Two moles of PCl₅ are placed in a 5.0 L container. Dissociation takes place according to the equation PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). At equilibrium, 0.40 mol of Cl₂ are present. Calculate the equilibrium constant (K_c) for this reaction under the conditions of this experiment.

Question

Quizplus · Accepted Answer

Initially, 2 moles of PCl5 are present in a 5.0 L container, so the initial concentration of PCl5 is 0.4 M. At equilibrium, 0.40 mol of Cl2 is present, meaning 0.40 mol of PCl5 has dissociated, resulting in 0.40 mol of PCl3 and 0.40 mol of Cl2. The equilibrium concentrations are: [PCl5] = 0.4 - 0.08 = 0.32 M, [PCl3] = 0.08 M, [Cl2] = 0.08 M. Kc = ([PCl3][Cl2])/[PCl5] = (0.08 * 0.08) / 0.32 = 0.020.

Two Moles of PCl5 Are Placed in a 5

Two Moles of PCl₅ Are Placed in a 5