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(C) Species$PCl_{5(g)}$$PCl_{3(g)}$$Cl_{2(g)}$Initial moles$1$$0$$0$Equilibrium moles$1 - \alpha$$\alpha$$\alpha$Total moles at equilibrium = $(1 - \a

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$\alpha = \left( \frac{K_P}{K_P + P} \right)^{1/2}$

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(C) Species$PCl_{5(g)}$$PCl_{3(g)}$$Cl_{2(g)}$Initial moles$1$$0$$0$Equilibrium moles$1 - \alpha$$\alpha$$\alpha$Total moles at equilibrium = $(1 - \alpha) + \alpha + \alpha = 1 + \alpha$.Partial pressures: $P_{PCl_5} = \frac{1 - \alpha}{1 + \alpha} P$,$P_{PCl_3} = \frac{\alpha}{1 + \alpha} P$,$P_{Cl_2} = \frac{\alpha}{1 + \alpha} P$.Equilibrium constant $K_P = \frac{P_{PCl_3} \cdot P_{Cl_2}}{P_{PCl_5}} = \frac{(\frac{\alpha}{1 + \alpha} P)(\frac{\alpha}{1 + \alpha} P)}{(\frac{1 - \alpha}{1 + \alpha} P)} = \frac{\alpha^2 P}{1 - \alpha^2}$.Rearranging: $K_P(1 - \alpha^2) = \alpha^2 P \implies K_P = \alpha^2(P + K_P) \implies \alpha^2 = \frac{K_P}{K_P + P}$.Therefore,$\alpha = \left( \frac{K_P}{K_P + P} \right)^{1/2}$.

Consider the following gas phase dissociation,$PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ with equilibrium constant $K_P$ at a particular temperature and at pressure $P$. The degree of dissociation $(\alpha)$ for $PCl_{5(g)}$ is

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