The hydrolysis of a sugar like cane sugar (sucrose) in dilute aqueous acid is FIRST ORDER with respect to sucrose, even though the rate law theoretically should involve water concentration too. Why is this so?

1. NCERT §3.2.3 explains the pseudo-first-order concept. 2. The full rate law for sucrose hydrolysis is rate $= k\,[\text{sucrose}][\mathrm{H_2O}]$ (second order overall). 3. In dilute aqueous solution, $[\mathrm{H_2O}] \approx 55\,\text{mol/L}$ — much larger than the sucrose concentration ($\sim 10^{-2}\,\text{mol/L}$) and essentially CONSTANT throughout the reaction. 4. So $k\,[\mathrm{H_2O}]$ can be lumped into an effective constant $k' = k\,[\mathrm{H_2O}]$, and the rate law becomes rate $= k'[\text{sucrose}]$ — first order in sucrose. 5. This is the PSEUDO-FIRST-ORDER trick: a reaction with one reactant in huge excess effectively reduces to the order of the limiting reactant. 6. Option A is false — water participates routinely. Option C is wrong — water is a reactant here, not a catalyst. Option D contradicts the experimental finding. _Source: NCERT Class 12 Chemistry Part 1, Ch 3, §3.2.3 (Pseudo first-order reactions, p. 8–9, e.g. sucrose inversion example)._