Question

A sample of PCl₅ was placed in a vessel with an original concentration of 0.40 mol/L. After the PCl₅ has decomposed and equilibrium has been established, 25% of the PCl₅ remains.

PCl₅(g) ---> PCl₃(g) + Cl₂(g)

What are the equilibrium concentrations of PCl₅, PCl₃, and Cl₂?

Answer 1

The equilibrium concentrations of PCl₅, PCl₃, and Cl₂ when 25% of the original PCl₅ remains are 0.10 mol/L, 0.30 mol/L, and 0.30 mol/L, respectively.

Step-by-step explanation:

To determine the equilibrium concentrations of PCl₅, PCl₃, and Cl₂, we start with the initial concentration of PCl₅ which is given as 0.40 mol/L. At equilibrium, 25% of PCl₅ means that 75% has decomposed.

Let's denote the initial concentration of PCl₅ as [PCl₅]i = 0.40 mol/L. The amount decomposed at equilibrium is 0.75 * [PCl₅]i, giving us the change in concentration of PCl₅, PCl₃, and Cl₂.

The decomposition reaction of PCl₅ is as follows:
PCl₅ (g) → PCl₃ (g) + Cl₂ (g).
This indicates that the concentration change for PCl₅ is -x, for PCl₃ is +x, and for Cl₂ is also +x.

Therefore, at equilibrium we have:
[PCl₅] = [PCl₅]i - (0.75 * [PCl₅]i) = 0.10 mol/L,
[PCl₃] = 0 + (0.75 * [PCl₅]i) = 0.30 mol/L,
[Cl₂] = 0 + (0.75 * [PCl₅]i) = 0.30 mol/L.