le chatelier’s principle

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, K = 10.38

K = [NO2]eq2

[N2O4]eq

Q =

K

[NO2]eq2

[N2O4]i

Q <

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]

K = [NO2]2

[N2O4] Q K>

decrease volume decrease nincrease volume increase n

Δn = 0 no effect of pressure

= (3.0)2

(0.87)

= 10.3 Q = (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)

Kp =

P(1.0 L)=

P = 119 atm

= (119)

/ 22= 242

(3/4.87)

KP 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

Calculationsreaction table ICE table

2HI (g)

Change

Equilibrium

H2 (g) + I2 (g)

Initial

at 453oC, at equilibrium,

calculate K

[HI] (M)[H2] (M) [I2] (M)

0.50 0.50 0.00

- x - x +2x

0.50 – x 0.50 – x 2x

K = [HI]2eq

[I2]eq[H2]eq

= (2x)2

(0.50–x)

= 0.50 – x x = 0.393

(0.786)2

(0.107)2

= = 54.3

[H2] = 0.107 M

(0.50–x)

Calculations

Change

Equilibrium

Initial

[H2] (M) [I2] (M) [HI] (M)

.623 .414 .224

-x +2x- x

.623 - x .414 - x .224 + 2x

2HI (g)H2 (g) + I2 (g)K = 54.3

Q = (.224)2

(.623) (.414)

= .195 < KK = 54.3 = (.224 + 2x)2

(.623 – x)(.414 – x)

50.3x2 - 57.2x+ 13.96 = 0ax2 bx c

x = -b ± b2 – 4ac2a

x = .782x = .355