MCQ
$\text{PCl}_5(\text{g})\rightleftharpoons\text{PCl}_3(\text{g}),$ if the degree of dissociation is a at equilibrium pressure$ 'p',$ then the equilibrium constant for the reaction is:
  • A
    $\text{K}_{\text{p}}=\frac{\alpha^2}{1+\alpha^2\text{P}}$
  • B
    $\text{K}_{\text{p}}=\frac{\alpha^2\text{P}^2}{1-\alpha^2}$
  • C
    $\text{K}_{\text{p}}=\frac{\alpha\text{P}^2}{1-\alpha^2}$
  • $\text{K}_{\text{p}}=\frac{\alpha^2\text{P}}{1-\alpha^2}$

Answer

Correct option: D.
$\text{K}_{\text{p}}=\frac{\alpha^2\text{P}}{1-\alpha^2}$
$\text{PCl}_5(\text{g})\rightleftharpoons\text{PCl}_3(\text{g})+\text{Cl}_2(\text{g})$
$\begin{matrix}\text{Initial}&1&0&0\\\text{At equilibrium}&1-\alpha&\alpha&\alpha\end{matrix}$
Total no. of moles $=1-\alpha+\alpha+\alpha=1+\alpha$
$\text{pPCl}_5=\frac{1-\alpha}{1+\alpha}$
$\text{pCl}_2=\text{pCl}_3=\frac{\alpha}{1+\alpha}$
$\text{K}_{\text{p}}=\frac{(\text{PCl}_3)(\text{Cl}_2)}{(\text{PCl}^-_3)}$
$=\frac{\frac{\alpha}{1+\alpha}\times\frac{\alpha}{1+\alpha}\times\text{p}^2}{\frac{1-\alpha}{1+\alpha}\times\text{p}}$
$=\frac{\alpha^2}{1-\alpha^2}\times\text{p}$

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