Question

If $8$ moles of $PC{l}_5$ heated in a closed vessel of $10$ litres capacity and $25\%$ of its disassociates into $PC{l}_3$ and $C{l}_2$ at the equilibrium then the value of Kp 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 $\frac{p}{15}$

Let $P_t$ is total pressure

$PC{l}_5\leftrightarrow PC{l}_3+C{l}_2$

$t=0$ $8$ $0$ $0$

$t=\left(eq\right)$ $8-2$ $2$ $2$ ($\because [PC{l}_5{]}_{eqm}=8-(25/100)\times 8=8-2=6$)

Total moles at equilibrium= $8-2+2+2=10$

$K_p=\frac{P_{PC{l}_3}\times P_{C{l}_2}}{P_{PC{l}_5}}=\frac{(\frac{2}{10}{P}_t\times \frac{2}{10}{P}_t)}{\frac{6}{10}{P}_t}=\frac{1}{15}{P}_t$