For the reaction ${\text{N}}_2{\text{O}}_4(g)\rightleftharpoons 2{\text{NO}}_2(g)$, the degree of dissociation $\alpha$ is related to the total pressure $P$ and equilibrium constant $K_p$ by which expression?

Solution

1. Initial moles are $1$ for ${\text{N}}_2{\text{O}}_4$ and $0$ for ${\text{NO}}_2$.
2. At equilibrium, moles are $(1-\alpha )$ for ${\text{N}}_2{\text{O}}_4$ and $2\alpha$ for ${\text{NO}}_2$, with total moles $=1+\alpha$.
3. Partial pressures are $p_{{\text{N}}_2{\text{O}}_4}=\frac{1-\alpha }{1+\alpha }P$ and $p_{{\text{NO}}_2}=\frac{2\alpha }{1+\alpha }P$.
4. The expression $$\begin{array}{rl}{K}_p& =\frac{(p_{{\text{NO}}_2}{)}^2}{p_{{\text{N}}_2{\text{O}}_4}}\\ & =\frac{(2\alpha P/(1+\alpha ){)}^2}{(1-\alpha )P/(1+\alpha )}\end{array}$$.
5. Simplifying yields $K_p=\frac{4{\alpha }^{2}P}{1-{\alpha }^2}$.
6. Hence the answer is (A).