It has been found for a chemical reaction with rise in temperature by 9 K the rate contant gets doubled . Assuming a reaction to be occuring at 300 K , the value of activation energy is found to be ____________ KJ `"mol"^(-1)` . (nearest integer)
(Given In 10 = 2.3 , R = 8.3 J`K^(-1) "mol"^(-1) , ` log 2 = 0.30)

Explanation

To find the activation energy (Ea) for the given chemical reaction, we can use the Arrhenius equation, which relates the rate constants at two different temperatures. The steps to solve the problem are as follows:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Initial temperature, \( T_1 = 300 \, K \)
- Rise in temperature, \( \Delta T = 9 \, K \)
- Final temperature, \( T_2 = T_1 + \Delta T = 300 + 9 = 309 \, K \)
- The rate constant doubles, so \( K_2 = 2K_1 \).

2. **Use the Arrhenius Equation**:
The Arrhenius equation in terms of two temperatures is given by:
\[ \log \frac{K_2}{K_1} = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

3. **Substitute the Known Values**:
Since \( K_2 = 2K_1 \), we can write:
\[ \log \frac{K_2}{K_1} = \log 2 \]
Given that \( \log 2 \approx 0.301 \), we substitute this into the equation:
\[ 0.301 = \frac{E_a}{2.303 \times 8.314} \left( \frac{1}{300} - \frac{1}{309} \right) \]

4. **Calculate the Temperature Difference**:
Calculate \( \frac{1}{T_1} - \frac{1}{T_2} \):
\[ \frac{1}{300} - \frac{1}{309} = \frac{309 - 300}{300 \times 309} = \frac{9}{92700} = \frac{1}{10300} \]

5. **Substitute and Rearrange**:
Now substitute this back into the equation:
\[ 0.301 = \frac{E_a}{2.303 \times 8.314} \times \frac{1}{10300} \]
Rearranging gives:
\[ E_a = 0.301 \times 2.303 \times 8.314 \times 10300 \]

6. **Calculate \( E_a \)**:
First calculate \( 2.303 \times 8.314 \):
\[ 2.303 \times 8.314 \approx 19.2 \]
Now multiply:
\[ E_a \approx 0.301 \times 19.2 \times 10300 \]
\[ \approx 0.301 \times 197376 \approx 59440.256 \, J/mol \]
Converting to kJ/mol:
\[ E_a \approx 59.44 \, kJ/mol \]
Rounding to the nearest integer gives:
\[ E_a \approx 59 \, kJ/mol \]

### Final Answer:
The value of activation energy \( E_a \) is approximately **59 kJ/mol**.