An equilibrium concentration from Kc

Worked example

In the Contact process, 2SO₂ + O₂ ⇌ 2SO₃ with Kc = 361 mol⁻¹ dm³. If [SO₂] = 0.228 mol dm⁻³ and [O₂] = 0.126 mol dm⁻³ at equilibrium, what is the concentration of SO₃?

1. 1. Kc from equilibrium concentrations: solve for [SO₃] = √(Kc[SO₂]²[O₂])

   $\text{Solve } K_\mathrm{c} = \prod_i [\mathrm{X}_i]^{\nu_i} \text{ for } [\mathrm{X}_{3}]$

   = 1.538 mol dm⁻³

Answer: [SO₃] = 1.54 mol dm⁻³

Method

Rearrange Kc = [SO₃]² ÷ ([SO₂]²[O₂]) to [SO₃] = √(Kc × [SO₂]² × [O₂]). Check: Kc here has units of dm³ mol⁻¹.

Common mistakes

• Kc = [SO₃]² ÷ ([SO₂]²[O₂]): square [SO₂], and take the square root at the end.

Formulas used

• Kc from equilibrium concentrations

  $K_\mathrm{c} = \dfrac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}$

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