In a 1 L closed container,4 moles of PCl_5 | vedclass

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(C) The dissociation reaction is: $PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)$.Initial moles: $PCl_5 = 4$,$PCl_3 = 0$,$Cl_2 = 0$.At equilibrium,$8

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$0.2$

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(C) The dissociation reaction is: $PCl_5(g) ightleftharpoons PCl_3(g) + Cl_2(g)$.Initial moles: $PCl_5 = 4$,$PCl_3 = 0$,$Cl_2 = 0$.At equilibrium,$80\%$ of $PCl_5$ remains undissociated,so the amount of $PCl_5$ dissociated is $20\%$ of $4 \ moles = 0.2 	imes 4 = 0.8 \ moles$.Moles at equilibrium: $PCl_5 = 4 - 0.8 = 3.2 \ moles$,$PCl_3 = 0.8 \ moles$,$Cl_2 = 0.8 \ moles$.Since the volume is $1 \ L$,the concentrations are: $[PCl_5] = 3.2 \ M$,$[PCl_3] = 0.8 \ M$,$[Cl_2] = 0.8 \ M$.The equilibrium constant $K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{0.8 	imes 0.8}{3.2} = \frac{0.64}{3.2} = 0.2$.

In a $1 \ L$ closed container,$4 \ moles$ of $PCl_5$ are heated. If $80\%$ of $PCl_5$ remains undissociated at equilibrium,what is the value of the equilibrium constant $(K_c)$?

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