Equilibrium concentration

Question

For the reaction: SO₂Cl₂(g) SO₂(g) + Cl₂(g), K = 2.99 ×10⁻⁷ at 500 K.

If a reaction mixture initially contains 0.35 mol SO₂Cl₂ in a 2.00 L flask, what is the equilibrium concentration of Cl₂ at 500 K?

Answer 1

Ice Table

Change moles to molarity 0.35 mol/2.00L = 0.175 M (the answer should have 2 significant figures. I am keeping 3 and will adjust at the end. You should not round up your numbers too soon)

SO₂Cl₂(g) SO₂(g) + Cl₂(g) K = 2.99 ×10⁻⁷ 

I 0.175 0 0

C -x +x +x

E 0.175-x x x

you can drop x from subtraction since the value of K is very small and the concentration of the products will be significantly smaller than the concentration of reactants. If you keep x you will have to solve a quadratic equation.

K = [SO₂][Cl₂]/SO₂Cl₂

2.99 ×10⁻⁷ =(x²)/0.175

x=√ (2.99 ×10⁻⁷x0.175)

x =2.29 x 10^-4

[Cl₂ ] = x = 2.3 x 10^-4 (with the correct number of significant figures)