1. Spontaneity, Entropy, and Free Energy
Chapter 17
Spontaneity, Entropy, and Free Energy

2. 17.1 Spontaneous Processes and Entropy
Entropy and the Second Law of Thermodynamics
The Effect of Temperature on Spontaneity
17.4 Free Energy
17.5 Entropy Changes in Chemical Reactions
17.6 Free Energy and Chemical Reactions
17.7 The Dependence of Free Energy on Pressure
17.8 Free Energy and Equilibrium
17.9 Free Energy and Work
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3. ex. 2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g)
(1) spontaneous process :
it occurs without outside intervention.
(2) spontaneous rxn & speed
 thermodynamics vs kinetics
the science of E transfer predict occur
study of the rates of rxn why somes are fast or slow? vs
ex. 2Na(s) + 2H2O(l) NaOH(aq) + H2(g)
Product favored ; reactant favored (spontaneous rxn)

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5. (3) why product - favored ?
DH <  energy  ? (exothermic)
How about H2O(s) → H2O(l) DH > 0 at 0℃
(4) something else !  Entropy ( S )
S = the driving force for a spontaneous process is an increase in the entropy of the universe.
= a measure of randomness or disorder of a system the great the randomness, the greater S.
= the number of arrangements  probability the higher probability, the higher S

6. The Microstates That Give a Particular Arrangement (State)
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7. ( S = 0 for a perfect crystal at 0 K) (6) DS = Sf - Si
(5) S  0 H , E ?
( S = 0 for a perfect crystal at 0 K)
(6) DS = Sf - Si
(7) generalization :
a) S(g) >> S(l) > S(s)
b) S : more complex molecules > simpler
(ex) : CH3 CH2 CH3 > CH3 CH3 > CH4
c) S : ionic solid (weaker force) > ionic solid (stronger force)
(ex) : NaF(s) > MgO(s)
d) liq and solid dissolve in a solvent DS > 0
gas dissolve in a solvent DS < 0

8. Predict the sign of S for each of the following, and explain:
Concept Check
Predict the sign of S for each of the following, and explain:
The evaporation of alcohol
The freezing of water
Compressing an ideal gas at constant temperature
Heating an ideal gas at constant pressure
Dissolving NaCl in water + –
a) + (a liquid is turning into a gas)
b) - (more order in a solid than a liquid)
c) - (the volume of the container is decreasing)
d) + (the volume of the container is increasing)
e) + (there is less order as the salt dissociates and spreads throughout the water)
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9. In any spontaneous process, there is always an increase in the entropy of the universe
DSuniv. > for spontaneous rxn
DSuniv. = DSsys. + DSsurr.
System: rxn R P
DSuniv. > 0 rxn →
DSuniv < 0 rxn ←
DSuniv = 0 rxn at equilibrium

10. a) sign ( ＋ or －) depend on “heat flow”
ex) H2O(l) ─→ H2O(g)
(1) System
(l) → (g) 18 ml (1 mol) → 31 L
DSsys. > 0
(2) Surrounding
a) sign ( ＋ or －) depend on “heat flow”
exothermic process  DSsurr > 0
endothermic process  DSsurr < 0

11. b) magnitude depend on the temp.
T  100 oC for rxn H2O(l) ─→ H2O(g)
(3) DSuniv = DSsys + DSsurr
const. T & P
qp = | DH |

12. The sign of ΔSsurr depends on the direction of the heat flow.
The magnitude of ΔSsurr depends on the temperature.
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13. DG < 0 R → P spontaneou DG > 0 R ← P nonspontaneous
G is a thermodynamic function that gives information about the spontaneity of the system.
G = H - TS (for system)
DG = DH - TDSsys (const. T & P)
DG < R → P spontaneou DG > R ← P nonspontaneous
DG = R  P equilibrium

14. DG = DH - TDSsys
DHo < 0
DSo > 0
DHo > 0
DSo < 0
at All temp.
at High temp.
spontaneous
at Low temp.
No Way !!
Table

15. A liquid is vaporized at its boiling point. Predict the signs of: w q
Concept Check
A liquid is vaporized at its boiling point. Predict the signs of: w q H S
Ssurr G
Explain your answers. – +
As a liquid goes to vapor, it does work on the surroundings (expansion occurs). Heat is required for this process. Thus, w = negative; q = H = positive. S = positive (a gas is more disordered than a liquid), and Ssurr = negative (heat comes from the surroundings to the system); G = 0 because the system is at its boiling point and therefore at equilibrium.
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16. Exercise
The value of Hvaporization of substance X is 45.7 kJ/mol, and its normal boiling point is 72.5°C.
Calculate S, Ssurr, and G for the vaporization of one mole of this substance at 72.5°C and 1 atm.
S = 132 J/K·mol
Ssurr = -132 J/K·mol
G = 0 kJ/mol
S = 132 J/K·mol
Ssurr = -132 J/k·mol
G = 0 kJ/mol
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17. Concept Check
Gas A2 reacts with gas B2 to form gas AB at constant temperature and pressure. The bond energy of AB is much greater than that of either reactant.
Predict the signs of:
H Ssurr S Suniv
Explain.
Since the average bond energy of the products is greater than the average bond energies of the reactants, the reaction is exothermic as written.
Thus, the sign of H is negative; Ssurr is positive; S is close to zero (cannot tell for sure); and Suniv is positive. –
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18.  4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
(1) ex.
 N2 (g) + 3H2(g) → 2NH3(g)
DS < 0
 H2(g) → 2H(g)
 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
(2) enthalpy (H) (at const. P)
DH determines a rxn exothermic
endothermic

19. (3) free energy (G) (at const. T & P)
spontaneous
DG determines a rxn nonspontaneous
equilibrium
(4) S → absolute values DST1→T2
S = 0 for a perfect crystal at 0 oK
3rd law of thermodynamics
(5) standard entropy So (298 K & 1 atm)

20. (1) DHo ： measured by calorimeter
(2) DGo ： standard free energy change
25 ℃ , 1 atm
 The more negative the value of DGo , the further a rxn will go to the right.
 Can be calculated by DGo = DHo - T DSo
ex. 17.9
 DGorxn can be calculated by using Hess’s law ex

21. thermodynamically unstable rxn ← when high temp.
 DGorxn can be calculated using Hess’s law
kinetically stable
thermodynamically unstable
rxn ← when high temp.

22.  DGof = standard free energy of formation
DGof = 0 for elements
ex & 17.12

23. A stable diatomic molecule spontaneously forms from its atoms.
Concept Check
A stable diatomic molecule spontaneously forms from its atoms.
Predict the signs of:
H° S° G°
Explain.
The reaction is exothermic, more ordered, and spontaneous.
Thus, the sign of H is negative; S is negative; and G is negative.
– – –
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24. Free Energy and Pressure
G = G° + RT ln(P) or
ΔG = ΔG° + RT ln(Q)
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25. (1) Slarge V > Ssmall V  Slow P > Shigh P G = Go + RT ln(P)
Go : the gas at the P = 1 atm
G : the gas at the P = P
ex. N2(g) + 3H2(g) → 2NH3(g)
DG = Gproducts – Greactants
= 2GNH3 – (GN2 + 3GH2)
where GNH3 = GoNH3 + RT ln(PNH3)
GN2 = ?
GH2 = ?

26. (2) The meaning of DG for a chemical rxn.
DG = (2GoNH3 – GoN2 – 3GoH2)
+ RT [2ln(PNH3) – ln(PN2) – 3ln(PH2)]
= DGo + RTln[(PNH3)2/ (PN2)(PH2)3]
DG = DGo + RTlnQ
(2) The meaning of DG for a chemical rxn.

27. The equilibrium point occurs at the lowest value of free energy available to the reaction system.
ΔG = 0 = ΔG° + RT ln(K)
ΔG° = –RT ln(K)
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28. (1) Fig & Fig. 17.9
A(g)  B (g)
at equilibrium : Gproducts = Greactants
 DG = GP – Gr = 0
& DG = DGo + RTlnQ
i.e. DG = DGo + RTlnk = 0 (equilibrium)
DGo = –RTlnk
 DGo = 0  k = 1
 DGo < 0  k > 1 products favor
 DGo > 0  k < 1 reactants favor

29. (2) Temp. dependence DGo = –RTlnK = DHo – TDSo y = mx + b
(ln K) vs. (1/T)

30. DGo = –33.3 kJ at 25oC ex. 17.14: N2 (g) + 3H2(g)  2NH3(g)
(a) PNH3 = 1.00, PN2 = 1.47 atm,
PH2 = 1.0010-2 atm
predict the rxn direction to reach equilibrium?
Solution : DG = DGo + RTlnQ

31. Solution : DG = DGo + RTlnQ DGo = DHo – TDSo
ex :
Calculate k for the follow rxn at 25 oC
4Fe (s) + 3O2(g)  2Fe2O3(s)
DHof (kJ/mol)
-826
DSof (J/kmol) 27
205 90
Solution : DG = DGo + RTlnQ
DGo = DHo – TDSo

32. ex.
Calculate k for
k = 3.7  10-6 at T = 900 K
k = ? at T = 550 K

33. Change in Free Energy to Reach Equilibrium
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34. Qualitative Relationship Between the Change in Standard Free Energy and the Equilibrium Constant for a Given Reaction
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35. Maximum possible useful work obtainable from a process at constant temperature and pressure is equal to the change in free energy.
wmax = ΔG
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36. All real processes are irreversible.
Achieving the maximum work available from a spontaneous process can occur only via a hypothetical pathway. Any real pathway wastes energy.
All real processes are irreversible.
First law: You can’t win, you can only break even.
Second law: You can’t break even.
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