Statement I: The inversion (hydrolysis) of cane sugar in acidic medium is a pseudo first order reaction.
Statement II: In the hydrolysis of cane sugar, water is present in large excess, so its concentration hardly changes during the reaction.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I.
In acidic medium, cane sugar (sucrose) reacts with water to give glucose and fructose:
\( \displaystyle \underset{\text{Cane sugar}}{\ce{C12H22O11}} + \ce{H2O} \xrightarrow{\ce{H+}} \underset{\text{Glucose}}{\ce{C6H12O6}} + \underset{\text{Fructose}}{\ce{C6H12O6}} \)
Statement I: The inversion (hydrolysis) of cane sugar in acidic medium is a pseudo first order reaction. — True. Both cane sugar and water take part, yet the rate law is Rate = k[C12H22O11]: the reaction behaves as first order.
Statement II: In the hydrolysis of cane sugar, water is present in large excess, so its concentration hardly changes during the reaction. — True. Water is the solvent and is in large excess; the small amount used up does not alter its concentration much.
Why Statement II explains Statement I?
Since [H2O] stays nearly constant, it does not change the rate, and only [C12H22O11] does.
A reaction that involves two reactants but follows a first order rate law for this reason is called pseudo first order.
NCERT (2026-27 edition):
3.3.3 Half-Life of a Reaction – “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.”