Question

The degree of dissociation of ${PCl}_5$ at 1 atm pressure is 0.2. Calculate the pressure at which ${PCl}_5$ is dissociated to 50%?

• A. $P=22.19atm$
• B. $P=0.1248atm$
• C. $P=2.165atm$
• D. $P=1.7atm$

Answer 1

Answer: (B)

$P=0.1248atm$

${PCl}_5\rightleftharpoons {PCl}_3+{Cl}_2$
1 0 0 Moles before dissociation
$(1-\alpha )$ $\alpha$ $\alpha$ Moles after dissociation
Given $\alpha =0.2$ 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,
$K_P=\frac{P{\alpha }^2}{1-{\alpha }^2}$
$0.0416=\frac{P\times {\left(0.5\right)}^2}{1-{\left(0.5\right)}^2}$
$\therefore P=0.1248atm$