The rate constant of a reaction increases by five times on increase in temperature from `27^@C` to `52^@C` . The value of activation energy in `kJ "mol"^(-1)` is ______ (Rounded-off to the nearest integer)

Answer

To find the activation energy (Ea) for the reaction given the change in the rate constant with temperature, we can use the Arrhenius equation in its logarithmic form:

\[ \log \left( \frac{k_2}{k_1} \right) = \frac{E_a}{2.303 R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]

### Step-by-Step Solution:

1. **Identify the given values:**
- The rate constant increases by five times: \( k_2 = 5k_1 \)
- Initial temperature \( T_1 = 27^\circ C = 300 \, K \) (convert to Kelvin by adding 273)
- Final temperature \( T_2 = 52^\circ C = 325 \, K \) (convert to Kelvin by adding 273)

2. **Substitute the values into the equation:**
\[ \log \left( \frac{5k_1}{k_1} \right) = \log(5) = 0.7 \]
Thus, we can rewrite the equation as:
\[ 0.7 = \frac{E_a}{2.303 \times 8.314} \left( \frac{1}{300} - \frac{1}{325} \right) \]

3. **Calculate \( \frac{1}{T_1} - \frac{1}{T_2} \):**
\[ \frac{1}{300} - \frac{1}{325} = \frac{325 - 300}{300 \times 325} = \frac{25}{97500} = \frac{1}{3900} \]

4. **Substitute this back into the equation:**
\[ 0.7 = \frac{E_a}{2.303 \times 8.314} \times \frac{1}{3900} \]

5. **Rearranging to solve for \( E_a \):**
\[ E_a = 0.7 \times 2.303 \times 8.314 \times 3900 \]

6. **Calculate the values:**
- Calculate \( 2.303 \times 8.314 \):
\[ 2.303 \times 8.314 \approx 19.175 \]
- Now calculate \( E_a \):
\[ E_a = 0.7 \times 19.175 \times 3900 \]
\[ E_a \approx 0.7 \times 19.175 \times 3900 \approx 52,271.69 \, J/mol \]

7. **Convert Joules to kJ:**
\[ E_a \approx 52.27 \, kJ/mol \]

8. **Round to the nearest integer:**
\[ E_a \approx 52 \, kJ/mol \]

### Final Answer:
The value of activation energy \( E_a \) is **52 kJ/mol**.