Why does a first-order curve flatten over time?
You will be able to: Use first-order decay and its logarithmic straight-line representation.
Why does a first-order curve flatten over time?
If a fixed fraction of the remaining reactant disappears each second, less absolute material is lost later because less remains. The curve bends even though the same rule operates throughout.
A useful starting point: What does a constant disappearance rate look like? →
Words and symbols before equations
- First order
- Disappearance rate proportional to the remaining concentration.
- Natural logarithm, ln
- Logarithm with base e, the inverse of the exponential function.
- Exponential decay
- Multiplicative decrease governed here by e^(−kt).
- Reference concentration
- A fixed concentration used to make a logarithm’s argument dimensionless.
What this picture assumes
One-species first-order disappearance −d[A]/dt=k[A], fixed k and volume. Logarithm uses [A]/(1 M), a dimensionless ratio. This k belongs to disappearance, not an unspecified normalized reaction rate.
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.
- At 10 s, [A]=0.48522 M; fraction remaining=0.60653. Log slope=−0.05 s⁻¹, independent of starting concentration.
- 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 −d[A]/dt=k[A], the solution is [A]t=[A]₀e^(−kt). As [A] falls, its rate k[A] also falls, so the concentration curve becomes less steep.
Taking logarithms gives ln([A]t/[A]₀)=−kt. A plot of ln([A]/1 M) versus time is linear with slope −k; the 1 M reference keeps the logarithm dimensionless.
For the same k, two samples with different starting concentrations lose the same fraction over a given interval, not the same number of moles.
This integrated law uses k for the named species’ disappearance. If a problem defines a normalized reaction rate with a coefficient, first translate that convention into the disappearance law.
A worked example, step by step
[A]₀=0.80 M and k=0.050 s⁻¹. Find concentration after 10 s.
- Use [A]t=[A]₀e^(−kt) for the stated first-order law.
- The exponent is −(0.050 s⁻¹)(10 s)=−0.50, dimensionless.
- [A]10=0.80e^(−0.50)≈0.485 M.
- About 60.7% remains; the absolute disappearance rate at that time is k[A]≈0.0243 M/s.
A first-order concentration curve is not a straight line. Its logarithmic concentration plot is.
Does the reactant lose the same concentration amount in each equal interval?
Compare with an explanation
No. It loses the same fraction; the absolute loss decreases as less remains.
Predict. Change one thing. Explain.
Change initial concentration while holding k fixed. Compare the concentration curve and log plot; explain why the log slope stays unchanged.
On narrow screens, swipe or scroll diagrams sideways to read all labels.
At 10 s, [A]=0.48522 M; fraction remaining=0.60653. Log slope=−0.05 s⁻¹, independent of starting concentration.
One-species first-order disappearance −d[A]/dt=k[A], fixed k and volume. Logarithm uses [A]/(1 M), a dimensionless ratio. This k belongs to disappearance, not an unspecified normalized reaction rate.
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 first-order sample falls from 0.60 M to 0.30 M in 20 s. Calculate k, predict concentration after another 20 s, and state the linear plot and slope sign.
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Compare with the answer and four-point rubric
- 1 point: k=ln(0.60/0.30)/20=ln 2/20≈0.0347 s⁻¹.
- 1 point: After another equal interval, concentration halves to 0.15 M.
- 1 point: ln([A]/1 M) versus time is linear.
- 1 point: Its slope is −k, negative for disappearance.
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 stays constant over equal intervals?
The fraction remaining, for fixed first-order k.
RECALL 2What is the first-order straight-line plot?
ln([A]/1 M) against time.
RECALL 3Why must kt be dimensionless?
It appears as an exponent and is formed from reciprocal-time k times time.
Revisit these tomorrow and a week later. Try a fresh problem and explain why the method applies.
Why does a first-order curve flatten over time?
- [A]t=[A]₀e^(−kt); ln([A]t/[A]₀)=−kt.
- Slope of ln([A]/1 M) vs t = −k; k units s⁻¹.
Remember: A first-order concentration curve is not a straight line. Its logarithmic concentration plot is.
Conditions: One-species first-order disappearance −d[A]/dt=k[A], fixed k and volume. Logarithm uses [A]/(1 M), a dimensionless ratio. This k belongs to disappearance, not an unspecified normalized reaction rate.
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.
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