Question

N2O4 (g) ⇌ 2NO2(g). KP = 0.10 at some Temperature. If 0.40 atm of N2O4 is placed in an evacuated flask, what are the pressures at equilibrium?

Answer 1

ICE Table:

                    p(N2O4)             p(NO2)            

initial             0.4                                   

change              -1x                 +2x               

equilibrium         0.4-1x              +2x               

Equilibrium constant expression is
Kp = p(NO2)^2/p(N2O4)
0.1 = (4*x^2)/((0.4-1*x))
0.04-0.1*x = 4*x^2
0.04-0.1*x-4*x^2 = 0
This is quadratic equation (ax^2+bx+c=0)
a = -4
b = -0.1
c = 4*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 = 0.65

roots are :
x = -0.1133 and x = 0.08828

since x can't be negative, the possible value of x is
x = 0.08828

At equilibrium:
p(N2O4) = 0.4-1x = 0.4-1*0.08828 = 0.31172 atm
p(NO2) = +2x = +2*0.08828 = 0.17656 atm

Answer:

p(N2O4) = 0.31 atm
p(NO2) = 0.18 atm