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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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
1Enter the elimination half-life in hours.
2The calculator applies k = ln(2) / t½.
3It calculates the remaining fraction after 1, 2, 3, 4, and 5 half-lives.
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.