Mg(OH)2 (s) Mg2+ + 2 OH-

K = [OH-]eq [Mg2+]eq

add H+ reacts with OH-

Q = [OH-]0 [Mg2+]0

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

Concentration

Fe3+ (aq) + SCN- (aq) FeSCN2+

add Fe(NO3)3 add reactant

add NaSCN add reactant

add C2O42- remove Fe2+

K = [FeSCN2+]

[Fe3+] [SCN-]

at equilibrium change

Q = [FeSCN2+]

[Fe3+] [SCN-]

Q K<

Q K>

ratef = kf [Fe3+] [SCN-]

Pressure

N2O4 (g) NO2 (g)2

add N2O4

at 25oC, Kc = 10.38

Kc = [NO2]eq2

[N2O4]eq

Qc =

Kc

[NO2]eq2

[N2O4]i

Qc <

increase P by adding reactant or product

Pressure

N2O4 (g) NO2 (g)2

decrease volume [N2O4] = mol N2O4

V

increase [N2O4]

[NO2] = mol NO2

V

increase [NO2]

Kc = [NO2]2

[N2O4] Qc Kc>

decrease volume decrease nincrease volume increase n

Δn = 0 no effect of pressure

= (3.0)2

(0.87)

= 10.3 Qc =(6.0)2 = 11.9

(1.74)

Pressure

N2O4 (g) NO2 (g)2 add inert gas

1.00 M Ar

2PNO

increase P = 3.0 M

2 4 PN O

= 0.87 M

= 3.0 mol/L

= 0.87 mol/L

at 298 K PV = nRT

P(1.0 L) = (3.87)(.08206)(298)

P = 95 atm

2PNO

[NO2][N2O4]

=(3/3.87)

= (.87/3.87)

K =

P(1.0 L)=

P = 119 atm

= (119)

/ 22= 242

(3/4.87)

K unchanged

(95) =73 atm

(95) =22atm

(73)2

(4.87)(.08206)(298)

=73atm

Temperature

treat heat reactant product

raising T adding heat as reactant

lowering T removing heat as product

endothermic exothermic

N2O4 (g) NO2 (g)2

ΔH > 0 ΔH < 0

ΔH > 0

ΔH < 0

ΔH = 58.0 kJ

changes K

heat + +heatheat