Question

The equlibrium constant K, for the following reaction is 2.00 x 10-2 at 613 K.

COCl2(g) CO(g) + Cl2(g)

An equilibrium mixture of the three gases in a 18.4 L container at 613 K contains 0.194 M COCl2, 6.22 x 10-2 M CO and 6.22 x10-2 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 7.62 L?

[COCl2]= _____________M

[CO]=____________M

[Cl2]=_____________M

Answer 1

COCl2(g) <----> CO(g) + Cl2(g)

Kc = [CO][Cl2]/[cocL2]

after decreasing volume

initial concentration of [COCl2] = 0.194*18.4/7.62 = 0.468 M

initial concentration of [cO] = 0.0622*18.4/7.62 =0.15 M

initial concentration of [CL2] = 0.0622*18.4/7.62 =0.15 M

generally after decreasing volume, partial pressures increases,so that equilibrium shifts towards less no of moles side.

0.02 = (0.15-x)^2/(0.468+x)

x = 0.0484

at equilibrium,

concentration of [COCl2] = 0.468+x = 0.468+0.0484 = 0.5164 M

concentration of [cO] = 0.15-x = 0.15-0.0484 = 0.1 M

concentration of [CL2] = 0.15-x = 0.15-0.0484 = 0.1 M