Question

Consider the following reaction: SO2Cl2(g)⇌SO2(g)+Cl2(g) Kc=2.99×10−7 at 227 ∘C If a reaction mixture initially contains 0.159 MSO2Cl2, what is the equilibrium concentration of Cl2 at 227 ∘C?

Answer 1

For the reaction :
SO2Cl2 = SO2 + Cl2 ,
The expression for Kc is :   
[..] denotes concentration in Molarity.

Here, initial concentration of SO2Cl2 = 0.159 M. Let x M of SO2Cl2 react to reach equilibrium. According to the equation, if x M of SO2Cl2 react, then x M of SO2 and x M of Cl2 will be formed.
Hence , final equilibrium concentrations are :
[SO2Cl2] = (0.159 - x) M
[Cl2] = [SO2] = x M.
And Kc = 2.99 x 10^-7 ( given) .
So, putting these values in the expression for Kc :
2.99 x 10^-7 =
Or, ( 2.99 x 10^-7)(0.159 - x) = x^2
Or, x^2 + ( 2.99 x 10^-7)x - (4.754 x 10^-8) = 0
This is a quadratic equation, and using the quadratic formula to solve it, x = - 0.000218 or x = 0.000218 . So, the positive solution is taken.
Hence, x = 0.000218.

Thus, equilibrium concentration of Cl2 = x M = 0.000218 M = 2.18 x 10^-4 M ( answer)