Question

$N_2$ and $H_2$ are taken in 1 : 3 molar ratio in a closed vessel to attained the following equilibrium $N_{2\left(g\right)}+3H_{2\left(g\right)}\rightleftharpoons 2N{H}_{3\left(g\right)}$ Find $K_p$ for reaction, if total pressure at equilibrium is 2P and $P_{N_2}$ is P/3.

• A. $\frac{1}{3P^2}$
• B. $\frac{4}{3P^2}$
• C. $\frac{4P^2}{3}$
• D. none

Answer 1

Answer: (B)

$\frac{4}{3P^2}$

The reaction of nitrogen with hydrogen can be shown as:

$N_2+3H_2\to 2N{H}_3$

At Initial P 3P

At equilibrium P-x P/3

From this

$P-x=\frac{P}{3}$

$x=\frac{2P}{3}$

$P-x=P-\frac{2P}{3}=\frac{P}{3}$

So pressure of the nitrogen at equilibrium is $\frac{P}{3}$ and for hydrogen is $P$

Now total pressure at equilibrium is $2P$

$P_{N_2}+P_{H_2}+P_{N{H}_3}=2P$

$\frac{P}{3}+P+P_{N{H}_3}=2P$

$P_{N{H}_3}=\frac{2P}{3}$

The equilibrium constant for pressure can be shown as

$K_P=\frac{[P_{N{H}_3}{]}^2}{\left[P_{N_2}\right][P_H{]}^3}$

$K_P=\frac{\left(4P^2/9\right)}{\left(P/3\right)\left(P^3\right)}$

$K_P=\frac{4}{3P^2}$