Milk turns sour at `40^(@)C` three times as faster as `0^(@)C`. Hence `E_(a)` in cal of process of turning of milk sour is `:`

Options

A. `(2.303xx2xx313xx273)/(40)log3`
B. `(2.303xx2xx313xx273)/(40)log.(1)/(3)`
C. `(2.303xx2xx40)/(273xx313)log3`
D. None of the above

Explanation

To solve the problem of determining the activation energy (Ea) for the process of milk turning sour, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- The rate of souring of milk at \(40^\circ C\) is three times faster than at \(0^\circ C\).
- This implies that the rate constant at \(40^\circ C\) (K2) is three times the rate constant at \(0^\circ C\) (K1):
\[ K_2 = 3K_1 \]

2. **Convert Temperatures to Kelvin:**
- Convert the temperatures from Celsius to Kelvin:
- \(T_1 = 0^\circ C = 273 \, K\)
- \(T_2 = 40^\circ C = 313 \, K\)

3. **Use the Arrhenius Equation:**
- The Arrhenius equation relates the rate constants and temperature:
\[ \log \left( \frac{K_2}{K_1} \right) = \frac{E_a}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
- Substitute \(K_2 = 3K_1\):
\[ \log(3) = \frac{E_a}{2.303 R} \left( \frac{1}{273} - \frac{1}{313} \right) \]

4. **Calculate the Difference in Inverse Temperatures:**
- Calculate \(\frac{1}{273} - \frac{1}{313}\):
\[ \frac{1}{273} - \frac{1}{313} = \frac{313 - 273}{273 \times 313} = \frac{40}{273 \times 313} \]

5. **Substitute Values into the Equation:**
- Using \(R = 2 \, \text{cal/(mol K)}\):
\[ \log(3) = \frac{E_a}{2.303 \times 2} \left( \frac{40}{273 \times 313} \right) \]

6. **Rearranging to Solve for Activation Energy (Ea):**
- Rearranging gives:
\[ E_a = 2.303 \times 2 \times 273 \times 313 \times \log(3) \times \frac{273 \times 313}{40} \]

7. **Calculating the Values:**
- Calculate \(\log(3) \approx 0.477\):
\[ E_a = 2.303 \times 2 \times 273 \times 313 \times 0.477 \times \frac{273 \times 313}{40} \]

8. **Final Calculation:**
- Compute the numerical value using a calculator to find \(E_a\) in calories.

### Final Answer:
- After performing the calculations, the value of \(E_a\) will be obtained.