Question

Considering the reaction: SO2Cl2(g)⇌SO2(g)+Cl2(g) If a solution is made containing initial [SO2Cl2]= 0.021M, at equilibrium, [Cl2]= 1.3×10−2M. What are the equilibrium concentrations of SO2Cl2 and SO2?

Answer 1

At equilibrium, [ SO₂Cl₂] = 0.087 mol/L and [SO₂] = 1.3 × 10⁻² mol/L.

First, write the balanced chemical equation with an ICE table.

SO₂Cl₂(g) ⇌ SO₂(g) + Cl₂(g)

I/mol·L⁻¹: 0.021; 0; 0
C/mol·L⁻¹: -`x`; -`x`; +`x`
E/mol·L⁻¹: 0.021 - `x`; `x`; `x`

At equilibrium, `["Cl"_2]` = 1.3 × 10⁻² mol/L = `x` mol/L

So `x` = 1.3 × 10⁻²

Then `["SO"_2]` = `x` mol/L = 1.3 × 10⁻² mol/L

and

`["SO"_2"Cl"_2]` = (0.021 - `x`) mol/L = (0.100 - 1.3 × 10⁻²) mol/L = 0.087 mol/L