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Elimination Rate Constant Calculator (k = ln2 / t½)

· Last updated：2026-08-04

This tool is for informational purposes only and does not constitute medical, financial, or legal advice; consult a qualified professional for advice.

Convert an elimination half-life to the first-order elimination rate constant: k = ln2 / t½ = 0.693 / t½. Also reports the fraction of drug remaining after 1/2/3/4/5 half-lives (50%, 25%, 12.5%, 6.25%, 3.125%), the number of half-lives required to reach steady state (≈4 for 94%, ≈5 for 97%), and the time to ~99% elimination (≈6.64 half-lives). Valid only for first-order (linear) elimination — NOT for zero-order or Michaelis-Menten kinetics (phenytoin, high-dose ethanol, high-dose aspirin). Not medical advice.

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Tool usage guide

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Key facts

Category
Health
Input types
number
Output type
json
Sample coverage
4
API ready
Yes

Overview

The Elimination Rate Constant Calculator converts a positive elimination half-life in hours into the first-order rate constant k using k = ln(2) / t½. It also reports the fraction remaining after 1 to 5 half-lives, estimated steady-state timing, and the time for approximately 99% elimination.

When to use

  • Calculate k or kel from a known elimination half-life.
  • Estimate how much drug remains after several half-lives.
  • Estimate steady-state timing or approximately 99% elimination under first-order kinetics.

How it works

  1. 1Enter the elimination half-life in hours.
  2. 2The calculator applies k = ln(2) / t½.
  3. 3It calculates the remaining fraction after 1, 2, 3, 4, and 5 half-lives.
  4. 4It estimates steady state at about 4 to 5 half-lives and approximately 99% elimination at 6.64 half-lives.

Use cases

  • Pharmacokinetic calculations using a known drug half-life.
  • Comparing how shorter and longer half-lives affect the elimination rate constant.
  • Estimating remaining drug fraction and approximate accumulation or washout timing.

Examples

1. Theophylline half-life calculation

Pharmacokinetics student

Background

A pharmacokinetics exercise uses a theophylline elimination half-life of 8 hours.

Problem

Calculate the first-order elimination rate constant and estimate steady-state and 99% elimination timing.

How to use

Enter 8 for Half-life t½ and select 4 decimal places.

halfLife: 8; decimalPlaces: 4

Outcome

The calculator returns k = 0.0866 h⁻¹. Steady state is estimated at 32–40 hours, and approximately 99% elimination at 53.1455 hours.

2. Faster elimination comparison

薬代謝研究者

Background

A compound has a shorter elimination half-life of 6 hours than the compound in a reference calculation.

Problem

Determine whether the shorter half-life produces a larger elimination rate constant and faster elimination timing.

How to use

Enter 6 for Half-life t½ and select 4 decimal places.

FAQ

What formula does the calculator use?

It uses k = ln(2) / t½, which is approximately 0.693 / t½.

What unit does k use?

Because the half-life is entered in hours, k is reported in h⁻¹.

What happens after five half-lives?

Approximately 3.125% remains, meaning about 96.875% has been eliminated.

How many half-lives are needed for steady state?

About 4 half-lives corresponds to approximately 94% and 5 half-lives to approximately 97%.

When is this calculator not appropriate?

It is for first-order, linear elimination only and is not appropriate for zero-order or Michaelis-Menten kinetics. It is not medical advice.

Hub

Pharmacokinetic Teaching and Dose-Model Checks

Explore pharmacokinetic formulas, model assumptions, and result consistency for teaching or QA without giving medication dosing or clinical advice.

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halfLife: 6; decimalPlaces: 4

Outcome

The calculator returns k = 0.1155 h⁻¹. Steady state is estimated at 24–30 hours, and approximately 99% elimination at 39.8591 hours.