Question

Nitrogen and hydrogen in the molar ratio $1:3$ was allowed to equilibrate. If $50\%$ of each reacts at the total pressure of $P$, then $K_p$ for the reaction, $N_2+3H_2\rightleftharpoons 2N{H}_3$ is

• A. $\frac{16}{3P^2}$
• B. $\frac{3P^2}{16}$
• C. $8P^2$
• D. $\frac{8}{P^2}$

Answer 1

Answer: (A)

Total no. of moles $=n-0.5n+3n-1.5n+n=3n$

$x_{N_2}=\frac{0.5n}{3n}=\frac{1}{6},x_{H_2}=\frac{1.5n}{3n}=\frac{1}{2},x_{N{H}_3}=\frac{n}{3n}=\frac{1}{3}$

$K_p=\frac{{\left(p_{N{H}_3}\right)}^2}{\left(p_{N_2}\right){\left(p_{H_2}\right)}^3}=\frac{{\left(\frac{1}{3}\cdot P\right)}^2}{\left(\frac{1}{6}P\right)\cdot {\left(\frac{1}{2}P\right)}^3}=\frac{1}{9}\times 6\times 8P^{-2}=\frac{16}{3P^2}$