Question

The following equilibrium 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
$2N{H}_3+\frac{5}{2}{O}_2\rightleftharpoons 2NO+3H_{2}O$, in terms of $K_1,K_2$ and $K_3$ is :

• A. $\frac{K_1{K}_2}{K_3}$
• B. $\frac{K_1{K}_3^2}{K_2}$
• C. $\frac{K_2{K}_3^3}{K_1}$
• D. $K_1{K}_2{K}_3$

Answer 1

Answer: (C)

$\frac{K_2{K}_3^3}{K_1}$

(1) $N_2+3H_2\rightleftharpoons 2N{H}_3$
(2) $N_2+O_2\rightleftharpoons 2NO$
(3) $H_2+\frac{1}{2}{O}_2\rightleftharpoons H_{2}O$
(4) $2N{H}_3+\frac{5}{2}{O}_2\rightleftharpoons 2NO+3H_{2}O$
(5) $=\left(2\right)+3\left(3\right)-\left(1\right)$
i.e. $K_4=\frac{K_2{K}_3^3}{K_1}$