Statement I: Hydrolysis of cane sugar is a first-order reaction

Statement II: Water is present in large excess during hydrolysis

Statement I and statement II are correct and statement II is correct explanation of statement I

The correct answer is option (1) → Statement I and statement II are correct and statement II is correct explanation of statement I

Analysis of Statement I

"Hydrolysis of cane sugar is a first-order reaction" (True)

The hydrolysis of cane sugar (sucrose) follows the chemical equation:

$\text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \overset{{\text{H}^+}}{\longrightarrow} \text{C}_6\text{H}_{12}\text{O}_6 \text{ (Glucose)} + \text{C}_6\text{H}_{12}\text{O}_6 \text{ (Fructose)}$

Experimentally, the rate of this reaction depends only on the concentration of sugar, making it behave like a first-order reaction.

Analysis of Statement II

"Water is present in large excess during hydrolysis" (True)

In an aqueous solution, the concentration of water is approximately 55.5 mol/L. Compared to the sugar concentration (which is usually very small), the amount of water used up during the reaction is negligible.

Why Statement II explains Statement I

The theoretical rate law for this reaction should be:

$\text{Rate} = k'[\text{C}_{12}\text{H}_{22}\text{O}_{11}][\text{H}_2\text{O}]$

However, because water is in large excess, its concentration remains constant throughout the reaction ($[\text{H}_2\text{O}] \approx \text{constant}$). We can combine this constant value with the rate constant $k'$ to create a new constant, $k$:

$\text{Rate} = k[\text{C}_{12}\text{H}_{22}\text{O}_{11}]$

Because the rate now depends only on the first power of the concentration of cane sugar, the reaction is called Pseudo First Order.

• Fact: The reaction involves two molecules (bimolecular).
• The "Trick": Because water is so abundant, it doesn't "limit" the speed of the reaction.
• Result: It behaves as if it is first-order. Statement II provides the physical reason for this behavior.