Question

The equilibrium constant for the reaction, $N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)$ is $k$, then the equilibrium constant for the equilibrium $2N{H}_3\left(g\right)\rightleftharpoons N_2\left(g\right)+3H_2\left(g\right)$ is:

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

Answer 1

The correct option is C $\frac{1}{k}$

$N_2\left(g\right)+3H_2\left(g\right)\rightleftharpoons 2N{H}_3\left(g\right)...\left(i\right)$
$2N{H}_3\left(g\right)\rightleftharpoons N_2\left(g\right)+3H_2\left(g\right)..\left(ii\right)$
Reaction $\left(ii\right)$ is inverse of reaction $\left(i\right)$,
Hence, the equilibrium constant for the reverse reaction will be $\frac{1}{k}$.