# Exercise 10 | Chemical Equilibrium

PCl(g) $⇄$ PCl(g) + Cl(g)     [KC = 0.040 M]

What is the equilibrium concentration of Cl2 in the system if the reaction was initiated with 0.40 mole of Cl2 and 0.40 mole of PCl3 in a 1.5 L container?

KC =

[Cl2]0 = [PCl3]0 = $\frac{0.40}{1.5}$ = 2.7 x 10-1 mol.L-1

From the stoichiometry of the chemical equation: 1 mole of Cl2 reacts with one mole of PCl3 to form one mole of PCl5:

[Cl2]eq = [Cl2]0 – x     with x = change in the number of moles of Cl2 per liter

[PCl3]eq = [PCl3]0 – x = [Cl2]0 – x

[PCl5]eq = x

KC =  =

KC x = [Cl2]02 – 2 [Cl2]0 x + x2

⇒ x2 – (2[Cl2]0 + KC) x + [Cl2]02 = 0

⇒ x = 0.40 or x = 0.18

So [Cl2]eq = [Cl2]0 – x = 0.27 - 0.40  < 0 ⇒ this is impossible

or [Cl2]eq = [Cl2]0 – x = 0.27 - 0.18 = 0.08 mol.L-1