Question

The following equilibria are given
$N_2+3H_2\rightleftharpoons 2N{H}_3,K_1$
$N_2+O_2\rightleftharpoons 2NO,K_2$
$H_2+\frac{1}{2}{O}_2\rightleftharpoons H_{2}O,K_3$
The equilibrium constant of the reaction, in terms of $K_1,K_2$ and $K_3$ is:
$2N{H}_3+\frac{5}{2}{O}_2\rightleftharpoons 2NO+3H_{2}O$

• A. $K=\frac{K_2\times K_3^2}{K_1}$
• B. $K=\frac{K_2^2\times K_3}{K_1}$
• C. $K=\frac{K_1\times K_2}{K_3}$
• D. $K=\frac{K_2\times K_3^3}{K_1}$

Answer 1

The correct option is D $K=\frac{K_2\times K_3^3}{K_1}$

If we reverse a reaction, then the new equilibrium constant becomes reciprocal to the older.

If we add two reactions, then equilibrium constants multiply to get a new equilibrium constant.

If we multiply the reaction by '2', then the equilibrium constant gets squared.

$N_2+3H_2\rightleftharpoons 2N{H}_3⟶K_{1}2N{H}_3\rightleftharpoons N_2+3H_2⟶\frac{1}{K_1}------\left(1\right)$

Multiplying equation (1) by 2.

$N_2+O_2\rightleftharpoons 2NO⟶K_2------\left(2\right)H_2+\frac{1}{2}{O}_2\rightleftharpoons H_{2}O⟶K_3$

Multiplying the above reaction by 3,

$3H_2+\frac{3}{2}{O}_2\rightleftharpoons 3H_{2}O⟶K_3^3------\left(3\right)$

Adding (1), (2) and (3) reactions,

$2N{H}_3+\frac{5}{2}{O}_2\rightleftharpoons 2NO+3H_{2}O$

Resultant equilibrium constant$=K^{\prime }=\frac{(K_3{)}^3\cdot K_2}{K_1}$