Question

The following system is at equilibrium with [N2O4] = 0.55 M and [NO2] = 0.25 M. If an additional 0.10 M NO2 is added to the reaction mixture with no change in volume, what will be the new equilibrium concentration of N2O4?

N2O4(g) <> 2 NO2(g)

Answer = 0.59 M

show work

Answer 1

Kc = [NO2]^2 / [N2O4]
= 0.25^2 / (0.55)
= 0.114

N2O4(g)   <------> 2 NO2(g)
0.55                                 0.35       (initial)
0.55+x                          0.35-2x (at equilibrium)

Kc = [NO2]^2 / [N2O4]
0.114 = (0.35-2x)^2 / (0.55+x)
0.0627 + 0.114*x = 0.1225 + 4*x^2 - 1.4*x
4*x^2 - 1.514*x + 0.0598 = 0

This is quadratic equation (ax^2+bx+c=0)
a = 4.0
b = -1.514
c = 0.0598

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 = 1.335

roots are :
x = 0.334 and x = 0.045

x can't be 0.334 as this will make the concentration negative.so,
x = 0.045

[N2O4] = 0.55 + x = 0.55 + 0.045 = 0.595 M

Answer: 0.595 M