Question

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

The equilibrium constant for the oxidation of $2$ mol of $N{H}_3$ to give NO is?

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

Answer 1

The correct option is B $K_2\cdot \frac{K_3^3}{K_1}$

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

$K=\frac{\left[NO{]}^2\right[H_{2}O{]}^3}{\left[N{H}_3{]}^2\right[O_2{]}^{5/2}}$

According to question for given reaction

$K_1=\frac{[N{H}_3{]}^2}{\left[N_2\right][H_2{]}^3}$

$K_2=\frac{[NO{]}^2}{\left[N_2\right]\left[O_2\right]}$

$K_3=\frac{\left[H_{2}O\right]}{\left[H_2\right][O_2{]}^{1/2}}$

$K=K_2\cdot \frac{K_3^3}{K_1}$ $=\frac{[NO{]}^2}{\left[N_2\right]\left[O_2\right]}$ $\times \frac{[H_{2}O{]}^3}{\left[H_2{]}^3\right[O_2{]}^{3/2}}\times \frac{\left[N_2\right][H_2{]}^3}{[N{H}_3{]}^2}$

$=\frac{\left[NO{]}^2\right[H_{2}O{]}^3}{\left[N{H}_3{]}^2\right[O_2{]}^{5/2}}$.