Question

The equilibrium constant, Kc, for the following reaction is 1.20×10-2 at 500 K. PCl5(g) PCl3(g) + Cl2(g) Calculate the equilibrium concentrations of reactant and products when 0.390 moles of PCl5(g) are introduced into a 1.00 L vessel at 500 K.

Answer 1

PCl5 (g) <-----> PCl3 (g) + Cl2 (g)

at equilibrium             0.390-X               X                    X

[PCl5] = 0.390-X) /1 , [PCl3] =[Cl2] = X/1

K = [PCl3] [Cl2] /[PCl5]

0.012 = X^2 / ( 0.390-X)

X^2 + 0.012X - 0.00468 = 0

X=0.06265

[PCl5] = 0.390-0.06265= 0.32735 M

[PCl3] =0.06265 M

[Cl2] = 0.06265 M