First-Order Reactions: Integrated Rate Law and Half-Life
First-order reactions are the most tested topic in Class 12 chemical kinetics. Radioactive decay and many decompositions follow this law. The key is to choose the right form of the equation and to keep the units clear. This page gives the formulas, the meaning of each symbol, and five numericals with every step shown.
The first-order rate law
For a reaction A → products with rate = k[A], the integrated forms are:
[A] = [A]0 e−kt
t½ = 0.693 / k
- [A]0 = initial concentration (or amount); [A] = concentration (or amount) after time t.
- k = rate constant, with unit time−1 (s−1, min−1).
- t½ = half-life. It does not depend on the starting concentration. This is the signature of first order.
Because the ratio [A]0/[A] appears, you may use any unit of amount (mol/L, grams, pressure units, or percentage) as long as you use the same unit for both. For gas-phase reactions where only pressure data is given, convert carefully using the stoichiometry.
Half-lives: a quick table
| Half-lives passed | Fraction left | Fraction reacted |
|---|---|---|
| 1 | 1/2 = 50% | 50% |
| 2 | 1/4 = 25% | 75% |
| 3 | 1/8 = 12.5% | 87.5% |
| 4 | 1/16 = 6.25% | 93.75% |
| 5 | 1/32 = 3.125% | 96.875% |
How to recognise a first-order reaction
- The rate is proportional to the concentration of one reactant, to the power one.
- A plot of ln[A] against time is a straight line with slope −k.
- The half-life stays the same however much reactant you start with.
- The unit of k is time−1, with no concentration term in it.
Order must be found from experiment or from the rate law given in the question. It cannot be read from the balanced equation alone. Radioactive decay and the decomposition of N2O5 are common first-order examples in textbooks.
Worked example 1: k from concentration data
A first-order reaction goes from 0.80 mol/L to 0.20 mol/L in 20 minutes. Find k and t½.
[A]0/[A] = 0.80 / 0.20 = 4.
k = ln 4 / 20 min = 1.3863 / 20 = 0.0693 min−1.
t½ = 0.693 / 0.0693 = 10.0 min.
Check without k: 0.80 → 0.40 → 0.20 takes two half-lives. Two half-lives = 20 min, so one half-life = 10 min. The two routes agree.
Worked example 2: percentage completion
A first-order reaction is 30% complete in 40 minutes. Find k, t½, and the time for 90% completion.
If 30% has reacted, 70% remains: [A]0/[A] = 100/70 = 1.4286.
k = ln(1.4286) / 40 = 0.35667 / 40 = 8.92 × 10−3 min−1.
t½ = 0.693 / 0.008917 = 77.7 min.
For 90% completion, 10% remains: ratio = 10. t = ln 10 / k = 2.3026 / 0.008917 = 258 min.
Quick check: 258 / 77.7 = 3.32 half-lives, and (1/2)3.32 = 0.10. Correct.
Worked example 3: amount left after n half-lives
A sample of 6.4 g of a substance decays by first-order kinetics with a half-life of 10 minutes. How much is left after 50 minutes?
Number of half-lives = 50 / 10 = 5.
Fraction left = (1/2)5 = 1/32.
Amount = 6.4 / 32 = 0.20 g.
Worked example 4: amount left after a general time
Using the reaction in Example 2 (k = 8.917 × 10−3 min−1), what percentage of the reactant is left after 100 minutes?
[A]/[A]0 = e−kt = e−(0.008917 × 100) = e−0.8917 = 0.410.
So 41.0% is left, and 59.0% has reacted.
Check with half-lives: 100 / 77.7 = 1.287 half-lives; (1/2)1.287 = 0.410. Same answer.
Worked example 5: the slope of a graph
A plot of ln[A] against time (in seconds) for a first-order reaction is a straight line with slope −4.6 × 10−3 s−1. Find k, the half-life, and the time for the concentration to fall to one-tenth.
k = −slope = 4.6 × 10−3 s−1.
t½ = 0.693 / 0.0046 = 150.7 s.
One-tenth left: t = ln 10 / k = 2.3026 / 0.0046 = 500.6 s, about 501 s.
Common mistakes
- Using the amount reacted instead of the amount left. In the formula, [A] is what remains.
- Mixing ln and log. If you use log10, the factor 2.303 must be present.
- Making the half-life depend on concentration. That is true for zero and second order, not for first order.
- Unit errors. If k is in min−1, time must be in minutes.
- Using 0.693 as k. It is the constant ln 2, used with t½.
- Calling 75% complete “half of two half-lives.” Two half-lives leave 25%, so 75% has reacted.
Exam relevance
| Question type | Starting point |
|---|---|
| Find k from two readings | k = (2.303/t) log([A]0/[A]) |
| Half-life to k, or k to half-life | t½ = 0.693/k |
| Time for x% completion | t = (2.303/k) log(100/(100 − x)) |
| Identify the order | Constant half-life, or straight ln[A] vs t graph |
All values here are practice numbers. Check your current board syllabus for which forms are required.
Check your half-life and decay arithmetic with the Half-Life calculator.
Open the Half-Life Calculator →Reaction mechanisms stick only with practice: attempt previous-year CSIR-NET questions and a timed quiz.
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