For the dissociation ${\text{PCl}}_5(g)\rightleftharpoons {\text{PCl}}_3(g)+{\text{Cl}}_2(g)$, the degree of dissociation $\alpha$ is related to the total pressure $P$ and equilibrium constant $K_p$ by which expression?

Solution

1. Initial moles: ${\text{PCl}}_5=1$, ${\text{PCl}}_3=0$, ${\text{Cl}}_2=0$.
2. Equilibrium moles: ${\text{PCl}}_5=1-\alpha$, ${\text{PCl}}_3=\alpha$, ${\text{Cl}}_2=\alpha$. Total moles = $1+\alpha$.
3. Partial pressures: $P_{{\text{PCl}}_5}=\frac{1-\alpha }{1+\alpha }P$, $P_{{\text{PCl}}_3}=\frac{\alpha }{1+\alpha }P$, $P_{{\text{Cl}}_2}=\frac{\alpha }{1+\alpha }P$.
4. $$\begin{array}{rl}{K}_p& =\frac{(\frac{\alpha }{1+\alpha }P)(\frac{\alpha }{1+\alpha }P)}{\frac{1-\alpha }{1+\alpha }P}\\ & =\frac{{\alpha }^{2}P}{1-{\alpha }^2}\end{array}$$.
5. Hence the answer is (A).