Inversion of cane sugar is a :

The correct answer is Option (2) → Pseudo first order reaction

A reaction whose true order is higher than one but which behaves as a first order reaction because one reactant is present in large excess is called a pseudo first order reaction.

In the inversion of cane sugar, sucrose is hydrolysed in the presence of acid. Water is the solvent and is present in such a large excess that its concentration hardly changes, so the rate depends only on the concentration of sucrose:

    \( \displaystyle \underset{\text{cane sugar}}{\ce{C12H22O11}}+\ce{H2O} \xrightarrow{\ce{H+}} \underset{\text{glucose}}{\ce{C6H12O6}}+\underset{\text{fructose}}{\ce{C6H12O6}} \)

    \( \displaystyle \text{Rate}=k[\ce{C12H22O11}] \)

So it is a pseudo first order reaction.

Why the other options are incorrect

• Option (1) Zero order reaction — Incorrect. The rate does depend on the sucrose concentration.

• Option (3) Second order reaction — Incorrect. It would be second order only if the water concentration changed appreciably, which it does not.

• Option (4) Third order reaction — Incorrect. Only two molecules take part, and only one concentration term appears in the rate law.

NCERT (2026-27 edition):
Temperature Dependence of the Rate 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. … The concentration of water does not get altered much during the course of the reaction. So, the reaction behaves as first order reaction. Such reactions are called pseudo first order reactions. Inversion of cane sugar is another pseudo first order reaction.”

Note: The same/similar question has been asked in different years by NTA as per the details given below: