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(B) For the reaction $PCl_5 \rightleftharpoons PCl_3 + Cl_2$,let the initial moles of $PCl_5$ be $1$.At equilibrium,moles are: $PCl_5 = (1 - \alpha)$,

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$0.05$

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(B) For the reaction $PCl_5 ightleftharpoons PCl_3 + Cl_2$,let the initial moles of $PCl_5$ be $1$.At equilibrium,moles are: $PCl_5 = (1 - \alpha)$,$PCl_3 = \alpha$,$Cl_2 = \alpha$,where $\alpha = 0.2$.Total moles at equilibrium $= 1 - \alpha + \alpha + \alpha = 1 + \alpha = 1.2$.Assuming volume $V = 1 \ L$,the concentrations are $[PCl_5] = 0.8$,$[PCl_3] = 0.2$,$[Cl_2] = 0.2$.$K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{0.2 	imes 0.2}{0.8} = \frac{0.04}{0.8} = 0.05$.

If the degree of dissociation for the reaction,$PCl_5 \rightleftharpoons PCl_3 + Cl_2$ is $20\%$ at $1 \ atm$ pressure,calculate $K_c$.

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