Why do the units of k depend on reaction order?
You will be able to: Calculate a rate constant and determine its units from the rate law.
Why do the units of k depend on reaction order?
A rate has units of concentration per time. Multiplying two concentrations produces concentration squared, so the rate constant must supply the missing units as well as the numerical scale.
A useful starting point: How do experiments reveal a rate law? →
Words and symbols before equations
- Rate constant, k
- Proportionality factor for a specified rate law and conditions.
- M
- mol/L, the concentration unit.
- s⁻¹
- Per second.
- Dimensional analysis
- Using units to check and organize a calculation.
What this picture assumes
Separate illustrative laws use numerical k=0.020 but different units. Comparing numerical k across orders does not rank physical reactions. Concentration is positive within the modeled regime.
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.
- Order 0 uses k units M/s; selected rate is 0.02 M/s. The three k values have unlike units and are separate illustrative laws.
- Check what the picture assumes below. Use the Explore task to predict one change before moving a control.
Connect the picture to the chemistry
Solve the measured law for k. For r = k[A]², divide r by [A]², including units: (M/s)/M² = M⁻¹ s⁻¹.
In general, an overall order p gives k units M^(1−p) s⁻¹ when rate is M/s. Zero-order k is M/s; first-order k is s⁻¹.
At fixed temperature and the same mechanism/conditions, changing concentration changes rate through the law while k remains the same. Changing temperature or catalytic pathway can change the relevant k.
The numerical size of k cannot directly rank rates of different orders because its units differ. Calculate rates at specified concentrations instead.
A worked example, step by step
A reaction has r = k[A]². At [A]=0.200 M, r=0.0120 M/s. Calculate k, then predict rate at 0.100 M at the same temperature.
- k = r/[A]² = 0.0120/(0.200)².
- k = 0.300 M⁻¹ s⁻¹ because (M/s)/M² = M⁻¹ s⁻¹.
- At 0.100 M, r = 0.300(0.100)² = 0.00300 M/s.
- Halving a second-order concentration quarters rate while k stays fixed.
Rate and rate constant are different quantities. k is not universally measured in s⁻¹.
Does k change as reactant is consumed at constant temperature in this model?
Compare with an explanation
No. The concentration dependence changes the rate; k remains fixed under the stated conditions.
Predict. Change one thing. Explain.
Switch among orders 0, 1 and 2. For each, inspect the matching k units and how concentration changes rate. Do not compare unlike k values as if their units were identical.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
Order 0 uses k units M/s; selected rate is 0.02 M/s. The three k values have unlike units and are separate illustrative laws.
Separate illustrative laws use numerical k=0.020 but different units. Comparing numerical k across orders does not rank physical reactions. Concentration is positive within the modeled regime.
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 r=k[A]²[B], concentrations are 0.10 M and 0.20 M, and rate is 0.0040 M/s. Calculate k with units and predict the rate when B is halved at fixed A and T.
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Compare with the answer and four-point rubric
- 1 point: Concentration product = (0.10)²(0.20) = 0.0020 M³.
- 1 point: k = 0.0040/0.0020 = 2.0.
- 1 point: Units are M⁻² s⁻¹.
- 1 point: Halving B halves rate to 0.0020 M/s while k remains fixed.
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 fixes k units?
Overall order and the chosen rate/concentration/time units.
RECALL 2What changes when concentration changes at fixed conditions?
The rate through the rate law, not k.
RECALL 3Why not compare different-order k values directly?
They have different units and concentration dependencies.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Why do the units of k depend on reaction order?
- k = r/([A]ᵐ[B]ⁿ).
- For overall order p, k units = M^(1−p) s⁻¹.
Remember: Rate and rate constant are different quantities. k is not universally measured in s⁻¹.
Conditions: Separate illustrative laws use numerical k=0.020 but different units. Comparing numerical k across orders does not rank physical reactions. Concentration is positive within the modeled regime.
Refresh Kid · AP Chemistry Unit 5 · Objectives 5.2.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 5.2, objective 5.2.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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