Kc from initial amounts and one equilibrium amount

Worked example

0.752 mol of hydrogen and 0.793 mol of iodine were sealed in a 5.00 dm³ vessel and heated: H₂(g) + I₂(g) ⇌ 2HI(g). At equilibrium the mixture contained 0.488 mol of HI. Calculate Kc.

1. 1. Multiply by a coefficient: 0.5 × n(HI) at equilibrium → H₂ (and I₂) used = ½ n(HI)

   = 0.2440 mol

2. 2. Subtract n(H₂) at start − H₂ (and I₂) used = ½ n(HI) → n(H₂) at equilibrium

   = 0.5080 mol

3. 3. Subtract n(I₂) at start − H₂ (and I₂) used = ½ n(HI) → n(I₂) at equilibrium

   = 0.5490 mol

4. 4. Concentration from moles and volume: solve for [H₂]

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

   = 0.1016 mol dm⁻³

5. 5. Concentration from moles and volume: solve for [I₂]

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

   = 0.1098 mol dm⁻³

6. 6. Concentration from moles and volume: solve for [HI]

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

   = 0.09760 mol dm⁻³

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

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

   = 0.8539

Answer: Kc = 0.854

Method

Making x mol HI uses x/2 mol each of H₂ and I₂. Equilibrium amounts: H₂ = a − x/2, I₂ = b − x/2, HI = x. Kc = [HI]² ÷ ([H₂][I₂]); with two moles of gas on each side the units (and the volume) cancel.

Common mistakes

• Forming 2 mol of HI uses only 1 mol each of H₂ and I₂: subtract half the HI formed.

Formulas used

• Concentration from moles and volume

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

• Kc from equilibrium concentrations

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

More equilibrium questions

• pH of a weak acid
• Buffer pH (Henderson–Hasselbalch)
• Kp from Kc
• Solubility from Ksp
• pH of a weak base
• Kc and its units for the Haber process
• Kc from equilibrium amounts and volume
• An equilibrium concentration from Kc
• Kp from equilibrium amounts and total pressure
• Kp and its units for the Haber process
• Ksp from a solubility in g dm⁻³
• Solubility with a common ion
• pH of benzoic acid from a weighed sample
• H₂ + I₂ ⇌ 2HI: equilibrium amounts