Integrated First-Order Problems

Introduction

A rate equation such as rate = k[A] tells you how fast a reaction is going at one instant. Most practical questions are different: how long until a drug level falls below its effective value, what fraction of a pollutant remains after a day, or how old a sample is. Answering them requires the integrated form of the rate law, which links concentration directly to time. For first-order reactions the result is especially neat, built on logarithms and a constant half-life.

Core explanation

The integrated law. For a first-order reaction, −d[A]/dt = k[A]. Separating variables and integrating from [A]₀ at t = 0 gives

ln([A]₀/[A]) = kt, or equivalently [A] = [A]₀e^(−kt)

A plot of ln[A] against t is a straight line with gradient −k; this is the standard graphical test for first-order kinetics.

Ratios, not absolute values. Only the ratio [A]₀/[A] appears, so any quantity proportional to concentration works: absorbance, mass, number of radioactive nuclei, or the partial pressure of a gas. There is no need to convert to mol dm⁻³.

Half-life. Setting [A] = [A]₀/2 gives ln 2 = kt½, so

t½ = ln 2/k ≈ 0.693/k

The half-life is independent of the starting concentration. After n half-lives the fraction remaining is (½)ⁿ: one quarter after two, one eighth after three, and so on. This constancy is a signature of first-order behaviour.

Three kinds of problem.

- Find k : from two concentrations at known times, k = ln([A]₀/[A])/t. - Find [A] at time t : [A] = [A]₀e^(−kt). - Find the elapsed time : t = ln([A]₀/[A])/k.

Unit consistency. k and t must use the same time unit. If k is in min⁻¹, time comes out in minutes.

Pressure data for gases. For A(g) → 2B(g) in a rigid vessel starting with pure A at pressure p₀, if x of A has reacted then p(A) = p₀ − x and p(total) = p₀ + x. So p(A) = 2p₀ − p(total), which can be substituted straight into the integrated law.

Base-10 logs. If using log₁₀, write log([A]₀/[A]) = kt/2.303. Mixing the two forms without the 2.303 factor is a common source of wrong answers.

Formulae

ln([A]₀/[A]) = kt; [A] = [A]₀e^(−kt); t½ = ln 2/k; fraction remaining after n half-lives = (½)ⁿ.

Step-by-step reasoning

1. Confirm that the reaction is first order (given, or from a constant half-life or linear ln plot). 2. Decide what is unknown: k, [A] or t. 3. Rearrange ln([A]₀/[A]) = kt for that unknown. 4. Keep time units consistent with k. 5. Check that the answer is sensible: concentration must fall, and time must be positive.

Visual explanation

On a plot of [A] against t, the curve decays smoothly, dropping by half every t½: 100%, 50%, 25%, 12.5%. On a plot of ln[A] against t, the same data form a straight line falling with slope −k. The logarithmic plot turns a curve that is hard to read into a line that is easy to measure.

Real-world analogy

A bank account losing a fixed percentage each year, rather than a fixed amount, behaves like first-order decay. Losing 10% a year always takes the same time to halve the balance, whether the balance started at a hundred or a million.

Real-world example

Radiocarbon dating uses the first-order decay of carbon-14, half-life about 5730 years. If a wooden artefact retains 25% of the carbon-14 of living wood, two half-lives have passed, giving an age of about 11 460 years.

Why?

Why is the first-order half-life independent of concentration? The rate is proportional to the amount present, so a larger sample decays faster in absolute terms but at the same fractional rate. The time for any fixed fraction, such as one half, to disappear is therefore the same.

Common misconception

"After two half-lives, nothing is left." Each half-life removes half of what remains, not half of the original amount. After two half-lives one quarter remains, and in principle the concentration never reaches exactly zero.

Worked example

Question: A first-order reactant falls from 0.0800 to 0.0500 mol dm⁻³ in 15.0 min. Find k, t½ and the time needed to reach 0.0100 mol dm⁻³.

Reasoning: k = ln(0.0800/0.0500)/15.0 = ln 1.60/15.0 = 0.470/15.0 = 0.0313 min⁻¹. t½ = 0.693/0.0313 = 22.1 min. To reach 0.0100 mol dm⁻³, the ratio is 8.00: t = ln 8.00/0.0313 = 2.079/0.0313 = 66.4 min. Check: 8 = 2³, so three half-lives, 3 × 22.1 = 66.3 min.

Answer: k = 0.0313 min⁻¹; t½ = 22.1 min; t = 66.4 min.

Quick check

1. A first-order reaction has k = 2.50 × 10⁻³ s⁻¹. What is its half-life? Answer: t½ = 0.693 ÷ 2.50 × 10⁻³ ≈ 277 s, about 4.6 minutes.

Exam focus

Write the integrated law before substituting, and state the unit of k. Recognise simple half-life multiples (½, ¼, ⅛) to save time, but use logarithms for non-integer fractions. Examiners often ask you to show that data are first order: demonstrate a constant half-life or a linear ln plot.

Advanced insight

Many reactions that are not truly first order behave as first order under pseudo-first-order conditions, with one reactant in large excess. Enzyme kinetics also approach first-order behaviour at low substrate concentrations. In pharmacology, drug elimination is often modelled as first order, and the dosing interval is set relative to the half-life so that concentrations stay within a therapeutic window.

Summary

For a first-order reaction, ln([A]₀/[A]) = kt and [A] = [A]₀e^(−kt). The half-life, t½ = ln 2/k, is independent of starting concentration, so after n half-lives a fraction (½)ⁿ remains. Any property proportional to concentration can be used, including gas partial pressures. Time and k must share units.

Practice questions

1. A reaction with k = 2.50 × 10⁻³ s⁻¹ runs for 600 s. What fraction of reactant remains? Answer: e^(−2.50 × 10⁻³ × 600) = e^(−1.50) ≈ 0.223, or about 22%. 2. How long does it take the same reaction to reach 10% of its initial concentration? Answer: t = ln 10/k = 2.303 ÷ 2.50 × 10⁻³ ≈ 921 s. 3. A reactant falls from 0.0800 to 0.0200 mol dm⁻³ in 46 min. Find t½ and k. Answer: One quarter remains, so two half-lives: t½ = 23 min and k = 0.693/23 ≈ 0.030 min⁻¹. 4. For A(g) → 2B(g) starting with pure A at 60.0 kPa, the total pressure is 80.0 kPa after 10.0 min. Find the partial pressure of A and k. Answer: p(A) = 2 × 60.0 − 80.0 = 40.0 kPa; k = ln(60.0/40.0)/10.0 = 0.405/10.0 ≈ 0.0405 min⁻¹.