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(A) The dissociation reaction is: $PCl_5(g) ⇌ PCl_3(g) + Cl_2(g)$ Initial moles: $2 \, mol$ of $PCl_5$,$0$ of $PCl_3$,$0$ of $Cl_2$. At equilibrium,$4

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$0.266$

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(A) The dissociation reaction is: $PCl_5(g) ⇌ PCl_3(g) + Cl_2(g)$ Initial moles: $2 \, mol$ of $PCl_5$,$0$ of $PCl_3$,$0$ of $Cl_2$. At equilibrium,$40\%$ of $PCl_5$ dissociates,so moles reacted $= 2 	imes 0.4 = 0.8 \, mol$. Moles at equilibrium: $n(PCl_5) = 2 - 0.8 = 1.2 \, mol$ $n(PCl_3) = 0.8 \, mol$ $n(Cl_2) = 0.8 \, mol$ Concentrations at equilibrium (Volume $= 2 \, L$): $[PCl_5] = 1.2 / 2 = 0.6 \, M$ $[PCl_3] = 0.8 / 2 = 0.4 \, M$ $[Cl_2] = 0.8 / 2 = 0.4 \, M$ $K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{0.4 	imes 0.4}{0.6} = \frac{0.16}{0.6} = 0.266$.

When $2 \, mol$ of $PCl_5$ is heated in a closed vessel of $2 \, L$ capacity,$40\%$ of $PCl_5$ dissociates into $PCl_3$ and $Cl_2$ at equilibrium. The value of the equilibrium constant $K_c$ will be:

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