Match List I with List II:
| List I | List II |
| (A) First order reaction | (I) \(C_{12}H_{22}O_{11} + H_2O \overset{H^+}{\longrightarrow} \underset{Glucose}{C_6H_{12}O_6} + \underset{Fructose}{C_6H_{12}O_6}\) |
| (B) Zero order reaction | (II) \(N_2O_5 \longrightarrow 2NO_2 + \frac{1}{2}O_2\) |
| (C) Pseudo first order reaction | (III) \(CH_3COOC_2H_5 + NaOH \longrightarrow CH_3COONa + C_2H_5OH\) |
| (D) Second order reaction | (IV) \(2NH_3\underset{\text{Pt catalyst}}{\overset{1130 K}{\longrightarrow}} N_2 (g) + 3H_2 (g)\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → A-II, B-IV, C-I, D-III
The order of a reaction is the sum of the powers of the concentrations in the experimental rate law. A reaction in which one reactant is in large excess behaves as first order and is called pseudo first order.
(A) First order reaction — This matches (II) N2O5 → 2NO2 + ½O2. The decomposition of \( \ce{N2O5} \) has the rate law rate = k[N2O5].
(B) Zero order reaction — This matches (IV) 2NH3 → N2(g) + 3H2(g) (1130 K, Pt catalyst). At high pressure the platinum surface is saturated with \( \ce{NH3} \), so the rate does not depend on its concentration: rate = k[NH3]0 = k.
(C) Pseudo first order reaction — This matches (I) C12H22O11 + H2O → C6H12O6 + C6H12O6 (H+). In the inversion of cane sugar water is present in large excess and its concentration hardly changes, so rate = k[C12H22O11].
(D) Second order reaction — This matches (III) CH3COOC2H5 + NaOH → CH3COONa + C2H5OH. In the saponification of an ester both reactants are taken in comparable amounts: rate = k[CH3COOC2H5][OH−].
Shortcut: (B)-(IV) appears only in Option (3).