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The acid-catalyzed hydrolysis of ethyl acetate follows first-order kinetics with respect to the ester, with rate constant k = 0.693 s⁻¹. Find the time (in seconds) required for 93.75% hydrolysis of the ester. (Use ln 2 = 0.693)

1. 1
2. 2
3. 3
4. 4

Correct answer: 4

Solution

First-order kinetics: fraction remaining = e^(-kt). After 93.75% hydrolysis, 6.25% = 0.0625 remains. 0.0625 = 1/16 = (1/2)⁴. So 4 half-lives pass. t_half = ln2 / k = 0.693 / 0.693 = 1 s. Total time = 4 * 1 = 4 s.

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