The rate constant for a second order reaction is given by `k=(5 times 10^11)e^(-29000k//T)`. The value of `E_a` will be:

Options

A. `275 kJ//mol e`
B. `250 kJ// mol e`
C. `241 kJ//mol e`
D. `225KJ//mol e`

Answer

To find the activation energy \( E_a \) for the given second-order reaction, we can use the Arrhenius equation, which relates the rate constant \( k \) to the temperature \( T \) and the activation energy \( E_a \):

\[ k = A e^{-\frac{E_a}{RT}} \]

Where:
- \( k \) is the rate constant,
- \( A \) is the pre-exponential factor,
- \( E_a \) is the activation energy,
- \( R \) is the universal gas constant (8.314 J/mol·K),
- \( T \) is the temperature in Kelvin.

### Step-by-Step Solution:

1. **Identify the given equation**:
We have the rate constant given as:
\[ k = (5 \times 10^{11}) e^{-\frac{29000}{T}} \]

2. **Compare with the Arrhenius equation**:
From the Arrhenius equation, we can see that:
\[ e^{-\frac{E_a}{RT}} = e^{-\frac{29000}{T}} \]
This implies:
\[ -\frac{E_a}{R} = -29000 \]

3. **Rearranging the equation**:
Multiply both sides by -1:
\[ \frac{E_a}{R} = 29000 \]

4. **Substituting the value of \( R \)**:
Now, substituting \( R = 8.314 \, \text{J/mol·K} \):
\[ E_a = 29000 \times R = 29000 \times 8.314 \]

5. **Calculating \( E_a \)**:
Performing the multiplication:
\[ E_a = 29000 \times 8.314 = 241106 \, \text{J/mol} \]

6. **Convert to kJ/mol**:
Since the question asks for \( E_a \) in kJ/mol, we convert:
\[ E_a = \frac{241106}{1000} = 241.106 \, \text{kJ/mol} \]

7. **Final Answer**:
Rounding to three significant figures, we get:
\[ E_a \approx 241 \, \text{kJ/mol} \]

### Conclusion:
The value of \( E_a \) is approximately **241 kJ/mol**.