Half-Life Across Reaction Orders

Introduction

Half-life is often introduced through radioactive decay, but chemical reactions can have several half-life patterns. For a stated disappearance law, it is the time required for concentration to fall from a chosen starting value to half that value. Zero-, first- and squared second-order laws give different formulas and different sequences of successive halving intervals.

Core explanation

For zero order, [A]t=[A]0−kt. Substituting [A]0/2 gives t₁/₂=[A]0/(2k). Half-life grows with initial concentration. Once the first half is consumed, the next halving starts with half as much A and takes half as long, assuming the same constant zero-order k remains valid.

For first order, [A]t=[A]0e^(−kt). Setting the ratio to one-half gives t₁/₂=ln2/k≈0.693/k. The starting concentration cancels. Every successive halving takes the same duration while k and conditions remain unchanged. Equal time intervals remove equal fractions, not equal molar amounts.

For the specific squared second-order law −d[A]/dt=k[A]², the reciprocal equation gives t₁/₂=1/(k[A]0). Half-life becomes shorter at higher initial concentration. After one halving, the next halving starts at half the concentration and takes twice as long as the first. This pattern is opposite the zero-order sequence.

Dimensions confirm each formula. Zero-order k has M/time, so [A]0/k gives time. First-order k has 1/time, so 1/k gives time. Second-order k has 1/(M time), so 1/(k[A]0) gives time. A formula that leaves concentration units behind is being applied to the wrong order or with inconsistent units.

The half-life is not a universal property of a chemical substance. It depends on reaction, conditions and sometimes starting concentration. Temperature changes k, and a catalyst can change the effective pathway and k. Even for first order, half-life is constant only under fixed conditions and within the law's valid range. A pseudo-first-order experiment has a constant observed half-life while the excess reagent remains effectively constant.

Measured half-lives can help identify order, but a few intervals are not definitive if data are noisy or a reaction becomes reversible. The definition also needs the starting point: a “second half-life” begins at the concentration remaining after the first one. For zero- and second-order reactions, its duration differs from the initial half-life.

Step-by-step reasoning

1. Identify the exact disappearance law and k units. 2. Substitute half the stated starting concentration into its integrated equation. 3. Derive or choose the matching half-life formula. 4. For successive halvings, use the new concentration as the next starting value. 5. Check unit cancellation and experimental validity.

Visual explanation

Draw three concentration-time curves marked at 1, 1/2, 1/4 and 1/8 of their starts. Under zero order, time gaps shrink; under first order, gaps are equal; under second order, gaps grow. Write each half-life formula below its curve.

Real-world analogy

Removing a fixed number of items per minute gives quicker later halvings, removing a fixed percentage gives equal halving times, and needing pairs of items to meet makes later halvings slower as the crowd thins.

Real-world example

In a lab, repeated concentration readings at half, quarter and eighth of an initial amount can suggest the order before a full fit. Equal spacing in time is characteristic of first-order behavior if conditions remain steady.

Why?

Why does second-order half-life double after each halving? The new starting concentration is half as large and t₁/₂=1/(k[A]start), so inverse proportionality doubles the interval.

Common misconception

“Half-life is always 0.693/k.” That expression belongs to first-order decay. Zero and squared second order also contain the starting concentration in their half-life formulas.

Worked example

Compare three reactions all starting at 0.40 M. Let zero-order k=0.020 M min⁻¹, first-order k=0.050 min⁻¹ and second-order k=0.50 M⁻¹ min⁻¹. Their initial half-lives are 0.40/(0.040)=10 min, 0.693/0.050=13.86 min and 1/(0.50×0.40)=5.0 min. These numbers cannot be compared as intrinsic substance properties because k units and mechanisms differ.

Quick check

1. Which simple order has concentration-independent half-life? Answer: First order, when k remains constant.

Exam focus

Memorize the formulas with their exact laws and units, then reason about successive intervals. Do not insert a second-order overall label into the one-species squared formula without checking the rate law.

Advanced insight

For a general power law −d[A]/dt=k[A]ⁿ with n≠1, integration gives a half-life proportional to [A]0^(1−n). The familiar zero- and second-order patterns are special cases of this broader scaling.

Summary

Zero-order half-life rises with starting concentration, first-order half-life is constant and squared second-order half-life falls as starting concentration rises. The sequence of successive halvings reveals these different concentration dependencies.

Practice questions

1. A first-order reaction has k=0.20 s⁻¹. What is its half-life? Answer: 0.693/0.20=3.47 s. 2. If a second-order starting concentration doubles at fixed k, what happens to initial half-life? Answer: It halves because t₁/₂=1/(k[A]0). 3. Does the second zero-order halving take longer or shorter than the first? Answer: Shorter; with half as much starting material and a constant absolute removal rate, it takes half the time.