Question

PCl₃(g)+Cl₂(g)⇄PCl₅(g) Kc=0.11 A 0.60mol sample of PCl₃(g) and a 0.70mol sample of Cl₂(g) are placed in a previously evacuated 1.0L rigid container, and the reaction represented above takes place. At equilibrium, the concentration of PCl₅(g) in the container is 0.040M.

Answer 1

The concentration of PCl₃(g) at equilibrium is approximately 0.34 M.

To determine the concentration of PCl₃ at equilibrium, we can use the given equilibrium constant (Kc) value, which is 0.11, and the concentrations of PCl₅ at equilibrium (0.040 M). The balanced chemical equation indicates that one mole of PCl₃ reacts with one mole of Cl₂ to produce one mole of PCl₅. Therefore, at equilibrium, the change in concentration of PCl₃ (x) and Cl₂ (x) will be equal.

Setting up an ICE (Initial, Change, Equilibrium) table, we start with the initial concentrations of PCl₃ and Cl₂, which are 0.60 M and 0.70 M, respectively. The change (x) represents the decrease in the concentrations of PCl₃ and Cl₂ and the increase in the concentration of PCl₅. At equilibrium, the concentrations are determined by adding the initial concentration and the change.

The equilibrium expression for the reaction is Kc= [PCl5​]/[PCl3​ ][Cl2​]. Substituting the known values into this expression and solving for x, we find x≈0.34. Therefore, at equilibrium, the concentration of PCl₃ is approximately 0.34 M. This calculation assumes that the reaction reaches equilibrium, and the reaction quotient Qc is close to the given equilibrium constant Kc.