Question

The following equilibrium constants are given:
$N_2\text{(g)}+H_2\text{(g)}\rightleftharpoons 2N{H}_3\text{(g)};K_1$
$N_2\text{(g)}+O_2\text{(g)}\rightleftharpoons 2NO\text{(g)};K_2$
$H_2\text{(g)}+\frac{1}{2}{O}_2\text{(g)}\rightleftharpoons H_{2}O\text{(g)};K_3$

The equilibrium constant for the oxidation of the $N{H}_3$​ by oxygen to give $NO$ is:
$2N{H}_3\text{(g)}+\frac{5}{2}{O}_2\text{(g)}\rightleftharpoons 2NO\text{(g)}+3H_{2}O\text{(g)}$

• 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

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

The given reactions are:
(I)$N_2\text{(g)}+H_2\text{(g)}\rightleftharpoons 2N{H}_3\text{(g)}$
$\therefore K_1=\frac{[N{H}_3{]}^2}{\left[N_2\right][H_2{]}^3}\left(\text{I}\right)$
(II)$N_2\text{(g)}+O_2\text{(g)}\rightleftharpoons 2NO\text{(g)}$
$\therefore K_2=\frac{[NO{]}^2}{\left[N_2\right]\left[O_2\right]}\left(\text{II}\right)$
(III)$H_2\text{(g)}+\frac{1}{2}{H}_2\text{(g)}\rightleftharpoons H_{2}O\text{(g)}$
$\therefore K_3=\frac{\left[H_{2}O\right]}{\left[H_2\right][O_2{]}^{1/2}}\left(\text{III}\right)$
Now, for the reaction,
$2N{H}_3\text{(g)}+\frac{5}{2}{O}_2\text{(g)}\rightleftharpoons 2NO\text{(g)}+3H_{2}O\text{(g)}$
Now, $K=\frac{\left[NO{]}^2\right[H_{2}O{]}^3}{\left[N{H}_3{]}^2\right[O_2{]}^{5/2}}$
$\Rightarrow \frac{\frac{[NO{]}^2}{\left[N_2\right]\left[O_2\right]}{\left(\frac{\left[H_{2}O\right]}{\left[H_2\right][O_2{]}^{1/2}}\right)}^3}{\frac{[N{H}_3{]}^2}{\left[N_2\right][H_2{]}^3}}$
$\Rightarrow K=\frac{K_2{K}_3^3}{K_1}$