Question

For the reaction $N_2{O}_4\rightleftharpoons 2N{O}_{2\left(g\right)}$, the degree of dissociation at equilibrium is $0.2$ at $1$ atm. Then $K_p$ will be?

• A. $1/2$
• B. $1/4$
• C. $1/6$
• D. $1/8$

Answer 1

The correct option is C $1/6$

$N_2{O}_4\rightleftharpoons 2N{O}_2$

$1-\alpha$ $2\alpha$

$P{N}_2{O}_4=\frac{P(1-\alpha )}{(1+\alpha )},P_{N{O}_2}=\frac{P\left(2\alpha \right)}{(1+\alpha )}$

$\Rightarrow K_P=\frac{P(N{O}_2{)}^2}{P_{N_2{O}_4}}=\frac{{\left(\frac{P\left(2\alpha \right)}{(1+\alpha )}\right)}^2}{\frac{P(1-\alpha )}{(1+\alpha )}}=\frac{P^{2}2{\alpha }^2}{P(1-\alpha )}$

$\Rightarrow K_P=4P{\alpha }^2$

$\Rightarrow \frac{K_P}{4P}={\alpha }^2$

$\Rightarrow \alpha =\sqrt{\frac{K_P}{4P}}$

Given $P=1,\alpha =0.2$

$\Rightarrow (0.2{)}^2=\frac{K_P}{4}$

$\Rightarrow K_P=\frac{1}{6}$ .