Integrated Rate Laws & Half-Life

You’ll be able to

Match zero-, first-, and second-order reactions to their integrated rate laws and linear plots
Use the first-order half-life relationship t½ = 0.693/k
Calculate concentration or time remaining after a whole number of half-lives

From rate law to concentration-vs-time

A rate law gives the instantaneous rate; an integrated rate law tells you the actual concentration at any time t. Each order integrates to a different equation — and, usefully, to a different straight-line plot. Finding which plot of the data is linear is how chemists confirm the order experimentally.

Integrated rate laws & linear plots

Zero: [A] = [A]₀ − kt (plot [A] vs t) | First: ln[A] = ln[A]₀ − kt (plot ln[A] vs t) | Second: 1/[A] = 1/[A]₀ + kt (plot 1/[A] vs t)

Whichever plot is a straight line reveals the order; the slope gives k (−k for zero/first order, +k for second order).

Half-life: the time to fall by half

The half-life (t½) is the time for a reactant to drop to half its concentration. For a first-order reaction, this time is constant — it does not depend on how much you start with — which is exactly why radioactive decay and many drug-clearance processes have a fixed half-life. Each successive half-life removes half of whatever remains.

First-order half-life

t½ = 0.693 / k

0.693 is ln 2. Because [A]₀ cancels out, first-order half-life is independent of starting concentration. (Zero order: t½ = [A]₀/2k; second order: t½ = 1/(k[A]₀) — these do depend on [A]₀.)

Worked example

A first-order reaction has a rate constant k = 0.0231 s⁻¹. Find its half-life. Then determine how long three half-lives take.

1.Use the first-order half-life formula: t½ = 0.693 / k.
2.Substitute k: t½ = 0.693 / 0.0231 s⁻¹.
3.Divide: t½ = 30. s.
4.For first order the half-life is constant, so three half-lives = 3 × 30. s = 90. s.

Answer: t½ = 30. s; three half-lives take 90. s

Checkpoint

A first-order reaction has a half-life of 20. min. What is its rate constant k?

Tip

Half-life questions with whole numbers of half-lives are just repeated halving — no calculator needed. After n half-lives the fraction remaining is (½)ⁿ: one half-life leaves 1/2, two leave 1/4, three leave 1/8, four leave 1/16.

Checkpoint

A first-order reactant starts at 80. g with a half-life of 10. min. How much remains after 30. min?

On the exam

A constant half-life is the signature of first order; a linear ln[A]-vs-t plot confirms it. If half-life instead grows as the reaction proceeds, suspect second order; if a plain [A]-vs-t plot is linear, it is zero order.

Answer the 2 checkpoints as you read.