A sample of milk splits after 60 min . At 300 K and after 40 min 400K when the population of lactobacillus acidophilus in it doubles . The activation energy (in KJ/mol) for this process is closest to (Given , `R = 8.3 J mol^(-1) K^(-1), " ln" (2/3) = 0.4 , e^(-3) = 4.0)`

Options

A. 39.8
B. 19.9
C. 3.98
D. 7.96

Answer

To find the activation energy (Ea) for the process described in the question, we can use the Arrhenius equation. The equation relates the rate constants of a reaction at two different temperatures to the activation energy. Here’s how we can solve the problem step by step:

### Step 1: Identify the given values
- Time at 300 K (T1) = 60 min
- Time at 400 K (T2) = 40 min
- R (gas constant) = 8.3 J/mol·K
- We need to find the activation energy (Ea).

### Step 2: Determine the rate constants (k1 and k2)
Since the population of lactobacillus acidophilus doubles, we can express the rate constants as:
- k1 corresponds to T1 (300 K) and is related to the time of 60 min.
- k2 corresponds to T2 (400 K) and is related to the time of 40 min.

Using the relationship between time and rate constant, we can say:
- k1 is inversely proportional to time, so \( k1 = \frac{1}{60} \)
- k2 is inversely proportional to time, so \( k2 = \frac{1}{40} \)

### Step 3: Set up the Arrhenius equation
The Arrhenius equation in terms of two temperatures is given by:
\[ \ln \left( \frac{k2}{k1} \right) = \frac{Ea}{R} \left( \frac{1}{T1} - \frac{1}{T2} \right) \]

### Step 4: Substitute the values into the equation
Substituting \( k1 \) and \( k2 \):
\[ \ln \left( \frac{1/40}{1/60} \right) = \ln \left( \frac{60}{40} \right) = \ln \left( \frac{3}{2} \right) \]
Using the provided value \( \ln \left( \frac{2}{3} \right) = -0.4 \), we can rewrite:
\[ \ln \left( \frac{3}{2} \right) = -\ln \left( \frac{2}{3} \right) = 0.4 \]

Now, substituting into the Arrhenius equation:
\[ 0.4 = \frac{Ea}{8.3} \left( \frac{1}{300} - \frac{1}{400} \right) \]

### Step 5: Calculate the temperature difference
Calculating \( \frac{1}{300} - \frac{1}{400} \):
\[ \frac{1}{300} - \frac{1}{400} = \frac{4 - 3}{1200} = \frac{1}{1200} \]

### Step 6: Substitute this back into the equation
Now substituting this back into the equation:
\[ 0.4 = \frac{Ea}{8.3} \cdot \frac{1}{1200} \]

### Step 7: Solve for Ea
Rearranging the equation to solve for Ea:
\[ Ea = 0.4 \cdot 8.3 \cdot 1200 \]

### Step 8: Calculate the value
Calculating:
\[ Ea = 0.4 \cdot 8.3 \cdot 1200 = 3984 \text{ J/mol} \]

### Step 9: Convert to kJ/mol
To convert to kJ/mol:
\[ Ea = \frac{3984}{1000} = 3.984 \text{ kJ/mol} \]

### Step 10: Round the answer
Rounding gives us:
\[ Ea \approx 4 \text{ kJ/mol} \]

### Final Answer
The activation energy for this process is closest to **4 kJ/mol**.