Question

Consider the following equilibrium in a closed container:
$PC{l}_5\left(g\right)\rightleftharpoons PC{l}_3\left(g\right)+C{l}_2\left(g\right)$
$PC{l}_5$ gas at a certain pressure is introduced in the container at $27{}^{o}C$. However, the total pressure at equilibrium at $207{}^{o}C$ was found to be double the initial value. The % dissociation of $PC{l}_5$ at $207{}^{o}C$ is: %

Answer 1

As per Gay-Lussac's law,
$\frac{P_1}{T_1}=\frac{P_2}{T_2}$
If,
$T_1=27{}^{o}C=300K\Rightarrow P_1=P\left(say\right)$
For $T_2=207{}^{o}C=480K\Rightarrow P_2=?$
$\therefore P_2=\left[\frac{T_2}{T_1}\right]\times P_1=\frac{480}{300}\times P$

$\begin{array}{cccccc}PC{l}_5\left(g\right)& \rightleftharpoons & PC{l}_3\left(g\right)& +& C{l}_2\left(g\right)& \\ P& & 0& & 0& \text{t=0 at}27{}^{o}C\\ \frac{480P}{300}& & 0& & 0& \text{t=0 at}207{}^{o}C\\ \frac{480P}{300}-P^{{}^{\prime }}& & P^{{}^{\prime }}& & P^{{}^{\prime }}& \text{t=t at}207{}^{o}C\end{array}$

$\therefore$ Total pressure at equilibrium at $207{}^{o}C$ is
$\frac{480P}{300}-P^{{}^{\prime }}+2P^{{}^{\prime }}$ = $2P$ (given)
$\therefore \frac{8P}{5}+P^{{}^{\prime }}=2P$
$\therefore P^{{}^{\prime }}=\frac{2P}{5}$
$\therefore$ % dissociation of $PC{l}_5$ at $207{}^{o}C$ = $\frac{P^{{}^{\prime }}}{\left[\frac{480P}{300}\right]}\times 100=\frac{\frac{2P}{5}}{\frac{8P}{5}}\times 100=25\%$