1. Chapter 17 Free Energy and Thermodynamics
Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro
Chapter 17 Free Energy and Thermodynamics
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2008, Prentice Hall

2. Reversible
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3. Comparing Potential Energy
The direction of spontaneity can be determined by comparing the potential energy of the system at the start and the end.
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4. Thermodynamics vs. Kinetics
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5. Diamond → Graphite
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7. 4 particles partition in 16 microstates 1 mole of particles partition in 2N (N~1023)
1/16
4/16
6/16
4/16
1/16

8. “There is a tide in the affairs of men, Which taken at the flood, leads on to fortune. Omitted, all the voyage of their life is bound in shallows and in miseries. On such a full sea are we now afloat. And we must take the current when it serves, or lose our ventures.”
William Shakspeare
Letters put together form words, in their specific organization,
convey timeless meaning!
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9. Increases in Entropy
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10. The 2nd Law of Thermodynamics
DSuniverse = DSsystem + DSsurroundings
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11. Entropy Change in State Change
when materials change state, the number of macrostates it can have changes as well
for entropy: solid < liquid < gas
because the degrees of freedom of motion increases solid → liquid → gas
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12. Entropy Change and State Change
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13. Ex. 17.2a – The reaction C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O(g) has DHrxn = kJ at 25°C. Calculate the entropy change of the surroundings.
Given:
Find:
DHsystem = kJ, T = 298 K
DSsurroundings, J/K
Concept Plan:
Relationships: DS
T, DH
Solution:
Check:
combustion is largely exothermic, so the entropy of the surrounding should increase significantly

14. DG, DH, and DS
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15. Ex. 17.3a – The reaction CCl4(g)  C(s, graphite) + 2 Cl2(g) has DH = kJ and DS = J/K at 25°C. Calculate DG and determine if it is spontaneous.
Given:
Find:
DH = kJ, DS = J/K, T = 298 K
DG, kJ
Concept Plan:
Relationships: DG
T, DH, DS
Solution:
Since DG is +, the reaction is not spontaneous at this temperature. To make it spontaneous, we need to increase the temperature.
Answer:

16. Ex. 17.3a – The reaction CCl4(g)  C(s, graphite) + 2 Cl2(g) has DH = kJ and DS = J/K. Calculate the minimum temperature it will be spontaneous.
Given:
Find:
DH = kJ, DS = J/K, DG < 0
T, K
Concept Plan:
Relationships: T
DG, DH, DS
Solution:
The temperature must be higher than 673K for the reaction to be spontaneous
Answer:

18. Relative Standard Entropies States
the gas state has a larger entropy than the liquid state at a particular temperature
the liquid state has a larger entropy than the solid state at a particular temperature
Substance
S°,
(J/mol∙K)
H2O (g)
70.0
H2O (l)
188.8
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19. Relative Standard Entropies Molar Mass
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20. Relative Standard Entropies Allotropes
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21. Relative Standard Entropies Molecular Complexity
Substance
Molar
Mass
S°,
(J/mol∙K)
Ar (g)
39.948
154.8
NO (g)
30.006
210.8
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22. Relative Standard Entropies Dissolution
Substance
S°,
(J/mol∙K)
KClO3(s)
143.1
KClO3(aq)
265.7
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23. Substance
S, J/molK
NH3(g)
192.8
O2(g)
205.2
NO(g)
210.8
H2O(g)
188.8
Ex –Calculate DS for the reaction 4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(l)
Given:
Find:
standard entropies from Appendix IIB
DS, J/K
Concept Plan:
Relationships: DS
SNH3, SO2, SNO, SH2O,
Solution:
Check:
DS is +, as you would expect for a reaction with more gas product molecules than reactant molecules

24. DGreaction = DHreaction – TDSreaction
Calculating DG
at 25C:
DGoreaction = SnGof(products) - SnGof(reactants)
at temperatures other than 25C:
assuming the change in DHoreaction and DSoreaction is negligible
DGreaction = DHreaction – TDSreaction
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26. Substance
DGf, kJ/mol
CH4(g)
-50.5
O2(g)
0.0
CO2(g)
-394.4
H2O(g)
-228.6
O3(g)
163.2
Ex –Calculate DG at 25C for the reaction CH4(g) + 8 O2(g)  CO2(g) + 2 H2O(g) + 4 O3(g)
Given:
Find:
standard free energies of formation from Appendix IIB
DG, kJ
Concept Plan:
Relationships:
DG
DGf of prod & react
Solution:

27. Ex. 17. 6 – The reaction SO2(g) + ½ O2(g)  SO3(g) has DH = -98
Ex – The reaction SO2(g) + ½ O2(g)  SO3(g) has DH = kJ and DS = J/K at 25°C. Calculate DG at 125C and determine if it is spontaneous.
Given:
Find:
DH = kJ, DS = J/K, T = 398 K
DG, kJ
Concept Plan:
Relationships:
DG
T, DH, DS
Solution:
Since DG is -, the reaction is spontaneous at this temperature, though less so than at 25C
Answer:

28. DG Relationships
if a reaction can be expressed as a series of reactions, the sum of the DG values of the individual reaction is the DG of the total reaction
DG is a state function
if a reaction is reversed, the sign of its DG value reverses
if the amounts of materials is multiplied by a factor, the value of the DG is multiplied by the same factor
the value of DG of a reaction is extensive
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29. Free Energy and Reversible Reactions
the change in free energy is a theoretical limit as to the amount of work that can be done
if the reaction achieves its theoretical limit, it is a reversible reaction
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30. Real Reactions
in a real reaction, some of the free energy is “lost” as heat
if not most
therefore, real reactions are irreversible
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31. DG under Nonstandard Conditions
DG = DG only when the reactants and products are in their standard states
there normal state at that temperature
partial pressure of gas = 1 atm
concentration = 1 M
under nonstandard conditions, DG = DG + RTlnQ
Q is the reaction quotient
at equilibrium DG = 0
DG = ─RTlnK
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34. DG = +46400 J + (8.314 J/K)(700 K)(ln 1.2 x 10-7)
Example - DG
Calculate DG at 427°C for the reaction below if the PN2 = 33.0 atm, PH2= 99.0 atm, and PNH3= 2.0 atm
N2(g) + 3 H2(g) ® 2 NH3(g)
Q =
PNH32
PN21 x PH23
(2.0 atm)2
(33.0 atm)1 (99.0)3 =
= 1.2 x 10-7
DH° = [ 2(-46.19)] - [0 +3( 0)] = kJ = J
DS° = [2 (192.5)] - [(191.50) + 3(130.58)] = J/K
DG° = J - (700 K)( J/K)
DG° = J
DG = DG° + RTlnQ
DG = J + (8.314 J/K)(700 K)(ln 1.2 x 10-7)
DG = J = -46 kJ

35. DGº and K Because DGrxn = 0 at equilibrium, then DGº = −RTln(K).
When K < 1, DGº is positive and the reaction is spontaneous in the reverse direction under standard conditions.
Nothing will happen if there are no products yet!
When K > 1, DGº is negative and the reaction is spontaneous in the forward direction under standard conditions.
When K = 1, DGº is 0 and the reaction is at equilibrium under standard conditions.

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37. Example - K
Estimate the equilibrium constant and position of equilibrium for the following reaction at 427°C
N2(g) + 3 H2(g) Û 2 NH3(g)
DH° = [ 2(-46.19)] - [0 +3( 0)] = kJ = J
DS° = [2 (192.5)] - [(191.50) + 3(130.58)] = J/K
DG° = J - (700 K)( J/K)
DG° = J
DG° = -RT lnK
J = -(8.314 J/K)(700 K) lnK
lnK = -7.97
K = e-7.97 = 3.45 x 10-4
since K is << 1, the position of equilibrium favors reactants

38. Why Is the Equilibrium Constant Temperature Dependent?
Combining these two equations
DG° = DH° − TDS°
DG° = −RTln(K)
It can be shown that
This equation is in the form y = mx + b.
The graph of ln(K) versus inverse T is a straight line with slope and y-intercept