1. Review Ssystem + Ssurroundings = Suniverse > 0
for spontaneous processes
Ssurroundings
= - Hsys T
energetic disorder
Ssystem
= qr T
positional disorder

2. - Ssystem + Ssurroundings = Suniverse > 0
for spontaneous processes
at 298 K
H2O(s)  H2O(l)
spontaneous
Ssurroundings
= - H T
H
a) >
b) <
Ssurr
< 0
Ssurr does not contribute to spontaneity

3. - + Ssystem + Ssurroundings = Suniverse for spontaneous processes
at 298 K
H2O(s)  H2O(l)
spontaneous
solid  liquid
increasing disorder S
> 0 S
positional disorder
S does contribute to spontaneity

4. - + Ssystem + Ssurroundings = Suniverse H2O(s)  H2O(l) S - H / T
solid  liquid
endothermic
positive
negative
At high T
a) spontaneous
b) non-spontaneous
Suniv
> 0
At low T
non-spontaneous
Suniv < 0

5. Suniverse= Ssystem+ Ssurroundings > 0
2NO2 (g)
 N2O4 (g)
NO2 - brown, toxic gas
N2O4 - colorless gas
Ssurr = - H / T H
= Hof N2O4
- 2 Hof NO2
= (9.66) - 2
(33.5)
-58 kJ mol-1 =
Ssurr 0
>
favors spontaneity

6. Suniverse= Ssystem+ Ssurroundings > 0- +
Suniverse= Ssystem+ Ssurroundings > 0
2NO2 (g)  N2O4 (g)
S =
Soproducts
- Soreactants
So (J mol-1 K-1)
N2O4
NO2
a) >
b) < So
N2O4 So
NO2
304.18
239.95 S
= [304.18
- (239.95)] 2
= J/mol K
2 mol gas  1 mol gas
non-spontaneous

7. Suniverse= Ssystem+ Ssurroundings > 0- +
Suniverse= Ssystem+ Ssurroundings > 0
2NO2 (g)  N2O4 (g) S
-H / T
J mol-1 K-1
+58 kJ mol-1
T(K)
At high T
a) spontaneous
b) non-spontaneous
At low T
spontaneous

8. G = H - TS Free Energy G
Ssystem+ Ssurroundings = Suniverse > 0
- H T
> 0
- T S
= Suniverse
H -
TS
= - TSuniverse
< 0
G 
H - TS G
= H
- TS

9. elements in standard states
G = H - TS
G < 0
spontaneous reaction
G > 0
non-spontaneous reaction
G = 0
equilibrium
G = wmax
maximum useful work
G is an extensive State function
Gof = 0
elements in standard states
Gorxn =
Gof products
- Gof reactants

10. G = H - TS calculate Go for: CH4 (g) + 2O2 (g)  CO2 (g)
+ 2H2O (l)
Gorxn =
[Gof CO2 (g) ] 2
(Gof H2O (l))
- [ ]
Gof CH4 (g)
+ 2(Gof O2 (g))
= kJ
a) >
b) <
Sorxn
Gorxn = Horxn - TSorxn
= [-892 kJ] -
[ ]
(298K)
(-242 J/K)
= -819 kJ

11. Rubber band Thermodynamics
State 1 = relaxed
State 2 = stretched
go from State 1 to State 2
What is sign of Go + -
What is sign of Ho
a) + b) -
What is sign of So
Go = Ho - TSo + - -

12. Go = Ho - TSo Ho So Go + - always positive - + always negative
always positive
- +
always negative
negative
a) high T
b) low T
+ +
positive
low T
a) high T
b) low T
- -
negative
positive
high T

13. Equilibrium Go = Ho - TS Go = 0 phase changes chemical reactions
TSo = 0
Ho
Ho
= So T
Hof prod -
Hof react
T =
Ho
= T
___________________________
Soprod -
Soreact
So
2NO2  N2O4
-58 kJ
= 331 K
J/K

14. Napoleon - 1812 Snwhite Sngrey tin buttons ΔHof (kJ/mol) So (J/mol K)
white tin
0.0
51.55
grey tin
-2.1
44.14
ΔHo =
-2.1
- 0.0
= -2.1 kJ
Snwhite
Sngrey 
ΔSo =
44.14
= -7.4 J/mol K
T =
-2100 J
= 283 K
= 10oC
-7.4 J/K
∆G298 = .105 kJ
∆G233 = kJ