Question 1 · Medium
A drug has a volume of distribution of 35 L and a clearance of 3.5 L/h. What is its elimination half-life?
- A0.1 hours
- B10 hours
- C6.9 hours
- D24 hours
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Answer: C · 6.9 hours
The elimination rate constant is k = CL / Vd = 3.5 / 35 = 0.1 per hour. The half-life is 0.693 / 0.1 = 6.93 hours. Steady state would be reached after about 5 half-lives, roughly 35 hours.
- A: 0.1 hours is a confusion with the rate constant.
- B: Ten hours is the reciprocal of k and ignores the 0.693.
- D: Twenty-four hours does not follow from the numbers.
Remember this
t1/2 = 0.693 x Vd / CL.