Question

For the reaction N2O4 <—> 2NO2

If [N2O4]initial= 0.25M and K= 4.5 what are the concentration of reactants and products at equilibrium?

Answer 1

ICE Table:

[N2O4] [NO2]

initial 0.25

change -1x +2x

equilibrium 0.25-1x +2x

Equilibrium constant expression is

Kc = [NO2]^2/[N2O4]

4.5 = (4*x^2)/((0.25-1*x))

1.125-4.5*x = 4*x^2

1.125-4.5*x-4*x^2 = 0

This is quadratic equation (ax^2+bx+c=0)

a = -4

b = -4.5

c = 1.125

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 38.25

roots are :

x = -1.336 and x = 0.2106

since x can't be negative, the possible value of x is

x = 0.2106

At equilibrium:

[N2O4] = 0.25-1x = 0.25-1* 0.2106 = 0.0394 M

[NO2] = +2x = +2* 0.2106 = 0.421 M

Answer:

[N2O4] = 0.0394 M

[NO2] = 0.421 M