1. Chapter 11 Spontaneous Change and Equilibrium
11-1 Enthalpy and Spontaneous Change
11-2 Entropy
11-3 Absolute Entropies and Chemical Reactions
11-4 The Second Law of Thermodynamics
11-5 The Gibbs Function
11-6 The Gibbs Function and Chemical Reactions
11-7 The Gibbs Function and the Equilibrium Constant
11-8 The Temperature Dependence of Equilibrium Constants
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2. Chapter 11 Spontaneous Change and Equilibrium
Second Law of Thermodynamics
In a real spontaneous process the Entropy of the universe (meaning the system plus its surroundings) must increase.
Suniverse > 0 (spontaneous process)
Suniverse = Ssystem + Ssurroundings > 0
if Suniverse = 0, then everything is in equilibrium
The 2nd Law of Thermodynamics profoundly affects how we look at nature and processes
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3. (conservation of energy)
Summarize a few Concepts
1st Law of Thermodynamics
In any process, the total energy of the universe remains unchanged: energy is conserved
A process and its reverse are equally allowed
Eforward = – Ereverse
(conservation of energy)
2nd Law of Thermodynamics
S, the entropy of a universe, increases in only one of the two directions of a reaction
Processes that decrease ΔS are impossible. Or improbable beyond conception
ΔSuniv > 0 Spontaneous
ΔSuniv = 0 Equilibrium
ΔSuniv < 0 Non-spontaneous
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4. (spontaneous process)
Gibbs Function
How are Enthalpy and Entropy related?
G = H - T•S
G has several names
Gibbs function
Gibbs free energy
Free Enthalpy
For the change in the Gibbs Energy of system, at constant Temperature and Pressure
ΔGsys = ΔHsys - T·ΔSsys
ΔGsystem = Δ Hsys - TΔSsystem = < 0
(spontaneous process)
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5. For a change at constant temperature and pressure
ΔGsys < 0 Spontaneous
ΔGsys = 0 Equilibrium
ΔGsys > 0 Non-spontaneous,
but the reverse is
spontaneous
From Earlier
ΔSuniv > 0 Spontaneous
ΔSuniv = 0 Equilibrium
ΔSuniv < 0 Non-spontaneous
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6. Typical example using Gibbs Free energy
Benzene, C6H6, boils at 80.1°C and ΔHvap = 30.8 kJ
a) Calculate ΔSvap for 1 mole of benzene
B) at 60°C and pressure = 1 atm does benzene boil?
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7. (equilibrium process)
Benzene, C6H6, boils at 80.1°C.
ΔHvap = 30.8 kJ
a) Calculate ΔSvap for 1 mole of benzene
B) at 60°C and pressure = 1 atm does benzene boil?
ΔGsys < 0 Spontaneous
ΔGsys = 0 Equilibrium
ΔGsys > 0 Non-spontaneous
ΔGsystem = Δ Hsys - TΔSsystem = 0
(equilibrium process)
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8. The Gibbs Function and Chemical Reactions
ΔG = ΔH - T·ΔS
ΔGf° is the standard molar Gibbs function of formation
Because G is a State Property, DGrxno for a general reaction
a A + b B → c C + d D
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9. Example 11-7 3NO(g) → N2O(g) + NO2(g)
Calculate ΔG° for the following reaction at K. Use Appendix D for additional information needed.
3NO(g) → N2O(g) + NO2(g)
Solution
From Appendix D
ΔGf°(N2O) = kJ mol-1
ΔGf°(NO2) = 51.29
ΔGf° (NO) = 86.55
ΔG°= 1(104.18)+1(51.29)-3(86.55)
ΔG°= − kJ therefore, spontaneous
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10. Effects of Temperature on ΔG°
For temperatures other than 298K
ΔG = ΔH - TΔS
Typically ΔH and ΔS are almost constant over a broad range
For above reaction, as Temperature increases ΔG becomes more positive, i.e., less negative.
Next Slide
ΔG has a strong temperature dependence
3NO (g) → N2O (g) + NO2 (g)
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11. G has a strong temperature dependence
3NO(g) → N2O(g) + NO2(g)
G has a strong temperature dependence
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12. For temperatures other than 298K or 25C ΔG = ΔH - T·ΔS
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13. For temperatures other than 298K or 25C ΔG = ΔH - T·ΔSB A C D
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14. For temperatures other than 298K or 25C ΔG = ΔH - T·ΔS
Case C
ΔH° > 0
ΔS° < 0
ΔG = ΔH - T·ΔS ΔG = (+) - T·(-) = positive
 ΔG > 0 or non-spontaneous at all Temp. B A
Case B
ΔH° < 0
ΔS° > 0
ΔG = ΔH - T·ΔS ΔG = (-) - T·(+) = negative
 ΔG < 0 or spontaneous at all temp. C D
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15. For temperatures other than 298K or 25C ΔG = ΔH - T·ΔS
The spontaneity of a reaction at different temperatures depends on the signs of Δ H° & ΔS°. When both signs are the same, the temperature determines the spontaneity.
For the next two cases, it is more interesting because there is a temperature at which
ΔG = 0
Then above or below a temperature the ΔG is either positive or negative
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17. For temperatures other than 298K or 25C ΔG = ΔH - T·ΔS
Case D
ΔH° < 0
ΔS° < 0
ΔG = ΔH - T·ΔS at a low Temp
ΔG = (-) - T·(-) = negative
 ΔG < 0 or spontaneous at low Temp. B A
Case A
ΔH° > 0
ΔS° > 0
ΔG = ΔH - T·ΔS ΔG = (+) - T·(+) at a High Temp
 ΔG < 0 or spontaneous at high Temp. C D
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18. The Gibbs Function and the Equilibrium Constant
What about non-standard states, other than 1 atm or a conc. [X] = 1 mol/L?
ΔG = ΔG° + RT ln Q
Where Q is the reaction quotient
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19. Where Q is the reaction quotient
ΔG = ΔG° + RT ln Q
Where Q is the reaction quotient
a A + b B ↔ c C + d D
If Q > K the rxn shifts towards the reactant side
The amount of products are too high relative to the amounts of reactants present, and the reaction shifts in reverse (to the left) to achieve equilibrium
If Q = K equilibrium
If Q < K the rxn shifts toward the product side
The amounts of reactants are too high relative to the amounts of products present, and the reaction proceeds in the forward direction (to the right) toward equilibrium
compare
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20. - ΔG ° = RT ln K ΔG = ΔG° + RT ln Q a A + b B ↔ c C + d D
Where Q is the reaction quotient
a A + b B ↔ c C + d D
If Q < K the rxn shifts towards the product side
If Q = K equilibrium
If Q > K the rxn shifts toward the reactant side
At Equilibrium conditions, ΔG = 0
- ΔG ° = RT ln K
NOTE: we can now calculate equilibrium constants (K) for reactions from standard ΔGf functions of formation
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21. Where Q is the reaction quotient
ΔG = ΔG° + RT ln Q
Where Q is the reaction quotient
a A + b B ↔ c C + d D
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22. Example 11-9 Found ΔG°= - 104.18 kJ mol -1
The ΔGr° for the following reaction at K was obtained in example Calculate the equilibrium constant for this reaction at 25C.
3NO(g) ↔ N2O(g) + NO2(g)
Strategy
Use - ΔG ° = RT ln K
Found ΔG°= kJ mol -1
from example 11-7
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23. Example 11-9 3NO(g) ↔ N2O(g) + NO2(g) Solution - ΔG ° = RT ln K
Use
- ΔG ° = RT ln K
Rearrange
Use ΔGr°= kJ mol-1
from Ex. 11-7
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24. Chapter 11 Spontaneous Change and Equilibrium
Examples / Exercises
11-2, 11-3, 11-4, 11-5, 11-6
11-7, 11-8, 11-9, 11-10, 11-11, 11-12
Problems WebCT
8abc 30abcd
44a 50abc
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