Half-life and first-order kinetics
The first-order integrated rate law, why its half-life is constant, k = 0.693 / t1/2, and worked examples from reaction kinetics and radioactive decay.
A half-life is the time it takes for half of something to be used up. For a first-order process, that time never changes. Start with 100 g or 1 mg, and the first half disappears in exactly as long as the next half, and the half after that. This makes first-order problems some of the most predictable on a chemistry exam, and they show up twice in General Chemistry II: once in kinetics and again in nuclear chemistry.
What "first order" means
A reaction is first order when its rate is proportional to the concentration of one reactant: rate = k[A]. Double [A] and the rate doubles. The rate constant k has units of inverse time (s^-1, min^-1, yr^-1).
Radioactive decay is always first order. The rate is k x N, where N is the number of radioactive nuclei present. Unlike ordinary reactions, a nuclide's half-life doesn't depend on temperature, pressure or its chemical environment.
The equations
Integrating rate = k[A] gives the first-order integrated rate law:
- ln[A]t = ln[A]0 - kt
- equivalently, [A]t = [A]0 e^(-kt)
- or, in half-lives, N = N0 (1/2)^(t / t1/2)
Set [A]t equal to half of [A]0 and solve for t, and you get the half-life:
t1/2 = ln 2 / k = 0.693 / k
Notice that [A]0 cancels out. That is the proof that a first-order half-life does not depend on the starting amount. A plot of ln[A] against time is a straight line with slope -k, which is how you recognize first-order data.
Example 1: whole half-lives (the shortcut)
100 g of carbon-14 (t1/2 = 5,730 years). How much is left after 11,460 years?
- Number of half-lives: 11,460 / 5,730 = 2
- Fraction left: (1/2)^2 = 1/4
- Amount left: 100 g x 1/4 = 25 g
A second one: an 80.0 mg sample goes through 4 half-lives. (1/2)^4 = 1/16 = 0.0625, and 80.0 mg x 0.0625 = 5.00 mg.
When the time is a whole number of half-lives, you never need a logarithm.
Example 2: k from the half-life, and back
Iodine-131 has a half-life of 8.02 days.
k = 0.693 / 8.02 days = 0.0864 per day
Going the other way: a nuclide with k = 0.0247 per year has t1/2 = 0.693 / 0.0247 = 28.1 years.
Example 3: a first-order reaction
A reaction is first order with k = 2.50 x 10^-3 s^-1 and [A]0 = 0.800 M. What is [A] after 300. s, and what is the half-life?
- ln[A]t = ln(0.800) - (2.50 x 10^-3)(300.) = -0.2231 - 0.750 = -0.9731
- [A]t = e^-0.9731 = 0.378 M
- t1/2 = 0.693 / (2.50 x 10^-3) = 277 s
As a sanity check, 300 s is just over one half-life, so [A] should be a little under half of 0.800 M. It is.
Example 4: time elapsed, not a whole number of half-lives
A wood sample has carbon-14 activity that is 62.5% of the activity in living wood. How old is it?
Solve N = N0 e^(-kt) for t:
t = (t1/2 / ln 2) x ln(N0 / N)
- N0 / N = 1 / 0.625 = 1.6
- ln(1.6) = 0.470
- t = (5,730 / 0.693) x 0.470 = 3,890 years (3 sig figs)
This is less than one half-life (5,730 years), which makes sense because more than half the carbon-14 remains.
Carbon-14 dating assumes the atmosphere's carbon-14 to carbon-12 ratio has stayed roughly constant, and that the organism stopped taking in carbon when it died. It works reliably up to about 50,000 years, after which too little carbon-14 is left to measure well.
Example 5: finding the half-life from data
A reactant falls from 0.800 M to 0.200 M in 40 minutes, and the reaction is first order. What is its half-life?
0.200 / 0.800 = 1/4 = (1/2)^2, so 2 half-lives have passed in 40 minutes. t1/2 = 40 / 2 = 20 minutes.
If the fraction isn't a clean power of 1/2, use t1/2 = t x ln 2 / ln(N0 / N).
How first order compares with zero and second order
| Order | Integrated law | Straight-line plot | Half-life |
|---|---|---|---|
| Zero | [A]t = [A]0 - kt | [A] vs t | [A]0 / 2k |
| First | ln[A]t = ln[A]0 - kt | ln[A] vs t | 0.693 / k |
| Second | 1/[A]t = kt + 1/[A]0 | 1/[A] vs t | 1 / (k[A]0) |
Only the first-order half-life is constant. Zero-order half-lives get shorter as the reaction proceeds, and second-order half-lives get longer. If an exam question says successive half-lives are equal, it is telling you the reaction is first order.
Mistakes to watch for
- Mixing time units. If t1/2 is in days, t must be in days, and k comes out per day.
- Using ln when the problem uses log. The integrated law uses the natural log. 0.693 is ln 2, not log 2.
- Applying (1/2)^n with a non-whole n and then rounding n. Either keep the decimal exponent or use the ln form.
- Using 0.693 / k for other orders. That formula belongs to first order only.
Check your work
The half-life calculator solves for the amount remaining, the time elapsed or the half-life, and reports k and the number of half-lives. Kinetics tells you how fast a reaction goes. Where it stops is the next unit, covered in ICE tables for equilibrium, step by step. Radioactive decay shows up again in the nuclear physics unit of College Physics 2, with the same equation written using the decay constant λ. For a full course map, see what's in General Chemistry II, and for the review schedule, what spaced repetition actually does.
Integrated rate laws and nuclear decay kinetics each have a chapter in Encodr's free General Chemistry II deck.
Half-Life Calculator
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