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A first-order reaction is 20% complete in 10 minutes. Find (i) the rate constant in min⁻¹ and (ii) the time (in min) for the reaction to reach 75% completion. [Use ln 10 = 2.3, ln 2 = 0.7]
- 0.0223, 69.3
- 0.0223, 62.1
- 0.0200, 69.3
- 0.0200, 62.1
Correct answer: 0.0223, 62.1
Solution
From 20% completion in 10 min, k = (1/10) ln(1/0.8) = 0.0223 min⁻¹. For 75% completion, t = (1/k) ln 4 = (2 ln2)/k = 1.4/0.0223 = 62.1 min.
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