The shape of the lesson
By the end of the lesson, learners can
- Lowering the barrier: Add a catalyst in the simulator and watch the activation energy line move left on the Boltzmann curve while the curve itself stays put.
- Same phase: homogeneous: Homogeneous catalysts take part in the reaction, forming an intermediate that is then used up, regenerating the catalyst.
- Different phase: heterogeneous: Heterogeneous catalysts are usually solids with gases or liquids passing over them.
- Catalyst: alternative route, lower Ea, not used up.
- Homogeneous: same phase, forms an intermediate.
- Heterogeneous: different phase; adsorb, react, desorb.
- No change to ΔH or equilibrium position.
One picture every idea returns to
A catalyst increases the rate of a reaction by providing an alternative route with a lower activation energy; it is chemically unchanged at the end. On the Boltzmann distribution, Ea moves left, so a larger proportion of molecules can react. ΔH is unchanged, and because forward and reverse rates rise equally, a catalyst does not move the position of equilibrium. A homogeneous catalyst is in the same phase as the reactants and usually forms an intermediate; a heterogeneous catalyst is in a different phase and works by adsorbing reactants onto its surface.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, A shortcut over the barrier.. A catalyst does not push molecules harder; it finds them an easier path. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Lowering the barrier | Add a catalyst in the simulator and watch the activation energy line move left on the Boltzmann curve while the curve itself stays put. Project tab 4 of the lab (Reaction kinetics lab, catalysts) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: Same phase: homogeneous | Homogeneous catalysts take part in the reaction, forming an intermediate that is then used up, regenerating the catalyst. | Pairs |
| 21 to 30 min | Teach and check: Different phase: heterogeneous | Heterogeneous catalysts are usually solids with gases or liquids passing over them. | Jigsaw |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 15 marks). Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Different phase: heterogeneous, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Jigsaw
Split the content into expert pieces. Expert groups master one piece each, then regroup so every home team has one expert per piece, and each expert teaches their piece to the team.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| A catalyst lowers the activation energy of the forward and reverse reactions, so it does not change ΔH, the position of equilibrium or the yield. | On a Boltzmann diagram for a catalysed reaction, do not redraw the curve; just move the E a line to the left. |
Homogeneous and heterogeneous catalysis: worksheet
6 questions, 15 marks. Written in the style of the papers; no past paper question is reproduced.
Distinguish between homogeneous and heterogeneous catalysis, with one example of each.
Draw a reaction pathway diagram for an exothermic reaction with and without a catalyst. Label ΔH and both activation energies.
Use the Boltzmann distribution to explain how a catalyst increases the rate of reaction.
Describe the stages by which a catalytic converter removes carbon monoxide and nitrogen monoxide, and write the equation.
Explain, with equations, how Fe2+ ions catalyse the reaction between peroxodisulfate and iodide ions.
Explain why adding a catalyst to the Haber process does not increase the equilibrium yield of ammonia.
Answer key and marking notes
- Homogeneous: the catalyst is in the same phase as the reactants, e.g. Fe2+(aq) in the reaction of S2O82− with I−.
- Heterogeneous: the catalyst is in a different phase, e.g. solid iron with gaseous N2 and H2 in the Haber process.
- Products below reactants, ΔH arrow downwards.
- Uncatalysed curve with a high hump labelled Ea.
- Catalysed curve with a lower hump labelled Ea(cat), same start and end levels.
- The catalyst provides a route with lower activation energy, so Ea moves to the left on the distribution.
- A greater proportion of molecules have energy ≥ Ea, so there are more successful collisions per unit time.
- CO and NO are adsorbed onto the Pt/Rh surface, weakening their bonds.
- They react on the surface and the products desorb.
- 2CO + 2NO → 2CO2 + N2.
- S2O82− + 2Fe2+ → 2SO42− + 2Fe3+.
- 2Fe3+ + 2I− → 2Fe2+ + I2.
- Fe2+ is regenerated; each step involves oppositely charged ions, so Ea is lower than for two repelling anions.
- The catalyst lowers the activation energy of the forward and reverse reactions equally.
- Both rates increase by the same factor, so the position of equilibrium does not change; equilibrium is simply reached faster.
8 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
How does a catalyst increase the rate of a reaction?
- Ait provides an alternative reaction pathway with a lower activation energy, and is not used up overall
- Bit gives the reactant molecules more kinetic energy, so more of them exceed the activation energy
- Cit makes the reaction more exothermic, so more energy is released to speed it up
- Dit increases the collision frequency by pushing the reactant particles closer together
The diagram shows the reaction pathway for a reaction with and without a catalyst. Which arrow represents the activation energy of the catalysed reaction?
- AB
- BA
- CC
- Dthe difference between A and B
For a reaction, ΔH = −92.0 kJ mol−1 and the uncatalysed activation energy is +180 kJ mol−1. A catalyst lowers the activation energy to +105 kJ mol−1. What is ΔH for the catalysed reaction, in kJ mol−1?
- A−92.0
- B−17.0
- C−167
- D+75.0
Which is an example of heterogeneous catalysis?
- Airon in the Haber process, N2(g) + 3H2(g) ⇌ 2NH3(g)
- BFe2+(aq) in the reaction between S2O82−(aq) and I−(aq)
- CH+(aq) from concentrated sulfuric acid in the esterification of ethanoic acid with ethanol
- DNO2(g) in the oxidation of SO2(g) to SO3(g) in the atmosphere
The graph shows the Boltzmann distribution for a gas mixture, with the activation energies of the uncatalysed and catalysed reactions marked. Why does the catalyst increase the rate?
- Athe curve is unchanged, but many more molecules have energy greater than or equal to the lower Ea(cat) (the shaded area)
- Bthe catalyst moves the curve to higher energy, so more molecules exceed Ea
- Cthe catalyst makes the peak higher, so more molecules have the most probable energy
- Dthe catalyst increases the total area under the curve, so there are more molecules to collide
In a catalytic converter, a platinum/rhodium surface catalyses 2CO(g) + 2NO(g) → 2CO2(g) + N2(g). In which order do the stages of heterogeneous catalysis occur?
- Aadsorption of reactants, weakening of bonds, reaction, desorption of products
- Badsorption of reactants, reaction, weakening of bonds, desorption of products
- Cweakening of bonds, adsorption of reactants, desorption of products, reaction
- Dadsorption of reactants, desorption of products, weakening of bonds, reaction
Iron is used as a catalyst in the Haber process, N2(g) + 3H2(g) ⇌ 2NH3(g). What is the effect of the catalyst on the equilibrium?
- Aequilibrium is reached more quickly, but the position of equilibrium and the yield of ammonia are unchanged
- Bthe position of equilibrium shifts to the right, so the yield of ammonia increases
- Cthe value of Kc increases, so more ammonia is present at equilibrium
- Donly the forward reaction is speeded up, so ammonia forms faster than it decomposes
The reaction S2O82−(aq) + 2I−(aq) → 2SO42−(aq) + I2(aq) is slow, but is catalysed by Fe2+(aq). Which statement explains the catalysis?
- AFe2+ is oxidised to Fe3+ by S2O82−, then Fe3+ oxidises I− and is reduced back to Fe2+; each step is between oppositely charged ions
- BFe2+ provides a solid surface on which S2O82− and I− are adsorbed close together
- CFe2+ is used up in the reaction, so it must be added in excess
- DFe2+ makes the reaction more exothermic, so the mixture warms up and reacts faster
Quiz key
A catalyst takes part in the reaction by an alternative route (mechanism) with a lower Ea, so at the same temperature a larger proportion of collisions have enough energy to succeed; it is regenerated, so it is not used up. It does not give the molecules more energy: the Boltzmann distribution is unchanged, and it is the Ea line that moves. ΔH depends only on the reactants and products, so a catalyst does not make the reaction more exothermic. A catalyst does not change the concentration or pressure of the reactants, so it does not change the collision frequency in that way.
The activation energy is the energy difference from the reactants to the highest point of the pathway. For the catalysed (dashed) route this is arrow B. Arrow A goes to the higher peak of the uncatalysed route, so it is the uncatalysed Ea. Arrow C is from reactants to products, which is ΔH and is the same for both routes. The difference between A and B is how much the catalyst lowers the activation energy, not the activation energy itself.
A catalyst changes the route between reactants and products, not the reactants or products themselves, so ΔH is unchanged at −92.0 kJ mol−1. −17.0 adds the 75 kJ mol−1 drop in Ea to ΔH, and −167 subtracts it; both wrongly assume that lowering the barrier shifts the product level. +75.0 is the amount by which Ea is lowered (180 − 105), which is not an enthalpy change of reaction.
In the Haber process the catalyst is solid iron and the reactants are gases, so they are in different phases: heterogeneous catalysis (like V2O5 in the Contact process and Pt/Pd/Rh in catalytic converters). Fe2+ and the ions it catalyses are all in aqueous solution, so that is homogeneous. H+ in esterification is in the same liquid phase as the acid and alcohol: homogeneous. NO2 and SO2 are both gases, so atmospheric NO2 is a homogeneous catalyst. (Enzymes are biological catalysts, usually acting in solution.)
A catalyst does not change the temperature, so the Boltzmann distribution is unchanged; it lowers the activation energy to Ea(cat). The area to the right of Ea(cat) (shaded) is much larger than the area to the right of Ea, so a much greater proportion of collisions succeed. Moving the curve to higher energy is what a temperature rise does, not a catalyst. The catalyst does not change the peak height, which would need a different temperature or number of molecules. The area is the total number of molecules, which a catalyst does not change.
Reactant molecules are first adsorbed onto active sites on the metal surface. Forming bonds to the surface weakens bonds within the reactant molecules (and holds them close together in a suitable orientation), so reaction happens with a lower Ea. The products then desorb, freeing the sites for more reactant. Bonds must be weakened before, not after, the reaction, because the weakening is what lowers Ea. Bonds are weakened by adsorption, so weakening cannot come first. Desorption releases the products, so it cannot happen before reaction has taken place.
The catalyst lowers Ea for the forward and the reverse reactions by the same amount, so both rates increase by the same factor. Equilibrium is reached sooner, but the position of equilibrium and the equilibrium yield do not change; in industry this is valuable because it allows a lower temperature to be used at an acceptable rate. The position does not shift to the right, since both rates change equally. Kc depends only on temperature, not on a catalyst. If only the forward reaction were speeded up, the equilibrium would shift, which would break the rule that a catalyst cannot change ΔH or the equilibrium constant.
Without a catalyst two negative ions must collide, and their repulsion gives a high Ea. With Fe2+: S2O82− + 2Fe2+ → 2SO42− + 2Fe3+, then 2Fe3+ + 2I− → 2Fe2+ + I2. Each step involves attraction between oppositely charged ions, so Ea is lower, and Fe2+ is regenerated. This is homogeneous catalysis: everything is in aqueous solution, so there is no solid surface for adsorption. The catalyst is not used up overall, so only a small amount is needed. ΔH for the overall reaction is unchanged by a catalyst.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.