`C_(12) H_(22)O_(11) + underset("excess") (H_(2)O) to underset("glucose")(C_(6)H_(12)O_(6) + underset("fructose")(C_(6)H_(12)O_(6)`
Rate law is expressed as

Options

A. `r = K[C_(12)H_(22)O_(11)][H_(2)O]`
B. `r = K[C_(12)H_(22)O_(11)]`
C. `r = K[H_(2)o]`
D. `r = K[C_(12)H_(22)O_(11)][H_(2)O]^(2-)`

Explanation

To derive the rate law for the reaction of sucrose (C₁₂H₂₂O₁₁) hydrolyzing into glucose (C₆H₁₂O₆) and fructose (C₆H₁₂O₆), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction is:
\[ C_{12}H_{22}O_{11} + \text{(excess)} H_2O \rightarrow C_6H_{12}O_6 + C_6H_{12}O_6 \]
Here, sucrose reacts with water to produce glucose and fructose.

2. **Determine the Rate Law Form**:
The general form of the rate law is:
\[ \text{Rate} = k [\text{Reactants}]^{\text{stoichiometric coefficients}} \]
In this case, the reactants are sucrose (C₁₂H₂₂O₁₁) and water (H₂O).

3. **Consider the Concentration of Water**:
Since water is present in excess, its concentration does not significantly change during the reaction. Therefore, it can be treated as a constant. This means we can omit it from the rate law.

4. **Write the Rate Law**:
The rate law will depend only on the concentration of sucrose:
\[ \text{Rate} = k' [C_{12}H_{22}O_{11}] \]
where \( k' \) is the modified rate constant that includes the concentration of water.

5. **Determine the Order of the Reaction**:
The stoichiometric coefficient of sucrose in the balanced equation is 1. Thus, the reaction is first order with respect to sucrose.

### Final Rate Law:
The final rate law for the hydrolysis of sucrose is:
\[ \text{Rate} = k [C_{12}H_{22}O_{11}] \]