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0.200 mol of PCl₅ is put in a 1.00 L flask and heated…

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$0.200$ mol of $\text{PCl}_5$ is put in a $1.00$ L flask and heated. It splits: $\text{PCl}_5\text{(g)} \rightleftharpoons \text{PCl}_3\text{(g)} + \text{Cl}_2\text{(g)}$ , with $K_c = 0.0400$ at this temperature. Find the equilibrium concentrations and the percentage of $\text{PCl}_5$ that splits.

Need a reminder? The key ideaShow

For $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$ : $K_c = \dfrac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$ . $K_p = K_c(RT)^{\Delta n}$ with $R = 0.0831$ L bar K $^{-1}$ mol $^{-1}$ . Pure solids and liquids are left out. $Q_c < K_c$ : forward; $Q_c > K_c$ : backward. $\Delta G^\circ = -2.303\,RT\log K$ .

The worked solution

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0.200 mol of PCl₅ is put in a 1.00 L flask and heated…

The worked solution

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