The rate constant is doubled when temperature increases from `27^(@)C` to `37^(@)C`. Activation energy in kJ is

Options

A. `34`
B. `54`
C. `100`
D. `50`

Explanation

To determine the activation energy (Ea) given that the rate constant (k) doubles when the temperature increases from 27°C to 37°C, we can use the Arrhenius equation, which is given by:

\[ k = A e^{-\frac{E_a}{RT}} \]

Where:
- \( k \) = rate constant
- \( A \) = pre-exponential factor
- \( E_a \) = activation energy
- \( R \) = universal gas constant (8.314 J/(mol·K))
- \( T \) = temperature in Kelvin

### Step-by-Step Solution:

1. **Convert Temperatures to Kelvin**:
- \( T_1 = 27°C = 27 + 273.15 = 300.15 K \)
- \( T_2 = 37°C = 37 + 273.15 = 310.15 K \)

2. **Set Up the Arrhenius Equation for Two Temperatures**:
- For \( T_1 \):
\[ k_1 = A e^{-\frac{E_a}{R T_1}} \]
- For \( T_2 \):
\[ k_2 = A e^{-\frac{E_a}{R T_2}} \]

3. **Since \( k_2 = 2k_1 \)**, we can write:
\[ 2k_1 = A e^{-\frac{E_a}{R T_2}} \]

4. **Divide the Two Equations**:
\[ \frac{k_2}{k_1} = \frac{A e^{-\frac{E_a}{R T_2}}}{A e^{-\frac{E_a}{R T_1}}} \]
\[ 2 = e^{-\frac{E_a}{R T_2}} / e^{-\frac{E_a}{R T_1}} \]
\[ 2 = e^{-\frac{E_a}{R T_2} + \frac{E_a}{R T_1}} \]

5. **Take the Natural Logarithm**:
\[ \ln(2) = -\frac{E_a}{R T_2} + \frac{E_a}{R T_1} \]
\[ \ln(2) = \frac{E_a}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \]

6. **Rearranging for Activation Energy (Ea)**:
\[ E_a = \frac{R \ln(2)}{\left(\frac{1}{T_1} - \frac{1}{T_2}\right)} \]

7. **Substituting the Values**:
- \( R = 8.314 \, \text{J/(mol·K)} \)
- \( T_1 = 300.15 \, K \)
- \( T_2 = 310.15 \, K \)

\[ \frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{300.15} - \frac{1}{310.15} \]

Calculate \( \frac{1}{300.15} \) and \( \frac{1}{310.15} \):
\[ \frac{1}{300.15} \approx 0.00332 \, K^{-1} \]
\[ \frac{1}{310.15} \approx 0.00322 \, K^{-1} \]
\[ \frac{1}{T_1} - \frac{1}{T_2} \approx 0.00332 - 0.00322 = 0.00010 \, K^{-1} \]

8. **Calculate Ea**:
\[ E_a = \frac{8.314 \cdot \ln(2)}{0.00010} \]
\[ \ln(2) \approx 0.693 \]
\[ E_a = \frac{8.314 \cdot 0.693}{0.00010} \approx 57500 \, J/mol \]
\[ E_a \approx 57.5 \, kJ/mol \]

### Final Answer:
The activation energy \( E_a \) is approximately **57.5 kJ/mol**.