Question

For the reaction; $N_{2\left(g\right)}+3H_{2\left(g\right)}\rightleftharpoons 2N{H}_{3\left(g\right)}$.

At $400K,K_p=41at{m}^{-2}$. Find the value of $K_p$ for each of the following reactions at the same temperature:

(i) $2N{H}_{3\left(g\right)}\rightleftharpoons N_{2\left(g\right)}+3H_{2\left(g\right)}$;

(ii) $\frac{1}{2}{N}_{2\left(g\right)}+\frac{3}{2}{H}_{2\left(g\right)}\rightleftharpoons N{H}_{3\left(g\right)}$;

(iii) $2N_{2\left(g\right)}+6H_{2\left(g\right)}\rightleftharpoons 4N{H}_{3\left(g\right)}$.

Answer 1

(i) $K_p=\frac{\left[H_2{]}^3\right[N_2]}{[N{H}_3{]}^2}={\left[\frac{[N{H}_3{]}^2}{\left[N_2\right][H_2{]}^3}\right]}^{-1}=(41{)}^{-1}=\frac{1}{41}=0.024at{m}^2$

$(\because \frac{[N{H}_3{]}^2}{\left[N_2\right][H_2{]}^3}=41)$

(ii) $K_p=\frac{\left[N{H}_3\right]}{\left[H_2{]}^{3/2}\right[N_2{]}^{1/2}}={\left[\frac{[N{H}_3{]}^2}{\left[H_2{]}^3\right[N_2]}\right]}^{1/2}=(41{)}^{1/2}=6.4at{m}^{-1}$

(iii) $K_p=\frac{[N{H}_3{]}^4}{\left[N_2{]}^2\right[H_2{]}^6}={\left[\frac{[N{H}_3{]}^2}{\left[N_2\right][H_2{]}^3}\right]}^2=(41{)}^2=1.681\times {10}^{3}at{m}^{-4}$.