What does a catalyst change on an energy diagram?
You will be able to: Explain catalytic acceleration while preserving reactant/product energy and equilibrium.
What does a catalyst change on an energy diagram?
A tunnel changes the route between two towns without changing the towns’ elevations. A catalyst likewise changes the reaction pathway rather than the chemical identities of the net starting and ending substances.
A useful starting point: How do you read several peaks and valleys? →
Words and symbols before equations
- Alternative pathway
- Different sequence of elementary events reaching the same net products.
- Catalyzed barrier
- Activation barrier along the catalyst-assisted route.
- Equilibrium composition
- Composition at which forward and reverse processes balance under specified conditions.
What this picture assumes
Schematic effective pathways: R=0, P=−20, uncatalyzed peak=80 and catalyzed peak=45 kJ/mol. Single humps are comparisons, not a claim that real catalysis has one step. Endpoints and equilibrium are unchanged.
Read the picture in three steps
- Read the species and labels first. Identify what each symbol and line represents. Read the units and fixed conditions before comparing quantities.
- Original forward barrier 80 kJ/mol; reverse 100 kJ/mol. ΔE=−20 kJ/mol. Add the alternative route to compare.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the chemistry
A catalyst can increase effective encounters and provide a route with lower relevant barriers. The new route may involve several steps and catalyst-bound intermediates.
For the same net reaction and conditions, reactant and product energy levels stay the same. Lowering a forward barrier also changes the reverse pathway; a catalyst does not change the equilibrium constant or make a different final equilibrium composition.
A reaction approaches equilibrium faster in both directions. A catalyst is not a fuel whose net consumption supplies extra reaction energy.
The simple overlay uses one effective hump per route to compare barriers. It does not imply every real catalytic mechanism has one elementary step or the same prefactor.
| Feature | Warming | Catalyst at fixed T |
|---|---|---|
| Energy distribution | Changes with temperature | Need not change |
| Pathway | Not necessarily changed | Alternative pathway |
| Barrier | Not lowered just by warming | Different effective barriers |
A worked example, step by step
Uncatalyzed R=0, TS=80, P=−20 kJ/mol. A schematic catalytic route has peak 45 with the same endpoints. Compare both forward/reverse barriers and ΔE.
- Uncatalyzed forward barrier is 80; reverse is 80−(−20)=100 kJ/mol.
- Catalyzed forward barrier is 45; reverse is 45−(−20)=65 kJ/mol.
- Both routes have ΔE=−20 kJ/mol.
- The route changes, but endpoint energies and equilibrium for the same net system do not. The overlay is schematic.
A catalyst does not lower reactant/product energy difference or guarantee more product at equilibrium.
Can a catalyst speed the reverse reaction too?
Compare with an explanation
Yes. The alternative pathway operates in reverse as well; it helps equilibrium be reached faster without shifting it.
Predict. Change one thing. Explain.
Toggle the alternative route at fixed endpoint energies. Compare both forward and reverse barriers and explain why the net energy difference is unchanged.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Original forward barrier 80 kJ/mol; reverse 100 kJ/mol. ΔE=−20 kJ/mol. Add the alternative route to compare.
Schematic effective pathways: R=0, P=−20, uncatalyzed peak=80 and catalyzed peak=45 kJ/mol. Single humps are comparisons, not a claim that real catalysis has one step. Endpoints and equilibrium are unchanged.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using concentration–time slopes, rate-law dependence, encounter geometry or the stated mechanism. Identify what the representation cannot tell you.
Apply the idea to a fresh problem Practice →Show what you understand.
Two original questions are a starting check, not proof of mastery. Explain your choice before revealing the answer.
Original written challenge
4 points · self-check · not an official AP questionA catalyst lowers an illustrative peak from 90 to 55 kJ/mol while R=0 and P=−15. Find old/new reverse barriers, state ΔE and explain why equilibrium yield need not increase.
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Compare with the answer and four-point rubric
- 1 point: Old reverse barrier=90−(−15)=105 kJ/mol.
- 1 point: New reverse barrier=55−(−15)=70 kJ/mol.
- 1 point: ΔE remains −15 kJ/mol.
- 1 point: The catalyst accelerates approach to the same equilibrium; it does not change the equilibrium constant.
Accept equivalent correct methods and explanations. This is a Refresh Kid teaching rubric, not an official AP scoring guideline.
Retrieve it before you reveal it.
RECALL 1What changes with catalysis?
The pathway and effective rates/barriers.
RECALL 2What happens to net endpoint energy difference?
It stays the same for the same reaction and conditions.
RECALL 3Does catalyst imply greater equilibrium yield?
No; it changes how quickly equilibrium is approached.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
What does a catalyst change on an energy diagram?
- Catalyst: altered mechanism and effective barriers; regenerated overall.
- Same net reaction: ΔE and equilibrium constant unchanged at fixed conditions.
Remember: A catalyst does not lower reactant/product energy difference or guarantee more product at equilibrium.
Conditions: Schematic effective pathways: R=0, P=−20, uncatalyzed peak=80 and catalyzed peak=45 kJ/mol. Single humps are comparisons, not a claim that real catalysis has one step. Endpoints and equilibrium are unchanged.
Refresh Kid · AP Chemistry Unit 5 · Objectives 5.11.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 5.11, objective 5.11.A. CED effective Fall 2024 and June 2026 clarifications checked September 16, 2026. Unit 5: Kinetics, Topics 5.1–5.11. Focused lesson names, examples, models and assessments are original Refresh Kid teaching materials, not additional official topics or official AP questions. Official corrections.
The model states its assumptions beside the diagram. Arrhenius calculations are not assessed in the current AP framework; temperature and activation energy are taught qualitatively here. Collection of intermediate-detection data is not assigned. Integrated rate laws explicitly use the monitored species’ disappearance constant, while event and normalized reaction rates are labeled separately. Pre-equilibrium models state their timescale assumptions and use free concentrations. Original illustrative data and geometry are not measured kinetics.
Teaching resources: The Organic Chemistry Tutor video titles/descriptions and topic coverage were checked for optional links; no claim is made to have watched every video. No creator scripts, examples, worksheets or artwork were copied. GitHub’s 3D website collection and its Three.js camera-control example informed the idea of controllable spatial inspection. Scientific diagrams, geometry and interactions here are original. The self-hosted Three.js runtime retains its MIT license. Camera rotation changes the view, not the chemistry.
Independent teacher review and observation of students remain pending. Implementation checks do not certify scientific accuracy, accessibility or learning effectiveness. This is a review edition.
Optional official resource: Released AP Chemistry questions and scoring guides. This archive contains questions across units; it is not an assignment of every question to this lesson.
The teaching sequence is informed by the IES learning guide; this exact implementation has not been evaluated with learners.
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