Question

In the process $N_2+3H_2\leftrightharpoons 2N{H}_3$, the initial concentration of nitrogen and hydrogen are $1$ mole per litre and $3$ moles per litre respectively. The equilibrium constant of the reaction is $x$. The value of $K_c$ for $2N{H}_3\leftrightharpoons N_2+3H_2$ will be:

• A. $x$
• B. $2x$
• C. $\frac{1}{x}$
• D. $3x$

Answer 1

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

When the first reaction $(N_2+3H_2\leftrightharpoons 2N{H}_3)$ is reversed, second reaction $(2N{H}_3\leftrightharpoons N_2+3H_2)$ is obtained.

In such a case, the value of the equilibrium constant for the second reaction will be the reverse of the value of the equilibrium constant for the first reaction.

Thus, $K=\frac{1}{K_1}=\frac{1}{x}$.