Half-Life in Drug Elimination: First-Order Kinetics and Repeated Doses

By Aniket Bhardwaj · 10 October 2026 · Updated 10 October 2026 · Physical Chemistry in Practice

How long does a medicine stay in the body, and why do doctors space doses at fixed intervals? Much of the answer is first-order kinetics, the same maths you use for radioactive decay. In pharmacokinetics, the half-life of a drug sets how fast its concentration falls and how it builds up when doses are repeated. This article connects the formula you know from physical chemistry to this practical field. It is a chemistry exercise, not medical advice: real dosing is decided by doctors using measured data.

A steady-state dose interval shows a 500 mg dose, a 666.7 mg peak, and a 166.7 mg trough after 8 h. — With a 4.0 h half-life, the 8 h interval leaves 0.25 of the peak at the trough.

The model: first-order elimination

For many drugs, at ordinary doses, the body removes a constant fraction of the drug per unit time. That is first-order kinetics. With a single compartment and an instant (intravenous) dose, the equations are:

C = C₀ e^−kt
t_½ = ln 2 / k = 0.693 / k
After n half-lives: C = C₀ × (½)ⁿ

Here C₀ is the starting concentration (mg/L), k is the elimination rate constant (per hour), and t is the time. The half-life t_½ does not depend on how much drug there is. That is what makes it useful.

Repeated doses and accumulation

If a second dose is given before the first has gone, drug builds up. Suppose a dose D is given every τ hours. Just before the next dose, the fraction left from earlier doses is e^−kτ. Adding up the geometric series gives the long-run (steady-state) results:

Accumulation factor R = 1 / (1 − e^−kτ)
Peak amount at steady state (just after a dose) = D / (1 − e^−kτ)
Trough amount (just before the next dose) = Peak × e^−kτ

Because e^−kτ = (½)^τ/t½, you can write the factor using the number of half-lives per dosing interval. If τ = t_½, then R = 1 / (1 − ½) = 2.

How long to reach steady state? After n half-lives of regular dosing, the amount reached is a fraction 1 − (½)ⁿ of the final steady-state level. After 4 half-lives that is 93.75 %. After 5 half-lives it is 96.9 %. This is why steady state is usually taken as about 4 to 5 half-lives. It depends only on the half-life, not on the dose.

Example 1: Concentration after several half-lives. A drug has t_½ = 6.0 h. The initial concentration is 80 mg/L.
Step 1: k = 0.693 / 6.0 = 0.1155 h⁻¹.
Step 2: After 18 h = 3 half-lives: C = 80 × (½)³ = 80 × 0.125 = 10 mg/L.
Step 3: Check with the exponential: C = 80 × e^{−0.1155 × 18} = 80 × e^−2.079 = 80 × 0.125 = 10 mg/L. ✓
Answer: 10 mg/L after 18 h.

Example 2: Time to fall to a target level. Using the same drug (k = 0.1155 h⁻¹, C₀ = 80 mg/L), how long until the concentration is 25 mg/L?
Step 1: t = (1/k) ln(C₀/C) = (1/0.1155) × ln(80/25).
Step 2: ln(3.2) = 1.1632.
Step 3: t = 1.1632 / 0.1155 = 10.07 h.
Step 4: Check: after 10 h, C = 80 × e^−1.155 = 25.2 mg/L. ✓
Answer: about 10.1 h.

Example 3: Steady-state amounts. A 500 mg dose is given every 8 h. The drug's half-life is 4.0 h. Find the accumulation factor, peak and trough at steady state.
Step 1: τ / t_½ = 8 / 4 = 2 half-lives per interval, so e^−kτ = (½)² = 0.25.
Step 2: R = 1 / (1 − 0.25) = 1.333.
Step 3: Peak = 500 × 1.333 = 666.7 mg in the body just after a dose.
Step 4: Trough = 666.7 × 0.25 = 166.7 mg just before the next dose.
Step 5: Check: trough + dose = 166.7 + 500 = 666.7 mg, which equals the peak. ✓
Answer: R = 1.33, peak ≈ 667 mg, trough ≈ 167 mg. Steady state takes about 4 to 5 half-lives, so about 16 to 20 h.

Example 4: What if the interval equals the half-life? A dose D is given every t_½.
Step 1: e^−kτ = ½, so R = 1 / (1 − ½) = 2.
Step 2: Peak = 2D and trough = 2D × ½ = D.
Answer: the body settles at an amount that swings between D and 2D. Longer intervals reduce accumulation; shorter intervals increase it.

The limits of this simple picture

Real drugs are often better described by several compartments (blood, tissues). Elimination may then look like two stages, a fast fall then a slower one.
Some drugs, when the enzymes that remove them become saturated, follow zero-order or mixed kinetics. The Michaelis–Menten model describes this, and the half-life is then no longer constant.
Oral doses are absorbed over time, so the peak is lower and later than the instant-dose model predicts.
Half-life differs between people and conditions. Do not apply numbers from a textbook example to a real treatment.

Common mistakes

Mixing up k and t_½. k = 0.693 / t_½. A longer half-life means a smaller k.
Thinking the drug is gone after two half-lives. After 2 half-lives, 25 % remains. After 5, about 3 % remains.
Using the wrong units. If k is per hour, t must be in hours.
Forgetting that the accumulation factor depends on the interval. It depends on τ / t_½, not on the size of the dose.
Claiming steady-state level depends on half-life only. The time to reach it depends on the half-life. The level depends on the dose, interval and half-life together.
Applying first-order maths to saturated, zero-order cases.

Exam relevance

Topic	Link to the syllabus
First-order kinetics and half-life	Core physical chemistry for JAM, GATE, CSIR-NET and CUET-PG
Geometric series for repeated doses	Appears as a numerical application; the key formula is 1/(1 − e^−kτ)
Enzyme saturation	Michaelis–Menten kinetics

Pharmacokinetics is an application, so the first-order maths is the part that carries over to your exam. Check your current official syllabus for the exact scope.

Check the half-life, rate constant and remaining amount for your own numbers after working the examples by hand.

Open the Half-Life Calculator →

Preparing for IIT-JAM, GATE, CSIR-NET or CUET-PG? ABC Chemistry offers batches for these exams online, for students across India.