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Integrated Rate Laws and Reaction Mechanisms

How does concentration change with time, how do you find the order from data, and what is a reaction mechanism?

AdvancedKineticsLast reviewed 5 October 2026
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What is it?

A rate law, such as rate =k[A]n= k[\ce{A}]^n, tells you how fast a reaction goes at one moment. An integrated rate law tells you the concentration at any time. Each order gives a different equation, and each can be rearranged into a straight line:

OrderIntegrated rate lawStraight-line plotSlopeHalf-life
0[A]=[A]0−kt[\ce{A}] = [\ce{A}]_0 - kt[A] against t−k-k[A]02k\dfrac{[\ce{A}]_0}{2k}
1ln⁡[A]=ln⁡[A]0−kt\ln[\ce{A}] = \ln[\ce{A}]_0 - ktln[A] against t−k-kln⁡2k\dfrac{\ln 2}{k}
21[A]=1[A]0+kt\dfrac{1}{[\ce{A}]} = \dfrac{1}{[\ce{A}]_0} + kt1/[A] against t+k+k1k[A]0\dfrac{1}{k[\ce{A}]_0}

Key idea

To find the order from concentration–time data, plot [A], ln[A] and 1/[A] against time. Only one plot is a straight line, and it tells you the order; its slope gives kk. A constant half-life is the signature of a first-order reaction.

Why does it matter?

  • Medicine. Drug levels in the blood usually fall by first-order kinetics, so doses are spaced by the drug’s half-life.
  • Shelf life. Manufacturers use integrated rate laws to predict how long medicines and foods stay within their limits.
  • Understanding reactions. The rate law is a clue to the mechanism: the sequence of simple steps by which a reaction actually happens.

How does it work?

1. Using the integrated rate laws

Each equation links four quantities: [A]0[\ce{A}]_0, [A][\ce{A}], kk and tt. Knowing any three, you can find the fourth. Make sure the units of kk match the time unit, and that ktkt has the right units: no units for first order (s⁻¹ × s), M⁻¹ for second order (M⁻¹ s⁻¹ × s).

2. Half-lives

  • First order: t1/2=ln⁡2/kt_{1/2} = \ln 2/k, independent of concentration. Each half-life halves what is left, just like radioactive decay.
  • Second order: t1/2=1/(k[A]0)t_{1/2} = 1/(k[\ce{A}]_0): each half-life is twice as long as the one before, because the concentration has halved.
  • Zero order: t1/2=[A]0/(2k)t_{1/2} = [\ce{A}]_0/(2k): each half-life is half as long as the one before.

3. Reaction mechanisms

Most reactions happen in several elementary steps. For an elementary step (only), the rate law follows from how many particles collide: a step A+B→\ce{A + B ->} has rate =k[A][B]= k[\ce{A}][\ce{B}].

  • The slowest step is the rate-determining step: the overall reaction cannot go faster than it.
  • A species made in one step and used up in a later one is an intermediate; it does not appear in the overall equation.
  • A proposed mechanism must (1) add up to the overall equation and (2) give a rate law that matches experiment.

For NOX2+CO→NO+COX2\ce{NO2 + CO -> NO + CO2}, experiments give rate =k[NOX2]2= k[\ce{NO2}]^2, with no CO. The accepted mechanism is:

NOX2+NOX2→NOX3+NO  (slow)NOX3+CO→NOX2+COX2  (fast)\begin{aligned} &\ce{NO2 + NO2 -> NO3 + NO}\ \ \text{(slow)} \\[2pt] &\ce{NO3 + CO -> NO2 + CO2}\ \ \text{(fast)} \end{aligned}

The slow first step involves two NOX2\ce{NO2} molecules, so rate =k[NOX2]2= k[\ce{NO2}]^2; CO only appears in the fast step, so it does not affect the rate. NOX3\ce{NO3} is the intermediate.

Think of it like this

A mechanism is like an assembly line. However fast the other workers are, the number of finished products per hour is set by the slowest worker (the rate-determining step). Speeding up a fast step changes nothing; speeding up the slow step speeds up everything.

More precisely

The integrated laws come from calculus: for first order, −d[A]dt=k[A]-\dfrac{d[\ce{A}]}{dt} = k[\ce{A}] integrates to ln⁡[A]=ln⁡[A]0−kt\ln[\ce{A}] = \ln[\ce{A}]_0 - kt. When a fast, reversible step comes before the slow step, the rate law can include species from that earlier step (the pre-equilibrium approach). A mechanism can never be proved, only supported: a mechanism that fits the data may still be replaced if new evidence appears.

Visualise it

Three graphs of the same data against time from 0 to 200 s. [A] against time is curved, so the reaction is not zero order. ln[A] against time is also curved, so it is not first order. 1/[A] against time is a straight line rising from 10 to 20 M⁻¹, so the reaction is second order. The slope of 1/[A] against t gives k = 0.0500 M⁻¹ s⁻¹.
Plot the data three ways: the straight-line plot identifies the order, and its slope gives k.
An energy profile for a two-step mechanism. From the reactants, the energy rises to a high first transition state (TS 1), labelled Ea step 1, slow; it falls to an intermediate in a shallow valley, rises to a lower second transition state (TS 2), labelled step 2, fast, and falls to the products. Step 1: NO2 + NO2 → NO3 + NO (slow). Step 2: NO3 + CO → NO2 + CO2 (fast). Rate = k[NO2] squared.
The step with the highest barrier is the rate-determining step; the intermediate sits in the valley between the two.

Worked example

Worked example: First order: concentration and time

Question: A first-order reaction has k=2.50×10−3 s−1k = 2.50 \times 10^{-3}\ \text{s}^{-1} and [A]0=0.500[\ce{A}]_0 = 0.500 M. Find (a) [A] after 300. s (b) the time for [A] to fall to 0.100 M (c) the half-life.

  1. (a) ln⁡[A]=ln⁡(0.500)−(2.50×10−3 s−1)(300. s)=−0.6931−0.750=−1.443\ln[\ce{A}] = \ln(0.500) - (2.50 \times 10^{-3}\ \text{s}^{-1})(300.\ \text{s}) = -0.6931 - 0.750 = -1.443, so [A]=e−1.443=[\ce{A}] = e^{-1.443} = 0.236 M
  2. (b) t=ln⁡([A]0/[A])k=ln⁡(0.500 M/0.100 M)2.50×10−3 s−1=1.6092.50×10−3 s−1=t = \dfrac{\ln([\ce{A}]_0/[\ce{A}])}{k} = \dfrac{\ln(0.500\ \text{M}/0.100\ \text{M})}{2.50 \times 10^{-3}\ \text{s}^{-1}} = \dfrac{1.609}{2.50 \times 10^{-3}\ \text{s}^{-1}} = 644 s
  3. (c) t1/2=0.69312.50×10−3 s−1=t_{1/2} = \dfrac{0.6931}{2.50 \times 10^{-3}\ \text{s}^{-1}} = 277 s

Worked example: Second order

Question: A second-order reaction has k=0.0400 M−1 s−1k = 0.0400\ \text{M}^{-1}\,\text{s}^{-1} and [A]0=0.200[\ce{A}]_0 = 0.200 M. Find [A] after 100. s and the first half-life.

  1. 1[A]=10.200 M+(0.0400 M−1 s−1)(100. s)=5.00 M−1+4.00 M−1=9.00 M−1\dfrac{1}{[\ce{A}]} = \dfrac{1}{0.200\ \text{M}} + (0.0400\ \text{M}^{-1}\,\text{s}^{-1})(100.\ \text{s}) = 5.00\ \text{M}^{-1} + 4.00\ \text{M}^{-1} = 9.00\ \text{M}^{-1}
  2. [A]=19.00 M−1=[\ce{A}] = \dfrac{1}{9.00\ \text{M}^{-1}} = 0.111 M
  3. t1/2=1(0.0400 M−1 s−1)(0.200 M)=t_{1/2} = \dfrac{1}{(0.0400\ \text{M}^{-1}\,\text{s}^{-1})(0.200\ \text{M})} = 125 s

Worked example: Finding the order from data

Question: [A] = 0.100, 0.0800, 0.0667, 0.0571 and 0.0500 M at t = 0, 50, 100, 150 and 200 s. Find the order and kk.

  1. ln[A] = −2.303, −2.526, −2.708, −2.862, −2.996: the differences shrink (−0.223, −0.182, −0.154, −0.134), so ln[A] is not linear: not first order.
  2. 1/[A] = 10.0, 12.5, 15.0, 17.5, 20.0 M⁻¹: it rises by exactly 2.5 M⁻¹ every 50 s, a straight line: second order.
  3. k=slope=2.5 M−150 s=k = \text{slope} = \dfrac{2.5\ \text{M}^{-1}}{50\ \text{s}} = 0.0500 M⁻¹ s⁻¹

Common mistake

Common mistake: Using the first-order half-life for every reaction

t1/2=0.693/kt_{1/2} = 0.693/k is only for first-order reactions. For second order the half-life depends on the starting concentration and grows as the reaction proceeds.

Common mistake: Writing a rate law from the overall equation

Orders come from experiment or from the slow elementary step, not from the coefficients of the overall equation. For NOX2+CO→NO+COX2\ce{NO2 + CO -> NO + CO2}, the rate law is k[NOX2]2k[\ce{NO2}]^2, not k[NOX2][CO]k[\ce{NO2}][\ce{CO}].

Common mistake: Mixing time units

If kk is in s⁻¹, time must be in seconds. Convert minutes or hours first, or ktkt will be wrong by a factor of 60 or 3600.

Notation note

  • [A]0[\ce{A}]_0 is the initial concentration; [A] or [A]t[\ce{A}]_t is the concentration at time tt.
  • “ln” is the natural logarithm (base e); ln 2 = 0.6931.
  • M⁻¹ s⁻¹ can also be written L mol⁻¹ s⁻¹.

Remember this

Remember this

  • Zero order: [A] linear in t. First order: ln[A] linear (slope −k). Second order: 1/[A] linear (slope +k).
  • Half-lives: zero [A]0/2k[\ce{A}]_0/2k; first ln⁡2/k\ln 2/k (constant); second 1/(k[A]0)1/(k[\ce{A}]_0).
  • The slow step is rate-determining; intermediates are made then used up; a mechanism must add up to the overall equation and match the rate law.

Test yourself

Check your understanding before moving on.

Flashcards

10 cards

  1. Question
    Zero-order integrated rate law and straight-line plot?
    Answer

    [A] = [A]₀ − kt; [A] against t is a straight line with slope −k.

  2. Question
    First-order integrated rate law and straight-line plot?
    Answer

    ln[A] = ln[A]₀ − kt; ln[A] against t is a straight line with slope −k.

  3. Question
    Second-order integrated rate law and straight-line plot?
    Answer

    1/[A] = 1/[A]₀ + kt; 1/[A] against t is a straight line with slope +k.

  4. Question
    Half-life for each order?
    Answer

    Zero: [A]₀/2k. First: ln 2/k (constant). Second: 1/(k[A]₀).

  5. Question
    How can you recognize first order from data?
    Answer

    Equal time intervals give equal fractions remaining, and the half-life is constant; ln[A] against t is straight.

  6. Question
    What happens to successive second-order half-lives?
    Answer

    Each is twice as long as the one before, because [A]₀ halves each time.

  7. Question
    What is an elementary step?
    Answer

    A single step in a mechanism; its rate law follows directly from the particles that collide.

  8. Question
    What is the rate-determining step?
    Answer

    The slowest step in a mechanism; it sets the overall rate.

  9. Question
    What is an intermediate?
    Answer

    A species formed in one step and used up in a later step; it does not appear in the overall equation.

  10. Question
    Two conditions for a valid mechanism?
    Answer

    The steps add up to the overall equation, and the predicted rate law matches the experimental one.

Quiz

7 questions

  1. Question 1EasyFor a first-order reaction, which plot gives a straight line?
    Show answer

    Answer: ln[A] against t

    ln[A] = ln[A]₀ − kt has the form y = c + mx, with slope −k.

  2. Question 2EasyA first-order reaction has a half-life of 20.0 min. What fraction remains after 60.0 min?
    Show answer

    Answer: 1/8

    60.0 min is 3 half-lives: (½)³ = 1/8.

  3. Question 3MediumFor a second-order reaction with k = 0.100 M⁻¹ s⁻¹ and [A]₀ = 0.500 M, what is the first half-life?
    Show answer

    Answer: 20.0 s

    t½ = 1/(k[A]₀) = 1/(0.100 M⁻¹ s⁻¹ × 0.500 M) = 20.0 s.

  4. Question 4MediumA plot of 1/[A] against time is a straight line with slope 0.0250 M⁻¹ s⁻¹. What is k?
    Show answer

    Answer: 0.0250 M⁻¹ s⁻¹

    For second order, the slope of 1/[A] against t equals +k, with units M⁻¹ s⁻¹.

  5. Question 5MediumA mechanism is: step 1 NO₂ + NO₂ → NO₃ + NO (slow); step 2 NO₃ + CO → NO₂ + CO₂ (fast). What is the rate law?
    Show answer

    Answer: rate = k[NO₂]²

    The slow step sets the rate, and it involves two NO₂ molecules.

  6. Question 6EasyIn the mechanism in the previous question, NO₃ is:
    Show answer

    Answer: an intermediate

    It is made in step 1 and used up in step 2, so it does not appear in the overall equation.

  7. Question 7HardSuccessive half-lives of a reaction are 50 s, 100 s and 200 s. What is the order?
    Show answer

    Answer: second

    Doubling half-lives are the signature of second order (t½ = 1/k[A]₀). First order gives constant half-lives; zero order gives shrinking ones.

Notes and downloads

References

  1. Brown, T. L.; LeMay, H. E., Jr.; Bursten, B. E.; Murphy, C. J.; Woodward, P. M.; Stoltzfus, M. W. Chemistry: The Central Science, 15th ed.; Pearson, 2022.

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