Half-life
The half-life of a first-order radioactive decay is 30 minutes. What percentage of the original reactant remains after 90 minutes?
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Solution
12.5%
Number of half-lives elapsed = total time / t₁/₂ = 90 / 30 = 3 half-lives. For a first-order process, the fraction remaining after n half-lives = (1/2)ⁿ. Therefore, fraction remaining = (1/2)³ = 1/8 = 0.125 = 12.5%. This is a key property of first-order reactions — the fraction decomposed in each successive half-life is always 50% of whatever amount is present at that moment.
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About This Question
- Subject
- chemistry
- Chapter
- chemical kinetics
- Topic
- half-life
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
12.5%
Number of half-lives elapsed = total time / t₁/₂ = 90 / 30 = 3 half-lives. For a first-order process, the fraction remaining after n half-lives = (1/2)ⁿ. Therefore, fraction remaining = (1/2)³ = 1/8 = 0.125 = 12.5%. This is a key property of first-order reactions — the fraction decomposed in each successive half-life is always 50% of whatever amount is present at that moment.
This medium difficulty chemistry question is from the chapter chemical kinetics, covering the topic of half-life. It appeared in the 2025 exam.
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