Question

The following equilibrium constants are given
$N_2+3H_2\rightleftharpoons 2N{H}_3;K_1$
$N_2+O_2\rightleftharpoons 2NO;K_2$
$H_2+1/2O_2\rightleftharpoons H_{2}O;K_3$
The equilibrium constant for the oxidation of the $N{H}_3$ by oxygen to give $NO$ is:

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

Answer 1

Answer: (B)

(I) $N_2+3H_2\rightleftharpoons 2N{H}_3;K_1=\frac{{\left[N{H}_3\right]}^2}{\left[N_2\right]{\left[H_2\right]}^3}$

(II) $N_2+O_2\rightleftharpoons 2NO;K_2=\frac{[NO{]}^2}{\left[N_2\right]\left[O_2\right]}$

(III) $H_2+\frac{1}{2}{O}_2⟶H_{2}O;K_3=\frac{\left[H_{2}O\right]}{\left[H_2\right]{\left[O_2\right]}^{1/2}}$

(II $+3\times$ III $-$ II) will give

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

$\therefore K=K_2\times K_3^3/K_1$