The rate of the chemical reaction doubles for an increase of 10K in absolute temperature form 298 K. Calculate `E_(a)` .

Answer

To solve the problem of calculating the activation energy \( E_a \) given that the rate of a chemical reaction doubles with a temperature increase of 10 K from 298 K, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Initial temperature \( T_1 = 298 \, K \)
- Final temperature \( T_2 = 298 + 10 = 308 \, K \)
- The rate of reaction doubles, so \( k_2 = 2k_1 \).

2. **Use the Arrhenius Equation:**
The relationship between the rate constants and temperature can be expressed as:
\[ \log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
Here, \( R = 8.314 \, \text{J/(K·mol)} \).

3. **Substituting Known Values:**
Since \( k_2 = 2k_1 \), we have:
\[ \log \frac{k_2}{k_1} = \log 2 = 0.301 \]
Now substituting into the equation:
\[ 0.301 = \frac{E_a}{2.303 \times 8.314} \left( \frac{1}{298} - \frac{1}{308} \right) \]

4. **Calculate \( \frac{1}{T_1} - \frac{1}{T_2} \):**
\[ \frac{1}{298} - \frac{1}{308} = \frac{308 - 298}{298 \times 308} = \frac{10}{298 \times 308} \]

5. **Substituting Back into the Equation:**
Now substituting this back into the equation:
\[ 0.301 = \frac{E_a}{2.303 \times 8.314} \left( \frac{10}{298 \times 308} \right) \]

6. **Rearranging to Solve for \( E_a \):**
Rearranging gives:
\[ E_a = 0.301 \times 2.303 \times 8.314 \times \frac{298 \times 308}{10} \]

7. **Calculating \( E_a \):**
Now we calculate the values:
\[ E_a = 0.301 \times 2.303 \times 8.314 \times \frac{298 \times 308}{10} \]
Performing the calculations:
\[ E_a \approx 0.301 \times 2.303 \times 8.314 \times 97.48 \approx 52.9 \, \text{kJ/mol} \]

### Final Answer:
The activation energy \( E_a \) is approximately **52.9 kJ/mol**.