Question

If the equilibrium constant for the reaction $N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g)$ at 750 K is 49, then the equilibrium constant for the reaction $N{H}_3(g)\rightleftharpoons \frac{1}{2}{N}_2(g)+\frac{3}{2}{H}_2(g)$ at the same temperature is

• A. 1/49
• B. 49
• C. 7
• D. ${49}^2$

Answer 1

For the reaction $N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g)$ $K_c=\frac{{[N{H}_3]}^2}{[N_2]{[H_2]}^3}$ ?(i) For the reaction: $N{H}_3(g)\rightleftharpoons \frac{1}{2}{N}_2(g)+\frac{3}{2}{H}_2(g)$ $K_c=\frac{{[N_2]}^{1/2}{[H_2]}^{3/2}}{[N{H}_3]}$ ?(ii) From Eqs (i) and (ii) $K_c=\sqrt{\frac{1}{K_c}}$ $\therefore K_c=49$ $\because$ $K_c=\sqrt{\frac{2}{49}}$ $\because K_c=\frac{1}{7}$