Kc from equilibrium amounts and volume

Worked example

Phosphorus(V) chloride dissociates: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). At equilibrium a 10.00 dm³ vessel contains 0.361 mol of PCl₅, 0.384 mol of PCl₃ and 0.220 mol of Cl₂. Calculate Kc and give its units.

1. 1. Concentration from moles and volume: solve for [PCl₅]

   $c = \dfrac{n}{V}$

   = 0.03610 mol dm⁻³

2. 2. Concentration from moles and volume: solve for [PCl₃]

   $c = \dfrac{n}{V}$

   = 0.03840 mol dm⁻³

3. 3. Concentration from moles and volume: solve for [Cl₂]

   $c = \dfrac{n}{V}$

   = 0.02200 mol dm⁻³

4. 4. Kc from equilibrium concentrations: solve for Kc

   $K_\mathrm{c} = \prod_i [\mathrm{X}_i]^{\nu_i}$

   = 0.02340 mol dm⁻³

Answer: Kc = 0.0234 mol dm⁻³

Method

Convert each amount to a concentration with c = n/V, then Kc = [PCl₃][Cl₂] ÷ [PCl₅]. Units: (mol dm⁻³)²/(mol dm⁻³) = mol dm⁻³. The volume only cancels when the moles of gas are the same on both sides.

Common mistakes

• Kc uses concentrations. With one mole on the left and two on the right, the volume does not cancel: divide each amount by V first.

Formulas used

• Concentration from moles and volume

  $c = \dfrac{n}{V}$

• Kc from equilibrium concentrations

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

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