Question

For the reaction $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$, the degree of dissociation at equilibrium is $0.2$ at $1$ atm pressure.

The equilibrium constant $K_p$ will be:

• A. $\frac{1}{2}$
• B. $\frac{1}{4}$
• C. $\frac{1}{6}$
• D. $\frac{1}{8}$

Answer 1

Answer: (C)

$\frac{1}{6}$

$\underset{\underset{1-\alpha }{1}}{N_2{O}_4}\rightleftharpoons \underset{\underset{2\alpha }{0}}{2N{O}_2}$

[Total moles $=1-\alpha +2\alpha =1+\alpha ]$

Partial pressure $\frac{1-\alpha }{1+\alpha }\times P\frac{2\alpha }{1+\alpha }\times P$

Given, $\alpha =0.2,P=1$

Partial pressure of $N_2{O}_4=\frac{1-0.2}{1+0.2}\times 1=\frac{0.8}{1.2}=\frac{2}{3}$

$p_{N{O}_2}=\frac{2\times 0.2}{1+0.2}\times 1=\frac{0.4}{1.2}=\frac{1}{3}$

$K_p=\frac{[p_{N{O}_2}{]}^2}{[p_{N_2{O}_4}{]}^2}=\frac{1\times 1\times 3}{3\times 3\times 2}=\frac{1}{6}$

Hence, the correct answer is $C$