Question

Calculate the equilibrium concentrations of N2O4 and NO2 at 25 ∘C in a vessel that contains an initial N2O4 concentration of 0.0365 M . The equilibrium constant Kc for the reaction N2O4(g)⇌2NO2(g) is 4.64×10−3 at 25 ∘C.

Express your answers using four decimal places separated by a comma.

Answer 1

ICE Table:

[N2O4] [NO2]

initial 0.0365

change -1x +2x

equilibrium 0.0365-1x +2x

Equilibrium constant expression is

Kc = [NO2]^2/[N2O4]

0.00464 = (4*x^2)/((3.65*10^-2-1*x))

1.694*10^-4-4.64*10^-3*x = 4*x^2

1.694*10^-4-4.64*10^-3*x-4*x^2 = 0

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

a = -4

b = -4.64*10^-3

c = 1.694*10^-4

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 = 2.731*10^-3

roots are :

x = -7.113*10^-3 and x = 5.953*10^-3

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

x = 5.953*10^-3

At equilibrium:

[N2O4] = 0.0365-1x = 0.0365-1*0.00595 = 0.03055 M

[NO2] = +2x = +2*0.00595 = 0.01191 M

0.0306 , 0.0119 M