`ArarrB, DeltaH= -10 KJ mol^(-1), E_(a(f))=50 KJ mol^(-1)`, then `E_(a)` of `BrarrA` will be

Options

A. `40 KJ mol^(-1)`
B. `50 KJ mol^(-1)`
C. `-50 KJ mol^(-1)`
D. `60 KJ mol^(-1)`

Explanation

To solve the problem, we need to find the activation energy for the backward reaction \( Br \rightarrow A \) using the given data:

1. **Given Data:**
- \( \Delta H = -10 \, \text{kJ mol}^{-1} \) (indicating an exothermic reaction)
- \( E_a(f) = 50 \, \text{kJ mol}^{-1} \) (activation energy for the forward reaction \( A \rightarrow B \))

2. **Understanding the Relationship:**
The relationship between the activation energies and the enthalpy change is given by the equation:
\[ \Delta H = E_a(f) - E_a(b) \]
where:
- \( E_a(f) \) is the activation energy for the forward reaction.
- \( E_a(b) \) is the activation energy for the backward reaction.

3. **Rearranging the Equation:**
To find \( E_a(b) \), we can rearrange the equation:
\[ E_a(b) = E_a(f) - \Delta H \]

4. **Substituting the Values:**
Now, we can substitute the known values into the equation:
\[ E_a(b) = 50 \, \text{kJ mol}^{-1} - (-10 \, \text{kJ mol}^{-1}) \]
This simplifies to:
\[ E_a(b) = 50 \, \text{kJ mol}^{-1} + 10 \, \text{kJ mol}^{-1} \]
\[ E_a(b) = 60 \, \text{kJ mol}^{-1} \]

5. **Conclusion:**
Therefore, the activation energy for the backward reaction \( Br \rightarrow A \) is:
\[ E_a(b) = 60 \, \text{kJ mol}^{-1} \]