Worked example 1
A second-order reaction has k = 0.250 M⁻¹ s⁻¹ and [A]0 = 0.500 M. Find [A] after 20.0 s.
Try it first: Work with reciprocals throughout and invert only at the very end.
0 of 2 steps revealed.
Integrated Rate LawsCourse level: AP Chemistry
Zero- and Second-Order Rate Laws and Graphical Analysis
What you'll be able to do: Identify reaction order from which plot is linear, and apply the zero- and second-order integrated rate laws and half-lives.
Best after: First-Order Integrated Rate Law and Half-Life
Introduction
Given a table of concentration against time and no other information, the order can be recovered by asking which of three plots gives a straight line. This lesson completes the set and compares all three orders side by side.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- Apply the zero- and second-order integrated rate laws
- Determine order by testing which of three plots is linear
- Compare half-life behaviour across the three orders
- Analyse errors in order assignment using half-life and slope evidence
Zero order
When the rate does not depend on concentration, concentration falls in a straight line until the reactant runs out. This happens when a surface or an enzyme is saturated, so adding more reactant cannot speed anything up.
For C = 0.600 M and k = 0.0150 M/s, [A] = 0.600 − 0.0150t predicts 0.300 M at 20.0 s and zero at 40.0 s. Do not continue the line into negative concentrations. In a surface-controlled experiment, the zero-order approximation may fail before complete depletion because the surface becomes unsaturated. A finite model intercept is not proof that the same kinetics holds all the way to zero.
[A]t = -kt + [A]0
Zero-order k carries units of M/s, unlike any other order.
Second order
For rate = k[A]², the reciprocal of concentration rises linearly with time. Note the sign: the slope of 1/[A] against t is +k, not -k, because the reciprocal grows as the concentration falls.
For C = 0.500 M, k = 0.250 M⁻¹ s⁻¹ and t = 20.0 s, first compute 1/[A] = 2.00 + 5.00 = 7.00 M⁻¹, then invert the complete result: [A] = 0.1429 M. Inverting the separate terms would be an algebra error. To find time instead, t = (1/[A]t − 1/C)/k; a smaller target concentration gives a positive reciprocal difference.
1/[A]t = kt + 1/[A]0
The three-plot test
Plot [A] against t, ln[A] against t, and 1/[A] against t. For data that follow one of these simple laws over a sufficient range, its transformed plot should be acceptably linear; noisy or short datasets can be ambiguous. Then read k from the slope, remembering the sign convention for each form.
These single-reactant formulas use −d[A]/dt = k[A]ⁿ at fixed conditions:
| Order | Integrated law | Linear ordinate versus t | Slope | Half-life from starting C |
|---|---|---|---|---|
| 0 | [A] = C − kt | [A] | −k | C/(2k) |
| 1 | ln([A]/C) = −kt | ln[A] | −k | ln(2)/k |
| 2 | 1/[A] = 1/C + kt | 1/[A] | +k | 1/(kC) |
The second-order row specifically describes a disappearance law proportional to [A]². A general law k[A][B] with unequal initial concentrations needs a different integrated expression. A stoichiometrically normalized rate constant can also carry a coefficient factor; state the convention used.
Compare correlation, not eyeballs alone. Over a short time window all three plots can look almost straight.
Half-lives as a second line of evidence
| Reaction order | Half-life behavior |
|---|---|
| Zero order | t(1/2) = [A]0/2k, which shortens as the reaction proceeds because each successive half-life starts from less material. |
| First order | constant. |
| Second order | t(1/2) = 1/(k[A]0), which lengthens as the reaction proceeds. |
Scroll horizontally if needed to read all columns.
Measuring two successive half-lives distinguishes the orders without any plotting.
zero: [A]0/2k first: 0.693/k second: 1/(k[A]0)
Error analysis
Claiming first order from a single half-life measurement is unsupported, because one half-life is consistent with any order. Claiming second order from a rising 1/[A] plot with visible curvature is also unsafe. A defensible assignment quotes both the linear plot and the trend in successive half-lives.
All smooth curves can look nearly straight over a short interval, and noisy transformed data may make several fits plausible. Inspect curvature and residuals, the range of concentrations and independent half-life evidence rather than claiming an exact order from two points. For first-order behavior, equal halving intervals support the log plot; for second order, consecutive halving intervals double. If a second-order sample takes 50 s to reach C/2, it takes another 100 s to reach C/4, giving 150 s total. Distinguish an additional interval from time measured since the start.
Use the transformed plot to get k
For a second-order reaction in A, 1/[A] = 1/[A]0 + kt, so a plot of reciprocal concentration against time has slope k in M⁻¹ s⁻¹. Two measured points can yield that slope without a graph.
The time for another halving then grows as concentration falls, unlike the constant first-order half-life.
Use reciprocal concentration only for the second-order integrated law.
Key ideas
Rule
Which plot is linear
[A] vs t means zero order, ln[A] vs t means first order, 1/[A] vs t means second order.
Rule
Slope signs
Zero and first order have slope -k; second order has slope +k.
Key concept
Half-life trend
Zero order half-lives shorten, first order stay constant, second order lengthen.
Key concept
Saturation
Zero-order behaviour usually signals a saturated catalyst, enzyme or surface.
Equation
Zero-order integrated law
[A]t = -kt + [A]0
- k = rate constant in M/s
Equation
Second-order integrated law
1/[A]t = kt + 1/[A]0
- k = rate constant in M⁻¹ s⁻¹
Equation
Second-order half-life
t(1/2) = 1 / (k[A]0)
- [A]0 = concentration at the start of that particular half-life
Worked examples
Worked example 2
Successive half-lives of a reaction are measured as 40 s, 80 s and 160 s. What is the order, and what is k if [A]0 = 0.100 M?
0 of 3 steps revealed.
Worked example 3
For a zero-order model with [A]₀ = 0.600 M and k = 0.0150 M/s, at what time does the extrapolated concentration reach zero? Assume this law remains valid up to that limit.
0 of 3 steps revealed.
Worked example 4
For a second-order process, [A] falls from 0.100 M to 0.0500 M in 50.0 s. Find k and when it reaches 0.0250 M.
Try it first: Convert each concentration to its reciprocal.
0 of 3 steps revealed.
Common mistakes
| Mistake | Why it's wrong | Check instead |
|---|---|---|
| Using slope = -k for the second-order plot. | 1/[A] increases with time, so its slope is positive and equals +k. | Match the sign to the specific integrated form being plotted. |
| Assuming a constant half-life without checking the order. | Constant half-life is a first-order property only. | Use [A]0/2k for zero order and 1/(k[A]0) for second order. |
| Judging linearity from a short time window. | All three plots look nearly straight over a small fraction of the reaction. | Use data spanning at least two half-lives before assigning an order. |
| Inverting each term separately in the second-order law. | 1/(a + b) is not 1/a + 1/b, so inverting mid-calculation corrupts the result. | Add reciprocals first, then invert once at the end. |
Scroll horizontally if needed to read all columns.
After the lesson · Question bank
Independent practice
Work through new problems on your own. The lesson's quick checks and worked examples stay above if you need to review a step.
33 topic questions across 3 difficulty levels.
Loading practice questions…
What you should now know
Zero order gives [A]t = -kt + [A]0, a straight line of [A] against t with slope -k and half-life [A]0/2k. Second order gives 1/[A]t = kt + 1/[A]0, a straight line of 1/[A] against t with slope +k and half-life 1/(k[A]0). Testing all three plots and keeping the one that is linear is the standard experimental route to reaction order, and the half-life behaviour, constant, shortening or lengthening, is an independent confirmation.
- Zero order: [A] vs t linear, slope -k, k in M/s
- First order: ln[A] vs t linear, slope -k
- Second order: 1/[A] vs t linear, slope +k
- Half-lives shorten, stay constant, or lengthen for zero, first and second order
- Assign order from a linear plot plus a half-life trend, not from one data pair
Your next step
Lesson: Not marked completed
Marking a lesson completed records that you finished reading it. Practice establishes mastery separately. Saved on this device.
Try a short set of different questions to check what you can do independently.
Watch and Learn
An optional video to support this lesson.
Loading video resources…
Supplementary learning resources
More resources for this topic →External resources belong to their respective creators. Chem Help is not affiliated with or endorsed by these providers. Links open the original provider; their terms apply.
Sources and further reading
- Chemistry 2e, Section 12.4: Integrated Rate LawsPaul Flowers, Klaus Theopold, Richard Langley, William R. Robinson · OpenStax, Rice University · Chapter 12.4View source
Lesson-specific rights and provenance need review.
The current online Chemistry 2e edition lists CC BY-NC-SA 4.0. The source version used for this lesson and its applicable rights remain unverified.
Recorded use: reference. This describes the stored citation, not verified authorship or reuse permission.
Recorded attribution / license claim (applicability unverified): Chemistry 2e, OpenStax, Rice University, licensed CC BY 4.0.
Source rights and review status - AP Chemistry Course and Exam DescriptionCollege Board · Unit 5: KineticsView source
Lesson-specific rights and provenance need review.
This College Board reference does not establish permission to reproduce its content. Any reproduction requires a verified applicable permission or other legal basis.
Recorded use: reference. This describes the stored citation, not verified authorship or reuse permission.
Source rights and review status - General Chemistry bookshelf (LibreTexts)LibreTexts, University of California, DavisView source
Lesson-specific rights and provenance need review.
The exact LibreTexts page and its license have not been established; the bookshelf citation does not grant permission for all linked material.
Recorded use: further reading. This describes the stored citation, not verified authorship or reuse permission.
Recorded attribution / license claim (applicability unverified): LibreTexts General Chemistry, licence varies by page, check the page footer.
Source rights and review status