Match the LIST-I with LIST-II.
| LIST-I Order of reaction | LIST-II Example |
| A. Zero order | I. Decomposition of hydrogen peroxide using iodide ion as catalyst in alkaline medium |
| B. First order | II. Hydrolysis of ethyl acetate in the presence of acid |
| C. Pseudo First order | III. The thermal decomposition of HI on gold surface |
| D. Second order | IV. Radioactive decay of unstable nuclei. |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-III, B-IV, C-II, D-I
The order of a reaction is the sum of the powers of the concentration terms in the experimental rate law. Standard examples of each order are:
(A) Zero order — This matches (III) The thermal decomposition of HI on gold surface. At high pressure the gold surface is fully covered with HI, so the rate does not depend on the concentration of HI.
\( \displaystyle \ce{2HI(g)} \xrightarrow{\text{Au}} \ce{H2(g) + I2(g)}\quad \text{Rate}=k \)
(B) First order — This matches (IV) Radioactive decay of unstable nuclei. The rate of decay is proportional to the number of nuclei present, so every radioactive decay is first order.
\( \displaystyle \ce{^{226}_{88}Ra -> ^{222}_{86}Rn + ^{4}_{2}He}\quad \text{Rate}=k[\ce{Ra}] \)
(C) Pseudo First order — This matches (II) Hydrolysis of ethyl acetate in the presence of acid. Water is present in such a large excess that its concentration hardly changes, so the reaction behaves as first order.
\( \displaystyle \ce{CH3COOC2H5 + H2O} \xrightarrow{\ce{H+}} \ce{CH3COOH + C2H5OH}\quad \text{Rate}=k'[\ce{CH3COOC2H5}] \)
(D) Second order — This matches (I) Decomposition of hydrogen peroxide using iodide ion as catalyst in alkaline medium. The rate is first order in \( \ce{H2O2} \) and first order in \( \ce{I^{-}} \), so the order of the reaction is 1 + 1 = 2.
\( \displaystyle \ce{2H2O2} \xrightarrow{\ce{I^-},\ \text{alkaline}} \ce{2H2O + O2}\quad \text{Rate}=k[\ce{H2O2}][\ce{I^-}] \)
Shortcut: (A)–(III) appears only in Option (3).