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 solve the problem of finding the activation energy for the process described, we can follow these steps:

### Step 1: Understand the relationship between rate constant and time
The rate constant \( k \) is inversely proportional to the time taken for the reaction to occur. Thus, we can express this as:
\[ k \propto \frac{1}{t} \]
This means we can write:
\[ k_1 = \frac{1}{t_1} \quad \text{and} \quad k_2 = \frac{1}{t_2} \]

### Step 2: 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:
- \( E_a \) is the activation energy,
- \( R \) is the gas constant (8.3 J/mol·K),
- \( T_1 \) and \( T_2 \) are the temperatures in Kelvin.

### Step 3: Substitute the known values
From the problem:
- \( t_1 = 60 \) min at \( T_1 = 300 \) K,
- \( t_2 = 40 \) min at \( T_2 = 400 \) K.

Thus, we can express \( k_1 \) and \( k_2 \):
\[ k_1 = \frac{1}{60} \quad \text{and} \quad k_2 = \frac{1}{40} \]

Now substituting into the Arrhenius equation:
\[ \ln \left( \frac{k_2}{k_1} \right) = \ln \left( \frac{1/40}{1/60} \right) = \ln \left( \frac{60}{40} \right) = \ln \left( \frac{3}{2} \right) \]

Using the provided value \( \ln \left( \frac{2}{3} \right) = -0.4 \):
\[ \ln \left( \frac{3}{2} \right) = -\ln \left( \frac{2}{3} \right) = 0.4 \]

### Step 4: Calculate the difference in reciprocal temperatures
Now we calculate:
\[ \frac{1}{T_2} - \frac{1}{T_1} = \frac{1}{400} - \frac{1}{300} \]
Finding a common denominator (1200):
\[ \frac{3 - 4}{1200} = -\frac{1}{1200} \]

### Step 5: Substitute into the Arrhenius equation
Now we substitute everything back into the Arrhenius equation:
\[ 0.4 = -\frac{E_a}{8.3} \left( -\frac{1}{1200} \right) \]
This simplifies to:
\[ 0.4 = \frac{E_a}{8.3 \times 1200} \]
Thus,
\[ E_a = 0.4 \times 8.3 \times 1200 \]

### Step 6: Calculate \( E_a \)
Calculating the right-hand side:
\[ E_a = 0.4 \times 9960 = 3984 \text{ J/mol} \]
To convert to kJ/mol:
\[ E_a = \frac{3984}{1000} = 3.984 \text{ kJ/mol} \]

### Final Answer
The activation energy \( E_a \) is approximately **3.98 kJ/mol**.

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