The equilibrium constant, K, for the following reaction is 1.34×10-2 at 601 K.
COCl2(g) CO(g) + Cl2(g)
An equilibrium mixture of the three gases in a 19.2 L container at 601 K contains 0.260 M COCl2, 5.90×10-2 M CO and 5.90×10-2 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 8.91 L?
[COCl2]= M
[CO] = M
[Cl2] = M
First, if (COCl2) = 0.260 at equilibrium it must have started at 0.260 M + 0.059 M = 0.319 M. With the pressure change the new concentration will be 0.319 x (19.2/8.91) = 0.687 M
Recalculate the ICE chart starting with this new value for COCl2.
..................COCl2 ==> CO + Cl2
I..................0.687............0........0
C...................-x................x.........x
E............0.687-x..............x..........x
Plug the E line into the Kc expression from above and solve for x, then evaluate each of the values in the E line. Le Chatelier's Principle tells you that the increase in pressure will shift the equilibrium to the left; i.e., to the side with fewer moles of gas so this new value for (COCl2) will be larger than the o.260 starting value if you convert back to the smaller pressure.
Post your work if you get stuck.