The rate constant of a reaction increases by 5% when its temperature is raised from `27^@c` to `28^@c.` The activation energy of the reaction is

Options

A. `36.6kJ//mol`
B. `16.6kJ//mol`
C. `46.6kJ//mol`
D. `26.6kJ//mol`

Explanation

To find the activation energy (EA) of the reaction given that the rate constant increases by 5% when the temperature is raised from 27°C to 28°C, we can use the Arrhenius equation in the form of the logarithmic relationship between the rate constants and temperatures.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial temperature (T1) = 27°C = 27 + 273 = 300 K
- Final temperature (T2) = 28°C = 28 + 273 = 301 K
- Rate constant at T1 (K1) = 100 (assuming an arbitrary value for simplicity)
- Rate constant at T2 (K2) = K1 + 5% of K1 = 100 + 5 = 105

2. **Use the Arrhenius Equation:**
The relationship between the rate constants and temperatures can be expressed as:
\[ \log \frac{K2}{K1} = \frac{EA}{2.303 R} \cdot \frac{(T2 - T1)}{T1 \cdot T2} \]

3. **Substitute Known Values:**
- \( K2 = 105 \)
- \( K1 = 100 \)
- \( R = 8.314 \, \text{J/mol K} \)
- \( T1 = 300 \, \text{K} \)
- \( T2 = 301 \, \text{K} \)

Substitute these values into the equation:
\[ \log \frac{105}{100} = \frac{EA}{2.303 \cdot 8.314} \cdot \frac{(301 - 300)}{300 \cdot 301} \]

4. **Calculate the Logarithm:**
\[ \log \frac{105}{100} = \log 1.05 \approx 0.02119 \]

5. **Calculate the Temperature Difference and Product:**
\[ T2 - T1 = 301 - 300 = 1 \, \text{K} \]
\[ T1 \cdot T2 = 300 \cdot 301 = 90300 \, \text{K}^2 \]

6. **Substitute Values into the Equation:**
\[ 0.02119 = \frac{EA}{2.303 \cdot 8.314} \cdot \frac{1}{90300} \]

7. **Rearranging to Solve for EA:**
\[ EA = 0.02119 \cdot 2.303 \cdot 8.314 \cdot 90300 \]

8. **Calculate EA:**
- First calculate \( 2.303 \cdot 8.314 \approx 19.1 \)
- Then calculate \( 0.02119 \cdot 19.1 \approx 0.404 \)
- Finally, \( EA \approx 0.404 \cdot 90300 \approx 36465.2 \, \text{J/mol} \approx 36.5 \, \text{kJ/mol} \)

9. **Final Result:**
The activation energy (EA) of the reaction is approximately **36.65 kJ/mol**.