Inversion of cane sugar in mild acidic conditions is a:

Options

A. Pseudo unimolecular reaction
B. Second order reaction
C. Termolecular reaction
D. Zero order reaction

Answer

To solve the question regarding the inversion of cane sugar in mild acidic conditions, we will analyze the reaction step by step:

### Step 1: Identify the Reaction
The inversion of cane sugar (sucrose) occurs when it is hydrolyzed in the presence of mild acid. The chemical equation for this reaction is:
\[ \text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_{12}\text{O}_6 + \text{C}_6\text{H}_{12}\text{O}_6 \]
This indicates that sucrose (C12H22O11) is converted into glucose (C6H12O6) and fructose (C6H12O6).

### Step 2: Determine the Type of Reaction
This reaction is a hydrolysis reaction, which is a type of reaction where water is used to break down a compound. In this case, it is the breakdown of sucrose into its constituent monosaccharides.

### Step 3: Analyze the Rate of Reaction
In the presence of water, which is typically in large excess, the reaction can be treated differently. The rate of the reaction can be expressed as:
\[ R = k \cdot [\text{sucrose}] \cdot [\text{water}] \]
However, since water is in large excess, its concentration remains relatively constant. Thus, we can combine the rate constant \( k \) with the concentration of water to form a new constant \( k' \):
\[ R = k' \cdot [\text{sucrose}] \]
This simplification leads us to conclude that the reaction behaves as a first-order reaction with respect to sucrose.

### Step 4: Classify the Reaction Order
Since the reaction appears to be a second-order reaction due to the presence of two reactants (sucrose and water), but water is in excess, we classify this reaction as a pseudo first-order reaction. This means that the reaction rate depends primarily on the concentration of sucrose.

### Conclusion
Thus, the inversion of cane sugar in mild acidic conditions is classified as a **pseudo first-order reaction**.

### Final Answer
The correct answer is: **Pseudo first-order reaction**.

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