A 0.210 mol sample of pcl5 (g) is injected into an empty 2.45 l reaction vessel held at 250 °c. calculate the concentrations of pcl5 (g) and pcl3 (g) at equilibrium.

When the balanced equation for this reaction is:

PCl5 (g) ↔ PCl3 (g) + Cl2 (g)

initial C (0.21/2.45) 0 0

change - X + X + X

equilibruimC (0.21/2.45) - X X X

So by substitution in Ka formula: when we have Ka at 250°C = 1.8 (must be given - missing in your question)

Ka = [PCl3][Cl2]/[PCl5]

1.8 = (X) (X) / (0.21/2.45) - X

1.8*0.086 - 1.8X = X^2

X^2 + 1.8X - 0.1548 = 0 by solving this equation

X = 0.082 mol

∴[PCl5] = (0.21/2.45) - X

= 0.0857 - 0.082 = 0.0037 mol

∴[PCl3] = X = 0.082 mol