1. Entropy, Free Energy, and Equilibrium
Chapter 16

2. Spontaneous Physical and Chemical Processes
A waterfall runs downhill
A lump of sugar dissolves in a cup of coffee
At 1 atm., water freezes below 00C and ice melts above 00C
Heat flows from a hotter object to a colder object
The expansion of a gas in an evacuated bulb
Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
18.2

3. spontaneous
nonspontaneous
18.2

4. Does a decrease in enthalpy mean a reaction proceeds spontaneously?
Spontaneous reactions
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH0 = kJ
H+ (aq) + OH- (aq) H2O (l) DH0 = kJ
H2O (s) H2O (l) DH0 = 6.01 kJ
NH4NO3 (s) NH4+(aq) + NO3- (aq) DH0 = 25 kJ
H2O
Spontaneous reactions could be either exothermic or endothermic.
Therefore, to predict the spontaneity of a reaction, another thermodynamic entity must be considered, i.e., entropy.
18.2

5. Entropy (S) is a measure of the randomness or disorder of a system.
DS = Sf - Si
If the change from initial to final results in an increase in randomness
Sf > Si
DS > 0
For any substance, the solid state is more ordered than the liquid state and the liquid state is more ordered than gas state
Ssolid < Sliquid << Sgas
H2O (s) H2O (l)
DS > 0
18.2

6. Thermodynamics
State functions are properties that are determined by the state of the system, regardless of how that condition was achieved.
energy, enthalpy, pressure, volume, temperature
, entropy
Potential energy of hiker 1 and hiker 2 is the same even though they took different paths.
6.7

8. At higher temperatures,
molecular motion increases
randomness increases.
entropy increases.

9. Microstates and Entropy
2 possible arrangements
50 % chance of finding the left empty

10. 4 possible arrangements
25% chance of finding the left empty
50 % chance of them being evenly dispersed

11. 4 possible arrangements
8% chance of finding the left empty
50 % chance of them being evenly dispersed

12. Processes that lead to an increase in entropy (DS > 0)
18.2

13. (a) Condensing water vapor Dec
How does the entropy of a system change for each of the following processes?
(a) Condensing water vapor
Dec
(b) Forming sucrose crystals from a supersaturated solution
Dec
(c) Heating hydrogen gas from 600C to 800C
Inc
(d) Subliming dry ice
Inc
18.2

14. (c) Sugar dissolves in water Inc
How does the entropy of a system change for each of the following processes?
Inc
(a) A solid melts
(b) A liquid freezes
Dec
(c) Sugar dissolves in water
Inc
(d) A vapor condenses to a liquid
Dec
18.2

15. First Law of Thermodynamics
Energy can be converted from one form to another but energy cannot be created or destroyed.
Second Law of Thermodynamics
The entropy of the universe increases in a spontaneous process and remains unchanged in an equilibrium process.
Spontaneous process:
DSuniv = DSsys + DSsurr > 0
Equilibrium process:
DSuniv = DSsys + DSsurr = 0
18.3

16. Entropy Changes in the System (DSsys)
The standard entropy of reaction (DS0 ) is the entropy change for a reaction carried out at 1 atm and 250C.
rxn
aA + bB cC + dD
DS0
rxn
dS0(D)
cS0(C) = [ + ] -
bS0(B)
aS0(A)
DS0
rxn
nS0(products) = S
mS0(reactants) -
What is the standard entropy change for the following reaction at 250C? 2CO (g) + O2 (g) CO2 (g)
S0(CO) = J/K•mol
S0(O2) = J/K•mol
S0(CO2) = J/K•mol
DS0
rxn
= 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]
DS0
rxn
= – [ ] = J/K•mol
18.3

18. Entropy Changes in the System (DSsys)
When gases are produced (or consumed)
If a reaction produces more gas molecules than it consumes, DS0 > 0.
If the total number of gas molecules diminishes, DS0 < 0.
If there is no net change in the total number of gas molecules, then DS0 may be positive or negative BUT DS0 will be a small number.
What is the sign of the entropy change for the following reaction? 2Zn (s) + O2 (g) ZnO (s)
The total number of gas molecules goes down, and a gas is converted to a solid, DS is negative.
18.3

19. Predict the sign of the entropy change for the following reactions?-
a) 3 O2 (g) O3 (g) +
b) 2 NaHCO3 (s) Na2CO3 (s) + H2O (l) + CO2 (g) -
c) N2 (g) + O2 (g) 2 NO (g) +
d) I2 (s) 2 I (g) -
e) 2 Hg (l) + O2 (g) 2 HgO (s)
18.3

21. Entropy Changes in the Surroundings (DSsurr)
Exothermic Process
DSsurr > 0
Endothermic Process
DSsurr < 0
Dssurr is proportional to - DHsys
DSsurr =
- DHsys T
18.3

22. Is the following reaction spontaneous at 25°C?
N2(g) + 3H2(g) 2NH3(g); DH° = kJ
DSuniv = DSsys + DSsurr > 0
Spontaneous process:
DSsys = -199 J/K
= [2 x S0(NH3) ] – [S0(N2) + 3 x S0(H2)]
DHsys = kJ
DSsurr =
- DHsys T =
- ( J)
298 K
= 311 J/K
DSuniv = DSsys + DSsurr = -199 J/K J/K = 112 J/K > 0
Process is spontaneous, but may be very slow.

23. Third Law of Thermodynamics
The entropy of a perfect crystalline substance is zero at the absolute zero of temperature.
18.3

24. For a constant-temperature process:
Gibbs Free Energy
Spontaneous process:
DSuniv = DSsys + DSsurr > 0
DSuniv = DSsys - > 0
- DHsys T
-TDSuniv = DHsys - TDSsys < 0
For a constant-temperature process:
Gibbs free energy (G)
DG = DHsys -TDSsys
DG < The reaction is spontaneous in the forward direction.
DG > The reaction is nonspontaneous as written. The
reaction is spontaneous in the reverse direction.
DG = The reaction is at equilibrium.
18.4

25. DG0 of any element in its stable form is zero.
The standard free-energy of reaction (DG0 ) is the free-energy change for a reaction when it occurs under standard-state conditions.
rxn
aA + bB cC + dD
DG0
rxn
dDG0 (D) f
cDG0 (C) = [ + ] -
bDG0 (B)
aDG0 (A)
DG0
rxn
nDG0 (products) f = S
mDG0 (reactants) -
Standard free energy of formation (DG0) is the free-energy change that occurs when 1 mole of the compound is formed from its elements in their standard states. f
DG0 of any element in its stable form is zero. f
18.4

28. 2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)
What is the standard free-energy change for the following reaction at 25 0C?
2C6H6 (l) + 15O2 (g) CO2 (g) + 6H2O (l)
DG0
rxn
nDG0 (products) f = S
mDG0 (reactants) -
DG0
rxn
6DG0 (H2O) f
12DG0 (CO2) = [ + ] -
2DG0 (C6H6)
DG0
rxn =
[12(–394.4) + 6(–237.2)] – [2(124.5)] = kJ
Is the reaction spontaneous at 25 0C?
DG0 = kJ
< 0
spontaneous
18.4

30. a) CaCO3 (s) CaO (s) + CO2 (g)
Calculate the standard free-energy change for the following reactions at 25 0C and determine whether the reactions are spontaneous.
a) CaCO3 (s) CaO (s) + CO2 (g)
b) 2 NaHCO3 (s) Na2CO3 (s) + H2O (l) + CO2 (g)
c) 2HgO (s) 2Hg (l) + O2 (g)
d) 2C2H4 (g) + 6O2 (g) CO2 (g) + 4H2O (l)
18.4

31. b) [-1044.4+ -237.2+ -394.4] - [ 2 NaHCO3 (s) ] =
Calculate the standard free-energy change for the following reactions at 25 0C and determine whether the reactions are spontaneous.
[( ) ] – [( ) ] =
b) [ ] - [ 2 NaHCO3 (s) ] =
c) [2 (0) + 0] - [2 (-58.49)] =
d) [4 (-394.4) + 4 (237.2) ] - [2C2H4 (g) + 0) ] =
18.4

32. DG = DH - TDS
18.4

33. Temperature and Spontaneity of Chemical Reactions
CaCO3 (s) CaO (s) + CO2 (g)
DH0 = kJ
DS0 = J/K
DG0 = DH0 – TDS0
At 25 0C, DG0 = kJ
nonspontaneous
T at which DG0 = 0 is when
the reaction starts to
become spontaneous
DG0 = 0 at 835 0C
18.4

34. To what temperature must MgCO3 must be heated to decompose it
To what temperature must MgCO3 must be heated to decompose it? Is this higher or lower than the temperature to decompose CaCO3?
MgCO3 (s) MgO (s) + CO2 (g)
DG0 = DH0 – TDS0 = 0
T = DH0
DS0
18.4

35. Gibbs Free Energy and Phase Transitions
At phase transitions, the system is at equilibrium (DG0 = 0)
DG0 = 0
= DH0 – TDS0
H2O (l) H2O (g)
DSvap = T
DHvap =
40.79 kJ
373 K
= 109 J/K
18.5

36. The molar heat of vaporization of argon is 6
The molar heat of vaporization of argon is 6.3 kJ/mol, and argon’s boiling point is -186°C. Calculate the entropy change of vaporization.
Ar (l) Ar (g)
DH vap
DSvap = T
DSvap = 6300 J
87 K
= 72.4 J/K
18.5

37. Gibbs Free Energy and Chemical Equilibrium
DG = DG0 + RT lnQ
R is the gas constant (8.314 J/K•mol)
T is the absolute temperature (K)
Q is the reaction quotient
At Equilibrium
DG = 0
Q = K
0 = DG0 + RT lnK
DG0 = - RT lnK
18.5

38. DG0 < 0
DG0 > 0
18.5

39. DG0 = - RT lnK
18.5

41. a) CaCO3 (s) CaO (s) + CO2 (g)
Using thermodynamic data, calculate the equilibrium constants for the following reactions at 25°C.
DG0 = - RT lnK
a) CaCO3 (s) CaO (s) + CO2 (g)
b) 2 NaHCO3 (s) Na2CO3 (s) + H2O (l) + CO2 (g)
c) 2HgO (s) 2Hg (l) + O2 (g)
d) 2C2H4 (g) + 6O2 (g) CO2 (g) + 4H2O (l)
18.5

43. BaF2 (s) Ba+2 (aq) + 2 F- (aq)
Calculate DG0 for the following process at 25°C. The Ksp of BaF2 is 1.7 x 10-6.
BaF2 (s) Ba+2 (aq) + 2 F- (aq)
Calculate DG0 for the following process at 25C. The Ksp of CaCO3 is 8.7 x 10-9.
CaCO3 (s) Ca+2 (aq) + CO3-2 (aq)
18.5

44. CaCO3 (s) Ca+2 (aq) + CO3-2 (aq)
Calculate the Ksp at 25°C for the following reaction, using standard free energy of formation (DG0f ) values.
CaCO3 (s) Ca+2 (aq) + CO3-2 (aq)
18.5