H2(g) + N2(g)
∆H°1 = ?
Hydrazine, N2H4, is a reactive intermediate (more on that in Ch. 12: Chemical Kinetics). Chemical reactions are far more complicated than discussed in Ch. 6 (stoichiometry).
The overall reaction and step 2 can be measured, but step 1 cannot be measured easily.
A fascinating book about Haber is:
Charles, Daniel; Master Mind: The Rise and Fall of Fritz Haber, The Nobel Laureate Who Launched the Age of Chemical Warfare; HarperCollins Publishers Inc., 2005.
2NH3(g)
N2H4(g) + H2(g)
∆H°2 = kJ
2NH3(g)
3H2(g) + N2(g)
∆H°1+2 = kJ
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29. Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Hess’s Law
∆H°1 + ∆H°2 = ∆H°1+2
∆H°1 = ∆H°1+2 - ∆H°2
= kJ - ( kJ) = 95.4 kJ
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30. Standard Heats of Formation
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Standard Heats of Formation
Standard Heat of Formation (∆H°f ): The enthalpy change for the formation of 1 mol of a substance in its standard state from its constituent elements in their standard states.
Standard states
CH4(g)
C(s) + 2H2(g)
∆H°f = kJ
Students need to remember solid, liquid, gases, and diatomics as discussed previously.
1 mol of 1 substance
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31. Standard Heats of Formation
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Standard Heats of Formation
Note the lack of anything that is 0.
Elements (standard state) are defined to have a standard enthalpy of formation of 0.
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32. Standard Heats of Formation
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Standard Heats of Formation
∆H° = ∆H°f (Products) - ∆H°f (Reactants)
cC + dD
aA + bB
∆H° = [c ∆H°f (C) + d ∆H°f (D)] - [a ∆H°f (A) + b ∆H°f (B)]
Using table values.
Products
Reactants
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33. Standard Heats of Formation
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Standard Heats of Formation
Using standard heats of formation, calculate the standard enthalpy of reaction for the photosynthesis of glucose (C6H12O6) and O2 from CO2 and liquid H2O.
C6H12O6(s) + 6O2(g)
6CO2(g) + 6H2O(l)
∆H° = ?
H° = [∆H°f (C6H12O6(s))] - [6 ∆H°f (CO2(g)) + 6 ∆H°f (H2O(l))]
Using table values.
The units for standard enthalpies of formation are kJ/mol while the unit for the standard enthalpy change is kJ.
Result is endothermic.
∆H° = [(1 mol)( kJ/mol)] -
[(6 mol)( kJ/mol) + (6 mol)( kJ/mol)]
= kJ
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34. Standard Heats of Formation
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Standard Heats of Formation
C6H12O6(s) + 6O2(g)
6CO2(g) + 6H2O(l)
∆H° = kJ
(1)
(2)
(3)
Why does the calculation “work”?
Reverse the “reaction” and reverse the sign on the standard enthalpy change.
CO2(g)
C(s) + O2(g)
∆H° = kJ
We need to see the two basic operations that can be done to a thermochemical equation.
The original reaction was exothermic (see the negative sign). Flipping the reaction changes it into an endothermic reaction which has a positive sign for the enthalpy change.
Becomes
C(s) + O2(g)
CO2(g)
∆H° = kJ
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35. Standard Heats of Formation
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Standard Heats of Formation
C6H12O6(s) + 6O2(g)
6CO2(g) + 6H2O(l)
∆H° = kJ
(1)
(2)
(3)
Why does the calculation “work”?
Multiply the coefficients by a factor and multiply the standard enthalpy change by the same factor.
C(s) + O2(g)
CO2(g)
∆H° = kJ
If you have 6 times as much “stuff,” you have 6 times the enthalpy change.
Becomes
6C(s) + 6O2(g)
6CO2(g)
∆H° = 6(393.5 kJ) = kJ
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36. Standard Heats of Formation
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Standard Heats of Formation
C6H12O6(s) + 6O2(g)
6CO2(g) + 6H2O(l)
∆H° = kJ
(1)
(2)
(3)
Why does the calculation “work”?
Reverse the “reaction” and reverse the sign on the standard enthalpy change.
H2O(l)
H2(g) O2(g) 2 1
∆H° = kJ
Becomes
H2(g) O2(g)
H2O(l) 2 1
∆H° = kJ
Copyright © 2010 Pearson Prentice Hall, Inc.

37. Standard Heats of Formation
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Standard Heats of Formation
C6H12O6(s) + 6O2(g)
6CO2(g) + 6H2O(l)
∆H° = kJ
(1)
(2)
(3)
Why does the calculation “work”?
Multiply the coefficients by a factor and multiply the standard enthalpy change by the same factor.
H2(g) O2(g)
H2O(l) 2 1
∆H° = kJ
If you have 6 times as much “stuff,” you have 6 times the enthalpy change.
Becomes
6H2(g) + 3O2(g)
6H2O(l)
∆H° = 6(285.8 kJ) = kJ
Copyright © 2010 Pearson Prentice Hall, Inc.

38. Standard Heats of Formation
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Standard Heats of Formation
C6H12O6(s) + 6O2(g)
6CO2(g) + 6H2O(l)
∆H° = kJ
(1)
(2)
(3)
Why does the calculation “work”?
C6H12O6(s)
6C(s) + 6H2(g) + 3O2(g)
∆H° = kJ
6C(s) + 6O2(g)
6CO2(g)
∆H° = kJ
Using table values.
6H2(g) + 3O2(g)
6H2O(l)
∆H° = kJ
C6H12O6(s) + 6O2(g)
6CO2(g) + 6H2O(l)
∆H° = kJ
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39. Bond Dissociation Energies
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Bond Dissociation Energies
Bond Dissociation Energy:
The amount of energy that must be supplied to break a chemical bond in an isolated molecule in the gaseous state and is thus the amount of energy released when the bond forms. or
Standard enthalpy changes for the corresponding bond-breaking reactions.
The first definition came from Chapter 5 and the second one from this chapter.
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40. Bond Dissociation Energies
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Bond Dissociation Energies
Note that all the bond dissociation energies are positive.
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41. Bond Dissociation Energies
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Bond Dissociation Energies
2HCl(g)
H2(g) + Cl2(g)
∆H° = D(Reactant bonds) - D(Product bonds)
∆H° = (DH-H + DCl-Cl) - (2DH-Cl)
∆H° = [(1 mol)(436 kJ/mol) + (1mol)(243 kJ/mol)] -
Reactants - products! This is opposite that for enthalpies of formation.
All BDE’s are positive. Add the values for the ones you have to break (reactants) and subtract the values for the ones you make (products, forming bonds releases energy).
This compares well to 2x(standard enthalpy of formation for HCl) = kJ/mol
(2mol)(432 kJ/mol)
= -185 kJ
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42. Fossil Fuels, Fuel Efficiency, and Heats of Combustion
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Fossil Fuels, Fuel Efficiency, and Heats of Combustion
CO2(g) + 2H2O(l)
CH4(g) + 2O2(g)
∆H°c = [∆H°f (CO2(g)) + 2 ∆H°f (H2O(l))] - [∆H°f (CH4(g))]
= [(1 mol)( kJ/mol) + (2 mol)( kJ/mol)] -
[(1 mol)(-74.8 kJ/mol)]
= kJ
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43. Fossil Fuels, Fuel Efficiency, and Heats of Combustion
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Fossil Fuels, Fuel Efficiency, and Heats of Combustion
CO2(g) + 2H2O(l)
CH4(g) + 2O2(g)
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44. An Introduction to Entropy
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
An Introduction to Entropy
Spontaneous Process: A process that, once started, proceeds on its own without a continuous external influence.
Entropy (S): The amount of molecular randomness in a system.
Spontaneous processes are
favored by a decrease in H (negative ∆H).
favored by an increase in S (positive ∆S).
Nonspontaneous processes are
favored by an increase in H (positive ∆H).
favored by a decrease in S (negative ∆S).
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45. Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
Copyright © 2010 Pearson Prentice Hall, Inc.

46. An Introduction to Free Energy
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
An Introduction to Free Energy
Gibbs Free Energy Change (∆G)
∆G = ∆H - T ∆S
Enthalpy of
reaction
Temperature
(Kelvin)
Entropy
change
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47. An Introduction to Free Energy
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
An Introduction to Free Energy
Gibbs Free Energy Change (∆G)
∆G = ∆H ∆S - T
∆G < 0 Process is spontaneous
∆G = 0 Process is at equilibrium
(neither spontaneous nor nonspontaneous)
∆G > 0 Process is nonspontaneous
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48. An Introduction to Free Energy
Chapter 8: Thermochemistry: Chemical Energy
11/24/2018
An Introduction to Free Energy
Copyright © 2010 Pearson Prentice Hall, Inc.