Chemistry 2e · Kinetics

Integrated Rate Laws

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On this page 9 sections

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 relates reaction rate to reactant concentration at one instant. An instead tells how a reactant's concentration changes over a measured time interval. It is the tool to use when a problem supplies an initial concentration, a concentration after some time, a rate constant, or a graph of concentration versus time.

For a reaction in which A disappears, the equation depends on its order. Experimental concentration–time data determine that order. With the correct form, calculate remaining reactant, elapsed time, or the rate constant k.

Why this matters

Integrated rate laws turn a changing chemical system into a testable pattern. They help chemists follow drug breakdown, pollutants, and reactions monitored by spectroscopy. A straight-line graph can reveal and provide k from its slope.

The college version

Core Concepts

Concentration, time, and reaction order

Let [A]₀ mean the concentration of A at time zero and [A]_t its concentration at time t. For a zero-order reaction, the rate does not depend on [A], so A is consumed by the same concentration amount in equal time intervals:

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

Thus, a plot of [A] versus t is linear. Its slope is -k, and k has units of mol L^-1 s^-1, often written M s^-1.

For a first-order reaction, the rate is proportional to [A]. The amount lost per interval decreases as less A remains:

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

Equivalently, ln([A]_t/[A]₀)=-kt. A plot of ln[A] versus t is linear, with slope -k; here k has units s^-1.

For a second-order reaction involving one reactant A, 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. The units of k are M^-1 s^-1. The positive slope is sensible: as concentration falls, its reciprocal rises.

Using plots as evidence

Transform the same data three ways: [A], ln[A], and 1/[A] versus time. The plot that is straight within experimental scatter identifies zero-, first-, or second-order behavior, respectively. Its slope supplies the appropriately signed k.

Units carry information about reaction order. Keep time units consistent: if k is in min^-1, use minutes rather than seconds.

Half-life patterns

The half-life, t_1/2, is the time required for [A] to decrease to one-half its current value. For first-order reactions,

$t 1/2 = 0.693 k$

so every halving takes the same time. Zero- and second-order half-lives depend on concentration: t_1/2=[A]₀/(2k) for zero order and t_1/2=1/(k[A]₀) for second order.

How It Works / Step-by-Step Process

Record concentrations at known times for the reactant being analyzed.
Test the three linearized plots or use the experimentally supplied order.
Choose the matching integrated rate law and preserve the units used for time and concentration.
Substitute known quantities, solve algebraically, and check that the sign and magnitude are chemically reasonable.
If half-life is requested, use the formula for that order rather than assuming the first-order expression.

Common Confusions

Common Confusion	Correct Understanding
“A larger k always has the same units.”	Units change with order; compare values only when the orders and units match.
“The coefficient in a balanced equation gives the order.”	That is true only for an elementary reaction; overall reactions require experimental data.
“A first-order reaction loses a constant amount each minute.”	It loses a constant fraction in equal intervals; the amount lost becomes smaller.
“Every half-life problem uses 0.693/k.”	That compact formula is only for first-order reactions.
“A positive second-order slope means concentration increases.”	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 passes. Different reactions use different recipes: some lose the same amount each minute, while others lose the same fraction. A graph helps us discover which recipe fits the experiment.

Worked example

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

Suppose A is found experimentally to react by first-order kinetics with k=0.230 min^-1. A sample begins at [A]₀=0.800 M. What concentration remains after 3.00 min?

Use ln([A]_t/[A]₀)=-kt:

$ln ([A] t /0.800)=-(0.230)(3.00)=-0.690$

Exponentiating gives [A]_t/0.800=e^-0.690 ≈ 0.502, so [A]_t ≈ 0.402 M. The answer is close to one-half because t_1/2=0.693/0.230=3.01 min. This agreement is a useful check, not an extra assumption: the constant half-life follows specifically from first-order kinetics.

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

Suppose the same reactant A follows second-order kinetics with k = 0.0445 M^-1min^-1, starting from [A]₀ = 0.250 M. How long does it take for [A] to fall to 0.100 M?

Write the integrated second-order law before substituting:

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

Substitute the known values:

$1 0.100 = 1 0.250 + (0.0445)t$

$10.0 M -1 = 4.00 M -1 + (0.0445 M -1 min -1)t$

$t = 10.0 - 4.00 0.0445 min = 134.8 min \approx 135 min$

The units cancel correctly: M^-1 divided by M^-1 min^-1 leaves minutes. Two checks apply: the time is positive, and it is shorter than the second-order half-life t_1/2 = 1/(k[A]₀) = 1/((0.0445)(0.250)) = 89.9 min — sensible, because 0.100 M is more than halfway down from 0.250 M.

Key takeaways

The straight graph identifies the order: [A] for zero order, ln[A] for first order, and 1/[A] for second order.
The slope is -k for zero- and first-order linear plots, but +k for the second-order reciprocal plot.
First-order reactions have a concentration-independent half-life; zero- and second-order reactions do not.
Use the order determined experimentally, not coefficients from an overall balanced equation.
A calculated concentration must be physically meaningful; a negative value signals that the zero-order equation has been extended beyond complete consumption.

Check yourself

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

Which plot is linear for a first-order reaction?
Show answer
A plot of ln[A] versus time; its slope is -k.
A graph of 1/[A] versus time is linear with positive slope. What order is the reaction, and what does the slope equal?
Show answer
It is second order in A, and the slope equals +k.
Why is the half-life of a first-order reaction unchanged by the starting concentration?
Show answer
Its integrated rate law leads to t_1/2=0.693/k, which contains no concentration term.
What is wrong with assigning reaction order directly from an overall balanced equation?
Show answer
Overall coefficients describe stoichiometry, whereas order is determined by the reaction mechanism and must usually be measured.
A zero-order calculation gives [A]_t=-0.10 M. How should that result be interpreted?
Show answer
Concentration cannot be negative. The reaction has reached completion before that time, so the zero-order equation no longer represents a real concentration.

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Study tools & related lessonsKey vocabulary · Related

Key vocabulary

integrated rate law: An equation connecting reactant concentration and elapsed time for a specified reaction order.
reaction order: The experimentally determined concentration dependence of a rate law.
rate constant, k: The proportionality constant for a rate law; its units depend on order.
half-life, t1/2: Time for a reactant concentration to become half of its starting or current value.
linear plot: A graph whose data fit a straight line; it can diagnose reaction order and give k from slope.
rate constant (k): Proportionality constant for a particular reaction under specified conditions

Sources & references

openstax.org — Chemistry 2e

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

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