Inversion of cane sugar is a pseudo first order reaction
Choose the correct reason from the statements given below.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → The concentration of water does not get altered during the reaction so the reaction behaves as a first order reaction
A reaction that is really of higher order but obeys a first order rate law is called a pseudo first order reaction. This happens when one reactant is present in such large excess that its concentration hardly changes during the reaction.
In the inversion of cane sugar, water is the solvent and is present in large excess:
\( \displaystyle \ce{C12H22O11 + H2O} \xrightarrow{\ce{H+}} \ce{C6H12O6}\,(\text{glucose}) + \ce{C6H12O6}\,(\text{fructose}) \)
The true rate law includes water, Rate \( = k'[\ce{C12H22O11}][\ce{H2O}] \). Because \( [\ce{H2O}] \) stays practically constant, it is absorbed into the constant:
\( \displaystyle \text{Rate} = k[\ce{C12H22O11}] \quad \text{where } k = k'[\ce{H2O}] \)
So the reaction behaves as first order because the concentration of water does not change appreciably.
Why the other options are incorrect
Option (1) \(H_2O\) is not involved in the reaction — Incorrect. Water is a reactant: one molecule of water is used for each molecule of sucrose hydrolysed.
Option (3) Order of the reaction with respect to \(H_2O\) is zero — Incorrect. The reaction is first order in water too; the dependence on water only cannot be seen because its concentration stays constant. A true zero order in water would mean water does not affect the rate at all.
Option (4) The rate law equation for the reaction is \(k[C_{12}H_{22}O_{11}]^{1/2}[H_2O]^{1/2}\) — Incorrect. The observed rate law is first order in sucrose, Rate = k\( \ce{[C12H22O11]} \); there are no half powers.
NCERT (2026-27 edition):
Half-Life of a Reaction – …
