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For the acid-catalysed hydrolysis of ethyl acetate, CH3COOC2H5(aq) + H2O ->[H+] CH3COOH(aq) + C2H5OH(aq), the observed rate law is rate = k[ester][H2O][H+]. With water in large excess (about 55.5 M), the reaction is pseudo-first-order in ester. If k = 1.386*10⁻³ M⁻² min⁻¹ and [H+] = 1.8 M, how many seconds will it take for the ester concentration to fall to half its initial value?

1. 300
2. 150
3. 600
4. 277

Correct answer: 300

Solution

With [H2O] ~ 55.5 M and [H+] = 1.8 M, k' = 1.386*10⁻³ * 55.5 * 1.8 = 0.1385 min⁻¹, so t(1/2) = 0.693/0.1385 = 5 min = 300 s.

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