Question

$N_2$ and $H_2$ in $1:3$ molar ratio are heated in a closed container having a catalyst. When the following equilibrium
$N_2(g{)}_3^{+}{H}_2(g)\rightleftharpoons 2N{H}_3(g)$ is attained, the total pressure is $10atm$ and mole fraction of $N{H}_3$ is $0.60$. The equilibrium constant $K_p$ for dissociation of $N{H}_3$ is

• A. $1.333at{m}^{-2}$
• B. $0.75at{m}^2$
• C. $0.75at{m}^{-2}$
• D. $1.333$ atm ${}^2$

Answer 1

Answer: (B)

$N_2(g)+3H_2(g)\rightleftharpoons 2N{H}_3(g)$
Mole fraction of $N{H}_3\left(X_{N{H}_3}\right)=0.6$
Mole fraction of $N_2$ and $H_2=1-0.6=0.4$
Total no. of mole of $N_2$ and $H_2=1+3=4$ Then,
$X_{N_2}=\frac{1}{4}\times 0.4=0.1$
$X_{H_2}=\frac{3}{4}\times 0.4=0.3$
Partial pressure of $N_2\left(P_{N_2}\right)=0.1\times 10=1atm$
Partial pressure of $H_2\left(P_{H_2}\right)$
$=0.3\times 10=3atm$
Partial pressure of $N{H}_3\left(P_{N{H}_3}\right)$
$=0.6\times 10=6atm$
Dissociation of ammonia
$2N{H}_3\rightleftharpoons N_2+3H_2$
$K_P=\frac{P_{N_2}\times P_{H_2}^3}{P_{N{H}_3}^{2H_2}}$
$=\frac{1\times (3{)}^3}{(6{)}^2}$
$=\frac{27}{36}=0.75at{m}^2$