H₂ + I₂ ⇌ 2HI: equilibrium amounts

C3L6 · UKChO Round 1 · IChO · unreviewed

H₂ + I₂ ⇌ 2HI in a sealed flask with Kc and initial amounts known; find the equilibrium amount of HI using the ICE helper.

Practise this with fresh numbers

Worked example

H₂(g) + I₂(g) ⇌ 2HI(g) has Kc = 54 at the temperature of the experiment. 0.75 mol of H₂ and 0.41 mol of I₂ are sealed in a 8.3 dm³ flask. How many moles of HI are present at equilibrium?

1. Concentration from moles and volume: solve for [H₂]₀
$c = \dfrac{n}{V}$
= 0.09036 mol dm⁻³
2. Concentration from moles and volume: solve for [I₂]₀
$c = \dfrac{n}{V}$
= 0.04940 mol dm⁻³
3. Equilibrium from initial concentrations (ICE): solve for ξ
$\text{Solve } K = \prod_i \left(\frac{[\mathrm{X}_i]_0 + \nu_i\,\xi}{c^\circ}\right)^{\nu_i} \text{ for } \xi$
= 0.04589 mol dm⁻³
4. Multiply by a coefficient: 2 × ξ → [HI]
= 0.09178 mol dm⁻³
5. Multiply [HI] × V → n(HI)
= 0.7618 mol

Answer: n(HI) = 0.762 mol

Method

Convert to concentrations, solve Kc = (2ξ)²/(([H₂]₀ − ξ)([I₂]₀ − ξ)) for ξ with every concentration positive, then n(HI) = 2ξV. (Δn_gas = 0, so the volume actually cancels.)

Common mistakes

ξ is the extent of reaction; each unit of ξ makes 2 HI, so [HI] = 2ξ.

Formulas used

Concentration from moles and volume
$c = \dfrac{n}{V}$
Equilibrium from initial concentrations (ICE)
$K = \prod_i \left(\frac{[\mathrm{X}_i]_0 + \nu_i\,\xi}{c^\circ}\right)^{\nu_i}$