Question

The degree of dissociation of $PC{l}_5\left(\alpha \right)$ obeying the equilibrium: $PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$ is related to the pressure at equilibrium by:

• A. $\alpha \propto P$
• B. $\alpha \propto \frac{1}{\sqrt{P}}$
• C. $\alpha \propto \frac{1}{P^2}$
• D. $\alpha \propto 1/P^4$

Answer 1

Answer: (B)

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

$PC{l}_5\left(g\right)\leftrightharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$

$1$ $0$ $0$

$1-\alpha$ $\alpha$ $\alpha$

$K_p=\frac{{\alpha }^2}{1-\alpha }$ $\left[\frac{P}{1+\alpha }\right]=\frac{{\alpha }^{2}P}{1-{\alpha }^2}a$

$⟹\alpha =\sqrt{\frac{K_p}{P}}⟹\alpha \propto \frac{1}{\sqrt{P}}$ (if $1-{\alpha }^2=1$)