1. Physical Equilibrium
ratevaporization
ratecondensation =
equilibrium
Go =
Ho - TSo = 0
Hof products
- Hof products
T =
Soproducts
- Soproducts
molecules with high K.E. leave liquid
vaporization
gas phase molecules collide with surface
condensation

2. 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
-20 F
∆G298 = .105 kJ
∆G244 = kJ

3. Chemical Equilibrium
kf [A][B]
= kr [C] A
+ B
 C
rate = kf
[A]
[B] kf
[C] =
= K A
+ B
rate = kr
[C]
C  kr
[A] [B]
equilibrium constant

4. Law of Mass Action aA
+ bB 
cC + dD
K =
[C]c
[D]d
[A]a
[B]b 2 O3 3 O2
at 300 K
Kc =
K = 2.6 x 1055
= [O2]  3
[O3] 2
if [O2] = 0.50 M
at equilibrium
2.6 x 1055 =
(0.50)3
[O3] =
6.9 x M
[O3]2

5. Law of Mass Action aA
+ bB 
cC + dD
K =
[C]c
[D]d
[A]a
[B]b 2 O3 3 O2
at 300 K
Kc =
K = 2.6 x 1055
= [O2]  3
[O3] 2 1 3 O2  2 O3
Kc =
= [O3] 2
2.6 x 1055
[O2]
large K
excess product 3
Kc = 3.5 x 10-56
small K 
excess reactant

6. Law of Mass Action aA
+ bB 
cC + dD
K =
[C]c
[D]d
K =
PCc
PDd
PAa
PBb
[A]a
[B]b 3 O2  O3 2
K =
3.5 x 10-56
( x 300 K)-1
Kc = 3.5 x 10-56
K =
1.4 x 10-57 Δn K
= M (mol/L)
gases P
K = Kc
(RT) n
= P
PV=nRT
Δn =
mol gas product - V RT
mol gas reactants

7. Law of Mass Action
CO (g)
+ H2O (g) 
H2 (g) +
CO2 (g)
at 830oC
equilibrium concentrations
[CO] = 0.20 M
(0.30)
(1.36)
K =
= 5.10
[H2O] = 0.40 M
(0.20)
(0.40)
[H2] = 0.30 M
[CO2] = 1.36 M
KC = K
homogeneous equilibrium
all in one phase
gas

8. Δn
K = Kc
(RT)
2 NO (g)
+ O2 (g)
 2NO2 (g)
at 230oC
equilibrium concentrations
[NO] = M
[O2] = M
Kc =
(15.5)2
= 6.44 x 105
[NO2] = 15.5 M
(0.0542)2(0.127)
K =
6.44 x 105
( )
(503 K) -1
= 1.56 x 104
K and Kc are
dimensionless
no units

9. Heterogeneous Equilibria
s, l, g
3 phases
CaCO3 (s) 
CaO (s) +
CO2 (g)
Kc = [CO2] [CaO]
= [CO2]
[CaCO3] L g g
mol
[CaO]
mol L
density
M.W. = ÷
constants
solids don’t appear in K
pure liquids don’t appear
H2O as a solvent

10. Predicting the direction of a reaction
H2 (g)
+ I2 (g) 
2 HI (g)
at 430oC
K = 54.3
start with:
0.243 mol H2
0.146 mol I2
1.0 L flask
1.98 mol HI
[HI]20 =
= 111
= Q
[H2]0
[I2]0
(0.243)
(0.146)
111
 54.3
not at equilibrium
0 = initial

11. Reaction Quotient, Q
Q =
[products]0
K =
[products]eq
[reactants]0
[reactants]eq
Q > K
reverse
reaction proceeds
Q = K
system at equilibrium
Q < K
forward
reaction proceeds

12. Reaction Quotient, Q
N2 (g)
+ 3H2 (g)
 2 NH3 (g)
at 200o C, Kc = 1085
[N2]0
= M
[H2]0
= 9.17 x 10-3 M
[NH3]0
= 1.83 x 10-4 M
Q =
(1.83 x 10-4)2
= 0.611
(9.17 x 10-3)3
(0.0711)

13. “insoluble” salts
Mg(OH)2 (s) 
Mg2+
+ 2 OH-
Ksp =
[Mg2+]eq
[OH-]2eq
= 1.02 x 10-11
1.02 x =
(x)
(2x) 2
= 4x3
x = 1.37 x 10-4
Q =
[OH-]20
[Mg2+]0
add H+
reacts with OH-