Worked example 1
A first-order reaction has k = 0.0250 s⁻¹. If [A]0 = 0.800 M, what is [A] after 60.0 s?
Try it first: Decide whether the ratio form or the logarithmic form is less work here.
0 of 3 steps revealed.
Chemical KineticsIntegrated Rate LawsContent level: Core 22 min
First-Order Integrated Rate Law and Half-Life
What you'll be able to do: Use the first-order integrated rate law and half-life relationship to find concentration, time or the rate constant.
Best after: Rate Laws and Reaction Order from Initial Rates
Introduction
A differential rate law answers how fast right now. An integrated rate law answers how much is left after a given time, which is the question that actually matters for radioactive decay, drug clearance and food spoilage.
These are recommended, not required. You can start this lesson at any time.
Learning objectives
- Apply the first-order integrated rate law to find [A], t or k
- Recognise first-order behaviour from a linear ln[A] versus t plot
- Use the constant half-life relationship in both directions
- Calculate the fraction remaining after a whole number of half-lives
Lesson
The first-order form
Integrating rate = k[A] over time gives a logarithmic relationship. Written as ln[A]t = -kt + ln[A]0 it is in y = mx + b form, so the graph of ln[A] against t is linear with slope -k and intercept ln[A]0.
ln[A]t = -kt + ln[A]0
The ratio form
Subtracting the intercept gives ln([A]t/[A]0) = -kt. This version is convenient whenever a question gives a fraction or percentage remaining rather than absolute concentrations, because the units cancel.
ln([A]t/[A]0) = -kt
Half-life is constant
Setting [A]t = [A]0/2 collapses the equation to t(1/2) = ln2/k = 0.693/k. The starting concentration cancels, so a first-order half-life never changes during the reaction. No other order behaves this way.
t(1/2) = 0.693 / k
Constant half-life is diagnostic. If successive half-lives are equal, the reaction is first order.
Counting half-lives
After n half-lives the fraction remaining is (1/2)ⁿ: 50 percent, 25 percent, 12.5 percent, 6.25 percent. When the elapsed time is a whole number of half-lives, counting is faster and safer than logarithms.
Check whether t divides evenly by t(1/2) before reaching for a calculator.
Where first-order kinetics shows up
Radioactive decay is strictly first order. So are many decompositions such as N₂O₅ and the elimination of many drugs from the bloodstream, which is why dosing intervals are quoted in half-lives.
Key ideas
Definition
Integrated rate law
An equation giving concentration as a function of time rather than rate as a function of concentration.
Definition
Half-life
The time required for the concentration of a reactant to fall to half its value.
Rule
First-order fingerprint
ln[A] versus t is linear and the half-life does not depend on the starting concentration.
Equation
Slope is -k
The slope of ln[A] against t equals minus the rate constant, so k is always positive.
Equation
First-order integrated rate law
ln[A]t = -kt + ln[A]0
- [A]t = concentration at time t
- [A]0 = initial concentration
- k = rate constant in s⁻¹
Equation
First-order half-life
t(1/2) = 0.693 / k
- t(1/2) = half-life in seconds
- 0.693 = ln 2
Worked examples
Worked example 2
The half-life of a first-order decomposition is 24.0 minutes. What percentage of the sample remains after 96.0 minutes?
0 of 3 steps revealed.
Worked example 3
A plot of ln[A] against time is a straight line with slope -3.2 × 10⁻³ s⁻¹. Find k and the half-life.
0 of 2 steps revealed.
Common mistakes
Reporting k as negative because the slope is negative.
Why it's wrong: The integrated law already contains the minus sign; k is a positive constant.
Check instead: Take k as the magnitude of the slope.
Assuming every reaction has a constant half-life.
Why it's wrong: Only first-order half-lives are independent of concentration; zero- and second-order half-lives change as the reaction runs.
Check instead: Confirm first order before using 0.693/k.
Mixing minutes and seconds between k and t.
Why it's wrong: The product kt must be dimensionless, so the time units must cancel.
Check instead: Convert so that the time unit in k matches the time unit of t.
Using log base 10 in place of ln.
Why it's wrong: The integrated law is derived with the natural logarithm, and using log10 introduces a factor of 2.303.
Check instead: Use ln, or multiply by 2.303 if a base-10 form is required.
Practice this skill
No practice questions are available for this topic yet. You can still practice the whole unit.
What you should now know
For a first-order reaction, ln[A]t = -kt + ln[A]0, so a plot of ln[A] against time is a straight line of slope -k. The half-life is t(1/2) = 0.693/k and is independent of the starting concentration, which is the defining fingerprint of first-order behaviour: every half-life removes half of whatever remains, so n half-lives leave a fraction (1/2)ⁿ.
- ln[A]t = -kt + ln[A]0 is the first-order integrated law
- A linear ln[A] versus t plot has slope -k
- t(1/2) = 0.693/k and does not depend on [A]0
- After n half-lives the fraction remaining is (1/2)ⁿ
- Keep time units consistent between k and t
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