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5.3.6Unit 5. Transition Metals and Organic Nitrogen Chemistry

Autocatalysis — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions

5 min read 6 parts Lesson 26 of 43 Unit quiz

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Part 1 of 6The Reaction

5.3.6 Autocatalysis
Equation
  • 2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 8H₂O + 10CO₂.
Conditions
  • The mixture is acidified and warmed to about 60 °C.
Observation
  • The purple colour fades very slowly at first and then suddenly speeds up.
Slow start
  • Both reactants are negative ions, so repulsion gives a high activation energy for the uncatalysed reaction.

Part 2 of 6The Autocatalyst

1
A small amount of Mn²⁺ forms by the slow uncatalysed reaction.
2
Mn²⁺ reacts with MnO₄⁻ to form Mn³⁺: 4Mn²⁺ + MnO₄⁻ + 8H⁺ → 5Mn³⁺ + 4H₂O.
3
Mn³⁺ then oxidises ethanedioate: 2Mn³⁺ + C₂O₄²⁻ → 2Mn²⁺ + 2CO₂.
4
Mn²⁺ is regenerated and also made in greater amounts, so the catalyst concentration increases as the reaction proceeds.
5
As Mn²⁺ builds up the reaction accelerates, until reactants are used up and the rate falls to zero.

Part 3 of 6Graphs

Graphs
Concentration-time
  • The concentration of MnO₄⁻ falls slowly at first, then steeply, then levels off, giving an S-shaped curve.
Rate-time
  • The rate rises to a maximum and then falls as the reactants are consumed.
Contrast
  • A normal reaction has its fastest rate at the start and decreases steadily.
Test
  • Adding Mn²⁺ at the start removes the initial slow period, which is evidence of autocatalysis.

Part 4 of 6Why the Reaction Is Slow Initially

Why the Reaction Is Slow Initially
Repulsion
  • The anions repel each other and rarely collide with sufficient energy.
Catalyst needed
  • The reaction goes by a faster route once the positive Mn²⁺ ion is present.
Temperature
  • Warming to about 60 °C gives more molecules enough energy to react, so the initial reaction is at least measurable.

Part 5 of 6Worked Example

Question Calculate the volume of CO₂ at room temperature and pressure (24 dm³ mol⁻¹) formed when 20.0 cm³ of 0.0100 mol dm⁻³ MnO₄⁻ reacts completely with excess ethanedioate.

  1. Moles MnO₄⁻ = 0.0200 dm³ x 0.0100 mol dm⁻³ = 2.00 x 10⁻⁴ mol.
  2. The ratio MnO₄⁻CO₂ is 2 : 10, so moles CO₂ = 5 x 2.00 x 10⁻⁴ = 1.00 x 10⁻³ mol.
  3. Volume = 1.00 x 10⁻³ x 24 = 0.024 dm³ = 24 cm³.

Answer 24 cm³ of CO₂

Part 6 of 6Exam Checklist & Tips

5.3.6 – Key Points for the Exam

  • Autocatalysis Catalysis by one of the products of the reaction.
  • Sigmoid curve An S-shaped graph, typical of autocatalytic reactions.
  • Ethanedioate The C₂O₄²⁻ ion.
  • Induction period The slow initial phase before the rate increases.

Common Mistakes

  • Saying the rate is highest at the start.
  • Forgetting that the catalyst is a product.
  • Writing the equation without balancing the charge.
  • Thinking that adding more MnO₄⁻ removes the slow period.

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