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?

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Answer 1 · 2014-09-14T16:26:48

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