Which graph identifies second-order disappearance?
You will be able to: Use reciprocal-concentration data and distinguish the three integrated-law plots.
Which graph identifies second-order disappearance?
A reaction can slow more strongly with falling concentration than first-order decay does. To identify the pattern, transform the same measured data rather than judging a curved graph by eye.
A useful starting point: Why does a first-order curve flatten over time? →
Words and symbols before equations
- Reciprocal concentration
- 1/[A], measured in M⁻¹.
- Second order in A
- Disappearance law proportional to [A]².
- Linearization
- Transforming measured values to test a predicted straight-line relation.
What this picture assumes
One-reactant law −d[A]/dt=(0.20 M⁻¹ s⁻¹)[A]². This k is a disappearance constant. The one-reactant integrated form is not automatically valid for arbitrary A+B kinetics.
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.
- [A]=0.25 M at 10 s. The reciprocal plot is linear with +0.20 M⁻¹ s⁻¹ slope. 1/[A]₀=2 M⁻¹.
- 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 stated law −d[A]/dt=k[A]², integration gives 1/[A]t = 1/[A]₀ + kt. Reciprocal concentration rises linearly while concentration itself falls.
The 1/[A] versus t slope is +k, not −k. k has units M⁻¹ s⁻¹ so kt has the reciprocal-concentration units of the vertical change.
Compare [A], ln([A]/1 M), and 1/[A] plots against the same times. Linear [A] supports zero order; linear log concentration supports first order; linear reciprocal concentration supports this second-order law.
Use multiple well-spaced data points and inspect residuals or deviations. Two points always define a line and cannot establish an order. These one-reactant forms are not automatically valid for every two-reactant second-order reaction.
| Order | Straight-line plot | Slope and k units |
|---|---|---|
| Zero | [A] versus t | −k; M/s |
| First | ln([A]/1 M) versus t | −k; s⁻¹ |
| Second in A | 1/[A] versus t | +k; M⁻¹ s⁻¹ |
A worked example, step by step
[A]₀=0.50 M and k=0.20 M⁻¹ s⁻¹ in −d[A]/dt=k[A]². Find [A] at 10 s.
- Calculate initial reciprocal concentration: 1/0.50 = 2.0 M⁻¹.
- Add kt = 0.20 × 10 = 2.0 M⁻¹.
- The final reciprocal is 4.0 M⁻¹.
- Invert at the end: [A]10 = 1/4.0 = 0.25 M. Do not mistake 4.0 M⁻¹ for concentration.
State the disappearance-rate convention. For an elementary 2A step, a normalized event-rate constant can differ by a factor of two from the k used here.
Can two concentration measurements prove second order?
Compare with an explanation
No. Multiple time points are needed because any two transformed points can lie on a straight line.
Predict. Change one thing. Explain.
Inspect the same second-order data in all three plot forms. Identify the straight one and explain its positive slope even though A is disappearing.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
[A]=0.25 M at 10 s. The reciprocal plot is linear with +0.20 M⁻¹ s⁻¹ slope. 1/[A]₀=2 M⁻¹.
One-reactant law −d[A]/dt=(0.20 M⁻¹ s⁻¹)[A]². This k is a disappearance constant. The one-reactant integrated form is not automatically valid for arbitrary A+B kinetics.
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 reciprocal-concentration plot rises from 2.0 to 5.0 M⁻¹ over 15 s and is linear across many measurements. Find k, initial/final concentrations and the supported disappearance law.
This response is not submitted or saved. Copy it before leaving.
Compare with the answer and four-point rubric
- 1 point: k=(5.0−2.0)/15=0.20 M⁻¹ s⁻¹.
- 1 point: Initial concentration=1/2.0=0.50 M.
- 1 point: Final concentration=1/5.0=0.20 M.
- 1 point: The data support −d[A]/dt=k[A]² under the stated one-reactant assumptions.
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 plot is linear for the stated second-order law?
1/[A] versus time.
RECALL 2Why does its graph rise?
As concentration decreases, its reciprocal increases.
RECALL 3Why inspect several data points?
Two points cannot discriminate among transformed linear fits.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Which graph identifies second-order disappearance?
- 1/[A]t = 1/[A]₀ + kt for −d[A]/dt=k[A]².
- Linear reciprocal plot slope +k; k units M⁻¹ s⁻¹.
Remember: State the disappearance-rate convention. For an elementary 2A step, a normalized event-rate constant can differ by a factor of two from the k used here.
Conditions: One-reactant law −d[A]/dt=(0.20 M⁻¹ s⁻¹)[A]². This k is a disappearance constant. The one-reactant integrated form is not automatically valid for arbitrary A+B kinetics.
Refresh Kid · AP Chemistry Unit 5 · Objectives 5.3.A · Review edition
Framework, scope and review status
Mapped to College Board CED, Topic 5.3, objective 5.3.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.
Want to work through this with a tutor?
Bring your question about Which graph identifies second-order disappearance? Your explanation and answers remain free to access.