What will be the concentrations of the three gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of 10.0 L?

Question

The equilibrium constant, K, for the following reaction is 2.21×10^-2 at 616 K.

1. COCl₂(g) CO(g) + Cl₂(g)

An equilibrium mixture of the three gases in a 17.7 L container at 616 K contains 0.157 M COCl₂,  5.89×10^-2 M CO and 5.89×10^-2 M Cl₂. What will be the concentrations of the three gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of 10.0 L?

[COCl2]	=	M
[CO]	=	M
[Cl2]	=	M

Answer 1

mols COCl2 = 0.157 mol/L x 17.7 L = 2.779 mols

mols CO = 5.89x10-2 mol/L x 17.7 L = 1.043 mols

mols Cl2 = 5.89x10-2 mol/L x 17.7 = 1.043 mols

COCl2(g) <===> CO(g) + Cl2(g) Keq = 2.21x10^-2

2.779...................1.043......1.043..........Initial

-x........................+x............+x...............Change

2.779-x.............1.043+x.....1.043+x......Equilibrium

2.21x10^-2 = (1.043+x)² / (2.779-x)

Solve for x and then to find concentrations after compression to 10.0 L perform the following calculations:

[COCl2] = 2.779-x / 10.0 = answer

[CO] = 1.043-x / 10.0 = answer

[Cl2] = 1.043-x / 10.0 = answer