First Order Reactions
A first-order reaction has a rate constant of 1.386 × 10⁻² min⁻¹. The time required for 75% completion of the reaction is:
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About This Question
- Subject
- chemistry
- Chapter
- chemical kinetics
- Topic
- first order reactions
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
100 min
For a first-order reaction, t₁/₂ = 0.693/k = 0.693/(1.386 × 10⁻²) = 50 min. After one half-life (50 min), 50% of the reactant remains. After two half-lives (100 min), 25% remains — meaning 75% has been consumed. Alternatively, using t = (2.303/k) × log([A]₀/[A]) = (2.303/0.01386) × log(100/25) = 166.2 × 0.6021 ≈ 100 min.
This medium difficulty chemistry question is from the chapter chemical kinetics, covering the topic of first order reactions. It appeared in the 2025 exam.
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