WBJEE · Chemistry · Chemical Equilibrium
Consider the following gas phase dissociation, \(\mathrm{PCl}_5(\mathrm{~g}) \rightleftharpoons \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})\) with equilibrium constant \(\mathrm{K}_{\mathrm{p}}\) at a particular temperature and at pressure \(P\). The degree of dissociation \((\alpha)\) for \(\mathrm{PCl}_5(\mathrm{~g})\) is
- A \(\alpha=\left(\frac{K_P}{K_P+P}\right)^{1 / 3}\)
- B \(\alpha=\left(\frac{\mathrm{K}_{\mathrm{P}}}{\mathrm{K}_{\mathrm{P}}+\mathrm{P}}\right)\)
- C \(\alpha=\left(\frac{\mathrm{K}_{\mathrm{P}}}{\mathrm{K}_{\mathrm{P}}+\mathrm{P}}\right)^{1 / 2}\)
- D \(\alpha=\left(\frac{\mathrm{K}_{\mathrm{P}}}{\mathrm{K}_{\mathrm{P}}+\mathrm{P}}\right)^2\)
Answer & Solution
Correct Answer
(C) \(\alpha=\left(\frac{\mathrm{K}_{\mathrm{P}}}{\mathrm{K}_{\mathrm{P}}+\mathrm{P}}\right)^{1 / 2}\)
Step-by-step Solution
Detailed explanation
\begin{array}{lccc} & \mathrm{PCl}_5(\mathrm{g}) &\rightleftharpoons & \mathrm{PCl}_3(\mathrm{g}) & + & \mathrm{Cl}_2(\mathrm{g}) \\ \text{initially} & 1 & & 0 & & 0 \\ \text{moles taken} & & & \\ \begin{array}{l} \text{moles of equilibrium} \end{array} & 1-\alpha & & \alpha & &…
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