How do warming and a catalyst affect equilibrium differently?
You will be able to: Distinguish temperature-dependent K from catalytic acceleration.
How do warming and a catalyst affect equilibrium differently?
Heating a reversible mixture can change its final color and composition. A catalyst at the same temperature may help it settle sooner without changing its equilibrium color.
A useful starting point: Why can compression change a gas equilibrium? →
Words and symbols before equations
- Endothermic forward reaction
- Forward conversion absorbs heat under the stated conditions.
- Exothermic forward reaction
- Forward conversion releases heat.
- Catalyst
- Provides faster pathways without changing the net equilibrium at fixed T.
- Temperature-dependent K
- The equilibrium constant belongs to a specified temperature.
What this picture assumes
Illustrative supplied constants: at 300 K, K=2; at 330 K, K=4 for the endothermic example and K=1 for the exothermic example. A ⇌ B total 1.00 M. These two examples are not a numerical van’t Hoff model. Catalyst holds T fixed; no timescale computed.
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.
- Forward reaction endothermic; T=300 K; supplied K=2. [B]eq=0.66667 M. Catalyst absent: same K and equilibrium composition at this T; rate behavior may differ.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the chemistry
Increasing temperature favors the endothermic direction. For an endothermic forward reaction K increases; for an exothermic forward reaction K decreases.
Heat is useful in a qualitative energy-accounting explanation, but it is not an extra concentration term in Q or K.
A catalyst changes pathways for both directions while preserving the equilibrium relation at fixed T. It does not change K or the equilibrium composition of the same closed inventory.
The model supplies illustrative K values at two temperatures; it does not derive them from a heat number. Predicting actual new K values requires additional data beyond a qualitative shift rule.
| Change | K at equilibrium | Reason |
|---|---|---|
| Temperature | Can change | New thermal conditions |
| Catalyst | Unchanged at fixed T | Both directions accelerated |
| Adding reactant | Unchanged at fixed T | Composition and Q change |
A worked example, step by step
For an endothermic A⇌B system, supplied K values are 2 at 300 K and 4 at 330 K. Total concentration is 1.00 M. Compare final B concentrations and the effect of a catalyst at 300 K.
- For A⇌B, [B]eq=CK/(1+K).
- At 300 K, [B]eq=2/3≈0.667 M.
- At 330 K, [B]eq=4/5=0.800 M: warming favors the endothermic products.
- A catalyst at 300 K leaves K=2 and [B]eq≈0.667 M; it changes the approach rate, not that endpoint.
A catalyst does not favor only the forward reaction or change K at fixed temperature.
If warming decreases K, which direction is exothermic?
Compare with an explanation
The forward direction is exothermic for that reaction convention.
Predict. Change one thing. Explain.
Switch the supplied forward reaction between endothermic and exothermic, then compare the two temperatures. Turn the catalyst on and check which displayed quantities remain unchanged.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Forward reaction endothermic; T=300 K; supplied K=2. [B]eq=0.66667 M. Catalyst absent: same K and equilibrium composition at this T; rate behavior may differ.
Illustrative supplied constants: at 300 K, K=2; at 330 K, K=4 for the endothermic example and K=1 for the exothermic example. A ⇌ B total 1.00 M. These two examples are not a numerical van’t Hoff model. Catalyst holds T fixed; no timescale computed.
Explain what you noticed: Answer the investigation prompt above. State one observation and explain it using relative rates, particle conservation, the Q/K comparison or the stated dissolution equilibrium. 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 reaction is exothermic forward. Explain the effect of warming on K and product preference, then explain why adding a catalyst at the original temperature has a different effect.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: Warming favors the endothermic reverse direction.
- 1 point: The forward K decreases.
- 1 point: The equilibrium expression becomes less product-favored under the new thermal conditions.
- 1 point: A catalyst at the original T does not change K or the equilibrium composition; it changes kinetics.
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 1Which change can alter K?
Temperature.
RECALL 2Is heat inserted into Q?
No.
RECALL 3Can Le Châtelier’s principle alone give a new numerical K?
No; numerical temperature-dependent data are needed.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do warming and a catalyst affect equilibrium differently?
- Warming favors the endothermic direction.
- A catalyst leaves K unchanged at fixed T.
Remember: A catalyst does not favor only the forward reaction or change K at fixed temperature.
Conditions: Illustrative supplied constants: at 300 K, K=2; at 330 K, K=4 for the endothermic example and K=1 for the exothermic example. A ⇌ B total 1.00 M. These two examples are not a numerical van’t Hoff model. Catalyst holds T fixed; no timescale computed.
Refresh Kid · AP Chemistry Unit 7 · Objectives 7.9.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 7.9, objective 7.9.A. CED effective Fall 2024 and June 2026 clarifications checked September 17, 2026. Unit 7: Equilibrium, Topics 7.1–7.12. 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. Converting between Kc and Kp and calculations for a dissolved species in equilibrium with its gas phase are excluded from assessed Unit 7 scope. Concentrations use mol/L and gas partial pressures use the stated pressure convention. Supplied constants are teaching data at fixed temperature unless otherwise specified. Ideal dilute-solution and ideal-gas approximations are stated. 3D views show inventories, not molecular trajectories, measured structures or proof of equilibrium from a single snapshot. Approximation checks are explicit; a small K alone does not justify neglecting every change.
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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