Question

Consider the reaction N2O4 (g) ? 2 NO2 (g). At equilibrium, a 2.00-L reaction vessel contains NO2 at a pressure of 0.269 atm and N2O4 at a pressure of 0.500 atm. The reaction vessel is then compressed to 1.00 L. What will be the pressures of NO2 and N2O4 once equilibrium is re-established?

Answer 1

Initial equilibrium pressure of,

[NO2] = 0.269 atm

[N2O4] = 0.5 atm

Kp = [NO2]^2/[N2O4]

     = (0.269)^2/(0.5)

     = 0.145

When the volume was reduced to 1 L,

new pressure of,

[NO2] = 0.269 atm x 2 L/1 L = 0.538 atm

[N2O4] = 0.5 atm x 2 L/1 L = 1 atm

let x be the change at equilibrium,

then,

        N2O4 <==> 2NO2

I         1.0             0.538

C       +x                -2x

E       1+x          0.538-2x

So,

0.145 = (0.538-2x)^2/(1-x)

0.145 - 0.145x = 4x^2 - 2.152x + 0.29

4x^2 - 2.007x + 0.145 = 0

x = 0.09 atm

The new equilibrium pressure for,

[NO2] = 0.538 - 2 x 0.09 = 0.358 atm

[N2O4] = 1.0 + 0.09 = 1.09 atm