Explain how a catalyst provides a lower-activation-energy route, show how it is regenerated, and interpret why an autocatalytic reaction first speeds up then slows down.
A catalyst changes the pathway, not the equilibrium
A catalyst increases reaction rate by providing an alternative pathway with lower activation energy and is regenerated overall. It may react during an individual step, so ‘does not take part in the reaction’ is misleading. Transition metals can bind reactants at surfaces or change oxidation state in solution, making alternative pathways available.
A heterogeneous catalyst is in a different phase from the reactants and reaction occurs at its surface. A homogeneous catalyst is in the same phase as the reactants and the pathway proceeds through intermediate species. Phase means a physically distinct region: two immiscible liquids are different phases even though both are liquids.
A catalyst does not change ΔH, ΔG°, E°cell, K or the equilibrium composition at a fixed temperature. It speeds approach to equilibrium by lowering barriers in both directions. It can permit an industrial process to run usefully at a lower temperature, but that changed operating temperature, not catalysis itself, changes equilibrium.
Adsorb, react, then desorb
CO and NO molecules adsorb onto active sites of a catalytic converter’s metal surface: bonds form between reactants and surface atoms. This changes electron distribution and weakens bonds within reactants, while holding reactants close in suitable orientations. At the surface, bonds rearrange to form CO₂ and N₂. These products desorb, freeing active sites for more molecules.
Adsorption is attachment at a surface; absorption means entry into the bulk of a material. A catalyst must bind reactants strongly enough to activate them but allow products to leave. A poison occupies active sites strongly and blocks their use. A high-area honeycomb support coated with small quantities of platinum-group metals exposes many sites without requiring a large mass of expensive metal.
The converter lowers carbon monoxide and nitrogen monoxide emissions through 2CO + 2NO → 2CO₂ + N₂. Carbon monoxide is oxidised and nitrogen monoxide reduced. Carbon dioxide is less acutely toxic than CO but remains a greenhouse gas, so ‘all pollutants become harmless’ is an overstatement. A cold converter works less effectively because the catalytic reaction still has an activation barrier.
Vanadium(V) oxide is reduced and regenerated
In the Contact process, V₂O₅ catalyses oxidation of SO₂ to SO₃. A simplified redox cycle shows V(V) in V₂O₅ reduced to V(IV) in V₂O₄ while sulfur(IV) becomes sulfur(VI). Oxygen then reoxidises V₂O₄ to V₂O₅. The catalyst’s oxidation state changes during the cycle and returns to its starting value.
Double the first equation before adding to the second. Two V₂O₅ consumed and two regenerated cancel, as do two V₂O₄ formed and consumed. The net equation is 2SO₂ + O₂ → 2SO₃. This cancellation is the evidence for regeneration; a catalytic scheme that leaves the catalyst permanently consumed is incomplete.
This is the required school-level oxide cycle. It links catalytic behaviour to accessible vanadium oxidation states; it is not a complete mechanistic description of every species in a working industrial catalyst. Use the model to account for atoms and electron transfer.
Iron(II) carries electrons between two reacting anions
The overall reaction S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂ is slow without a catalyst partly because the reacting ions both have negative charge and repel on approach. Dissolved Fe²⁺ provides a different route: peroxodisulfate oxidises Fe²⁺ to Fe³⁺, then Fe³⁺ oxidises iodide and regenerates Fe²⁺. All reactants and catalyst are in the aqueous phase, so this is homogeneous catalysis.
Each stage involves attraction between oppositely charged ions and a different electron-transfer pathway. Avoid treating electrostatic attraction alone as a universal proof of a lower barrier; the observed catalysis and viable intermediate cycle support the mechanism. Fe³⁺ can also initiate the same cycle because the steps can begin in the other order when the necessary reactants are present.
Using E°(S₂O₈²⁻/SO₄²⁻) = +2.01 V, E°(Fe³⁺/Fe²⁺) = +0.77 V and E°(I₂/I⁻) = +0.54 V gives +1.24 V for the first step and +0.23 V for the second. The iron potential lies between the two reactant couples, making both redox stages thermodynamically feasible. This does not by itself prove a fast catalytic rate: kinetics must still permit the steps.
A reaction product can accelerate its own formation
Acidified manganate(VII) oxidises ethanedioate to carbon dioxide and is reduced to Mn²⁺. Initially the uncatalysed reaction is slow. As Mn²⁺ accumulates, it provides a faster route through Mn³⁺ intermediates, so the rate increases. Because the catalyst is also a product of the overall reaction, this is autocatalysis.
One simplified sequence is 4Mn²⁺ + MnO₄⁻ + 8H⁺ → 5Mn³⁺ + 4H₂O, followed by 2Mn³⁺ + C₂O₄²⁻ → 2Mn²⁺ + 2CO₂. Multiply the first by two and the second by five before adding. Ten Mn³⁺ cancel, and ten Mn²⁺ are regenerated after eight were used: two Mn²⁺ remain as net products, matching the overall reaction.
The rate eventually falls because reactants are depleted, despite the high catalyst concentration. For a plot of [MnO₄⁻] against time, expect an initially shallow downward gradient, then a steeper descent, then a shallow approach to completion. The magnitude of the gradient is the rate of manganate consumption. Adding a small known amount of Mn²⁺ at the start shortens or removes the initial slow period.
Follow the purple ion’s absorbance at a suitable wavelength using a blank and a calibration where needed. Keep temperature, starting concentrations, total volume and optical path length controlled when comparing runs. Absorbance can follow concentration without withdrawing samples, but bubbles, dirty cuvettes and excessively high absorbance can distort results. Label a drawn curve as schematic unless it is backed by actual measurements.
Build a connected comparison
To compare platinum in a converter with aqueous Mn²⁺, start with the shared lower-activation-energy route and regeneration. Then explain the physical distinction: gases interact at a solid surface, whereas dissolved ions react through soluble intermediates. Link adsorption, weakened bonds and desorption in the first case; link product formation, rising catalyst concentration and acceleration in the second.
June 2023 9CH0/01 Q8 assessed this comparison through linked reasoning as well as content. Its report highlights confusion between adsorption and absorption and omitted explanations of the initial slow anion–anion reaction. A list of unrelated catalyst facts does not show how the contrasting mechanisms produce the observed behaviour.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.
Q1. Why is ‘a catalyst never reacts’ an inaccurate definition?Show answer
It can form bonds or change oxidation state in individual steps but is regenerated overall. Add all mechanism steps and cancel the catalyst/intermediates to check this. Its pathway lowers activation energy without changing the reaction’s equilibrium constant.
Q2. Explain the sequence by which a converter turns CO and NO into CO₂ and N₂.Show answer
CO and NO adsorb at metal active sites, forming surface bonds that weaken their internal bonds and position them for reaction. New bonds form to produce CO₂ and N₂, which desorb and release the sites. Overall: 2CO + 2NO → 2CO₂ + N₂.
Q3. Identify the vanadium oxidation-number changes in the Contact-process cycle and derive the net equation.Show answer
V changes +5 → +4 in SO₂ + V₂O₅ → SO₃ + V₂O₄, then +4 → +5 in 2V₂O₄ + O₂ → 2V₂O₅. Double the first step and add; vanadium oxides cancel to give 2SO₂ + O₂ → 2SO₃.
Q4. For iron-catalysed iodide/peroxodisulfate, why is a catalyst couple between +0.54 V and +2.01 V useful?Show answer
Its reduced form can be oxidised by peroxodisulfate and its oxidised form can oxidise iodide, giving positive emfs for both stages. Fe³⁺/Fe²⁺ at +0.77 V gives +1.24 V and +0.23 V. This establishes thermodynamic compatibility, while actual catalysis also requires sufficiently fast steps.
Q5. Why does acidified MnO₄⁻/C₂O₄²⁻ first speed up and later slow down? Predict the effect of initially added Mn²⁺.Show answer
Mn²⁺ forms as a product and catalyses a faster route, so increasing catalyst concentration initially increases rate. Later reactant depletion dominates and rate falls. Initially added Mn²⁺ reduces the induction period; it does not increase the equilibrium yield by changing K.
Sources
Sources and examiner guidance (reviewed 9 October 2026)
- Pearson Edexcel 9CH0 specification — Issue 3 — Topic 15A/B, printed pp.33–35; UK A-Level content and Core Practical 12.
- Chemrevise — UK Edexcel Topic 15 — Pages 1–13 reviewed as a secondary coverage check; original Finesse teaching and questions.
- Pearson 9CH0/01 mark scheme — June 2023 — Q7(b–c), Q8; PDF pp.22–24 and 26–27. Electronic configurations, aqua-ion acidity and catalytic explanations.
- Pearson 9CH0/01 examiner report — June 2023 — Q7(b–c), Q8, pp.26–30 and 33–35. Question-specific evidence, not a universal checklist of mark allocations.
- Pearson 9CH0/01 question paper — June 2023 — Q8, pp.18–19, reviewed for the actual catalyst comparison and extended-response context.
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