Question

The equilibrium constant Kc for the reaction

N2O4 <—> 2 NO2 at 25degrees C is 170. Suppose 13.3g of N2O4 is placed in a 4.000-L flask at 25 C.

Calculate the percentage of the original N2O4 that is dissociated?

Answer 1

Molar mass of N2O4,
MM = 2*MM(N) + 4*MM(O)
= 2*14.01 + 4*16.0
= 92.02 g/mol


mass(N2O4)= 13.3 g

use:
number of mol of N2O4,
n = mass of N2O4/molar mass of N2O4
=(13.3 g)/(92.02 g/mol)
= 0.1445 mol
volume , V = 4.000 L


use:
Molarity,
M = number of mol / volume in L
= 0.1445/4
= 0.03613 M

ICE Table:

                    [N2O4]              [NO2]             

initial             0.03613                               

change              -1x                 +2x               

equilibrium         0.03613-1x          +2x               

Equilibrium constant expression is
Kc = [NO2]^2/[N2O4]
170.0 = (4*x^2)/((0.03613-1*x))
6.1421-170*x = 4*x^2
6.1421-170*x-4*x^2 = 0
This is quadratic equation (ax^2+bx+c=0)
a = -4
b = -1.7*10^2
c = 6.142

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 = 2.9*10^4

roots are :
x = -42.54 and x = 3.61*10^-2

since x can't be negative, the possible value of x is
x = 3.61*10^-2

% dissociation = x*100/c
= (3.61*10^-2)*100 / 0.03613
= 99.9 %

Answer: 99.9 %