Lacto bacillus has generation time 60 min. at 300 K and 40 min. at 400 K. Determine activation energy in `(kJ)/(mol)` . (`R = 8.3 J K^ (–1) mol^(–1)` ) `[ log_e (2/3) = - 0.4 ]`

Answer

To determine the activation energy (Ea) for the growth of Lactobacillus at two different temperatures, we can use the Arrhenius equation in its logarithmic form. Here’s a step-by-step solution:

### Step 1: Understand the Arrhenius Equation
The Arrhenius equation can be expressed in logarithmic form as:
\[ \log K = \log A - \frac{E_a}{2.303RT} \]
Where:
- \( K \) is the rate constant,
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin.

### Step 2: Set Up the Equation for Two Temperatures
For two different temperatures \( T_1 \) and \( T_2 \) with corresponding generation times \( t_1 \) and \( t_2 \), we can write:
\[ \log \frac{K_2}{K_1} = -\frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
In this case, \( K_1 \) and \( K_2 \) can be related to the generation times as:
\[ K_1 \propto \frac{1}{t_1} \quad \text{and} \quad K_2 \propto \frac{1}{t_2} \]
Thus, we can express the equation as:
\[ \log \frac{t_2}{t_1} = -\frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

### Step 3: Substitute Given Values
Given:
- \( t_1 = 60 \) minutes at \( T_1 = 300 \) K
- \( t_2 = 40 \) minutes at \( T_2 = 400 \) K

Substituting these values into the equation:
\[ \log \frac{40}{60} = -\frac{E_a}{2.303 \times 8.314} \left( \frac{1}{300} - \frac{1}{400} \right) \]

### Step 4: Simplify the Logarithm
Calculating the logarithm:
\[ \log \frac{40}{60} = \log \frac{2}{3} = -0.4 \]
Thus, we have:
\[ -0.4 = -\frac{E_a}{2.303 \times 8.314} \left( \frac{1}{300} - \frac{1}{400} \right) \]

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

### Step 6: Substitute and Rearrange to Solve for \( E_a \)
Now substituting back into the equation:
\[ -0.4 = -\frac{E_a}{2.303 \times 8.314} \left( \frac{1}{1200} \right) \]
Rearranging gives:
\[ E_a = 0.4 \times 2.303 \times 8.314 \times 1200 \]

### Step 7: Calculate \( E_a \)
Calculating \( E_a \):
\[ E_a = 0.4 \times 2.303 \times 8.314 \times 1200 \]
Calculating the numerical values:
\[ E_a \approx 0.4 \times 2.303 \times 8.314 \times 1200 \approx 3.984 \text{ kJ/mol} \]

### Final Answer
The activation energy \( E_a \) is approximately:
\[ \boxed{3.984 \text{ kJ/mol}} \]