Question

8 g mol of $PC{l}_5$ is heated in a closed vessel of 10 L capacity and 25% of it dissociates into $PC{l}_3$ and $C{l}_2$ at the equilibrium, then value of $K_p$ will be equal to:

• A. $\frac{P}{30}$
• B. $\frac{P}{15}$
• C. $\frac{2}{3P}$
• D. $\frac{3}{2P}$

Answer 1

The correct option is B $\frac{P}{15}$

$\begin{array}{cccccc}& PC{l}_5\left(g\right)& \rightleftharpoons & PC{l}_3\left(g\right)& +& C{l}_2\left(g\right)\\ \text{initially}& 8& & 0& & 0\\ \text{At eq.}& (8-8\alpha )& & 8\alpha & & 8\alpha \end{array}$
Degree of dissociation $\left(\alpha \right)$ = 25% or 0.25

Total moles $=8-8\alpha +8\alpha +8\alpha$
$\text{Total moles}=8+8\alpha =8+8\times \frac{25}{100}=10$

$X_{PC{l}_5}=\frac{8-8\alpha }{10},X_{PC{l}_3}=\frac{8\alpha }{10},X_{C{l}_2}=\frac{8\alpha }{10}$
$P_{PC{l}_5}=\frac{6}{10}\times P,P_{PC{l}_3}=\frac{2}{10}\times P_{C{l}_2}=\frac{2}{10}\times P$
$K_p=\frac{P_{PC{l}_3}\times P_{C{l}_2}}{P_{PC{l}_5}}=\frac{(\frac{2}{10}\times P)\times (\frac{2}{10}\times P)}{(\frac{6}{10}\times P)}$
$=\frac{4P}{60}=\frac{P}{15}$