Question

Given equilibrium constant for $N_2 + 3H_2 \rightleftharpoons 2NH_3$ is K

Answer 1

(i) - (a), (ii) - (b)

Answer 2

Let the given reaction be: $N_2 + 3H_2 \rightleftharpoons 2NH_3$ The equilibrium constant is given as $K$. The expression for $K$ is: $K = \frac{[NH_3]^2}{[N_2][H_2]^3}$

For the reaction in Column I (i): $NH_3 \rightleftharpoons \frac{1}{2}N_2 + \frac{3}{2}H_2$ Let its equilibrium constant be $K_i$. The expression for $K_i$ is: $K_i = \frac{[N_2]^{1/2}[H_2]^{3/2}}{[NH_3]}$ We can see that $K_i$ is the reciprocal of the square root of $K$: $K_i = \left(\frac{[NH_3]^2}{[N_2][H_2]^3}\right)^{-1/2} = K^{-1/2} = \frac{1}{\sqrt{K}}$ Thus, (i) matches with (a).

For the reaction in Column I (ii): $\frac{1}{2}N_2 + \frac{3}{2}H_2 \rightleftharpoons NH_3$ Let its equilibrium constant be $K_{ii}$. The expression for $K_{ii}$ is: $K_{ii} = \frac{[NH_3]}{[N_2]^{1/2}[H_2]^{3/2}}$ We can see that $K_{ii}$ is the square root of $K$: $K_{ii} = \left(\frac{[NH_3]^2}{[N_2][H_2]^3}\right)^{1/2} = K^{1/2} = \sqrt{K}$ Thus, (ii) matches with (b).