Question

The degree of dissociation of $PC{l}_5\left(g\right)$ obeying the equilibrium, $PC{l}_5\rightleftharpoons PC{l}_3+C{l}_2$, is approximately 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. $\alpha \propto \frac{1}{P^4}$

Answer 1

The correct option is A $\alpha \propto P$

${PCl}_5\rightleftharpoons {PCl}_3+{Cl}_2$

Initial $1$ $0$ $0$

equilibrium $1-x$ $x$ $x$

Total number of moles at equilibrium $=1-x+x+x=1+x$

We know that partial pressure is the product of mole fraction and the total pressure. Therefore

$P_{{PCl}_5}=\frac{(1-x)P}{(1+x)}$

$P_{{PCl}_3}=\frac{xP}{1+x}$

$P_{{PCl}_2}=\frac{xP}{1+x}$

We know that

$K_P=\frac{P_{{PCl}_3}.P_{{Cl}_2}}{P_{{PCl}_5}}$

$K_P=\frac{x^2{P}^2}{{(1+x)}^2}(1+x)(1+x).\frac{1}{P}$

$K_P=\frac{x^{2}P}{1-x^2}$ when $X<<1$

$K_P=x^{2}P$

This equation can be used to predict the influence of pressure on this equilibrium.