How do you remove an intermediate from a rate law?
You will be able to: Use a fast pre-equilibrium relation to eliminate an intermediate before a slow step.
How do you remove an intermediate from a rate law?
A fast reversible assembly step keeps making and undoing an intermediate while a slower step uses it. The intermediate amount is then related approximately to the available starting particles.
A useful starting point: How does a slow first step predict the rate law? →
Words and symbols before equations
- Pre-equilibrium
- Approximation that an earlier reversible step stays near equilibrium while a slower step removes its intermediate.
- Forward/reverse rate constants
- k₁ and k₋₁ for the two directions of the early step.
- Concentration-form equilibrium ratio, Kc
- Ratio of rate constants used here with explicit concentration units.
- Intermediate elimination
- Substitution that removes an intermediate from the final observed-rate expression.
What this picture assumes
Fast 2NO ⇌ N₂O₂ pre-equilibrium, then slow N₂O₂+O₂ → 2NO₂. Kc=2.0 M⁻¹ and k₂=0.50 M⁻¹ s⁻¹. Free concentrations, not analytical starting totals. Fast reverse relaxation must dominate slow depletion.
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.
- Free NO=0.1 M; O₂=0.2 M. Intermediate=0.02 M; normalized rate=0.002 M/s; NO₂ appearance=0.004 M/s.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the chemistry
Consider the supplied model 2NO ⇌ N₂O₂ (fast pre-equilibrium), then N₂O₂+O₂ → 2NO₂ (slow, elementary). The slow event rate is r=k₂[N₂O₂][O₂].
The rapid forward and reverse event rates are approximately equal: k₁[NO]²≈k₋₁[N₂O₂]. Rearranging gives [N₂O₂]≈(k₁/k₋₁)[NO]²=Kc[NO]².
Substitute into the slow law: r≈k₂Kc[NO]²[O₂]. The final expression uses free reactant concentrations rather than the unmeasured intermediate.
This requires the reversible step to relax much faster than the slow consumption step. Fast does not mean irreversible or complete. Kc here is a concentration-form ratio with units M⁻¹; thermodynamic equilibrium constants use standardized activities.
A worked example, step by step
For this supplied model, Kc=2.0 M⁻¹, [NO]=0.10 M, [O₂]=0.20 M and k₂=0.50 M⁻¹ s⁻¹. Find intermediate concentration and normalized net reaction rate.
- [N₂O₂]≈Kc[NO]²=2.0(0.10)²=0.020 M.
- Use the slow step: r=0.50(0.020)(0.20)=0.0020 M/s.
- The same result follows from r≈k₂Kc[NO]²[O₂].
- The net reaction is 2NO+O₂ → 2NO₂; NO₂ appearance is 2r=0.0040 M/s. Concentrations are free concentrations, not initial analytical totals.
Do not leave an intermediate in the final observable rate law when the supplied pre-equilibrium relation allows its elimination.
Does fast pre-equilibrium mean all NO becomes N₂O₂?
Compare with an explanation
No. Both directions operate, with their rates nearly balancing; the intermediate depends on the equilibrium relation.
Predict. Change one thing. Explain.
Keep O₂ fixed and double free NO. Predict the intermediate concentration and rate factors. Then vary O₂ alone, keeping the approximation assumptions in mind.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Free NO=0.1 M; O₂=0.2 M. Intermediate=0.02 M; normalized rate=0.002 M/s; NO₂ appearance=0.004 M/s.
Fast 2NO ⇌ N₂O₂ pre-equilibrium, then slow N₂O₂+O₂ → 2NO₂. Kc=2.0 M⁻¹ and k₂=0.50 M⁻¹ s⁻¹. Free concentrations, not analytical starting totals. Fast reverse relaxation must dominate slow depletion.
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 questionFor fast A+B ⇌ I and slow I+B → AB₂, derive an intermediate relation and overall rate law, give the net equation, and state the timescale assumption.
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Compare with the answer and four-point rubric
- 1 point: [I]≈(k₁/k₋₁)[A][B] from the fast reversible elementary step.
- 1 point: r≈k₂[I][B]=k₂(k₁/k₋₁)[A][B]².
- 1 point: Net equation A+2B → AB₂.
- 1 point: The reversible first step must re-equilibrate much faster than I is drained by the slow step.
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 1Why introduce pre-equilibrium?
To relate a fast intermediate to reactant concentrations before the slow step.
RECALL 2What gets substituted?
The intermediate concentration in the slow-step law.
RECALL 3What must be checked?
That the fast reversible step remains near equilibrium on the slower reaction timescale.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
How do you remove an intermediate from a rate law?
- Fast 2A ⇌ I: [I]≈(k₁/k₋₁)[A]².
- Slow I+B: r≈k₂(k₁/k₋₁)[A]²[B].
Remember: Do not leave an intermediate in the final observable rate law when the supplied pre-equilibrium relation allows its elimination.
Conditions: Fast 2NO ⇌ N₂O₂ pre-equilibrium, then slow N₂O₂+O₂ → 2NO₂. Kc=2.0 M⁻¹ and k₂=0.50 M⁻¹ s⁻¹. Free concentrations, not analytical starting totals. Fast reverse relaxation must dominate slow depletion.
Refresh Kid · AP Chemistry Unit 5 · Objectives 5.9.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 5.9, objective 5.9.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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