Question

The degree of dissociation of $P{Cl}_5\left(\alpha \right)$ obeying the equilibrium $P{Cl}_5\rightleftharpoons P{Cl}_3+{Cl}_2$ 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)

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

$P{Cl}_5\rightleftharpoons P{Cl}_3+{Cl}_2$
$\alpha =$ the degree of dissociation of $P{Cl}_5$
$P=$ total equilibrium pressure.
Suppose initially, 1 mole of $P{Cl}_5$ is present.
$\alpha$ moles of $P{Cl}_5$ will dissociate to form $\alpha$ moles of $P{Cl}_3$ and $\alpha$ moles of ${Cl}_2$.
$(1-\alpha )$ moles of $P{Cl}_5$ will remain at equilibrium.
Total number of moles $=(1-\alpha )+\alpha +\alpha =(1+\alpha )$
Mole fraction of $P{Cl}_5=\frac{(1-\alpha )}{(1+\alpha )}$
Mole fraction of $P{Cl}_3=\frac{\left(\alpha \right)}{(1+\alpha )}$
Mole fraction of ${Cl}_2=\frac{\left(\alpha \right)}{(1+\alpha )}$
Partial pressure of $P{Cl}_5=\frac{(1-\alpha )}{(1+\alpha )}P$
Partial pressure of $P{Cl}_3=\frac{\left(\alpha \right)}{(1+\alpha )}P$
Partial pressure of ${Cl}_2=\frac{\left(\alpha \right)}{(1+\alpha )}P$
$K_p=\frac{P_{PCl3}{P}_{Cl2}}{P_{PCl5}}$
$\therefore K_p=\frac{\frac{\alpha }{1+\alpha }P\times \frac{\alpha }{1+\alpha }P}{\frac{1-\alpha }{1+\alpha }P}$
$\therefore K_p=\frac{{\alpha }^{2}P}{1-{\alpha }^2}$
Assume $1-{\alpha }^2=1$ as the degree of dissociation is small.
$K_p={\alpha }^{2}P$
$\therefore \alpha =\sqrt{\frac{K_p}{P}}$
$\alpha \propto \frac{1}{\sqrt{P}}$