Activation energy of the reaction `AtoB+38KCal` is 20KCal.
Then activation energy of reaction `BtoA` will be :-

Options

A. 20 KCal
B. `-20 KCal`
C. 18 KCal
D. 58KCal

Explanation

To solve the problem, we need to find the activation energy of the reverse reaction (B to A) given the activation energy of the forward reaction (A to B) and the enthalpy change (ΔH) of the reaction.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Activation energy for the forward reaction (A to B) = 20 kcal
- Enthalpy change (ΔH) for the reaction (A to B) = -38 kcal (since the reaction is exothermic)

2. **Understand the Relationship:**
- For a reaction, the relationship between the activation energies and the enthalpy change is given by the equation:
\[ \Delta H = E_a (\text{forward}) - E_a (\text{backward}) \]
- Here, \(E_a (\text{forward})\) is the activation energy for the forward reaction (A to B), and \(E_a (\text{backward})\) is the activation energy for the reverse reaction (B to A).

3. **Substitute the Known Values:**
- We can rearrange the equation to find the activation energy for the backward reaction:
\[ E_a (\text{backward}) = E_a (\text{forward}) - \Delta H \]
- Substituting the known values:
\[ E_a (\text{backward}) = 20 \text{ kcal} - (-38 \text{ kcal}) \]

4. **Calculate the Activation Energy for the Backward Reaction:**
- Simplifying the equation:
\[ E_a (\text{backward}) = 20 \text{ kcal} + 38 \text{ kcal} = 58 \text{ kcal} \]

5. **Conclusion:**
- The activation energy for the reaction B to A is 58 kcal.

### Final Answer:
The activation energy of the reaction B to A is **58 kcal**.

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