What is the order of hydrolysis of methyl acetate in the acidic medium?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Pseudo first order
Acid hydrolysis of methyl acetate involves ester and water:
\( \ce{CH3COOCH3 + H2O} \xrightarrow{\ce{H+}} \ce{CH3COOH + CH3OH} \)
Rate = k′\( \ce{[CH3COOCH3][H2O]} \). Water is present in such a large excess that its concentration hardly changes during the reaction, so Rate = k[CH3COOCH3]. The reaction is truly bimolecular but behaves as a first order reaction: a pseudo first order reaction.
Why the other options are incorrect
Option (1) Zero-order — Incorrect. The rate does depend on the ester concentration.
Option (3) First order — Incorrect. It only looks first order because water is in excess; it is not a true first order reaction.
Option (4) Second order — Incorrect. It would be second order only if water were not in large excess.
NCERT (2026-27 edition):
Half-Life of a Reaction – “… The order of a reaction is sometimes altered by conditions. There are many reactions which obey first order rate law although they are higher order reactions. Consider the hydrolysis of ethyl acetate which is a chemical reaction between ethyl acetate and water. In reality, it is a second order reaction and concentration of both ethyl acetate and water affect the rate of the reaction. But water is taken in large excess for hydrolysis, therefore, concentration of water is not altered much during the reaction. Thus, the rate of reaction is affected by concentration of ethyl acetate only. … Such reactions are called pseudo first order reactions.”
Note: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| – | – | – | 02-06 S3 31-05 S1 | – |