Question

The equilibrium constants of the following are:
$N_2+3H_2\rightleftharpoons 2N{H}_3;K_1$
$N_2+O_2\rightleftharpoons 2NO;K_2$
$H_2+\frac{1}{2}{O}_2\to H_{2}O;K_3$

The equilibrium constant $\left(K\right)$ of the reaction:
$2N{H}_3+\frac{5}{2}{O}_2\stackrel{K}{\rightleftharpoons }2NO+3H_{2}O$, will be:

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

Answer 1

The correct option is C $\frac{K_2{K}_3^3}{K_1}$

$N_2+3H_2\rightleftharpoons 2N{H}_3;K_1$ ..... (1)
$N_2+O_2\rightleftharpoons 2NO;K_2$.....(2)
$H_2+\frac{1}{2}{O}_2\to H_{2}O;K_3$ ..... (3)

Multiply (3) equation by 3, we get

$3H_2+\frac{3}{2}{O}_2\to 3H_{2}O;K_3^3$ --- (4)
Add equation (2) and (4) and then subtract equation (1)

$2N{H}_3+\frac{5}{2}{O}_2\stackrel{K}{\rightleftharpoons }2NO+3H_{2}O$
$\therefore K=\frac{K_2{K}_3^3}{K_1}$