Rate constant from total pressure

UKChO Round 1 · unreviewed

Gas-phase first-order decomposition followed by total pressure; deduce the reactant pressure first.

Worked example

Pure N₂O₅ (50.7 kPa) decomposes at constant volume and temperature: 2N₂O₅ → 4NO₂ + O₂ (first order). After 21 min the total pressure is 94.1 kPa. Find k in s⁻¹.

1. Weighted sum of 1.6666666666666667·p(N₂O₅) at start − 0.6666666666666666·p(total) → p(N₂O₅) at time t = (5p₀ − 2p)/3
= 21.77 kPa
2. First-order integrated rate law: solve for k
$k = \dfrac{1}{t}\ln\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]}$
= 6.711 × 10⁻⁴ s⁻¹

Answer: k = 6.71 × 10⁻⁴ s⁻¹

Method

If x of N₂O₅ has reacted, the mixture holds p₀ − x of N₂O₅, 2x of NO₂ and x/2 of O₂: total p = p₀ + 1.5x. So p(N₂O₅) = p₀ − x = (5p₀ − 2p)/3. Partial pressures are proportional to concentrations at constant V and T, so k = ln(p₀/p(N₂O₅)) ÷ t.

Common mistakes

The rise in pressure is not the N₂O₅ left. If x kPa of N₂O₅ react, the total rises by 1.5x, so p(N₂O₅) = p₀ − (p − p₀)/1.5.

Formulas used

First-order integrated rate law
$\ln\dfrac{[\mathrm{A}]_0}{[\mathrm{A}]} = kt$

Practise this with fresh numbers

Rate constant from total pressure

Worked example

Method

Common mistakes

Formulas used

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