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