Chemistry: Atoms First 2e · Kinetics

Integrated Rate Laws

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In 30 seconds
Why this matters
The college version
Eli explains
Worked example
Key takeaway
Check yourself
Study tools
Sources & references

In 30 seconds

A rate law describes how fast a reaction proceeds at a single moment, expressed in terms of reactant concentrations. An answers a different question: given that a reaction started at some concentration, how much reactant remains after a certain amount of time? It is obtained by integrating (hence the name) the differential rate law, so that concentration appears as a function of elapsed time rather than as an instantaneous rate.

The form of the integrated rate law depends on the — zero, first, or second — and that order must come from experiment, not from the coefficients of a balanced equation. With the correct form in hand, you can solve for a remaining concentration, an elapsed time, or the rate constant k. The same equations also generate half-life expressions, which are especially useful for describing processes such as radioactive decay and drug elimination.

Why this matters

Integrated rate laws turn a changing chemical system into something predictable. Pharmacists use first-order kinetics to estimate how much of a medication remains in the body between doses; forensic and environmental chemists use them to predict how long a spilled pollutant will persist; and astronomers use the first-order decay of carbon-14 or potassium-40 to date ancient materials. In the lab, a researcher monitoring a reaction by spectroscopy can determine both the reaction order and the rate constant from a single straight-line graph. Understanding which integrated equation applies — and what its slope means — is a core skill for any chemistry exam and for interpreting real kinetic data.

The college version

Core Concepts

Concentration as a function of time

Let [A]₀ be the concentration of reactant A at time zero and [A]_t its concentration at time t. For a zero-order reaction, the rate is independent of [A], so equal amounts of A disappear in equal time intervals:

$[A] t = [A] 0 - kt$

A plot of [A] versus t is a straight line with slope -k, and k has units of concentration per time, such as mol L^-1s^-1 or M s^-1.

For a first-order reaction, the rate is proportional to [A]. Each equal time interval removes the same fraction of what remains, so the amount consumed per interval shrinks as the reaction proceeds:

$ln [A] t = ln [A] 0 - kt$

A plot of ln[A] versus t is linear with slope -k; here k carries units of s^-1 (or min^-1, h^-1, and so on).

For a second-order reaction in a single reactant, the rate is proportional to [A]²:

$1 [A] t = 1 [A] 0 + kt$

A plot of 1/[A] versus t is linear with slope +k, and k has units of M^-1s^-1. The positive slope makes sense: as [A] falls, its reciprocal rises.

Using linear plots as evidence

The same concentration–time data can be plotted three ways: [A], ln[A], or 1/[A] against time. Whichever plot is a straight line (within experimental scatter) identifies the order: linear [A] means zero order, linear ln[A] means first order, and linear 1/[A] means second order. The slope of that line supplies the rate constant with the correct sign. Always keep time units consistent — if k is in min^-1, express times in minutes, not seconds.

Half-life patterns

The half-life, t_1/2, is the time needed for the concentration to fall to one-half of its current value. For a first-order reaction:

$t 1/2 = 0.693 k$

Because this expression contains no concentration term, every halving takes exactly the same amount of time — a hallmark of first-order behavior. Zero- and second-order half-lives do depend on concentration: t_1/2 = [A]₀/(2k) for zero order and t_1/2 = 1/(k[A]₀) for second order, so each successive half-life is shorter (zero order) or longer (second order) than the last.

How It Works / Step-by-Step Process

Record the reactant's concentration at several known times (or obtain the data from an instrument such as a spectrometer).
Identify the order by testing the three linear plots, or use an order supplied by the experiment.
Select the matching integrated rate law and keep time and concentration units consistent.
Substitute the known values, solve algebraically, and check that the result is physically sensible (positive time, positive concentration).
If a half-life is requested, use the half-life expression for that order rather than assuming the first-order formula.

Common Confusions

Do Not Confuse	With	Difference
The straight-line plot for first order	The plot for second order	ln[A] vs. t is linear (slope -k) for first order; 1/[A] vs. t is linear (slope +k) for second order.
Losing a constant amount each interval	Losing a constant fraction each interval	Zero-order reactions lose a constant amount; first-order reactions lose a constant fraction, so the amount shrinks.
The first-order half-life formula 0.693/k	Half-lives of other orders	Zero-order half-life is [A]₀/(2k) and second-order is 1/(k[A]₀); both depend on concentration.
Coefficients in the balanced equation	Reaction order	Coefficients are stoichiometry; order comes from the mechanism and must be measured.
A positive slope on the second-order plot	Concentration increasing	The plotted quantity is 1/[A], which rises as [A] falls.

Eli explains

The same idea, in plain words

Explain it like I’m 10

An integrated rate law is a recipe for tracking how much of a chemical is left as time goes by. Different reactions follow different recipes: some lose the same amount every minute, and others lose the same fraction. Drawing a graph of the data tells us which recipe the reaction is actually using.

Worked example

Worked example 1: concentration remaining after a set time (first order)

The decomposition of hydrogen peroxide in dilute solution follows first-order kinetics with k = 0.0346 min^-1. A sample begins at [H₂O₂]₀ = 1.25 M. What concentration remains after 40.0 min?

Write the integrated first-order law before substituting:

$ln [H 2 O 2] t = ln [H 2 O 2] 0 - kt$

Substitute the known values:

$ln [H 2 O 2] t = ln (1.25) - (0.0346 min -1)(40.0 min) = 0.2231 - 1.384 = -1.161$

Exponentiate both sides:

$[H 2 O 2] t = e -1.161 \approx 0.313 M$

Check with the half-life: t_1/2 = 0.693/0.0346 = 20.0 min. Since 40.0 min is exactly two half-lives, 1.25 M → 0.625 M → 0.313 M. The agreement confirms the calculation. Note that the units of k (min^-1) required time in minutes; had the problem supplied seconds, they would need converting first.

Worked example 2: time to reach a target concentration (second order)

Dimerization of butadiene, 2 C₄H₆ → C₈H₁₂, is second order in butadiene with k = 0.250 M^-1s^-1. Starting from [C₄H₆]₀ = 0.500 M, how long until the concentration reaches 0.125 M?

Write the integrated second-order law first:

$1 [C 4 H 6] t = 1 [C 4 H 6] 0 + kt$

Substitute:

$1 0.125 = 1 0.500 + (0.250 M -1 s -1)t$

$8.00 M -1 = 2.00 M -1 + (0.250 M -1 s -1)t$

$t = 8.00 - 2.00 0.250 s = 24.0 s$

Dimensional analysis confirms the units: M^-1 divided by M^-1s^-1 leaves seconds. As a sanity check, the second-order half-life is t_1/2 = 1/(k[A]₀) = 1/((0.250)(0.500)) = 8.0 s; reaching 0.125 M (two halvings) takes longer than one half-life but less than the naive doubling of 16 s, because second-order half-lives lengthen as concentration falls — the computed 24.0 s is consistent with that pattern.

Key takeaways

The linear plot identifies the order: [A] versus t → zero order, ln[A] versus t → first order, 1/[A] versus t → second order.
The slope is -k for zero- and first-order plots but +k for the second-order reciprocal plot.
Only first-order reactions have a concentration-independent half-life, t_1/2 = 0.693/k.
Reaction order is an experimental quantity; never assign it from the coefficients of an overall balanced equation.
Units of k encode the order: M s^-1, s^-1, and M^-1s^-1 for zero, first, and second order, respectively.
A calculated negative concentration means the reaction finished before that time; the integrated equation no longer describes a real concentration.

Check yourself

5 review questions from the chapter. Try each one, then open the answer.

Which plot would you make to test whether a reaction is first order, and what would the slope equal?
Show answer
Plot ln[A] versus time; a straight line indicates first order, and its slope is -k.
A plot of 1/[A] versus time is linear with a positive slope. What is the reaction order, and what does the slope equal?
Show answer
Second order in A; the slope equals +k.
Why is the half-life of a first-order reaction independent of the starting concentration?
Show answer
The first-order integrated law gives t_1/2 = 0.693/k, which contains no concentration term, so the time for each halving is fixed.
A reaction has k = 0.0500 M^-1s^-1. What is its order, and what are the units telling you?
Show answer
The units M^-1s^-1 are characteristic of a second-order reaction (one reactant, or overall second order).
A zero-order calculation returns [A]_t = -0.05 M. How should you interpret this result?
Show answer
A concentration cannot be negative. The reaction reached completion before that time, so the zero-order equation is no longer physically meaningful there.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

Integrated rate law: An equation relating reactant concentration to elapsed time for a given reaction order.
Reaction order: The experimentally determined dependence of rate on each reactant's concentration.
Rate constant k: The proportionality constant in a rate law; its units change with order.
Half-life t1/2: The time for a concentration to drop to half of its starting or current value.
Linear plot: A graph of transformed concentration data that fits a straight line.

Sources & references

openstax.org — Chemistry Atoms First 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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