Question

At 473K, partially dissociated vapours of ${PCl}_5$ are 62 times as heavy as $H_2$. Calculate the degree of dissociation of ${PCl}_5$.

• A. $\alpha =19.367$
• B. $\alpha =3.4684$
• C. $\alpha =1.37$
• D. $\alpha =0.6814$

Answer 1

The correct option is D $\alpha =0.6814$

${PCl}_5\rightleftharpoons {PCl}_3+{Cl}_2$
$1$ $0$ $0$ Moles before dissociation
$(1-\alpha )$ $\alpha$ $\alpha$ Moles after dissociation

Given = $0.2$ at $1$ atm pressure
$K_P=\frac{n_{{PCl}_3}\times n_{{Cl}_2}}{n_{{PCl}_5}}\times {\left[\frac{P}{\sum n}\right]}^{△n}$
$=\frac{\alpha .\alpha }{(1+\alpha )}\left[\frac{P}{1+\alpha }\right]=\frac{P{\alpha }^2}{1-{\alpha }^2}=\frac{1\times {\left(0.2\right)}^2}{1-{\left(0.2\right)}^2}$
$K_P=0.0416atm$
Again when $\alpha$ is desired at 0.5, $K_P$ remains constant and thus,
$K_{SP}=\frac{P{\alpha }^2}{1-{\alpha }^2}$
$0.0416=\frac{P\times {\left(0.5\right)}^2}{1-{\left(0.5\right)}^2}$
$\therefore P=0.1248atm$