Question

The equilibrium constant, K, for the following reaction is 1.20×10-2 at 500 K. PCl5(g) PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 1.00 L flask at 500 K contains 0.200 M PCl5, 4.90×10-2 M PCl3 and 4.90×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if 3.12×10-2 mol of Cl2(g) is added to the flask?

[PCl5] =

[PCl3] =

[Cl2] =  

Answer 1

Given K = 1.2 x 10^-2 = 0.012

equillibrium conc of PCl₅ = 0.2 M since the volume is 1 L . number of mole at equilibrium = 0.200 mole

similarly equliibrium conc of PCl₃ = 4.9 x 10^-2 M = 0.049 M , number of mole at equillibrium = 0.049 mole

            equillibrium conc of Cl₂ = 4.9 x 10^-2 M = 0.049 M, number of mole at equillibrium = 0.049 mole

This suggest that the initial mole of PCl₅ = 0.200 + 0.049 = 0.249

The given equation is

    initial mole 0.249 0    0                           

PCl₅                            PCl₃           +           Cl₂

mole at equillibrium             0.200                                   0.049                       0.049

mole of Cl₂ added                                                                                      0.0312

mole after addition             0.200                                      0.049                      0.08.02

mole after re-establishing

the equillibrium                 0.200 + x                           0.049 -x           0.0802 -x

Since according to lechatlier principle, addition of Cl₂ will shift the reaction in the reverse direction. Hence in the re established equillibrium the concentration of PCl₅ will increase.

Since the total volume is 1 Litre, the number of mole = mol/L

                   K = [PCl3] [Cl2] / [PCl5]

             0.012 = [0.049 -x] [0.0802 -x] / [0.2+x]

          0.0024 + 0.012x = x² - 0.1292 x +0.00393

        x² - 0.1412x + 0.00153 = 0

     This is a quadratic equation in x and when solving gives x =[0.1412 (0.01993 - 0.00612) ] /2

                                                                                         = [0.1412 0.1175]/2

                                                                                         = 0.12935 or 0.01185

                                    The value of x= 0.12935 is not possible as it will exceed the initial mole of PCl₅ .

                                      Hence x = 0.01185

Therefore the concentrations after the re established equillibrium is

[PCl5] = 0.2 + 0.01185)

           = 0.21185 M = 0.212 M

[PCl₃] = 0.049 - 0.01185

          = 0.03715 M = 3.72 x 10^-2 M

[Cl]₂ = 0.0802 - 0.01185

       = 0.06835 M = 6.84 x 10^-2 M