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 approximately related to the pressure at equilibrium by :(given $\alpha <<1$)

• 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: (A)

$\alpha \propto P$

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

initial $1$ $0$ $0$

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

$\alpha =$ degree of dissociation

$K_P=\frac{\left[P_{{Cl}_2}\right]\left[P_{{PCl}_3}\right]}{\left[P_{{PCl}_5}\right]}=\frac{\left(\frac{\alpha }{1+\alpha }\right)P\left(\frac{\alpha }{1+\alpha }\right)P}{\left(\frac{1-\alpha }{1+\alpha }\right)P}$

$K_p=\frac{{\alpha }^{2}P}{(1-\alpha )(1+\alpha )}$

total moles $=1-\alpha +\alpha +\alpha =(1+\alpha )$

$K_p=\left(\frac{{\alpha }^2}{1-{\alpha }^2}\right)P$

$\alpha =\sqrt{\frac{K_p}{P}}$ if $1-{\alpha }^2=1$

$K_p$ is constant.

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