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, $N _{2}+3 H _{2} \rightleftharpoons 2 NH _{3}$ is

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

Answer 1

Answer: (A)

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

$x_{ N _{2}}=\frac{0.5 n}{3 n}=\frac{1}{6}, x_{ H _{2}}=\frac{1.5 n}{3 n}=\frac{1}{2}, x_{ NH _{3}}=\frac{n}{3 n}=\frac{1}{3}$

$K_{p}=\frac{\left(p_{ NH _{3}}\right)^{2}}{\left(p_{ N _{2}}\right)\left(p_{ H _{2}}\right)^{3}}=\frac{\left(\frac{1}{3} \cdot P\right)^{2}}{\left(\frac{1}{6} P\right) \cdot\left(\frac{1}{2} P\right)^{3}}=\frac{1}{9} \times 6 \times 8 P^{-2}=\frac{16}{3 P^{2}}$