Question

The degree of dissociation of $P C l_{5}(\alpha)$ obeying the equilibrium $PCl _{5} \rightleftharpoons PCl _{3}+ Cl _{2}$ is related to the equilibrium pressure by

• A. $\alpha \propto \frac{1}{P_4}$
• B. $\alpha \propto \frac{1}{\sqrt{P}}$
• C. $\alpha \propto \frac{1}{P^2}$
• D. $\alpha \propto P$

Answer 1

Answer: (B)

$\alpha \propto \frac{1}{\sqrt{P}}$

Answer 2

$\underset{1-\alpha}{PCl _{5}} \rightleftharpoons \underset{\alpha}{PCl _{3}}+ \underset{\alpha}{Cl _{2}}$
$\therefore K_{p}=\frac{\frac{\alpha}{1+\alpha} P \times \frac{\alpha}{1+\alpha} P}{\frac{1-\alpha}{1+\alpha} P}=\frac{\alpha^{2} P}{1-\alpha^{2}}$
or $K_{p}=\alpha^{2} P$
$\therefore \alpha=\sqrt{\frac{K_{p}}{P}}$ when $1-\alpha^{2}=1$