Question

Calculate the equilibrium concentrations of N2O4 and NO2 at 25 ∘C in a vessel that contains an initial N2O4 concentration of 0.0507 M . The equilibrium constantKc for the reaction N2O4(g)⇌2NO2(g) is 4.64×10−3 at 25 ∘C.

Answer 1

N₂O₄(g)    ⇌      2NO₂(g)

0.0507                 0                                  initial

0.0507 - x           2x                           at equilibrium

Kc = [NO₂]² / [N₂O₄]

0.00464 = (2x)² / (0.0507 - x)

4x² + 0.00464x - 0.000235 = 0

Using quadratic equation,

x = [- 0.00464 +-(0.000215 + 0.00376)^1/2] / 8

x = 0.00903 M

So, at equilbrium,

[N₂O₄] = 0.0507 - 0.00903 = 0.0417 M

[NO₂] = 2*0.00903 = 0.0181 M