Question

The following equilibrium constants are given

$N_2+3H_2\rightleftharpoons 2N{H}_3;K_1$

$N_2+O_2\rightleftharpoons 2NO;K_2$

$H_2+\frac{1}{2}{O}_2⟶H_{2}O;K_3$

The equilibrium constant for the oxidation of $N{H}_3$ by oxygen to give $NO$ is:

• A. $\frac{K_1{K}_2}{K_3}$
• B. $\frac{K_2{K}_3^3}{K_1}$
• C. $\frac{K_{21}{K}_3^2}{K_1}$
• D. $\frac{K_2^2{K}_3}{K_1}$

Answer 1

The correct option is B

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

Reverse the first equation:

$2N{H}_3\to N_2+3H_2;\frac{1}{K_1}$

Next up, multiply the second equation by 3:

$3H_2+\frac{3}{2}{O}_2{K}_3^3$

Add the two resulting ones, and we get :

$N_2+O_2\to 2NO{K}_2$
$\underset{–}{\overline{2N{H}_3+\frac{5}{2}{O}_2\to 2NO+3H_{2}OK=\frac{K_2{K}_3^3}{K_1}}}$