The rate constant for a first order reaction becomes six times when the temperature is raised from 350 K to 400 K. Calculate the activation energy for the reaction
`[R=8.314JK^(-1)mol^(-1)]`

Options

A. `4.17 kJmol^(-1)`
B. `41.7 kJmol^(-1)`
C. `417.0 kJmol^(-1)`
D. `4170 kJmol^(-1)`

Answer

To solve the problem, we will use the Arrhenius equation, which relates the rate constants of a reaction at two different temperatures to the activation energy (Ea) of the reaction.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Initial temperature, \( T_1 = 350 \, K \)
- Final temperature, \( T_2 = 400 \, K \)
- The rate constant at \( T_2 \) is six times that at \( T_1 \): \( K_2 = 6 K_1 \)
- Gas constant, \( R = 8.314 \, J \, K^{-1} \, mol^{-1} \)

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

3. **Substituting Known Values:**
Substitute \( K_2 = 6 K_1 \) into the equation:
\[ \log(6) = \frac{E_a}{2.303 \times 8.314} \left( \frac{1}{350} - \frac{1}{400} \right) \]

4. **Calculate \( \log(6) \):**
\[ \log(6) \approx 0.778 \]

5. **Calculate the Temperature Difference:**
First, calculate \( \frac{1}{350} - \frac{1}{400} \):
\[ \frac{1}{350} = 0.002857 \quad \text{and} \quad \frac{1}{400} = 0.0025 \]
\[ \frac{1}{350} - \frac{1}{400} = 0.002857 - 0.0025 = 0.000357 \]

6. **Substituting Back into the Equation:**
Now substitute everything back into the equation:
\[ 0.778 = \frac{E_a}{2.303 \times 8.314} \times 0.000357 \]

7. **Calculate \( 2.303 \times 8.314 \):**
\[ 2.303 \times 8.314 \approx 19.1 \]

8. **Rearranging to Find \( E_a \):**
Rearranging gives:
\[ E_a = \frac{0.778 \times 19.1}{0.000357} \]

9. **Calculate \( E_a \):**
\[ E_a \approx \frac{14.85}{0.000357} \approx 41680.7 \, J/mol \approx 41.68 \, kJ/mol \]

10. **Final Answer:**
The activation energy \( E_a \) for the reaction is approximately:
\[ E_a \approx 41.7 \, kJ/mol \]