The degree of dissociation of PCl_{5(g)} at 1 | vedclass

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(A) Given,Total pressure $P = 16.8 \ bar$ and degree of dissociation $\alpha = 0.4$.For the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2

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$3.2 \ bar$

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(A) Given,Total pressure $P = 16.8 \ bar$ and degree of dissociation $\alpha = 0.4$.For the reaction $PCl_{5(g)} ightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,the equilibrium moles are $(1-\alpha)$,$\alpha$,and $\alpha$ respectively.Total moles at equilibrium $= 1-\alpha + \alpha + \alpha = 1+\alpha$.The partial pressures are $p_{PCl_5} = \frac{1-\alpha}{1+\alpha} P$,$p_{PCl_3} = \frac{\alpha}{1+\alpha} P$,and $p_{Cl_2} = \frac{\alpha}{1+\alpha} P$.The expression for $K_p$ is $K_p = \frac{p_{PCl_3} 	imes p_{Cl_2}}{p_{PCl_5}} = \frac{\alpha^2 P}{1-\alpha^2}$.Substituting the values: $K_p = \frac{(0.4)^2 	imes 16.8}{1-(0.4)^2} = \frac{0.16 	imes 16.8}{1-0.16} = \frac{0.16 	imes 16.8}{0.84} = \frac{2.688}{0.84} = 3.2 \ bar$.

The degree of dissociation of $PCl_{5(g)}$ at $16.8 \ bar$ and $127 \ ^oC$ is $0.4$. The value of $K_p$ for the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ is

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