Question

The equilibrium constant, K, for the following reaction is 1.42×10-2 at 504 K.

PCl5(g) --> PCl3(g) + Cl2(g)

An equilibrium mixture of the three gases in a 10.9 L container at 504 K contains 0.279 M PCl5, 6.29×10-2 M PCl3 and 6.29×10-2 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 4.95 L?

[PCl5] = M

[PCl3] = M

[Cl2] = M

Answer 1

PCl5(g) --> PCl3(g) + Cl2(g)

Moles of PCl5 = Molarity*V in L = 0.279*10.9 = 3.0411

Moles of PCl3 = 0.685

Moles of Cl2 = 0.685

Kc = [Cl2] [PCl3] / [PCl5] = 1.42*10^-2

When volume is compressed to 4.95 L

[PCl5] = Moles / 4.95 = 0.614 M

[PCl3] = 0.138 M

[Cl2] = 0.138 M

Q = [Cl2] [PCl3] / [PCl5] = 0.138^2 / 0.614 = 0.031

Q > Kc

When Q>K, there are more products than reactants. To decrease the amount of products, the reaction will shift to the left and produce more reactants. For Q>K:

PCl5(g) --> PCl3(g) + Cl2(g)

0.614 0.138 0.138

0.614+x 0.318-x 0.318-x

Kc = [Cl2] [PCl3] / [PCl5] = [0.38-x]^2 / [0.614+x] = 0.0142

0.00872 + 0.0142x = 0.1444 - 0.76x + x^2

x^2 - 0.7742x + 0.13568 = 0

x = 0.268

[PCl5] = 0.614+0.268 = 0.882 M

[PCl3] = 0.318-0.268 = 0.05 M

[Cl2] = 0.318-0.268 = 0.05 M