Question

Consider the following equation: N2O4(g) ⇄ 2 NO2(g) Kc = 5.8 × 10-3 If the initial concentration of N2O4(g) = 0.040 M and the initial concentration of NO2(g) is 0 M, what is the equilibrium concentration of N2O4(g)?

Answer 1

Answer : The  concentration of $N_2O_4$ at equilibrium is, 0.03977 M

Explanation :  

The given balanced chemical reaction is,

                          $N_2O_4(g)\rightleftharpoons 2NO_2(g)$

Initial conc.         0.04            0

At eqm. conc.    (0.04-x)       (2x)

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

Now put all the given values in this expression, we get :

$5.8\times 10^{-3}=\frac{x}{(0.04-x)}$

By solving the term 'x', we get :

x = 0.00023 M

Concentration of $NO_2$ at equilibrium = 2x M = 2 × 0.00023 = 0.00046 M

Concentration of $N_2O_4$ at equilibrium = (0.04 - x) M = (0.04 - 0.00023) = 0.03977 M

Therefore, the concentration of $N_2O_4$ at equilibrium is, 0.03977 M