Question

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

• A. $P/30$
• B. $P/15$
• C. $2/3P$
• D. $3/2P$

Answer 1

Answer: (B)

$P/15$

${PCl}_5\left(g\right)\rightleftharpoons {PCl}_3+{Cl}_{2}10L$ vessel

$25$% dissociated moles of ${PCl}_5=\frac{8}{208}=\frac{1}{26}$

$\alpha =0.25$

${PCl}_5\rightleftharpoons {PCl}_3+{Cl}_2$

At $t=0$ $\frac{1}{26}$ $0$ $0$

At $\frac{1}{26}(1-\alpha )$ $\frac{\alpha }{26}$ $\alpha /26$

$t=eq$

Let $P$ be total pressure at equilibrium.

$P^{{Cl}_2}=\frac{\frac{\alpha }{26}}{\frac{1}{26}(1+\alpha )}P\Rightarrow P/5$

$P_{{PCl}_3}=P/5$

$P_{{PCl}_5}=3P/5$ $K_P=\frac{P/5.P/5}{3P/5}\Rightarrow \frac{P}{15}$