The degree of dissociation of $\mathrm{PCl}_{5}(\alpha)$ obeying the equilibrium $\mathrm{PCl}_{5}…

The degree of dissociation of $\mathrm{PCl}_{5}(\alpha)$ obeying the equilibrium $\mathrm{PCl}_{5} ightleftharpoons \mathrm{PCl}_{3}+\mathrm{Cl}_{2}$ is related to the equilibrium pressure by
  1. $\alpha \propto \frac{1}{\mathrm{P}^{4}}$
  2. $\alpha \propto \frac{1}{\sqrt{\mathrm{P}}}$
  3. $\alpha \propto \frac{1}{\mathrm{P}^{2}}$
  4. $\alpha \propto \mathrm{P}$

Solution

$\mathrm{PCl}_{5} ightleftharpoons \mathrm{PCl}_{3}+\mathrm{Cl}_{2}$
$\begin{array}{lll}1-\alpha & \alpha & \alpha\end{array}$
$\therefore \mathrm{K}_{\mathrm{p}}=\frac{\frac{\alpha}{1+\alpha} \mathrm{P} \times \frac{\alpha}{1+\alpha} \mathrm{P}}{\frac{1-\alpha}{1+\alpha} \mathrm{P}}=\frac{\alpha^{2} \mathrm{P}}{1-\alpha^{2}}$
or, $\mathrm{K}_{\mathrm{p}}=\alpha^{2} \mathrm{P} \quad \therefore \alpha=\sqrt{\frac{\mathrm{K}_{\mathrm{p}}}{\mathrm{P}}}$ when $1-\alpha^{2}=1$

Asked in: JEE-TOPICTESTS-CHEMISTRY

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