The inverison of cane sugar proceeds with half life of `500 min` at `pH 5` for any concentration of sugar. However, if `pH = 6`, if the half life changes to `50 min`. The rate law expresison for the sugar inverison can be written as

Options

A. `r = k["sugar"]^(2)[H]^(6)`
B. `r = k["sugar"]^(1)[H]^(0)`
C. `r = k["suagr"]^(0)[H^(o+)]^(6)`
D. `r = k["sugar"]^(0)[H^(o+)]^(1)`

Explanation

To derive the rate law expression for the inversion of cane sugar, we can follow these steps:

### Step 1: Understand the Reaction
The inversion of cane sugar (sucrose) involves its conversion into glucose and fructose. This reaction occurs in an acidic medium, and we need to analyze how the pH affects the reaction rate.

### Step 2: Identify the Given Data
- At pH 5, the half-life (t₁/₂) is 500 minutes.
- At pH 6, the half-life (t₁/₂) is 50 minutes.

### Step 3: Relate Half-Life to Concentration
For a first-order reaction, the half-life is given by the formula:
\[ t_{1/2} = \frac{0.693}{k} \]
However, in this case, the reaction is influenced by the concentration of hydrogen ions (H⁺) as well.

### Step 4: Write the Rate Law Expression
The rate law can be expressed as:
\[ \text{Rate} = k [\text{sugar}]^m [\text{H}^+]^n \]
Where:
- \( [\text{sugar}] \) is the concentration of sucrose.
- \( [\text{H}^+] \) is the concentration of hydrogen ions.
- \( m \) is the order with respect to sugar.
- \( n \) is the order with respect to hydrogen ions.

### Step 5: Analyze the Effect of pH on H⁺ Concentration
At pH 5, the concentration of H⁺ is:
\[ [\text{H}^+] = 10^{-5} \, \text{M} \]
At pH 6, the concentration of H⁺ is:
\[ [\text{H}^+] = 10^{-6} \, \text{M} \]

### Step 6: Relate Half-Lives to H⁺ Concentration
From the relationship of half-lives and H⁺ concentration, we can write:
\[ \frac{t_{1/2, \text{initial}}}{t_{1/2, \text{final}}} = \left(\frac{[\text{H}^+]_{\text{initial}}}{[\text{H}^+]_{\text{final}}}\right)^{1-n} \]

Substituting the values:
\[ \frac{500}{50} = \left(\frac{10^{-5}}{10^{-6}}\right)^{1-n} \]
This simplifies to:
\[ 10 = 10^{1-n} \]

### Step 7: Solve for n
From the equation:
\[ 10 = 10^{1-n} \]
This implies:
\[ 1 - n = 1 \]
Thus,
\[ n = 0 \]

### Step 8: Determine the Order with Respect to Sugar
Since the order with respect to sugar is typically first-order for such reactions, we can conclude:
- \( m = 1 \)
- \( n = 0 \)

### Final Rate Law Expression
The rate law expression for the inversion of cane sugar is:
\[ \text{Rate} = k [\text{sugar}]^1 [\text{H}^+]^0 \]
This simplifies to:
\[ \text{Rate} = k [\text{sugar}] \]