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 the container is $380$ torr. If equilibrium constant of the above equilibrium is $0.667atm$, then the 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

Answer: (B)

$\frac{1}{2}$

$N_2{O}_4\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)$
1 0
$1-x$ $2x$

$K_P=\frac{p_{N{O}_2}^2}{p_{N_2{O}_4}}=\frac{0.5\times (2x/1+x{)}^2}{(1-x)(1+x)}$ (380 torr=0.5 atm)

$0.667-0.667x^2=2x^2$

$x=0.5$