Question

If the value of equilibrium constant $K_c$ for the reaction, $N_2+3H_2\rightleftharpoons 2N{H}_3$ is $7$. The equilibrium constant for the reaction $2N_2+6H_2\rightleftharpoons 4N{H}_3$ will be :

• A. $49$
• B. $7$
• C. $14$
• D. $28$

Answer 1

The correct option is A $49$

$N_2+3H_2\rightleftharpoons 2N{H}_3$, given that $K_c=7$

$2N_2+6H_2\rightleftharpoons 4N{H}_3$

If the first reaction is multiplied by 2, the equilibrium constant becomes square of the first reaction $K_c$.

So, $K_c^{\prime }=K_c^2=7^2=49$