Half-Lives and Reaction Order

Introduction

The half-life, t½, is the time taken for a reactant's concentration to halve. It is a single number that is easy to read from a graph, yet it carries a powerful clue about order. For a first-order process the half-life never changes; for other orders it shifts in a predictable way as concentration falls. Watching how the half-life behaves is therefore one of the quickest ways to identify the kinetic form of a reaction.

Core explanation

Set [A] = [A]₀/2 in each integrated rate law and solve for t.

Zero order. From [A] = [A]₀ − kt:

t½ = [A]₀ ÷ 2k

The half-life is proportional to the starting concentration. Each successive half-life is half as long as the one before, because the constant rate removes a smaller absolute amount each time.

First order. From ln[A] = ln[A]₀ − kt:

t½ = ln 2 ÷ k ≈ 0.693 ÷ k

The half-life is independent of concentration. It takes the same time to go from 100% to 50% as from 50% to 25%. This constant half-life is the fingerprint of first-order kinetics and explains why radioactive isotopes are characterised by a single half-life.

Second order. From 1/[A] = 1/[A]₀ + kt:

t½ = 1 ÷ k[A]₀

The half-life is inversely proportional to starting concentration. Each successive half-life is twice as long as the previous one, because as the reactant thins out, encounters become rarer.

General nth order (n ≠ 1). t½ is proportional to [A]₀^(1−n). Taking logarithms, log t½ = constant + (1 − n) log[A]₀. A plot of log t½ against log[A]₀ from runs at different starting concentrations has a gradient of 1 − n, giving the order directly, even if it is fractional.

Two practical methods. First, from one curve, read successive half-lives: constant (first order), halving (zero order) or doubling (second order). Second, compare half-lives from separate runs with different [A]₀. Both use the same idea: the way t½ depends on concentration reveals n.

Beyond half-lives. Any fractional lifetime, such as the time to 75% or 90% conversion, depends on [A]₀ in the same way as t½. For first order, t₇₅ is exactly 2t½; for second order it is 3t½. Such ratios provide a check that does not require knowing k.

Formulae

Zero order: t½ = [A]₀/2k. First order: t½ = ln 2/k. Second order: t½ = 1/(k[A]₀).

Step-by-step reasoning

1. On a concentration–time graph, mark [A]₀, [A]₀/2, [A]₀/4 and [A]₀/8. 2. Read the time at each mark and calculate successive half-lives. 3. Compare them: equal, shrinking by half or doubling. 4. Assign the order and calculate k from the appropriate half-life formula.

Visual explanation

On a first-order decay curve, draw horizontal lines at halving concentrations and drop vertical lines to the time axis: the vertical lines are evenly spaced. On a second-order curve the spacing doubles each time, and on a zero-order line it halves.

Real-world analogy

A first-order half-life is like a bouncing ball that always rises to half its previous height after an equal interval of time, regardless of how high it started. Second order is like trying to find a partner in an emptying room: the fewer people left, the longer each pairing takes.

Real-world example

Carbon-14 decays by first-order kinetics with a half-life of about 5730 years. Because the half-life does not depend on how much ¹⁴C remains, archaeologists can compare the ¹⁴C fraction in an ancient sample with that in living matter and calculate its age.

Why?

Why is the first-order half-life constant? The rate is proportional to [A], so the fraction of molecules reacting per unit time is fixed at k. Halving always requires the same time, whatever the absolute amount present.

Common misconception

"Every reaction has a fixed half-life." Only first-order reactions do. Quoting a single half-life for a second-order reaction is meaningless unless the starting concentration is also stated.

Worked example

Question: In one run the successive half-lives are 40 s, 80 s and 160 s, with [A]₀ = 0.050 mol dm⁻³. Identify the order and find k.

Reasoning: Each half-life doubles, so the reaction is second order. Using t½ = 1/(k[A]₀): k = 1/(40 × 0.050) = 0.50.

Answer: Second order; k = 0.50 mol⁻¹ dm³ s⁻¹.

Quick check

1. A reaction's half-life stays at 12 minutes as the concentration falls. What is the order, and what is k? Answer: First order, with k = 0.693 ÷ 12 min ≈ 0.058 min⁻¹.

Exam focus

State clearly that a constant half-life proves first order. Remember to convert time units consistently and to use ln 2 ≈ 0.693. When given half-lives from several starting concentrations, look at whether t½ rises, falls or stays constant as [A]₀ increases.

Advanced insight

The half-life method assumes a simple single-reactant rate law. If products catalyse or inhibit the reaction, or the reaction is reversible, the half-life drifts in ways that mimic a different order. Comparing half-lives from separate runs with fresh reagents avoids product effects that contaminate successive half-lives on one curve.

Summary

Half-life depends on starting concentration as [A]₀^(1−n). Zero-order half-lives shrink, first-order half-lives are constant at ln 2/k, and second-order half-lives lengthen as 1/(k[A]₀). Successive half-lives on one curve, or half-lives from runs at different starting concentrations, therefore identify the kinetic order and yield k.

Practice questions

1. A zero-order reaction has [A]₀ = 0.80 mol dm⁻³ and k = 0.020 mol dm⁻³ s⁻¹. Find t½. Answer: t½ = 0.80 ÷ (2 × 0.020) = 20 s. 2. Doubling [A]₀ halves the half-life. What is the order? Answer: Second order, since t½ is inversely proportional to [A]₀. 3. A first-order reaction has t½ = 30 s. What fraction of reactant remains after 90 s? Answer: 90 s is three half-lives, so (1/2)³ = 1/8 remains. 4. A log t½ against log[A]₀ plot has a gradient of −0.5. Find the order. Answer: 1 − n = −0.5, so n = 1.5.