Question

4) The equilibrium constant for the reaction shown below is 4.63 x 10^-3 at 527^oC.

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

a)Write the equilibrium constant expression in terms of the chemical concentrations.

b)What is the equilibrium constant for the reverse reaction?

c)If 0.50 mol of COCl₂ is placed in a 2.0 L flask, what will the concentration of all of the reactants and products be when the reaction reaches equilibrium at 527^oC?

Answer 1

The reaction shown below:

            COCl₂(g) ó CO(g) + Cl₂(g)

1. The equilibrium constant expression in terms of the chemical concentrations for this reaction is as follows:

Keq = [CO] [Cl2] / [COCl2]

1. The equilibrium constant expression in terms of the chemical concentrations for the reverse reaction of this given reaction is as follows:

Keq = [COCl2]/ [CO] [Cl2]

1. To calculate the concentration of all of the reactants and products be when the reaction reaches equilibrium at 527^oC first calculate molarity of COCl₂ as follows:

Molarity of COCl₂ = 0.50 mol / 2.0 L= 0.25 M

Now make the ICE table as follows:

           COCl₂(g) ó CO(g) + Cl₂(g)

I           0.25 M             0              0

C         -x                    +x           +x

E          0.25M-x          x          x

Given that, Keq =4.63 x 10^-3

Then;

Keq = [CO] [Cl2] / [COCl2]

4.63 x 10^-3 =    x*x] / 0.25-x

if 0.25-x=0.25then;

x²= 4.63 x 10^-3 *0.25

x²= 1.16 x 10^-3

x=0.03

Hence [CO] =[Cl2] =0.03 M at equilibrium and [COCl2] = 0.25-0.034=0.22 M