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(D) The equilibrium reaction is: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$Given volume $V = 1000\, cm^3 = 1\, L$.The equilibrium concentr

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$74$

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(D) The equilibrium reaction is: $PCl_{5(g)} ightleftharpoons PCl_{3(g)} + Cl_{2(g)}$Given volume $V = 1000\, cm^3 = 1\, L$.The equilibrium concentrations are:$[PCl_5] = 2\, mol / 1\, L = 2\, M$$[PCl_3] = 2\, mol / 1\, L = 2\, M$$[Cl_2] = 3\, mol / 1\, L = 3\, M$Calculate $K_C$:$K_C = \frac{[PCl_3][Cl_2]}{[PCl_5]} = \frac{2 	imes 3}{2} = 3$Calculate $K_P$ using the relation $K_P = K_C(RT)^{\Delta n_g}$:Here,$\Delta n_g = (1 + 1) - 1 = 1$,$R = 0.0821\, L \cdot atm \cdot K^{-1} \cdot mol^{-1}$,and $T = 27 + 273 = 300\, K$.$K_P = 3 	imes (0.0821 	imes 300)^1 = 3 	imes 24.63 = 73.89 \approx 74\, atm$.

In a closed container of $1000\, cm^3$,$2\, mol$ of $PCl_5$,$2\, mol$ of $PCl_3$,and $3\, mol$ of $Cl_2$ are found to be at equilibrium at $27\, ^oC$. Then $K_P$ for the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ at $27\, ^oC$ is $.....$ $atm$.

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