Question

For the reaction shown here, Kc = 0.513 at 500 K.
N2O4(g)⇌2NO2(g)

Part A

If a reaction vessel initially contains an N2O4 concentration of 5.50×10⁻² M at 500 K, what are the equilibrium concentrations of N2O4 and NO2 at 500 K?

[N2O4], [NO2] =

Answer 1

ICE Table:

[N2O4] [NO2]

initial 0.055

change -1x +2x

equilibrium 0.055-1x +2x

Equilibrium constant expression is

kc = [NO2]^2/[N2O4]

0.513 = (4*x^2)/((5.5*10^-2-1*x))

2.822*10^-2-0.513*x = 4*x^2

2.822*10^-2-0.513*x-4*x^2 = 0

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

a = -4

b = -0.513

c = 2.822*10^-2

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 = 0.7146

roots are :

x = -0.1698 and x = 4.154*10^-2

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

x = 4.154*10^-2

At equilibrium:

[N2O4] = 0.055-1x = 0.055-1*0.04154 = 0.01346 M

[NO2] = +2x = +2*0.04154 = 0.08309 M

1.35*10^-2 M, 8.31*10^-2 M