Question

The degree of dissociation of $PCl_{5}\left(\right.g\left.\right)$ for the equilibrium $PCl_{5}\left(g\right)\rightleftharpoons PCl_{3}\left(g\right)+Cl_{2}\left(\right.g\left.\right)$ is approximately related to the pressure at equilibrium $(P)$ by the relation $\left[\right.\alpha \ll1\left]\right.$

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

Answer 1

Answer: (B)

$\therefore K_{P}=\frac{\left(\frac{\alpha P}{1 + \alpha }\right) \left(\frac{\alpha }{1 + \alpha } P\right)}{\left(\frac{1 - \alpha }{1 + \alpha }\right) P}$
$=\frac{\left(\alpha \right)^{2}}{1 - \left(\alpha \right)^{2}}P\approx\left(\alpha \right)^{2}P\left(\alpha \ll 1\right)$
Neglecting ' $\alpha ^{2}$ '
$K_{P}=\alpha ^{2}P$
$\therefore \alpha =\frac{\sqrt{K_{P}}}{\sqrt{P}}$