Question

A sample of pure PCl₅ was introduced into an evacuated vessel at 473 K. After equilibrium was attained, the concentration of PCl₅ was found to be 0.5 × 10^–1 mol L^–1. If the value of K_c is 8.3 × 10^–3, what are the concentrations of PCl₃ and Cl₂ at equilibrium?

\[\ce{PCl5 (g) ⇌ PCl3 (g) + Cl2(g)}\]

Answer 1

Let the concentrations of both PCl₃ and Cl₂ at equilibrium be x mol L^–1. The given reaction is:

	PCl5(g)	↔	PCl3(g)	+	Cl2(g)
At equilibrium	0.5 × 10-1 mol L-1		x mol L-1		x mol L-1

 It is given that the value of equilibrium constant, `"K"_"C"`  is `8.3 xx 10^(-3)`

Now we can write the expression for equilibrium as:

`(["PCl"_2]["Cl"_2])/["PCl"_5] = "K"_"C"`

`=> (x xx x)/(0.5 xx 10^(-1)) = 8.3 xx 10^(-1)`

`=> x^2 = 4.15 xx 10^(-4)`

`=> x = 2.04 xx 10^(-2)`

= 0.0204

= 0.02 (approximately)

Therefore at equilibrium.

`["PCl"_3] = ["Cl"_2] = 0.02 " mol L"^(-1)`