Match List I with List II
| LIST I | LIST II |
| A. Decomposition of H2O2 | I. Second order reaction |
| B. Decomposition of H2 in the presence of Hg and light | II. Pseudo first order reaction |
| C. Hydrolysis of ester by alkali | III. First order reaction |
| D. Hydrolysis of ester in excess of acid | IV. Photosensitisation |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-III, B-IV, C-I, D-II
(A) Decomposition of H2O2 — This matches (III) First order reaction. Rate = k[H2O2]: \( \ce{2H2O2 -> 2H2O + O2} \).
(B) Decomposition of H2 in the presence of Hg and light — This matches (IV) Photosensitisation. \( \ce{H2} \) does not absorb the light itself; mercury vapour absorbs it and passes the energy on to \( \ce{H2} \), which then dissociates into atoms. Mercury acts as a photosensitiser.
(C) Hydrolysis of ester by alkali — This matches (I) Second order reaction. In saponification the ester and \( \ce{OH^-} \) are present in comparable amounts, so the rate depends on both: rate = k[ester][OH−].
\( \ce{CH3COOC2H5 + NaOH -> CH3COONa + C2H5OH} \)
(D) Hydrolysis of ester in excess of acid — This matches (II) Pseudo first order reaction. Water is present in large excess, so its concentration hardly changes and the rate depends on [ester] only: rate = k′[ester].
\( \ce{CH3COOC2H5 + H2O} \xrightarrow{\ce{H^+}} \ce{CH3COOH + C2H5OH} \)
Note: The question is from beyond the NCERT textbook.