Milk turns sour at `40^(@)C` three times faster as at `0^(@)C`. The energy of activation for souring of milk is:

Options

A. `4.693 kcal`
B. `2.6 kcal`
C. `6.6 kcal`
D. None of these

Explanation

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

### Step 1: Understand the relationship between the rate constants
Given that the souring of milk at `40°C` is three times faster than at `0°C`, we can denote the rate constants at these two temperatures as \( k_1 \) and \( k_2 \):
- \( k_2 = 3k_1 \)

### Step 2: Convert temperatures to Kelvin
We need to convert the temperatures from Celsius to Kelvin:
- \( T_1 = 0°C = 273 K \)
- \( T_2 = 40°C = 313 K \)

### Step 3: Use the Arrhenius equation
The Arrhenius equation is given by:
\[ k = A e^{-\frac{E_a}{RT}} \]
Taking the ratio of the rate constants at two different temperatures gives us:
\[ \frac{k_2}{k_1} = \frac{A e^{-\frac{E_a}{RT_2}}}{A e^{-\frac{E_a}{RT_1}}} \]
This simplifies to:
\[ \frac{k_2}{k_1} = e^{-\frac{E_a}{RT_2} + \frac{E_a}{RT_1}} = e^{\frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right)} \]

### Step 4: Take the natural logarithm
Taking the natural logarithm of both sides gives:
\[ \ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]
Substituting \( k_2 = 3k_1 \):
\[ \ln(3) = \frac{E_a}{R} \left(\frac{1}{273} - \frac{1}{313}\right) \]

### Step 5: Calculate the difference in reciprocals of temperatures
Calculating \( \frac{1}{273} - \frac{1}{313} \):
\[ \frac{1}{273} \approx 0.003663 \quad \text{and} \quad \frac{1}{313} \approx 0.003194 \]
Thus:
\[ \frac{1}{273} - \frac{1}{313} \approx 0.003663 - 0.003194 = 0.000469 \]

### Step 6: Substitute values into the equation
Now substituting the values into the equation:
\[ \ln(3) \approx 1.0986 \]
And using \( R = 2 \, \text{cal/(mol K)} \):
\[ 1.0986 = \frac{E_a}{2} \times 0.000469 \]

### Step 7: Solve for \( E_a \)
Rearranging gives:
\[ E_a = \frac{1.0986 \times 2}{0.000469} \]
Calculating this:
\[ E_a \approx \frac{2.1972}{0.000469} \approx 4687.65 \, \text{cal/mol} \approx 4.69 \, \text{kcal/mol} \]

### Final Answer
The energy of activation for the souring of milk is approximately **4.69 kcal/mol**.

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