Common Registration Assessment, Part 1

Clearance from half-life and volume of distribution

Three original questions linking half-life, volume and clearance, with the working shown and the usual traps pointed out.

Published 4 October 2026. Independent revision material, not GPhC questions, not for patient care.

Vials and test tubes in a laboratory rack
Photo from Pexels, used under the Pexels licence. It is a stock image and does not show the exercise.

For a drug with first-order elimination, clearance is the elimination rate constant multiplied by the volume of distribution: CL = k × Vd, where k = 0.693 ÷ half-life.

The exam moves in both directions: given half-life and volume to find clearance, or given clearance and volume to find half-life. Keep every unit in hours and litres until you convert at the end.

The method

  1. Write k = 0.693 ÷ t½ with the half-life in hours.
  2. Multiply k by the volume of distribution in litres to get clearance in L/h.
  3. If the volume is quoted per kg, multiply by the weight first.
  4. Convert L/h to mL/min only if the question asks for it: × 1,000 ÷ 60.
  5. For the reverse, k = CL ÷ Vd, then t½ = 0.693 ÷ k.

Three practice questions

Work each one on paper first, then open the answer. All drugs and patients are fictional.

Question 1

Drug X has a half-life of 8 hours and a volume of distribution of 50 L. What is its clearance in L/h to two decimal places?

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Answer: 4.33 L/h

Working: k = 0.693 ÷ 8 = 0.0866 per hour. CL = 0.0866 × 50 = 4.33 L/h.

Question 2

Drug Y has a half-life of 6 hours and a volume of distribution of 0.7 L/kg. For a 70 kg patient, what is the clearance in mL/min to the nearest whole number?

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Answer: 94 mL/min

Working: Vd = 0.7 × 70 = 49 L. k = 0.693 ÷ 6 = 0.1155 per hour. CL = 0.1155 × 49 = 5.66 L/h. 5.66 × 1,000 ÷ 60 = 94 mL/min.

Question 3

Drug Z has a clearance of 6 L/h and a volume of distribution of 40 L. What is its half-life to one decimal place?

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Answer: 4.6 hours

Working: k = CL ÷ Vd = 6 ÷ 40 = 0.15 per hour. t½ = 0.693 ÷ 0.15 = 4.62 hours, which rounds to 4.6 hours.

How this is examined

Questions often hide the volume behind a per-kg figure and a body weight. Resolve the actual volume in litres before applying CL = k × Vd.

Where marks are lost

  • Using 0.7 instead of 0.693 for ln 2.
  • Multiplying clearance by volume when the question asks for the rate constant: k = CL ÷ Vd.
  • Converting to mL/min too early and carrying rounded figures through the rest of the working.

Frequently asked questions

Where does 0.693 come from?

It is the natural logarithm of 2. For first-order elimination the rate constant is ln 2 divided by the half-life.

Does clearance change with dose?

For first-order drugs clearance is constant regardless of dose or concentration, which is why the same value of k applies at every point in the decay.

Sources

The questions above are original and use fictional drugs. PreRegExamPrep is not affiliated with or endorsed by the General Pharmaceutical Council.

Practise until the method is automatic

Try 15 free questions with worked answers. No sign-up required.

More calculation topics

See also the formula sheet, the eight sample questions and the approved calculator page.