Question

Consider the reaction: N2(g) 2 O2(g)N2O4(g) Write the equilibrium constant for this reaction in terms of the equilibrium constants, Ka and Kb, for reactions a and b below: a.) N2O4(g) 2NO2(g) Ka b.) 1/2 N2(g) O2(g) NO2(g) Kb

Answer 1

Answer : The equilibrium constant for this reaction is, $K=\frac{(K_b)^2}{K_a}$

Explanation :

The given main chemical reaction is:

$N_2(g)+2O_2(g)\rightarrow N_2O_4(g)$;  $K$

The intermediate reactions are:

(1) $N_2O_4(g)\rightarrow 2NO_2(g)$;  $K_a$

(2) $\frac{1}{2}N_2(g)+O_2(g)\rightarrow NO_2(g)$;  $K_b$

We are reversing reaction 1 and multiplying reaction 2 by 2 and then adding both reaction, we get:

(1) $2NO_2(g)\rightarrow N_2O_4(g)$;  $\frac{1}{K_a}$

(2) $N_2(g)+2O_2(g)\rightarrow 2NO_2(g)$;  $(K_b)^2$

Thus, the equilibrium constant for this reaction will be:

$K=\frac{1}{K_a}\times (K_b)^2$

$K=\frac{(K_b)^2}{K_a}$

Thus, the equilibrium constant for this reaction is, $K=\frac{(K_b)^2}{K_a}$