Question

If $K_1$ is the equilibrium constant for the reaction $2N{H}_3\rightleftharpoons N_2+3H_2$, then equilibrium constant for the reaction, $\frac{1}{2}{N}_2+\frac{3}{2}{H}_2\rightleftharpoons N{H}_3$ will be

• A. $\sqrt{K_1}$
• B. $K_1^2$
• C. $\frac{1}{\sqrt{K_1}}$
• D. $\frac{1}{\sqrt{K_1^2}}$

Answer 1

The correct option is C $\frac{1}{\sqrt{K_1}}$

$2N{H}_3\to N_2+3H_2$ $K_{eq}$ = $K_1$

$N{H}_3\to \frac{1}{2}{N}_2+\frac{3}{2}{H}_2$ $K_{eq}={K_1}^{1/2}$

$\frac{1}{2}{N}_2+\frac{3}{2}{H}_2\to N{H}_3$ $K_{eq}={K_1}^{-1/2}$ = $\frac{1}{{\sqrt{K}}_1}$