Question

One mole of N2O4 is added to a 1.00L container at equilibrium consisting of a mixture of gases, .20 mole NO2 and 2.00 mole N2O4 at equilibrium and 50 degrees Celsius.

a. Calculate the equilibrium constant value at equilibrium before stress is applied.

b. Use ICE table. Determine the concentrations at the reestablished equilibrium. Thank you.

Answer 1

the reaction:

N2O4 = 2NO2

the Kc:

Kc = [NO2]^2/[N2O4]

find initial concentrations:

[N2O4] = 2/1 = 2 M

[NO2] = 0.2/1 = 0.2 M

calculate Kc

Kc = [NO2]^2/[N2O4] = (0.2^2)/(2) = 0.02

after addint 1 M of N2O4:

ICE TABLE:

(I)

intially (after equilbirum is interrupted)

[N2O4] = 2+1 = 3 M

[NO2] = 0.2 M

(C) the change in concentrations:

[N2O4] = -x

[NO2] = +2x

(E) in equilibrium:

[N2O4] = 3 - x

[NO2] = 0.2+2x

substitute this in K

Kc = [NO2]^2/[N2O4]

0.02 = (0.2+2x)^2 / (3-x)

0.02 *3 - 0.02 x = 0.2*0.2 + 2*0.2x + 4x^2

4x^2 + (2*0.2+0.02)x +  0.2*0.2 -0.02 *3 = 0

4x^2 + 0.42x -0.02 = 0

x = 0.03556

[N2O4] = 3 - x = 3-0.03556 = 2.96444

[NO2] = 0.2+2x = 0.2+2*0.03556 = 0.27112

proof of K = 0.27112^2/ (2.96444) = 0.0247959 which is pretyt near to 0.02