Concentration Changes Over Time
Choose concentration, logarithmic or reciprocal graphs and interpret model limits and half-life.
AP Chemistry Topic 5.3 · CED printed p. 96, 97 / PDF p. 103, 104. This teaching, practice and feedback were added for the AP edition; original author material remains separately traceable.
Before you start
- Balanced equations and stoichiometric ratios
- Molarity and concentration versus amount
- Bond rearrangement and energy conservation
- Slopes, logarithms and reciprocal graphs
A curved [A]-versus-time trace tells you the rate changes, but its shape alone can be hard to diagnose. Test the appropriate transformed plots. Here we use the disappearance convention for a single monitored reactant under conditions where the simple law applies. A constant slope in [A] versus t supports zero order. Compare multiple points, fit residuals, and axes with units; two points always form a line and cannot by themselves establish an order.
For first order, . A plot of ln([A]/1 M) versus t is linear with slope −k; dividing by a reference unit makes the logarithm dimensionless. In ordinary AP notation, ln[A] means using consistent numerical concentration units. Equal time intervals remove equal fractions, so concentration drops quickly at first and more slowly later. The line is in the log plot, not the raw concentration plot.
For second order in A, . The reciprocal plot is linear with positive slope k, measured in M−1 s−1. Raw concentration decreases, but its reciprocal increases. A straight log plot and a straight reciprocal plot are different models; inspect data across a meaningful concentration range to distinguish them.
Zero order: [A]t =[A]0 −kt and raw slope −k, with k in M s−1. First order: log slope −k, with k in s−1. Second order: reciprocal slope +k, with k in M−1 s−1. Choose the model before substituting numbers. A zero-order extrapolation stops when reactant runs out; negative predicted concentration is a failure of extending the model, not a physical result.
First-order half-life is . Every halving takes the same time because the fractional disappearance rate is constant. After two half-lives, one quarter remains; this does not mean two identical amounts were removed. For zero and second order, half-life depends on the starting concentration, so constant half-life is specific evidence for first-order behavior.
Radioactive decay illustrates first-order kinetics: each nucleus has a constant probability of decaying per time, and the sample’s average number decaying per time is proportional to the number still present. Chemical temperature and catalysts do not ordinarily change the nuclear decay constant. This is a mathematical connection, not a claim that nuclear decay rearranges ordinary chemical bonds. Use supplied half-life or k data; no collection or handling of radioactive material is needed.
Let's work through it
For a first-order disappearance of A, [A]0 =0.0800 M and [A]60s =0.0400 M; multiple measurements establish a linear log plot.
- One halving occurs in 60.0 s, so k =0.693/60.0 =0.0116 s−1. Equivalently k =−ln(0.0400/0.0800)/60.0.
- At 180 s, three half-lives have passed: [A] =0.0800(½)3 =0.0100 M.
- A log-plot slope is negative because concentration falls, but k is positive. A nuclear first-order sample with the same half-life would have the same fraction remaining, even though its microscopic process differs.
For a zero-order regime, [A] falls from 0.0500 to 0.0400 M in 20.0 s and to 0.0300 M in another 20.0 s.
- Constant absolute losses make the raw concentration plot linear. k =0.0100/20.0 =5.00×10⁻⁴ M s⁻¹.
- The straight line [A] =0.0500−kt reaches zero at 100 s; do not extrapolate past depletion.
In a second-order regime, [A] is 0.100 M initially and 0.0500 M at 100 s; additional measurements establish a straight reciprocal plot.
- Initial and final reciprocals are 10.0 and 20.0 M⁻¹.
- k =(20.0−10.0)/100 =0.100 M⁻¹ s⁻¹, the positive reciprocal slope.
- At 200 s the reciprocal is 30.0 M⁻¹, giving [A] =0.0333 M. This is not another halving, so half-life is not constant.
A supplied isotope has half-life 5.00 h and starts with 48.0 μg undecayed isotope.
- In 15.0 h, three half-lives pass; remaining isotope is 48.0(½)³ =6.00 μg.
- The fraction follows the same first-order mathematics as the chemical example. The microscopic change is nuclear decay, so ordinary chemical catalysis does not provide a new nuclear pathway.
Make the picture move
Interactive concept model
Use data to identify a rate law.
Predict first. Compare the initial-rate rows. What changes when initial concentration doubles? Then compare the half-life dependence in the two scenarios.
rate = k[A]; [A]ₜ = [A]₀ exp(−kt)
- First-order concentration
- Labeled selected or reference state
On a small screen, scroll inside the graph or diagram to read its labels. The values and explanation also appear below.
For the specified first-order law, doubling initial [A] multiplies initial rate by 2. At 20 s, [A] = 0.147 mol L⁻¹.
Read the data and comparisons
| Initial [A] (mol L⁻¹) | Initial rate (mol L⁻¹ s⁻¹) | Rate relative to selected initial rate |
|---|---|---|
| 0.2 | 0.01 | 0.5 |
| 0.4 | 0.02 | 1 |
| 0.8 | 0.04 | 2 |
Macroscopic / data. Initial rate is a concentration change per unit time; it differs from the concentration itself.
Particle view. Changing particle concentration can change encounter frequency, but a rate-law exponent alone does not prove a molecular mechanism.
Symbolic view. The supplied exponent is 1; k units are s⁻¹.
Why it works. Order is established from rate evidence, not read from an overall balanced equation. The table compares otherwise identical hypothetical initial-rate trials. The curve follows the integrated law for disappearance of A; the plotted curve is not the table of initial rates. A straight ln([A]/1 M) versus time line has slope −k for first order; a straight 1/[A] versus time line has slope +k for second order. The concentration reference makes the logarithm dimensionless without changing its slope.
Conditions and limits. Hypothetical single-reactant disappearance with unit stoichiometric coefficient, constant volume and temperature (298 K), constant k = 0.05 s⁻¹, no reverse reaction over this interval. Supplied ideal data are model outputs, not measurements; rates and concentrations are rounded to three significant digits.
Try it from memory
Pause & retrieve
Try it before you peek.
Write your answer and the reason from memory. Then compare with the worked answer. Your self-assessment is a study decision, not an automatically graded score.
Question 1
Which plot should be straight for second-order disappearance of A?
See every recall question with its answer
These references preserve each prompt and its answer for direct evidence links. Try the interactive questions above before opening the worked reasoning.
Retrieval 1
Which plot should be straight for second-order disappearance of A?
Worked answer and reason
1/[A] versus time, with positive slope k and concentration-reciprocal units on the vertical axis.
The integrated law is linear in the reciprocal.
Retrieval 2
Is half of the original amount removed in every first-order half-life?
Worked answer and reason
No. Half of what remains is removed each time; after two half-lives one quarter of the original remains.
First order gives constant fractional rather than constant absolute loss.
Catch the tempting mistake
- Any decreasing concentration line is first order.
- A straight raw concentration line supports zero order. First order makes the log plot straight.
- Half-life must double when starting amount doubles.
- First-order half-life is independent of starting amount under the same conditions.
Put it to work
Give the chemical reason as well as the calculation. Worked solutions are on a separate page.
- Unit 5 problem 6 · ap application · data
- Unit 5 problem 7 · ap application · data
- Unit 5 problem 8 · ap application · data
- Unit 5 problem 9 · ap application · multiple
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For the teacher
- Require a graph choice, a slope sign and k units before arithmetic.
- If students fit only two points, give a third or fourth point that distinguishes the competing models.
Source objectives and scope
Source edition: ap-chemistry-ced-fall2024-local-b5dfe867. See this topic in the course outline →
- 5.3.A
- Identify the rate law expression of a chemical reaction using data that show how the concentrations of reaction species change over time.
- 5.3.A.1
- The order of a reaction can be inferred from a graph of concentration of reactant versus time.
- 5.3.A.2
- If a reaction is first order with respect to a reactant being monitored, a plot of the natural log (ln) of the reactant concentration as a function of time will be linear.
- 5.3.A.3
- If a reaction is second order with respect to a reactant being monitored, a plot of the reciprocal of the concentration of that reactant versus time will be linear.
- 5.3.A.4
- The slopes of the concentration versus time data for zeroth, first, and second order reactions can be used to determine the rate constant for the reaction. Zeroth order: EQN: [A]t − [A]0 = −kt First order: EQN: ln[A]t − ln[A]0 = −kt Second order: EQN: 1/[A]t − 1/[A]0 = kt
- 5.3.A.5
- Half-life is a critical parameter for first order reactions because the half-life is constant and related to the rate constant for the reaction by the equation: EQN: t1/2 = 0.693/k.
- 5.3.A.6
- Radioactive decay processes provide an important illustration of first order kinetics.
Science references for this teaching
- College Board AP Chemistry Course and Exam Description, Fall 2024 — Exact unit learning requirements and exclusions; supplied PDF is pinned by the integration catalog.
- OpenStax Chemistry 2e: Integrated Rate Laws — Scientific check: linearized zero/first/second-order forms and first-order half-life.
Glossary and deeper reading · 4 refreshers
These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.
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Stoichiometry
Stoichiometry uses a balanced chemical equation to relate amounts of reactants and products. Equation coefficients provide mole ratios rather than direct mass ratios. For example, forming water from hydrogen and oxygen consumes two moles of hydrogen molecules per mole of oxygen molecules.
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Molarity
Molarity is the amount of a solute in moles divided by the final solution volume in liters. It describes concentration rather than the total amount of solute. For example, 0.10 mole dissolved to make 0.50 liter of solution gives a molarity of 0.20 mole per liter.
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Reaction half-life
A reaction half-life is the time needed for a reactant's concentration to fall to half its starting value under the specified model. A first-order reaction has a half-life independent of its starting concentration, while other orders generally do not. For example, consecutive equal time intervals halve a first-order reactant's concentration repeatedly.
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Catalyst
A catalyst participates in a reaction pathway and is regenerated overall. It can increase the reaction rate by providing a different pathway without changing the reaction's equilibrium constant at fixed temperature. For example, a catalyst can help a mixture reach equilibrium sooner without changing the final equilibrium composition.