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?

Question

Jeremiah Adams · Accepted Answer

[SO₂Cl₂] = 0.087 mol/L and [SO₂] = 1.3 × 10⁻² mol/L are the values at equilibrium. First, use an ICE table to write the balanced chemical equation. g = SO₂Cl₂(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 additionally #["SO"_2"Cl"_2]# = (0.021 - #x#) mol/L = (0.100 - 1.3 × 10⁻²) mol/L = 0.087 mol/L

Avery Adams · Answer

undefined To find the equilibrium concentrations of SO2Cl2 and SO2, we need to use the equilibrium constant expression $ K_{eq} = \frac{[SO2][Cl2]}{[SO2Cl2]} $. Given the initial concentration of SO2Cl2 and the equilibrium concentration of Cl2, we can rearrange the expression and solve for the equilibrium concentration of SO2.

$$ K_{eq} = \frac{[SO2][Cl2]}{[SO2Cl2]} $$

$$ [SO2] = \frac{K_{eq} \times [SO2Cl2]}{[Cl2]} $$

Substitute the given values into the equation:

$$ [SO2] = \frac{K_{eq} \times 0.021M}{1.3 \times 10^{-2}M} $$

Given that the equilibrium constant $ K_{eq} $ for the reaction is not provided, we cannot directly calculate the equilibrium concentration of SO2. We would need the value of $ K_{eq} $ to proceed with the calculation.