2NO + O₂: pre-equilibrium with bimolecular steps
UKChO Round 2 · IChO · unreviewed
Fast pre-equilibrium 2NO ⇌ N₂O₂ then slow N₂O₂ + O₂ → 2NO₂; rate constants and initial concentrations known. Find d[NO₂]/dt.
Worked example
Mechanism: 2NO ⇌ N₂O₂ (fast; k₁ = 1.2 × 10⁸ dm³ mol⁻¹ s⁻¹, k₋₁ = 3.0 × 10¹⁰ s⁻¹); N₂O₂ + O₂ → 2NO₂ (slow; k₂ = 14000 dm³ mol⁻¹ s⁻¹). Calculate d[NO₂]/dt at [NO] = 0.0041 mol dm⁻³, [O₂] = 3.1 × 10⁻⁴ mol dm⁻³.
1. Divide k₁ ÷ k₋₁ → K₁
= 0.004000 mol⁻¹ dm³
2. Multiply K₁ × k₂ → K₁k₂
= 5.600 × 10⁻⁵ m⁶ s⁻¹ mol⁻²
3. Raise to a power: ([NO])^2 → [NO]²
= 1.681 × 10⁻⁵ mol² dm⁻⁶
4. Multiply K₁k₂ × [NO]² → K₁k₂[NO]²
= 0.05648 min⁻¹
5. Multiply K₁k₂[NO]² × [O₂] → rate of step 2
= 2.918 × 10⁻⁷ mol dm⁻³ s⁻¹
6. Multiply by a coefficient: 2 × rate of step 2 → d[NO₂]/dt
= 5.836 × 10⁻⁷ mol dm⁻³ s⁻¹
Answer: d[NO₂]/dt = 5.84 × 10⁻⁷ mol dm⁻³ s⁻¹
Method
K₁ = k₁/k₋₁ gives [N₂O₂] = K₁[NO]². The slow step has rate k₂[N₂O₂][O₂] = K₁k₂[NO]²[O₂], and it forms two NO₂, so d[NO₂]/dt = 2K₁k₂[NO]²[O₂]. Overall third order.
Common mistakes
- That is the rate of the slow step. Each N₂O₂ + O₂ event makes two NO₂, so d[NO₂]/dt is twice it.
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