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

SECTION 1

Historical Context & Motivation

The concept of half-life originated not in chemistry but in the study of radioactive decay, where physicists in the late nineteenth and early twentieth centuries observed that unstable nuclei disintegrate following remarkably predictable exponential kinetics. Ernest Rutherford and Frederick Soddy coined the term around 1903 to describe the time required for half of a radioactive sample to undergo transformation. Chemists soon recognized that the same mathematical framework governs a broad class of chemical reactions whose rates depend on the concentration of a single reactant raised to the first power—reactions now classified as first-order. This realization unified nuclear physics and chemical kinetics under a single exponential decay law and provided experimentalists with a powerful, concentration-independent clock for characterizing reaction speed.

1850s • Wilhelmy's Pioneering Kinetics

Ludwig Wilhelmy quantified the acid-catalyzed inversion of sucrose, producing one of the first measured rate laws and demonstrating exponential concentration decay over time.

1903 • Rutherford & Soddy Define Half-Life

Rutherford and Soddy introduced the half-life concept for radioactive decay, noting that each radionuclide possesses a characteristic, constant half-life regardless of the amount present.

1920s • First-Order Kinetics Formalized

Chemists such as Hinshelwood and Lindemann extended exponential decay models to gas-phase decomposition reactions, firmly establishing the integrated first-order rate law and its associated half-life expression.

1949 • Libby's Radiocarbon Dating

Willard Libby applied the first-order half-life of carbon-14 (≈ 5730 years) to develop radiocarbon dating, earning the 1960 Nobel Prize and demonstrating the immense practical reach of half-life concepts.

Modern Era • Pharmacokinetics & Beyond

Today, first-order half-life governs drug elimination models in pharmacology, pollutant degradation in environmental science, and shelf-life determinations in food chemistry.

A central question motivates this lesson: given a reaction that follows first-order kinetics, how do we relate the half-life to the rate constant, and how can we exploit the constancy of that half-life to predict concentrations at arbitrary times, determine rate constants from experimental observations, and compare the speeds of different reactions? Answering these questions requires us to derive and apply the half-life relationship that emerges naturally from the integrated first-order rate law.

SECTION 2

Core Principles & Definitions

Before diving into derivations, it is essential to anchor several foundational ideas. A first-order reaction is one whose instantaneous rate is directly proportional to the concentration of one reactant; doubling that concentration doubles the rate. The proportionality constant linking rate to concentration is the rate constant k, which carries units of reciprocal time (s⁻¹, min⁻¹, etc.) for a first-order process. The half-life (t₁/₂) is defined as the time required for the reactant concentration to decrease to exactly one-half of its value at any chosen starting point. Crucially, for first-order kinetics—and only for first-order kinetics—this half-life is a constant that does not depend on the initial concentration, a property that distinguishes first-order processes from zero-order and second-order reactions.

First-Order Rate Law

Rate = k[A]. The reaction rate at any instant is proportional to the current concentration of reactant A. This yields exponential decay when integrated over time.

Integrated Rate Law

ln[A] = ln[A]₀ − kt. A plot of ln[A] versus time is linear with slope −k and y-intercept ln[A]₀, providing a diagnostic test for first-order behavior.

Constant Half-Life

t₁/₂ = ln 2 / k ≈ 0.693 / k. Because this expression contains only the rate constant, the half-life is independent of [A]₀—a unique fingerprint of first-order kinetics.

Successive Halving

After n half-lives, the remaining fraction of reactant is (1/2)ⁿ. This geometric progression lets you quickly estimate concentrations without a calculator.

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KEY TAKEAWAY

Think of a first-order half-life like a photocopier that shrinks every document to 50 % of its current size each time you press the button. It does not matter whether the original document was poster-sized or index-card-sized—the time per button press (half-life) stays the same. That constancy is the hallmark of first-order kinetics: the fraction consumed per unit time is fixed, even though the absolute amount consumed per unit time steadily decreases as the reactant is depleted.

SECTION 3

Visual Explanation — Exponential Decay Curve

The exponential decay curve shows [A] decreasing by half during each successive half-life interval. The cyan dashed line marks the first half-life (t₁/₂), the violet dashed line marks 2t₁/₂, and the pink dashed line marks 3t₁/₂. Notice that the time interval between consecutive half-life markers is identical—this constancy is the defining signature of first-order kinetics.

The diagram above captures the essential behavior of a first-order process. Starting at [A]₀, the concentration drops to ½[A]₀ after one half-life, to ¼[A]₀ after two half-lives, and to ⅛[A]₀ after three. Each successive half-life removes the same fraction (50 %) of whatever remains, even though the absolute amount consumed decreases with each interval. After ten half-lives, only about 0.1 % of the original reactant survives—(½)¹⁰ ≈ 0.00098. The shape of the curve is a classic exponential decay, asymptotically approaching zero but never quite reaching it within a finite time.

SECTION 4

Mathematical Framework

The half-life expression for a first-order reaction is not an independent postulate; it follows directly from the integrated rate law. Starting from the differential rate law, we integrate and then impose the half-life boundary condition to obtain the key formula.

DIFFERENTIAL RATE LAW

−d[A]/dt = k[A]

where [A] is the instantaneous concentration, k is the first-order rate constant (s⁻¹), and t is time.

INTEGRATED RATE LAW

ln[A] = ln[A]₀ − kt ⟺ [A] = [A]₀ e^(−kt)

Separation of variables and integration from t = 0 to t yields the logarithmic form on the left and the equivalent exponential form on the right. [A]₀ denotes the initial concentration at t = 0.

To derive the half-life, set [A] = ½[A]₀ at t = t₁/₂ in the integrated rate law:

HALF-LIFE DERIVATION

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

Because ln(½) = −ln 2 ≈ −0.693, the [A]₀ terms cancel completely. The result, t₁/₂ = 0.693 / k, depends only on k. This cancellation is unique to first-order kinetics.

CONCENTRATION AFTER n HALF-LIVES

[A] = [A]₀ × (½)ⁿ where n = t / t₁/₂

When the elapsed time equals a whole number of half-lives, this formula provides an exact result. For non-integer n, the exponential form [A] = [A]₀ $e^{−kt}$ is more appropriate, but this expression is useful for quick mental estimates.

ℹ

Why Only First-Order?

For zero-order reactions the half-life is t₁/₂ = [A]₀ / (2k), and for second-order reactions it is t₁/₂ = 1 / (k[A]₀). Both depend on [A]₀, so their half-lives change as the reaction proceeds. The cancellation of [A]₀ in the first-order derivation is what makes its half-life a true constant of the reaction rather than a function of the starting conditions.

SECTION 5

Half-Life Across Reaction Orders

Appreciating the special nature of first-order half-life is easier when it is set alongside the corresponding expressions for other common reaction orders. The following diagram and table highlight the key contrasts.

The zero-order half-life (amber line) increases linearly with [A]₀; the first-order half-life (cyan dashed line) is a horizontal line, independent of [A]₀; and the second-order half-life (pink curve) decreases as [A]₀ increases (inversely proportional). This graphical comparison instantly reveals the unique constancy of the first-order half-life.

Reaction Order	Half-Life Expression	Depends on [A]₀?	Behavior as Reaction Proceeds
Zero-order	t₁/₂ = [A]₀ / (2k)	Yes — directly proportional	Half-life shortens as [A] drops
First-order	t₁/₂ = ln 2 / k	No — independent	Half-life remains constant
Second-order	t₁/₂ = 1 / (k[A]₀)	Yes — inversely proportional	Half-life lengthens as [A] drops

Summary of half-life expressions for the three most common reaction orders.

An experimentalist can exploit these contrasts as a diagnostic tool. If successive half-life intervals measured during a single experiment remain constant, the reaction is almost certainly first-order. If each successive half-life is shorter, zero-order kinetics may be at play; if each is longer, second-order kinetics is likely. This observational strategy complements the graphical method of plotting ln[A] vs. t and checking for linearity.

SECTION 6

Worked Example

The decomposition of dinitrogen pentoxide (N₂O₅) in the gas phase at 45 °C is a classic first-order reaction:

2 N₂O₅(g) → 4 NO₂(g) + O₂(g)

The rate constant at 45 °C is k = 5.1 × 10⁻⁴ s⁻¹. If the initial concentration is [N₂O₅]₀ = 0.0240 mol/L, determine (a) the half-life, (b) the concentration after 1500 s, and (c) the time required for 90 % of the N₂O₅ to decompose.

N₂O₅ Decomposition at 45 °C

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Step 1 — Calculate the half-life

Apply the first-order half-life formula: t₁/₂ = ln 2 / k = 0.6931 / (5.1 × 10⁻⁴ s⁻¹).

t₁/₂ ≈ 1359 s (≈ 22.6 min)
2
Step 2 — Find [N₂O₅] at t = 1500 s

Use the exponential form: [A] = [A]₀ $e^{−kt}$ . Substitute: [N₂O₅] = 0.0240 × $e^{−5.1 × 10⁻⁴ × 1500}$ . Calculate the exponent: −5.1 × 10⁻⁴ × 1500 = −0.765. Then $e^{−0.765}$ = 0.4653.

[N₂O₅] ≈ 0.0240 × 0.4653 = 0.0112 mol/L
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Step 3 — Verify with half-life reasoning

1500 s is slightly more than one half-life (1359 s), so we expect [N₂O₅] to be somewhat less than ½ × 0.0240 = 0.0120 mol/L. Our calculated value of 0.0112 mol/L is indeed slightly below 0.0120, confirming internal consistency.
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Step 4 — Time for 90 % decomposition

If 90 % has decomposed, 10 % remains: [A] = 0.10 × [A]₀. Substitute into the integrated law: ln(0.10 × [A]₀) = ln[A]₀ − kt. This simplifies to ln(0.10) = −kt, giving t = −ln(0.10) / k = 2.3026 / (5.1 × 10⁻⁴ s⁻¹).

t ≈ 4515 s (≈ 75.3 min, or about 3.32 half-lives)
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Step 5 — Interpret

Since (½)^3.32 ≈ 0.10, it takes roughly 3.32 half-lives for 90 % consumption, regardless of the initial concentration. This is a general result for first-order reactions and illustrates why pharmacologists use the rule of thumb that ≈ 3.3 half-lives are needed to eliminate 90 % of a drug.

SECTION 7

Strengths, Limitations & Applications

The first-order half-life relationship is one of the most versatile tools in kinetics, but its applicability hinges on the reaction genuinely being first-order. Below is a comparison of its strengths and limitations.

Strengths	Limitations
Constant half-life enables rapid estimation of concentrations at any time without solving exponential equations.	Assumes the reaction is strictly first-order over the entire concentration range studied. Deviations at very high or very low concentrations (e.g., saturation kinetics) invalidate the formula.
Direct link between t₁/₂ and k allows experimentalists to determine rate constants from a single half-life measurement.	Temperature dependence of k (via the Arrhenius equation) means the half-life changes with temperature. A quoted t₁/₂ is only valid at the specified temperature.
Applicable across disciplines: radioactive decay, pharmacokinetics, pollutant degradation, enzyme kinetics (pseudo-first-order conditions).	Many real reactions only appear first-order under pseudo-first-order conditions (one reagent in large excess). If that excess is consumed, the apparent order and half-life change.
The successive-halving rule [(½)ⁿ] gives quick mental arithmetic estimates of remaining reactant.	For reversible reactions approaching equilibrium, the simple first-order half-life applies only to the initial rate regime, not near equilibrium where the back reaction becomes significant.

Strengths and limitations of the first-order half-life relationship.

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KEY TAKEAWAY

In pharmacology, the biological half-life of a drug determines dosing schedules. A drug with t₁/₂ ≈ 6 hours that must remain above a therapeutic threshold will require dosing roughly every 6 hours; after about 5 half-lives (30 hours with no additional dose), the drug is considered effectively eliminated (only ≈ 3 % remains). This practical example underscores why the concentration-independent constancy of first-order half-life is not merely an academic curiosity but a cornerstone of applied science.

SECTION 8

Connection to Advanced Theory

The first-order half-life formula is the starting point for several deeper investigations in kinetics and physical chemistry. Understanding how it connects to more advanced concepts will help you navigate upper-division and graduate-level material.

Introductory Concept	Advanced Extension
t₁/₂ = ln 2 / k (constant at fixed T)	Combining with the Arrhenius equation k = A $e^{−Eₐ/RT}$ yields t₁/₂ = (ln 2 / A) $e^{Eₐ/RT}$ , showing exponential temperature dependence of half-life.
Single-step first-order decay	Consecutive first-order reactions (A → B → C) produce transient intermediates whose concentrations are governed by coupled exponential terms and the Bateman equations.
Pseudo-first-order approximation	Michaelis–Menten enzyme kinetics transitions from pseudo-first-order (low [S]) to zero-order (high [S]) behavior, so the effective half-life of substrate conversion is concentration-dependent at saturation.
Discrete half-life counting	In pharmacokinetics, multi-compartment models assign different half-lives to distribution (α) and elimination (β) phases, reflecting the complexity of drug transport between body compartments.

Introductory vs. advanced perspectives on first-order half-life.

As you progress through physical chemistry and biochemistry, you will encounter situations where the simple t₁/₂ = ln 2 / k relation is embedded within larger kinetic schemes. The ability to recognize when a process reduces to effective first-order behavior—and when it does not—is a powerful analytical skill that builds directly on the foundations established in this lesson.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

A colleague claims that doubling the initial concentration of a first-order reactant will double its half-life. Is this claim correct? Explain your reasoning by referencing the half-life formula.

Show answer

The claim is incorrect. For a first-order reaction, t₁/₂ = ln 2 / k, which contains no concentration term. Because [A]₀ cancels during the derivation (ln(½[A]₀/[A]₀) = ln ½, independent of [A]₀), the half-life is determined solely by the rate constant k. Doubling [A]₀ has no effect on t₁/₂; the reaction simply starts with more material, each half-life still reducing the remaining amount by exactly 50 %. This concentration independence is the hallmark that distinguishes first-order kinetics from zero-order (where t₁/₂ ∝ [A]₀) and second-order (where t₁/₂ ∝ 1/[A]₀).

Question 2: BASIC CALCULATION

A first-order reaction has a rate constant of 2.80 × 10⁻³ min⁻¹. Calculate the half-life in minutes.

Show answer

Apply t₁/₂ = ln 2 / k = 0.6931 / (2.80 × 10⁻³ min⁻¹) = 247.5 min ≈ 248 min. This is a direct substitution into the first-order half-life equation. The result tells us that every 248 minutes, the concentration of the reactant drops to half of whatever value it had at the start of that interval. In hours, this corresponds to approximately 4.13 hours.

Question 3: INTERMEDIATE

The first-order decomposition of a certain pesticide in soil has a half-life of 18.0 days. If a field initially contains 120 mg/kg of the pesticide, how many days are required for the concentration to fall below 15.0 mg/kg?

Show answer

First determine the rate constant: k = ln 2 / t₁/₂ = 0.6931 / 18.0 days = 0.03851 day⁻¹. Then use the integrated rate law: ln([A]/[A]₀) = −kt, so t = −ln([A]/[A]₀) / k = −ln(15.0/120) / 0.03851 = −ln(0.125) / 0.03851 = 2.0794 / 0.03851 ≈ 54.0 days. Alternatively, note that 15.0/120 = 1/8 = (½)³, so exactly 3 half-lives are needed: 3 × 18.0 = 54.0 days. Both methods confirm that 54 days are required for the pesticide concentration to decrease from 120 to 15.0 mg/kg.

Question 4: APPLIED

A patient receives an intravenous dose of a drug whose elimination follows first-order kinetics with t₁/₂ = 4.0 hours. The minimum effective plasma concentration is 2.0 μg/mL, and the peak plasma concentration immediately after injection is 32 μg/mL. How many hours after injection will the drug first drop below the therapeutic threshold?

Show answer

The fraction remaining at the threshold is 2.0/32 = 0.0625 = (½)⁴, so the concentration reaches 2.0 μg/mL after exactly 4 half-lives: t = 4 × 4.0 h = 16.0 h. Alternatively, compute k = ln 2 / 4.0 = 0.1733 h⁻¹, then t = −ln(2.0/32) / 0.1733 = −ln(0.0625) / 0.1733 = 2.7726 / 0.1733 ≈ 16.0 h. The drug drops below the therapeutic threshold 16 hours after injection, at which point the patient would need a new dose. This type of calculation is foundational in clinical pharmacokinetics for designing dosing regimens.

Question 5: CRITICAL THINKING

An experimentalist monitors a reaction and records the following successive half-life values: 42 min, 84 min, 170 min. Are these data consistent with a first-order reaction? If not, what reaction order do they suggest, and how could the half-life formula for that order explain the observed trend?

Show answer

For a first-order reaction, all successive half-lives should be equal. Here, the half-life roughly doubles each time (42 → 84 → 170 min), which is inconsistent with first-order kinetics. The data suggest second-order kinetics. For a second-order reaction, t₁/₂ = 1/(k[A]₀). After one half-life, the concentration halves, so the next half-life is 1/(k × ½[A]₀) = 2/(k[A]₀), which is twice the first. After the second half-life the concentration is ¼[A]₀, giving the third half-life as 4/(k[A]₀), four times the first. The observed pattern (42, 84, ~168) approximately matches this doubling pattern (the slight deviation to 170 reflects experimental uncertainty). To confirm, one could plot 1/[A] vs. t and check for linearity, the diagnostic for second-order kinetics. This problem illustrates how monitoring successive half-lives serves as a practical method for determining reaction order.

SUMMARY

Lesson Summary

Summary

For a first-order reaction, the half-life is given by t₁/₂ = ln 2 / k ≈ 0.693 / k, a quantity that depends only on the rate constant k and is therefore independent of the initial concentration [A]₀. This constancy arises because the [A]₀ terms cancel during derivation from the integrated first-order rate law ln[A] = ln[A]₀ − kt. After n half-lives, the remaining fraction of reactant is (½)ⁿ, enabling rapid mental estimation of concentrations.

The concentration-independent half-life distinguishes first-order kinetics from zero-order (t₁/₂ ∝ [A]₀) and second-order (t₁/₂ ∝ 1/[A]₀) processes, making it a powerful diagnostic tool for reaction order. Applications span radioactive decay, pharmacokinetics, and environmental degradation. To convert between half-life and rate constant, simply use k = ln 2 / t₁/₂ or t₁/₂ = ln 2 / k—the single most important relationship in this lesson.