Can a mechanism produce a fractional reaction order?
You will be able to: Derive a square-root concentration dependence from a fast dissociation equilibrium.
Can a mechanism produce a fractional reaction order?
If one particle quickly splits into two reactive pieces, the concentration of each piece need not be proportional to the original species. Solving the equilibrium relation can introduce a square root.
A useful starting point: How do you remove an intermediate from a rate law? →
Words and symbols before equations
- Fractional order
- A noninteger concentration exponent in a rate law.
- Free concentration
- Concentration of the specified species actually present, not its initial total before equilibration.
- Square root
- Number which, multiplied by itself, gives the stated quantity.
What this picture assumes
Fast N₂O₄ ⇌ 2NO₂, then slow NO₂+CO → NO+CO₂. Kc=0.040 M and k₂=0.50 M⁻¹ s⁻¹. v denotes NO appearance; normalized N₂O₄+2CO → 2NO+2CO₂ rate is v/2. Pre-equilibrium and free concentrations assumed.
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 N₂O₄=0.25 M; CO=0.2 M. NO appearance=0.01 M/s; normalized N₂O₄+2CO → 2NO+2CO₂ rate=0.005 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
For the supplied fast pre-equilibrium N₂O₄ ⇌ 2NO₂, equal forward/reverse rates give [NO₂]²≈Kc[N₂O₄]. Thus [NO₂]≈√(Kc[N₂O₄]).
If the supplied slow elementary step is NO₂+CO → NO+CO₂, the NO formation rate is v=k₂[NO₂][CO]. Substitution gives v≈k₂√Kc[N₂O₄]^(1/2)[CO].
Quadrupling free N₂O₄ at fixed CO doubles this rate. A fractional order can emerge from ordinary elementary steps; it does not require half a molecule to collide.
Two slow events per one forward dissociation give the net equation N₂O₄+2CO → 2NO+2CO₂. The normalized net rate is v/2 because v here denotes NO appearance. The fast step must remain near equilibrium.
A worked example, step by step
Kc=0.040 M, free [N₂O₄]=0.250 M, [CO]=0.200 M and k₂=0.50 M⁻¹ s⁻¹. Find [NO₂] and NO formation rate in the supplied model.
- [NO₂]²≈0.040 × 0.250 = 0.0100 M².
- [NO₂]≈√0.0100 = 0.100 M.
- v(NO)=0.50(0.100)(0.200)=0.0100 M/s.
- The normalized rate for N₂O₄+2CO → 2NO+2CO₂ is 0.00500 M/s; distinguish the two rate definitions.
Use free equilibrium concentrations. Treating an initial total as free N₂O₄ without a material-balance calculation can be wrong.
Does a half-order rate law require a half molecule in an elementary collision?
Compare with an explanation
No. The fractional power can arise when an equilibrium relationship is substituted into an elementary-step law.
Predict. Change one thing. Explain.
Quadruple free N₂O₄ while CO is fixed. Compare the NO₂ and NO-formation rate factors, then explain why a half-order does not mean a half-particle event.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Free N₂O₄=0.25 M; CO=0.2 M. NO appearance=0.01 M/s; normalized N₂O₄+2CO → 2NO+2CO₂ rate=0.005 M/s.
Fast N₂O₄ ⇌ 2NO₂, then slow NO₂+CO → NO+CO₂. Kc=0.040 M and k₂=0.50 M⁻¹ s⁻¹. v denotes NO appearance; normalized N₂O₄+2CO → 2NO+2CO₂ rate is v/2. Pre-equilibrium and free concentrations assumed.
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 X₂ ⇌ 2X and slow X+Y → XY, derive v(XY), predict the effect of quadrupling free X₂, give the net equation and distinguish normalized net rate.
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Compare with the answer and four-point rubric
- 1 point: [X]≈√(Kc[X₂]); v(XY)=k₂√(Kc[X₂])[Y].
- 1 point: Quadrupling free X₂ doubles v at fixed Y.
- 1 point: Net X₂+2Y → 2XY.
- 1 point: The normalized net rate is v(XY)/2; the stated v measures product appearance.
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 1Where can fractional order come from?
Eliminating intermediates through pre-equilibrium relationships.
RECALL 2What does the square root act on?
Kc times the free parent-species concentration.
RECALL 3Why label the rate explicitly?
Species appearance and normalized reaction rate can differ by a coefficient.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Can a mechanism produce a fractional reaction order?
- Fast A₂ ⇌ 2A: [A]≈√(Kc[A₂]).
- Slow A+B: v≈k₂√(Kc[A₂])[B].
Remember: Use free equilibrium concentrations. Treating an initial total as free N₂O₄ without a material-balance calculation can be wrong.
Conditions: Fast N₂O₄ ⇌ 2NO₂, then slow NO₂+CO → NO+CO₂. Kc=0.040 M and k₂=0.50 M⁻¹ s⁻¹. v denotes NO appearance; normalized N₂O₄+2CO → 2NO+2CO₂ rate is v/2. Pre-equilibrium and free concentrations assumed.
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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