Question

At ${250}^{o}C$ and $1$ atmospheric pressure, the vapour density of $P{Cl}_5$ is $57.9$.

Calculate:

(i) $K_p$ for the reaction, $P{Cl}_5\rightleftharpoons P{Cl}_3\left(g\right)+{Cl}_2\left(g\right)$ at ${250}^{o}C$

(ii) the percentage dissociation when pressure is doubled.

Answer 1

(i) Molecular mass of $P{Cl}_5=208.5$
Vapour density, $D=\frac{208.5}{2}=104.25$
Observed vapour density, $d=57.9$
Degree of dissociation, $\alpha =\frac{D-d}{d}=0.80$
$P{Cl}_5\rightleftharpoons P{Cl}_3\left(g\right)+{Cl}_2\left(g\right)$
At equilibrium: $1-\alpha$ $\alpha$ $\alpha$
$(1-0.8)$ $0.80$ $0.80$
Total number of moles$=1+\alpha =1.80$
Partial pressure of $P{Cl}_5=\frac{0.2}{1.80}\times 1=\frac{1}{9}$
Partial pressure of $P{Cl}_3=\frac{0.80}{1.80}\times 1=\frac{4}{9}$
Partial pressure of ${Cl}_2=\frac{0.80}{1.80}\times 1=\frac{4}{9}$
So, $K_p=\frac{p_{P{Cl}_3}\times p_{{Cl}_2}}{p_{P{Cl}_5}}=1.78$
(ii) Let the degree of dissociation be $\alpha$.
At equilibrium.
$p_{P{Cl}_5}=\left(\frac{1-\alpha }{1+\alpha }\right).P=\frac{1-\alpha }{1+\alpha }\times 2$
$p_{P{Cl}_3}=\left(\frac{\alpha }{1+\alpha }\right).P=\frac{\alpha }{1+\alpha }\times 2$
$p_{{Cl}_2}==\left(\frac{\alpha }{1+\alpha }\right).P=\frac{\alpha }{1+\alpha }\times 2$
$K_p=\frac{\frac{\alpha }{1+\alpha }\times 2\times \frac{\alpha }{1+\alpha }\times 2}{\frac{1-\alpha }{1+\alpha }\times 2}=1.78$
or ${\alpha }^2=\frac{0.89}{1.89}$
$\alpha =0.686$

Thus, $P{Cl}_5$ is $68.6$% dissociated.