Question

In a container equilibrium
$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
is attained at ${25}^{o}C.$ The total equilibrium pressure in container is 380 torr. If equilibrium constant of above equilibrium is 0.667 atm, then degree of dissociation of $N_2{O}_4$ at this temperature will be:

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{2}{3}$
• D. $\frac{1}{4}$

Answer 1

The correct option is B $\frac{1}{2}$

$\begin{array}{cccc}& N_2{O}_4& \rightleftharpoons & 2N{O}_2\\ \text{initial}& 1& & \\ \text{At equilibrium}& 1-\alpha & & 2\alpha \end{array}$

Total moles at equilibrium $=1-\alpha +2\alpha =1+\alpha$
$\because$ We know that
$\text{Partial pressure}=\text{total pressure}\times \text{mole fraction}$
$\therefore$ Partial pressure of $N_2{O}_4=\frac{(1-\alpha )}{(1+\alpha )}\cdot P$
Partial pressure of $N{O}_2=\frac{\left(2\alpha \right)}{(1+\alpha )}\cdot P$
$\therefore K_P=\frac{[P_{N{O}_2}{]}^2}{P_{N_2{O}_4}}=\frac{{[\frac{2\alpha }{1+\alpha }\cdot P]}^2}{\left(\frac{1-\alpha }{1+\alpha }\right)\cdot P}=\frac{\frac{4{\alpha }^2}{(1+\alpha {)}^2}\cdot P^2}{\frac{(1-\alpha )}{(1+\alpha )}P}$
$\Rightarrow K_P=\frac{4{\alpha }^2.P}{(1-\alpha )(1+\alpha )}=\frac{4{\alpha }^2}{(1-{\alpha }^2)}\times \frac{1}{2}$
{P = 380 torr = 0.5 atm}
$\Rightarrow 0.667=\frac{2}{3}=\frac{4{\alpha }^2}{(1-{\alpha }^2)}\cdot \frac{1}{2}$
$\Rightarrow \frac{4{\alpha }^2}{(1-{\alpha }^2)}=\frac{4}{3}or3{\alpha }^2=1-{\alpha }^2$
$\Rightarrow {\alpha }^2=\frac{1}{4},\alpha =\frac{1}{2}$
Therefore, the correct option is (b).