Question

For the reaction $N_{2}O_{4}\rightleftharpoons2NO_{2 \left(g\right)}$ , the degree of dissociation of $N_{2}O_{4}$ is $0.2$ at $1atm.$ Then the $K_{p}$ of $2NO_{2}\rightleftharpoons N_{2}O_{4}$ is

Answer 1

$\text{K}_{\text{p}}=\frac{\left[\text{P}_{\text{NO}_{\text{2}}}\right]^{2}}{\text{P}_{\text{N}_{2} \text{O}_{\text{4}}}}$

$\text{ K}_{\text{p}}=\frac{\text{4} \alpha ^{2}}{1 - \alpha ^{2}}\text{p}$

$\Rightarrow \mathrm{K}_{\mathrm{p}}=\frac{4 \times(0.2)^2}{1-(0.2)^2}=\frac{0.16}{0.96}$

$\text{ for}\text{ }2\text{NO}_{2}\overset{}{\rightleftharpoons}\text{N}_{\text{2}}\text{O}_{4}; \, \text{K}_{\text{p}}=\frac{0.96}{0.16}=6$