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.
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Chemical KineticsIntegrated Rate LawsContent level: Challenge 26 min
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
Lesson
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.
[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.
1/[A]t = kt + 1/[A]0
The three-plot test
Plot [A] against t, ln[A] against t, and 1/[A] against t. Exactly one will be acceptably linear, and that identifies the order. Then read k from the slope, remembering the sign convention for each form.
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
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. 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.
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?
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Worked example 3
For a zero-order reaction, [A]0 = 0.600 M and k = 0.0150 M/s. How long until the reactant is completely consumed?
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Common mistakes
Using slope = -k for the second-order plot.
Why it's wrong: 1/[A] increases with time, so its slope is positive and equals +k.
Check instead: Match the sign to the specific integrated form being plotted.
Assuming a constant half-life without checking the order.
Why it's wrong: Constant half-life is a first-order property only.
Check instead: Use [A]0/2k for zero order and 1/(k[A]0) for second order.
Judging linearity from a short time window.
Why it's wrong: All three plots look nearly straight over a small fraction of the reaction.
Check instead: Use data spanning at least two half-lives before assigning an order.
Inverting each term separately in the second-order law.
Why it's wrong: 1/(a + b) is not 1/a + 1/b, so inverting mid-calculation corrupts the result.
Check instead: Add reciprocals first, then invert once at the end.
Practice this skill
No practice questions are available for this topic yet. You can still practice the whole unit.
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
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
Chemistry 2e, OpenStax, Rice University, licensed CC BY 4.0. License