Pharmacokinetics

Elimination Rate Constant (kₑ) Calculator

Finds the first-order elimination rate constant from a half-life, from clearance and Vd, or from a dose and C₀.

Elimination rate constant

0.1732h⁻¹

Rapid elimination – multiple daily doses

Inputs

Half-life t½ (hours)

The elimination half-life, e.g. from a drug monograph.

Try an example (fills the half-life method)

Working

kₑ = 0.693 ÷ 40.1732h⁻¹

Calculated t½4.00hours

Fraction eliminated per hour (kₑ × 100)17.3%

Elimination curve

Percentage of the drug remaining over the first 10 hours.

kₑ = 0.693 / t½

kₑ = CL / Vd

Vd = Dose / C₀ then kₑ = CL / Vd

Key relationships between the parameters:

t½ = 0.693 / kₑ

CL = kₑ × Vd

AUC = Dose / (kₑ × Vd)

0.693 is ln 2. In first-order elimination the concentration falls exponentially, C = C₀ · e^(−kₑ·t), which is what the curve above plots as a percentage of C₀. First-order kinetics per the MSD Manual.

Worked example: kₑ from an IV bolus dose, C₀ and clearance

In a pharmacokinetics practical, Drug X is given as a single 500 mg IV bolus. The plasma concentration back-extrapolated to time zero is 12.5 mg/L, and total clearance was found to be 5 L/h. Find the elimination rate constant and the half-life.

Dose (IV bolus): 500 mg
C₀ (back-extrapolated): 12.5 mg/L
Clearance CL: 5 L/h
Method: From dose and C₀

Volume of distribution
Vd = Dose ÷ C₀ = 500 mg ÷ 12.5 mg/L = 40 L
Elimination rate constant
kₑ = CL ÷ Vd = 5 L/h ÷ 40 L = 0.1250 h⁻¹
Half-life
t½ = 0.693 ÷ kₑ = 0.693 ÷ 0.125 h⁻¹ = 5.544 h ≈ 5.54 h
Fraction lost in the first hour
kₑ × 100 = 12.5 % (approximation) exact: 1 − e^(−0.125 × 1) = 0.1175 → 11.8 %

kₑ = 0.1250 h⁻¹, so t½ ≈ 5.54 h (Vd = 40 L).

About 12 % of the drug in the body is removed each hour, and the level halves roughly every 5.5 hours, so it takes about 4–5 half-lives (22–28 h) for a single dose to be almost completely eliminated. This is an educational calculation on illustrative values for a one-compartment drug, not a patient-specific dosing decision.

Common mistakes

Using log₁₀ 2 (0.301) instead of ln 2 (0.693). First-order elimination is exponential in base e, so kₑ = 0.693 ÷ t½; 0.301 gives a value 2.3 times too small.

Mixing time units. A half-life in minutes gives kₑ in min⁻¹; it must be multiplied by 60 before comparing it with a kₑ in h⁻¹.

Mixing volume units. CL in mL/min with Vd in L gives a number that is neither h⁻¹ nor min⁻¹ — convert CL to L/h (× 60 ÷ 1000) first.

Using a Vd in L/kg as if it were total litres. Multiply by body weight before dividing CL by it.

Reading kₑ × 100 as the exact percentage lost per hour. It is a small-kₑ approximation; the exact fraction is 1 − e^(−kₑ), which is noticeably smaller once kₑ is above about 0.1 h⁻¹.

Using the first measured concentration as C₀. C₀ is the value back-extrapolated to time zero on a log-concentration plot, not the first sample.

Assumptions and limitations

Assumes first-order (linear) elimination. Drugs with saturable metabolism, such as high-dose phenytoin or ethanol, have no single kₑ.

Assumes a one-compartment body. For drugs with a distinct distribution phase, Dose ÷ C₀ underestimates the true volume, and the kₑ here describes only the terminal phase.

The dose and C₀ method is valid only for an IV bolus. After an oral dose, absorption and bioavailability distort both C₀ and Vd.

kₑ is a population or study value; disease (renal or hepatic impairment), age and interacting drugs change it in individual patients.

Guide: Pharmacokinetic Parameters: AUC, Clearance, Half-Life and BioavailabilityThe concept in full, with practice questions.Volume of Distribution CalculatorWork out Vd on its own, in L and L/kg, before you divide clearance by it.Clearance CalculatorGet CL from a dose and AUC, or from an infusion at steady state.Accumulation Index CalculatorSee how kₑ and the dosing interval decide how much a repeated dose builds up.

References

Ducharme MP, Shargel L, editors. Shargel and Yu's Applied Biopharmaceutics & Pharmacokinetics. 8th ed. New York: McGraw Hill; 2022.
Rowland and Tozer's Clinical Pharmacokinetics and Pharmacodynamics: Concepts and Applications. 5th ed. Philadelphia: Wolters Kluwer; 2019.

Common questions

Reciprocal time — here h⁻¹, because half-life is entered in hours and clearance in L/h. If you enter a half-life in minutes the answer is in min⁻¹; multiply by 60 to convert to h⁻¹.

Only approximately, and only when kₑ is small. The exact fraction lost in one hour is 1 − e^(−kₑ). For kₑ = 0.1 h⁻¹ that is 9.5% rather than 10%; for larger kₑ the gap widens.

Because kₑ = CL / Vd. With the same clearance, a drug spread through a larger volume has a smaller share of itself in the blood passing the liver and kidneys at any moment, so it leaves more slowly.

No. Zero-order drugs are removed at a constant amount per hour, so there is no constant kₑ or half-life. Use the order-of-kinetics calculator to compare the two patterns.

Prepared by the PharmaWallah student team. Verify results with your textbook or instructor. Report an error · Editorial policy