In the start of summer, a given sample of milk turns sour at room temperature `(27^(@)C)` in `48` hours. In a refrigerator at `2^(@)C`, milk can be stored there times longer before it sours.
The activation energy of the souring of milk is `(kJ mol^(-1))`

Options

A. `30.210`
B. `30.146`
C. `30.0`
D. `35.126`

Explanation

To find the activation energy (Ea) for the souring of milk, we can use the Arrhenius equation and the information provided in the question. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We know that:
- At room temperature (27°C or 300 K), milk turns sour in 48 hours.
- In a refrigerator at 2°C (or 275 K), milk can be stored three times longer before it sours, which means it lasts 144 hours.

### Step 2: Set Up the Rate Constant Ratio
Since the time taken for the reaction is inversely proportional to the rate constant (k), we can express the relationship between the rate constants at the two temperatures:
\[ \frac{k_2}{k_1} = \frac{t_1}{t_2} = \frac{48}{144} = \frac{1}{3} \]
This implies:
\[ k_2 = \frac{1}{3} k_1 \]

### Step 3: Use the Arrhenius Equation
The Arrhenius equation relates the rate constants at two different temperatures:
\[ \ln \left(\frac{k_2}{k_1}\right) = -\frac{E_a}{R} \left(\frac{1}{T_2} - \frac{1}{T_1}\right) \]
Where:
- \( R \) is the gas constant (8.314 J/mol·K)
- \( T_1 = 300 \, K \) (27°C)
- \( T_2 = 275 \, K \) (2°C)

### Step 4: Substitute Known Values
Substituting the values we have:
\[ \ln \left(\frac{1}{3}\right) = -\frac{E_a}{8.314} \left(\frac{1}{275} - \frac{1}{300}\right) \]
Calculating \( \ln \left(\frac{1}{3}\right) \):
\[ \ln \left(\frac{1}{3}\right) = -\ln(3) \approx -1.0986 \]

### Step 5: Calculate the Temperature Difference
Now calculate \( \frac{1}{275} - \frac{1}{300} \):
\[ \frac{1}{275} - \frac{1}{300} = \frac{300 - 275}{275 \times 300} = \frac{25}{82500} = \frac{1}{3300} \]

### Step 6: Rearranging the Equation
Now we can rearrange the equation to solve for \( E_a \):
\[ -1.0986 = -\frac{E_a}{8.314} \cdot \frac{1}{3300} \]
Multiplying both sides by \(-8.314 \cdot 3300\):
\[ E_a = 1.0986 \cdot 8.314 \cdot 3300 \]

### Step 7: Calculate \( E_a \)
Calculating the right side:
\[ E_a \approx 1.0986 \cdot 8.314 \cdot 3300 \approx 30.147 \, kJ/mol \]

### Final Answer
Thus, the activation energy for the souring of milk is approximately:
\[ E_a \approx 30.147 \, kJ/mol \]