For an exothermic reaction `ArarrB`, the activation energy is `65kJmol^(-1)` and enthalpy of reaction is `42kJmol^(-1)`. The activation energy for the reaction `BrarrA` will be:

Options

A. `23kJmol^(-1)`
B. `107kJmol^(-1)`
C. `65kJmol^(-1)`
D. `42kJmol^(-1)`

Explanation

To solve the problem, we need to determine the activation energy for the reverse reaction \( B \rightarrow A \) given the activation energy for the forward reaction \( A \rightarrow B \) and the enthalpy change of the reaction.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Activation energy for the forward reaction (\( E_a \) forward): \( 65 \, \text{kJ/mol} \)
- Enthalpy change of the reaction (\( \Delta H \)): \( 42 \, \text{kJ/mol} \)

2. **Understand the Relationship**:
For an exothermic reaction, the relationship between the activation energies and the enthalpy change can be expressed as:
\[ E_a \text{ (backward)} = E_a \text{ (forward)} + \Delta H \]
This equation states that the activation energy for the reverse reaction is equal to the activation energy for the forward reaction plus the enthalpy change.

3. **Substitute the Values**:
Now, substitute the known values into the equation:
\[ E_a \text{ (backward)} = 65 \, \text{kJ/mol} + 42 \, \text{kJ/mol} \]

4. **Calculate the Activation Energy for the Backward Reaction**:
\[ E_a \text{ (backward)} = 107 \, \text{kJ/mol} \]

5. **Final Answer**:
The activation energy for the reaction \( B \rightarrow A \) is \( 107 \, \text{kJ/mol} \).