Question

In a container equilibrium $N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$ is attained at ${25}^{\circ }C$. The total equilibrium pressure in container is $380torr$. If equilibrium constant of above equilibrium is $0.667atm$, 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 A $\frac{1}{3}$

$N_2{O}_4⟺2N{O}_2$

Initially P 0

Equilibrium p-$\alpha$ 2$\alpha$

$\alpha$ is the degree of dissociation

$K_P=\frac{\left[P\right(N{O}_2){]}^2}{\left[P\right(N_2{O}_4\left)\right]}$

$\left[P\right(N{O}_2\left)\right]$=2$\alpha$

$\left[P\right(N_2{O}_4\left)\right]$=P-$\alpha$

$0.667=\frac{(2\alpha {)}^2}{0.5-\alpha }\alpha =0.3$ $380$ torr=$0.5$ atm=1/3