1. Mg(OH)2 (s) 
Mg2+
+ 2 OH-
K =
[OH-]eq
[Mg2+]eq
Q =
[OH-]0
[Mg2+]0
add H+
reacts with OH-

2. Le Chatelier’s Principle
equilibrium
balance
forward reaction
reverse reaction
changes in experimental conditions
disturb balance
equilibrium shifts
counteract disturbance
concentration
pressure
(gas phase)
temperature

3. Concentration
Fe3+ (aq)
+ SCN- (aq)
FeSCN2+ 
add Fe(NO3)3
add reactant Q
< K
add NaSCN
add reactant
add C2O42-
remove Fe2+ Q
> K
at equilibrium
change
K =
[FeSCN2+]
Q =
[FeSCN2+]
[Fe3+]
[SCN-]
[Fe3+]
[SCN-]
ratef = kf
[Fe3+]
[SCN-]

4. Pressure
N2O4 (g)  2
NO2 (g)
at 25oC, Kc = 10.38
increase P
by adding reactant or product
Kc =
[NO2]eq2
add N2O4
Qc =
[NO2]eq2
[N2O4]eq
[N2O4]i Qc
< Kc

5. Pressure
N2O4 (g)  2
NO2 (g)
decrease volume
[N2O4]
= mol N2O4
increase [N2O4] V
[NO2]
= mol NO2
increase [NO2] V
(6.0)2
= 11.9
Kc =
[NO2]2
= (3.0)2
Qc =
= 10.3
(1.74)
[N2O4]
(0.87) Qc
> Kc
decrease volume
decrease n
increase n
increase volume
Δn = 0
no effect of pressure

6. Pressure N2O4 (g)  2 NO2 (g) add inert gas 1.00 M Ar [NO2] = 3.0 M
= 3.0 mol/L
increase P
[N2O4]
= 0.87 M
= 0.87 mol/L
PV = nRT
at 298 K
P(1.0 L)=
(4.87)
(.08206)
(298) P
(1.0 L)
= (3.87)
(.08206)
(298)
P = 119 atm
P = 95 atm 2
PNO
=(3/3.87)
(95)
=73 atm 2
PNO =
(3/4.87)
(119)
=73atm 2 4
PN O
= (.87/3.87)
(95)
=22atm K
unchanged
K =
(73)2
/ 22
= 242

7. Temperature
changes K
heat
heat +
N2O4 (g)  2
NO2 (g)
+heat
ΔH = 58.0 kJ
treat heat
reactant
product
endothermic
exothermic
ΔH > 0
ΔH < 0
raising T
adding heat as reactant
ΔH > 0
lowering T
removing heat as product
ΔH < 0