Question

Sulfuryl chloride decomposes to sulfur dioxide and chlorine by the following reaction: SO2Cl2(g) ⇌ SO2(g) + Cl2(g) Kc = 0.045 at 648 K If an initial concentration of 6.76 x 10-2 M SO2Cl2 is allowed to equilibrate, what is the equilibrium concentration of Cl2?

Answer 1

ICE Table:

[SO2Cl2] [SO2] [Cl2]

initial 0.0676 0 0

change -1x +1x +1x

equilibrium 0.0676-1x +1x +1x

Equilibrium constant expression is

Kc = [SO2]*[Cl2]/[SO2Cl2]

0.045 = (1*x)(1*x)/((6.76*10^-2-1*x))

0.045 = (1*x^2)/(6.76*10^-2-1*x)

3.042*10^-3-4.5*10^-2*x = 1*x^2

3.042*10^-3-4.5*10^-2*x-1*x^2 = 0

This is quadratic equation (ax^2+bx+c=0)

a = -1

b = -4.5*10^-2

c = 3.042*10^-3

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 1.419*10^-2

roots are :

x = -8.207*10^-2 and x = 3.707*10^-2

since x can't be negative, the possible value of x is

x = 3.707*10^-2

At equilibrium:

[Cl2] = x = 3.707*10^-2

Answer: 3.71*10^-2