Question

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

PCl₅(g) ⇌ PCl₃(g) + Cl₂(g)

Answer 1

K_C = \(\frac{x+x}{0.5\times 10^{-1}}\)

8.3 x 10^-3 = \(\frac{x^2}{0.5\times 10^{-1}}\)

or, x² = 8.3 x 10^-3 x 0.5 x 10^-1 = 4.15 x 10^-4

x = 2.04 x 10^-2

[PCl₅] = 0.02 mol^-1

[Cl₂] = 0.02 mol L^-1