Question

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

• A. $1.333\, atm ^{-2}$
• B. $0.75 \,atm ^{2}$
• C. $0.75 \,atm ^{-2}$
• D. $1.333$ atm $^{2}$

Answer 1

Answer: (B)

$N_{2}(g)+3 H_{2}(g) \rightleftharpoons 2 N H_{3}(g)$
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 $N_{2}\left(P_{N_{2}}\right)=0.1 \times 10=1 atm$
Partial pressure of $H_{2}\left(P_{H_{2}}\right)$
$=0.3 \times 10=3\, atm$
Partial pressure of $N H_{3}\left(P_{N H_{3}}\right)$
$=0.6 \times 10=6\, atm$
Dissociation of ammonia
$2 NH _{3} \ce{<=>}N _{2}+3 H _{2}$
$K_{P}=\frac{P_{N_{2}} \times P_{H_{2}}^{3}}{P_{N H_{3}}^{2 H_{2}}}$
$=\frac{1 \times(3)^{3}}{(6)^{2}}$
$=\frac{27}{36}=0.75 atm ^{2}$