1. Chapter 15 - Spontaneity, Entropy, and Free Energy
1 Spontaneous Processes
2 The Isothermal Expansion and Compression of an Ideal Gas
3 The Definition of Entropy
4 Entropy and Physical Changes
5 Entropy and the Second Law of Thermodynamics
6 The Effect of Temperature on Spontaneity
7 Free Energy
8 Entropy Changes in Chemical Reactions
9 Free Energy and Chemical Reactions
10 The Dependence of Free Energy on Pressure
11 Free Energy and Equilibrium
12 Free Energy and Work (skip)
13 Reversible and Irreversible Processes: A Summary (skip)
14 Adiabatic Processes (skip)

2. Spontaneous Processes
Kinetics vs Thermo
Thermodynamics predicts direction and “driving force”.
Kinetics predicts speed (rate).
Spontaneous Processes
Occur on some timescale (maybe slowly) without outside intervention (examples: a battery will discharge, a hot cup of coffee will cool to ambient temperature).
All spontaneous processes proceed toward “states” (macrostates) with the greatest number of accessible microstates.

3. Microstates and Macrostates:
An available microstate describes a specific detailed microscopic configuration (molecular rotations, translations, vibrations, electronic configuration) that a system can visit in the course of its fluctuations.
A macrostate describes macroscopic properties such as temperature and pressure.
For a gas at constant T: the number of available microstates increases with volume. For gas, liquid or solid, the number of available microstates increases with T (the number of available vibrational microstates, electronic microstates, etc. increases with T). When you heat anything, you increase the number of available microstates.
When a liquid vaporizes, the number of available microstates increases.
When a liquid freezes, the number of available microstates decreases.

4. What is a spontaneous process?
probable
not probable
This is not a spontaneous process.
The reverse process (going from right to left) is spontaneous.
a) A gas will spontaneously expand to fill the available space.
b) There is a ‘driving’ force that causes a gas to spontaneously expand to fill a vacuum.
c) The entropy of the universe increases with a gas expands to fill a vacuum.
a=b=c

5. Why is this not a spontaneous process?
probable
not probable
There are more available microstates on the left hand side than on the right hand side.
If a system gains degrees of freedom (more constituents, more room to move, more available quantum states, more available rotational, vibrational, translational or electronic states), then it gains entropy.
A spontaneous process increases entropy (but you must consider both the system and the surroundings)

6. probable
not probable
The probability of finding both molecules on the left side is ¼
(this is one available microstate out of four possible microstates that will give this arrangement)
The probability of finding one molecule on the each side is ½.
(this are two possible microstates out of four possible microstates that will give this arrangement)
The probability of finding both molecules on the right side is ¼
(this is one microstate out of four possible microstates)
The probability of occurrence of a particular arrangement (state) depends on the number of ways (microstates) in which that arrangement can be achieved. All microstates are equally probable.

7. You will see that arrangement III is most probable.
There are three possible arrangements of four molecules in two chambers.
The arrangement with the greatest number of microstates is most probable. Label the molecules a,b,c,d and count the microstates.
You will see that arrangement III is most probable.

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9. 2 molecules Probability of finding 2 molecules on the same side is 1/4

10. Definitions of Entropy “S”
Entropy is related to probability
If a system has several available macrostates, it will spontaneously proceed to the one with the largest number of available microstates.
The macrostate with the greatest probability (largest number of available microstates) has the highest entropy.
When you heat something you increase its entropy.
S = kB ln Ω Joules/Kelvin
Kb = Boltzmann’s constant, the gas constant per molecule (R/NA)
Ω = the number of available microstates of a given state
∆S = q/T J / mol-K

11. Ludwig Boltzmann (1844-1906) Highlights Moments in a Life
Established the logarithmic connection between entropy and probability in his kinetic theory of gases.
The Boltzmann constant (k or kB) is the physical constant relating temperature to energy.
Moments in a Life
Suffered from bipolar disorder and depression
Ironically, in Max Planck’s Nobel Prize speech in 1918, it was pointed out that Boltzmann never introduced the constant k, Planck did.
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12. Twice the number of microstates
One He in the gas phase expands from volume V1 to 2V1
Ω2 = 2Ω1
Twice the number of microstates
If 1 mole of He (instead of 2 He)
and the gas expands from V1 to V2
The change in entropy of a gas is dependent on the change in volume of the gas
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13. The isothermal expansion of an ideal gas.
Isothermal – system and surroundings maintain constant temperature.
ΔE = 0 = q + w
then q = – w
Consider only reversible and irreversible processes
For a reversible, cyclic process both the system and the surroundings are returned exactly to their original positions.
Cyclic expansion-compression process “work is converted to heat”
Work → Heat
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14. The isothermal expansion of an ideal gas.
∆E=0 (energy of a perfect gas depends only on T)
∆E= w + q
w = -q
This important relationship entropy (determined by number of available microscopic states) is related to a macroscopic properties of heat and temperature.

15. ↓ ΔS(A) < 0 (cools) ΔS(B) > 0 (heats) |ΔS(A)| > |ΔS(B)|
Brick A (warm)
Brick B (warm)
w(A)
q(A)
ΔE(A)
ΔH(A)
ΔS(A)
W(B)
Q(B)
ΔE(B)
ΔH(B)
ΔS(B) ↓
Brick A (cold)
Brick B (hot)
ΔS(A) < 0 (cools) ΔS(B) > 0 (heats)
|ΔS(A)| > |ΔS(B)|
ΔS(uni) = ΔS(A) + ΔS(B)
ΔS(uni) < 0
This is not a spontaneous process.

16. Entropy and Physical Change
Temperature Dependence of Entropy:
Cp and Cv are is the heat capacities of the system.
ΔS(T1 to T2) here should be written ΔSsys(T1 to T2)

17. Example
Calculate the change in entropy that occurs when a sample containing 1.00 mol of water (ice) is heated from – 20 °C to +20°C at 1 atm pressure. The molar heat capacities of H2O (s) and H2O (l) are 38.1 J K-1mol-1 and 75.3 J K-1mol-1 respectively and the enthalpy of fusion (melting) is 6.01 kJ mol-1 at 0°C.
Solution
ΔS from 253K to 273K = n Cp ln(T2/T1) = (1.00)(38.1)ln (273/253) = 2.90 J/K
ΔS phase change from liq to gas = qrev/T = ΔHfus/T = (6010/273) = 22.0 J/K
Δfrom 273K to 293K = n Cp ln(T2/T1) = (1.00)(75.3)ln (293/273) = 5.3 J/K
Total ΔS = ΔS1 + ΔS2 + ΔS3

18. First Law of Thermodynamics
The change in the internal energy of a system is equal to the work done on it plus the heat transferred to it. The Law of Conservation of Energy
E = q + w
Second Law of Thermodynamics
For a spontaneous process the Entropy of the universe (meaning the system plus its surroundings) increases.
Suniverse > 0
Third Law of Thermodynamics
In any thermodynamic process involving only pure phases at equilibrium, the entropy change,  S, approaches zero at absolute zero temperature; also the entropy of a crystalline substance approaches zero at 0K.
S = 0 at 0 K
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19. 0 =ΔEforward + ΔEreverse (Energy is conserved in both directions)
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 by the first law
0 =ΔEforward + ΔEreverse
(Energy is conserved in both directions)
2nd Law of Thermodynamics
Processes that increase ΔSuniverse are spontaneous.
ΔSuniv > 0 Spontaneous Forward
ΔSuniv = 0 At Equilibrium
ΔSuniv < 0 Spontaneous Reverse
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20. Suniverse = Ssystem + Ssurroundings
The sign of ΔSsur depends on the direction of the heat flow.
The magnitude of ΔSsur depends on the temperature
This is ΔH of the system.
If the reaction is exothermic, ΔH has a negative sign and ΔSsurr is positive
If the reaction is endothermic, ΔH has a positive sign and ΔSsurr is negative

21. Ssystem + Ssurroundings = Suniverse

22. Ssolid < S liquid < Sgas
Summary of Entropy
Entropy is a quantitative measure of the number of microstates available to the molecules in a system. It is a measure of the number of ways in which energy or molecules can be arranged.
Entropy is the degree of randomness or disorder in a system
The Entropy of all substances is positive
Ssolid < S liquid < Sgas
ΔSsys is the Entropy Change of the system
ΔSsur is the Entropy Change of the surroundings
ΔSuni is the Entropy Change of the universe
S has the units J K-1mol-1

23. Josiah Willard Gibbs (1839-1903)
Highlights
Devised much of the theoretical foundation for chemical thermodynamics.
Established the concept free energy
Moments in a Life
1863 Yale awarded him the first American Ph.D. in engineering
Book: Equilibrium of Heterogeneous Substances, deemed one of the greatest scientific achievements of the 19th century.
Will never be famous like Michael Jackson.
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24. Gibbs Free Energy ΔG = ΔHsys - TΔSsys
Allows us to focus on the system only, without considering the surroundings.
ΔG = - TΔSuni
G is called
Gibbs Function, or
Gibbs Free Energy, or
Free Energy.

25. Free Energy
ΔG < 0 Spontaneous
ΔG = 0 Equilibrium
ΔG > 0 Spontaneous Reverse
Entropy
ΔSuniv > 0 Spontaneous Forward
ΔSuniv = 0 Equilibrium
ΔSuniv < 0 Spontaneous Reverse

26. Benzene, C6H6, boils at 80.1°C (at 1 atm) and ΔHvap = 30.8 kJ
Gibbs Free energy
Benzene, C6H6, boils at 80.1°C (at 1 atm) and ΔHvap = 30.8 kJ
a) Calculate ΔSvap for 1 mole of benzene at 60°C and pressure = 1 atm.

27. Benzene, C6H6, boils at 80°C at 1 atm. ΔHovap = 30.8 kJ
a) Calculate ΔSvap for 1 mole of benzene Start with ΔGvap=ΔHvap-TΔSvap at the boiling point, ΔGvap = 0 so ΔHvap = TbΔSvap

28. Benzene, C6H6, boils at 80.1°C (at 1 atm) and ΔHvap = 30.8 kJ
Gibbs Free energy
Benzene, C6H6, boils at 80.1°C (at 1 atm) and ΔHvap = 30.8 kJ
b) Does benzene spontaneously boil at 60°C?

29. Effects of Temperature on ΔG°
ΔG° = ΔH° - TΔS°
Typically ΔH and ΔS are almost constant over a broad range
For the reaction above, as Temperature increases ΔG becomes more positive, i.e., less negative.
3NO (g) → N2O (g) + NO2 (g)

30. Effects of Temperature on ΔG

31. For temperatures other than 298K or 25C ΔG = ΔH - T·ΔSB A D C

32. 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. D C

33. 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. D C

35. ΔSrxn° = ΣS°products – ΣS°reactants
Entropies of Reaction
ΔSrxn° = ΣS°products – ΣS°reactants
ΔSrxn° is the sum of products minus the sum of the reactants, for one mole of reaction (that is what ° means)
For a general reaction
a A + b B → c C + d D
Appendix 4 tabulates standard molar entropy values, S° in units JK-1mol-1

36. 2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g)
Example
Calculate ΔSr° at K for the reaction
2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g)
(b) Calculate ΔS° of the system when g of H2S(g) reacts with excess O2(g) to give SO2(g) and H2O(g) and no other products at K

37. 2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g)
Calculate ΔSr° at K for the reaction
2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g)
Solution
(a) Look up each S° of formation [Note this is for “one mole of the reaction” as written: i.e. 2 moles of H2S, 3 moles of O2, etc]
ΔSrxn°= 2S°(SO2(g) ) + 2S°(H2O(g)) -2S°(H2S(g) ) - 3S°(O2(g))
ΔSrxn°= 2(248) + 2(189) -2(206) - 3(205)
= – 153 JK-1mol-1

38. 2H2S(g) + 3O2(g) → 2SO2(g) +2H2O(g)
(a) ΔSrxn= -153 JK-1mol-1
(b) Calculate ΔS° when 26.7 g of H2S(g) reacts with excess O2(g) to give SO2(g) and H2O(g) and no other products at K
Solution:

39. Free Energy and Chemical Reactions
ΔG = ΔH - T·ΔS
ΔGf° is the standard molar Gibbs function of formation
Because G is a State Property, for a general reaction
a A + b B → c C + d D

40. Calculate ΔG° for the following reaction at 298. 15K
Calculate ΔG° for the following reaction at K. Use Appendix 4 for additional information needed.
3NO(g) → N2O(g) + NO2(g)
Solution From Appendix 4
ΔGf°(N2O) = 104 kJ mol-1
ΔGf°(NO2) = 52
ΔGf° (NO) = 87
ΔG°= 1(104) + 1(52) – 3(87)
ΔG°= − 105 kJ therefore, spontaneous
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41. Where Q is the reaction quotient
The Dependence of Free Energy on Pressure
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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42. At Equilibrium conditions, ΔG = 0
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 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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43. Use ΔG°= - 105 kJ mol -1 (from previous)
Calculate the equilibrium constant for this reaction at 25C.
3NO(g) ↔ N2O(g) + NO2(g)
Strategy
Use - ΔG ° = RT ln K
Use ΔG°= kJ mol -1 (from previous)

44. 3NO(g) ↔ N2O(g) + NO2(g) Solution ΔGrxn°= – 105 kJ mol-1
Use – ΔG ° = RT ln K
Rearrange
ΔGrxn°= – 105 kJ mol-1

45. 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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46. The Temperature Dependence of Equilibrium Constants
Where does this come from?
Recall ΔG = ΔH - T·ΔS
Divide by RT, then multiply by -1
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47. A plot of y = mx + b or ln K vs. 1/T
Notice that this is y = mx + b the equation for a straight line
A plot of y = mx + b or
ln K vs. 1/T
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48. If we have two different Temperatures and K’s (equilibrium constants)or
Now given ΔH and T at one temperature, we can calculate K at another temperature, assuming that ΔH and ΔS are constant over the temperature range
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49. ΔT = − i m K J.H. van’t Hoff (1852-1901) van’t Hoff Factor (i)
The Person Behind the Science
J.H. van’t Hoff ( )
Highlights
Discovery of the laws of chemical dynamics and osmotic pressure in solutions
his work led to Arrhenius's theory of electrolytic dissociation or ionization
The Van't Hoff equation in chemical thermodynamics relates the change in temperature to the change in the equilibrium constant given the enthalpy change.
Moments in a Life
1901 awarded first Noble Prize in Chemistry
van’t Hoff Factor (i)
ΔT = − i m K
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50. The reaction 2 Al3Cl9 (g) → 3 Al2Cl6 (g)
Has an equilibrium constant of 8.8X103 at 443K and a ΔHr°= 39.8 kJmol-1 at 443K. Estimate the equilibrium constant at a temperature of 600K.
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51. P = mg/A V
Infinite-step expansion
2 step expansion
6 step expansion
Expansion (V2 > V1): Work flows out of the system and the Work sign is negative
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53. Compression (V2 < V1): Work is put into system,
P = mg/A V
Compression (V2 < V1): Work is put into system,
ln(V2/V1) is negative and the Work is positive