The ratio of rate constant at `27^(@)C` and `37^(@)C` is `Q_(10)`. What should be the energy of activation of a reaction for which `Q_(10) = 2.5` ?

Options

A. `71 kJ`
B. `212 kJ`
C. `35 kJ`
D. `12.1 kJ`

Explanation

To solve the problem, we will use the Arrhenius equation and the concept of the temperature coefficient \( Q_{10} \).

### Step-by-Step Solution:

1. **Understanding the Given Information**:
We are given that the ratio of the rate constants at two temperatures (27°C and 37°C) is \( Q_{10} = 2.5 \).

2. **Convert Temperatures to Kelvin**:
- \( T_1 = 27°C = 27 + 273 = 300 \, K \)
- \( T_2 = 37°C = 37 + 273 = 310 \, K \)

3. **Use the Arrhenius Equation**:
The Arrhenius equation in logarithmic form for two different temperatures is given by:
\[ \log \left( \frac{K_2}{K_1} \right) = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) \]
Here, \( K_1 \) and \( K_2 \) are the rate constants at temperatures \( T_1 \) and \( T_2 \) respectively.

4. **Substituting \( Q_{10} \) into the Equation**:
Since \( Q_{10} = \frac{K_2}{K_1} \), we can write:
\[ \log(2.5) = \frac{E_a}{2.303R} \left( \frac{1}{300} - \frac{1}{310} \right) \]

5. **Calculate the Difference in Inverses of Temperatures**:
\[ \frac{1}{300} - \frac{1}{310} = \frac{310 - 300}{300 \times 310} = \frac{10}{93000} = \frac{1}{9300} \]

6. **Substituting Values**:
Now, substituting this back into the equation:
\[ \log(2.5) = \frac{E_a}{2.303R} \cdot \frac{1}{9300} \]

7. **Calculate \( \log(2.5) \)**:
Using a calculator or logarithm table:
\[ \log(2.5) \approx 0.39794 \]

8. **Substituting the Value of R**:
The gas constant \( R \) in kJ/(mol·K) is approximately \( 8.314 \times 10^{-3} \, kJ/(mol·K) \).

9. **Rearranging to Find \( E_a \)**:
Rearranging the equation gives:
\[ E_a = \log(2.5) \cdot 2.303R \cdot 9300 \]

10. **Calculating \( E_a \)**:
Substituting the values:
\[ E_a = 0.39794 \cdot 2.303 \cdot 8.314 \times 10^{-3} \cdot 9300 \]
\[ E_a \approx 71 \, kJ/mol \]

### Final Answer:
The energy of activation \( E_a \) is approximately **71 kJ/mol**.