First-Order Reaction Kinetics and Half-Life
In a first-order reaction the rate is proportional to the concentration of one reactant, rate = k[A]; the integrated rate law is ln[A] = ln[A]0 − kt, and the half-life, t½ = 0.693/k, is constant and independent of the starting concentration. In brief: if k = 0.0231 min−1, the half-life is 30.0 min, so 0.800 M falls to 0.400 M after 30 min, 0.200 M after 60 min and 0.100 M after 90 min.
First-order kinetics is the most commonly met in chemistry and in related sciences. It describes radioactive decay, many decompositions, the elimination of drugs from the body and the unimolecular reactions of molecules. This article presents the equations, shows how to use them, gives a table of the fraction remaining after each half-life and works through problems on finding k, concentrations and times.
What is a first-order rate law?
For the reaction A → products, with first order in A:
rate = −d[A]/dt = k[A]
The rate is directly proportional to the concentration, so it decreases as the reactant is consumed. The rate constant k has the units of s−1 (or min−1, h−1), which means that the value is independent of the concentration unit used.
What are the integrated forms?
Integrating the rate law gives three equivalent forms:
ln[A] = ln[A]0 − ktln([A]0 / [A]) = kt[A] = [A]0 e−kt
A plot of ln[A] against t is a straight line with a slope of −k. This is the graphical test for first order. Because the equations contain a ratio of concentrations, any quantity proportional to concentration (pressure, absorbance, activity or the mass of a sample) can be used instead of molarity.
What is the half-life?
The half-life t½ is the time required for the concentration to fall to half of its value. Setting [A] = [A]0/2 gives ln 2 = kt½, so:
t½ = ln 2 / k = 0.693 / k
The half-life is a constant for a given first-order reaction: it takes the same time to fall from 1.0 M to 0.5 M as from 0.10 M to 0.05 M. This is a distinguishing property of first-order reactions, and the reason for the use of the half-life in nuclear decay and in pharmacology.
Fraction remaining after n half-lives
| Number of half-lives | Time | Fraction remaining | Percent remaining |
|---|---|---|---|
| 0 | 0 | 1 | 100 % |
| 1 | t½ | 1/2 | 50 % |
| 2 | 2t½ | 1/4 | 25 % |
| 3 | 3t½ | 1/8 | 12.5 % |
| 4 | 4t½ | 1/16 | 6.25 % |
| 5 | 5t½ | 1/32 | 3.13 % |
| 7 | 7t½ | 1/128 | 0.78 % |
| 10 | 10t½ | 1/1024 | 0.098 % |
Worked examples
Example 1: half-life and remaining concentration
A first-order reaction has a rate constant of k = 0.0231 min−1. The initial concentration is 0.800 M.
- Half-life: t½ = 0.693 / 0.0231 = 30.0 min.
- After 90 min (3 half-lives): [A] = 0.800 / 23 = 0.100 M.
- Check with the integrated law: ln([A]0/[A]) = 0.0231 × 90 = 2.079, so [A]0/[A] = e2.079 = 8.00 and [A] = 0.100 M.
Example 2: finding k from two concentrations
The concentration of a reactant falls from 0.500 M to 0.300 M in 100 s. Then k = ln(0.500/0.300) / 100 s = 0.5108 / 100 = 5.11 × 10−3 s−1, and t½ = 0.693 / 5.11 × 10−3 = 136 s.
Example 3: time for 90 % conversion
How long does it take for the reaction in Example 1 to be 90 % complete? Then [A] = 0.10[A]0, so t = ln(10) / k = 2.303 / 0.0231 = 99.7 min. About 3.3 half-lives are needed.
Example 4: caffeine in the body
The elimination of caffeine is roughly first order, with a half-life of about 5 hours in healthy adults (the range is 3–7 hours). After a coffee with 200 mg of caffeine, the amount left after 15 hours (3 half-lives) is 200 / 8 = 25 mg. After 10 hours (2 half-lives) it is 50 mg, which is why late-afternoon coffee can affect sleep.
Where do first-order processes occur?
- Radioactive decay: carbon-14 has a half-life of 5730 years, and iodine-131 of 8.02 days.
- Unimolecular decompositions: N2O5 → NO2 + O2, and the isomerization of cyclopropane to propene.
- Pharmacokinetics: the elimination of most drugs from the body at normal doses.
- Pseudo-first-order reactions: bimolecular reactions in which one reactant is in such a large excess that its concentration is almost constant, as in the hydrolysis of an ester in dilute aqueous solution.
Who studied first-order decay?
Ludwig Wilhelmy’s work on the conversion of sucrose in 1850 followed a first-order rate law, probably the earliest example. Ernest Rutherford and Frederick Soddy showed in 1902 that radioactive decay follows an exponential law, which is mathematically identical, and Rutherford introduced the idea of half-life in 1907. The concept was then taken over in chemistry and pharmacology.
Which mistakes are common?
- Using the half-life formula for other orders: t½ = 0.693/k is only for first-order reactions.
- Forgetting the natural logarithm: ln and log10 differ by a factor of 2.303.
- Mixing time units: k in min−1 goes with time in minutes.
- Assuming that the reaction ends after two half-lives: 25 % is still left.
- Adding half-lives of different concentration steps wrongly: each one is the same length, so use n × t½.