Question

The equilibrium constant, K, for the following reaction is 1.29×10-2 at 600 K. COCl2(g) CO(g) + Cl2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 600 K contains 0.244 M COCl2, 5.61×10-2 M CO and 5.61×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.44×10-2 mol of CO(g) is added to the flask?

[COCl2] = M

[CO] = M

[Cl2] = M

Answer 1

i) The given reaction is

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

Equillibrium expresssion ,K_c , for this equillibrium is as follows

K_c = [CO][Cl2]/[COCl2]

K_c value is 1.29×10^-2

ii) Make ICE table   

Concentration	[COCl2]	[CO]	[Cl2]
Initial concentration	0.244	0.0561+0.0344 = 0.0905	0.0561
change in concentration	- x	+ x	+ x
Equillibrium concentration	0.244 -x	0.0905 + x	0.0561 + x

iii) From ICE table we know that

(( 0.0905 + x)(0.0561 +x))/(0.244 -x )= 1.29×10^-2

solving for x

x = - 0.01319

iv) Concentration at equillibrium

[COCl2] = 0.244 -(- 0.01319) = 0.2572M

[CO] = 0.0905 - 0.01319 = 0.0773M

[Cl2] = 0.0561 - 0.01319 = 0.0429M