First-Order Half-Life — Use half-life relationships for first-order reactions - Chemistry 2

SECTION 1

Historical Context & Motivation

Understanding how quickly a substance breaks down has fascinated scientists for centuries. Whether tracking the decay of a radioactive element or studying how a drug leaves the bloodstream, researchers needed a simple way to describe the speed of disappearance. The concept of half-life — the time it takes for exactly half of a substance to react or decay — became one of the most powerful and intuitive tools in all of science. Its story weaves together nuclear physics, physical chemistry, and pharmacology into a single elegant idea.

1898 • Rutherford Observes Exponential Decay

Ernest Rutherford studied the radioactivity of thorium and noticed that its intensity dropped by half in equal time intervals. He coined the term "half-life" to describe this constant halving period, laying the groundwork for first-order decay mathematics.

1913 • Soddy and Isotope Theory

Frederick Soddy showed that radioactive elements transform into different isotopes through decay. Each isotope has its own characteristic half-life, reinforcing the idea that half-life is an intrinsic property of first-order processes.

1920s • Chemical Kinetics Adopts Half-Life

Physical chemists realized that many chemical reactions — not just nuclear decay — follow first-order kinetics. They borrowed the half-life concept from nuclear physics and applied it to reaction rates, drug metabolism, and pollutant breakdown.

1949 • Carbon-14 Dating Developed

Willard Libby introduced radiocarbon dating using the 5,730-year half-life of carbon-14. This application of first-order half-life revolutionized archaeology, geology, and environmental science.

Modern Era • Half-Life in Medicine and Environment

Today, pharmacologists use half-life to design drug dosing schedules, and environmental scientists use it to predict how long pollutants persist. The same first-order math applies across all these fields.

The central question that drove these discoveries was elegant in its simplicity: if the rate of a reaction depends only on how much reactant is present, what does that tell us about the time needed for half of it to disappear? The answer — that this time is constant no matter how much you start with — is both surprising and incredibly useful.

SECTION 2

Core Principles & Definitions

Before diving into the math, you need to understand what makes a reaction first-order and why its half-life behaves so differently from reactions of other orders. A first-order reaction is one in which the rate depends on the concentration of only one reactant, raised to the first power. As the reactant concentration drops, the rate drops proportionally — the reaction naturally slows itself down in a very predictable way.

First-Order Rate Law

The rate equals the rate constant (k) multiplied by the reactant concentration: rate = k[A]. As [A] decreases, the rate decreases proportionally.

Half-Life (t₁/₂)

The half-life is the time required for the reactant concentration to fall to exactly one-half of its initial value. For first-order reactions, this time does not change.

Constant Half-Life

In a first-order process, t₁/₂ is independent of the starting concentration. Whether you begin with 1.0 M or 0.001 M, each successive halving takes the same amount of time.

Rate Constant (k)

The rate constant k has units of s⁻¹ (or min⁻¹) for first-order reactions. A larger k means a shorter half-life and faster decay.

Exponential Decay

First-order reactions follow an exponential decay pattern. The concentration–time curve is a smooth, ever-flattening curve that never truly reaches zero.

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KEY TAKEAWAY

Think of a first-order half-life like a phone battery that always loses half its remaining charge every hour. If you start at 100%, after one hour you have 50%. After another hour you have 25%, then 12.5%, and so on. The time between each halving is always the same — one hour — regardless of how much charge remains. That's the defining feature of first-order kinetics: the half-life stays constant no matter the starting amount.

SECTION 3

Visual Explanation — Exponential Decay Curve

The exponential decay curve shows how the concentration [A] decreases over time. Notice that each half-life interval (marked by the colored dashed lines) has the same width along the time axis. After one half-life the concentration drops to ½[A]₀, after two half-lives to ¼[A]₀, and after three half-lives to ⅛[A]₀.

The diagram above is the signature shape of a first-order reaction. The curve drops steeply at first — when the concentration is high, the reaction is fast. As the concentration falls, the rate slows down, and the curve flattens. The crucial observation is that the three vertical dashed lines are evenly spaced along the time axis. Each interval represents one half-life, and every half-life the concentration drops by exactly half. This equal spacing only occurs in first-order reactions; for second-order reactions, each successive halving takes longer and longer.

SECTION 4

Mathematical Framework

The mathematics of first-order half-life starts from the integrated rate law for a first-order reaction. By setting the final concentration equal to half the initial concentration, you can derive a beautifully simple formula that links half-life directly to the rate constant. Let's build up to it step by step.

FIRST-ORDER RATE LAW

Rate = k[A]

k = rate constant (s⁻¹), [A] = concentration of reactant A at time t.

INTEGRATED RATE LAW (FIRST-ORDER)

ln[A]ₜ = −kt + ln[A]₀

[A]ₜ = concentration at time t, [A]₀ = initial concentration, k = rate constant, t = elapsed time. This equation can also be written as ln([A]ₜ/[A]₀) = −kt.

To derive the half-life formula, you substitute [A]ₜ = ½[A]₀ into the integrated rate law. When t equals the half-life (t₁/₂), exactly half the original reactant remains.

DERIVATION

ln(½[A]₀ / [A]₀) = −k·t₁/₂ → ln(½) = −k·t₁/₂ → −0.693 = −k·t₁/₂

The [A]₀ terms cancel. Since ln(½) = −0.693, dividing both sides by −k gives the half-life formula.

FIRST-ORDER HALF-LIFE FORMULA

t₁/₂ = 0.693 / k

t₁/₂ = half-life (in seconds, minutes, etc.), k = first-order rate constant (s⁻¹, min⁻¹, etc.). Notice that [A]₀ does not appear — the half-life is independent of the initial concentration.

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Where does 0.693 come from?

The number 0.693 is the natural logarithm of 2 (ln 2 ≈ 0.6931). It appears because we are solving for the time at which the concentration is halved — that's a factor of 2. You don't need to memorize why; just remember that t₁/₂ = 0.693 / k for any first-order process.

SECTION 5

Tracking Concentration Through Multiple Half-Lives

One of the most practical skills in working with first-order half-life is tracking how much reactant remains after several half-lives. Rather than plugging into the exponential equation every time, you can use a simple pattern: after n half-lives, the fraction remaining is (½)ⁿ. The table and diagram below make this pattern clear.

Number of Half-Lives (n)	Fraction Remaining	Percent Remaining	Percent Reacted
0	1	100%	0%
1	1/2	50.0%	50.0%
2	1/4	25.0%	75.0%
3	1/8	12.5%	87.5%
4	1/16	6.25%	93.75%
5	1/32	3.13%	96.87%

Concentration remaining after n half-lives for a first-order reaction

Each bar represents the percentage of reactant remaining after n half-lives. The height of each bar is exactly half the previous one, visually reinforcing the constant halving pattern of first-order decay.

A useful shortcut: after about 7 half-lives, less than 1% of the original reactant remains (0.78% to be exact). This is why pharmacologists often say a drug is effectively eliminated from your body after roughly 5–7 half-lives. The same logic applies to chemical reactions and radioactive decay.

SECTION 6

Worked Example

Let's work through a complete problem that shows how to move between the rate constant, half-life, and remaining concentration.

Decomposition of N₂O₅

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Step 1 — Read the Problem

The decomposition of dinitrogen pentoxide (N₂O₅) in the gas phase is a first-order reaction with a rate constant k = 5.1 × 10⁻⁴ s⁻¹ at 45 °C. If the initial concentration of N₂O₅ is 0.160 M, find (a) the half-life, and (b) the concentration remaining after 25.0 minutes.
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Step 2 — Calculate the Half-Life

Use the first-order half-life formula: t₁/₂ = 0.693 / k. Substitute: t₁/₂ = 0.693 / (5.1 × 10⁻⁴ s⁻¹).

t₁/₂ = 1,359 s ≈ 22.6 min
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Step 3 — Convert Time and Count Half-Lives

The question asks about t = 25.0 min. Convert t₁/₂ to minutes for easy comparison: t₁/₂ ≈ 22.6 min. The number of half-lives elapsed: n = 25.0 min / 22.6 min ≈ 1.11 half-lives. Since n is not a whole number, we'll use the integrated rate law for an exact answer.
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Step 4 — Apply the Integrated Rate Law

Convert 25.0 min to seconds: 25.0 × 60 = 1,500 s. Use ln[A]ₜ = −kt + ln[A]₀. Substituting: ln[A]ₜ = −(5.1 × 10⁻⁴)(1500) + ln(0.160) = −0.765 + (−1.833) = −2.598. Solve for [A]ₜ: [A]ₜ = e⁻²·⁵⁹⁸.

[A]ₜ = 0.0744 M
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Step 5 — Verify with the Half-Life Logic

After exactly 1 half-life (22.6 min), the concentration would be 0.160 / 2 = 0.080 M. Since 25.0 min is slightly more than one half-life, our answer of 0.0744 M is slightly less than 0.080 M. This makes sense — the answer checks out!

✓ Answer confirmed: t₁/₂ ≈ 22.6 min and [N₂O₅] at 25.0 min ≈ 0.0744 M

SECTION 7

Comparing Half-Lives Across Reaction Orders

The constant half-life of first-order reactions is special. Other reaction orders have half-lives that depend on the initial concentration. Understanding these differences helps you identify reaction order from experimental data and avoid common mistakes.

Feature	Zero-Order	First-Order	Second-Order
Rate Law	Rate = k	Rate = k[A]	Rate = k[A]²
Half-Life Formula	t₁/₂ = [A]₀ / 2k	t₁/₂ = 0.693 / k	t₁/₂ = 1 / k[A]₀
Depends on [A]₀?	Yes — increases with [A]₀	No — independent of [A]₀	Yes — decreases with [A]₀
Successive Half-Lives	Get shorter over time	Stay the same	Get longer over time
Units of k	M·s⁻¹	s⁻¹	M⁻¹·s⁻¹

How half-life behavior varies with reaction order

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KEY TAKEAWAY

If you're given experimental data and notice that each successive half-life is the same length, you've found strong evidence that the reaction is first-order. If the half-lives get progressively longer, the reaction is likely second-order. If they get shorter, it may be zero-order. This is one of the quickest ways to determine reaction order from raw data.

SECTION 8

Connection to Advanced Theory

The first-order half-life equation you've learned here is a gateway to more advanced topics in chemical kinetics, nuclear chemistry, and even pharmacology. Understanding how this basic formula extends into more complex scenarios will prepare you for AP Chemistry and beyond.

This Lesson	Advanced Extension
t₁/₂ = 0.693 / k (single step)	Multi-step mechanisms with rate-determining steps; pseudo-first-order kinetics in complex reactions
Constant half-life for a single reactant	Pharmacokinetic models: drug absorption, distribution, metabolism, and excretion (ADME) all modeled with half-lives
Exponential decay: [A]ₜ = [A]₀e⁻ᵏᵗ	Arrhenius equation links k to temperature: k = Ae⁻ᴱᵃ/ᴿᵀ, so half-life changes with temperature
Radioactive decay half-life (nuclear)	Decay chains where one product is itself radioactive with its own half-life (secular equilibrium)

How first-order half-life connects to advanced topics

One particularly important extension is the Arrhenius equation, which shows that the rate constant k increases as temperature rises. Since t₁/₂ = 0.693 / k, a higher temperature means a shorter half-life — reactions speed up at higher temperatures. This concept is used in food science (why refrigeration slows spoilage), materials engineering, and environmental chemistry.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A first-order reaction has a half-life of 20 minutes. If you double the initial concentration of the reactant, what happens to the half-life? Explain your reasoning.

Show answer

The half-life remains 20 minutes. For a first-order reaction, the half-life formula is t₁/₂ = 0.693 / k. Since neither [A]₀ nor any concentration term appears in this equation, doubling the initial concentration has no effect on t₁/₂. This independence from starting concentration is the defining characteristic that sets first-order half-life apart from zero-order and second-order reactions.

Question 2: BASIC CALCULATION

A certain first-order reaction has a rate constant k = 2.5 × 10⁻³ s⁻¹. Calculate the half-life of this reaction.

Show answer

Apply t₁/₂ = 0.693 / k = 0.693 / (2.5 × 10⁻³ s⁻¹) = 277 seconds, which is approximately 4.6 minutes. The units of k (s⁻¹) cancel with the denominator, leaving the half-life in seconds. A larger rate constant would give a shorter half-life, indicating a faster reaction.

Question 3: INTERMEDIATE

The first-order decomposition of hydrogen peroxide (H₂O₂) has a half-life of 11.1 hours at a certain temperature. If you start with a 0.500 M solution, what concentration remains after 33.3 hours?

Show answer

First, determine the number of half-lives: n = 33.3 h / 11.1 h = 3.00 half-lives. After 3 half-lives, the fraction remaining is (½)³ = 1/8. Therefore [A]ₜ = 0.500 M × (1/8) = 0.0625 M. You can verify by halving step by step: 0.500 → 0.250 → 0.125 → 0.0625 M. This approach works beautifully when the elapsed time is an exact multiple of the half-life.

Question 4: APPLIED

A medication follows first-order elimination from the body. If 400 mg is administered and only 25 mg remains after 12 hours, determine the half-life and the rate constant k for elimination.

Show answer

Find the fraction remaining: 25 mg / 400 mg = 1/16 = (½)⁴, so exactly 4 half-lives have elapsed in 12 hours. Therefore t₁/₂ = 12 h / 4 = 3.0 hours. Now calculate k: k = 0.693 / t₁/₂ = 0.693 / 3.0 h = 0.231 h⁻¹. This means the body eliminates about 23.1% of the remaining drug each hour. Pharmacists use this kind of calculation to determine how often a patient should take a dose to maintain a therapeutic level.

Question 5: CRITICAL THINKING

A scientist measures the half-life of a reaction at two different initial concentrations. At [A]₀ = 0.80 M, the half-life is 45 seconds. At [A]₀ = 0.40 M, the half-life is 90 seconds. Is this reaction first-order? Justify your answer, identify the likely reaction order, and calculate the rate constant.

Show answer

This reaction is not first-order. A first-order half-life would remain constant at both concentrations. Here, when [A]₀ was halved (from 0.80 to 0.40 M), the half-life doubled (from 45 to 90 s). This inverse relationship between t₁/₂ and [A]₀ is characteristic of a second-order reaction, where t₁/₂ = 1 / (k[A]₀). Using the first data point: 45 s = 1 / (k × 0.80 M), so k = 1 / (45 × 0.80) = 0.028 M⁻¹s⁻¹. Verifying with the second data point: t₁/₂ = 1 / (0.028 × 0.40) = 89.3 s ≈ 90 s. This confirms second-order kinetics. This problem illustrates why testing multiple concentrations is essential for correctly identifying reaction order.

SUMMARY

Lesson Summary

Summary

A first-order reaction has a rate that depends on the concentration of one reactant raised to the first power. Its half-life is given by the formula t₁/₂ = 0.693 / k, where k is the rate constant. The defining feature of first-order kinetics is that the half-life is independent of the initial concentration — each successive halving of the reactant takes the same amount of time. After n half-lives, the fraction of reactant remaining is (½)ⁿ.

You can use the integrated rate law ln[A]ₜ = −kt + ln[A]₀ to find concentrations at any time, and you can convert between k and t₁/₂ using the half-life formula. Comparing half-life behavior across zero-order, first-order, and second-order reactions is one of the most reliable methods for determining reaction order from experimental data. These principles apply broadly to radioactive decay, drug elimination, and chemical decomposition.