Heat of neutralisation of a strong dibasic acid in dilute solution by NaOH is nearly :

Options

A. `-27.4` kcal/eq
B. 13.7 kcal/mol
C. `-13.7`kcal/eq
D. `-13.7` kcal/mol

Answer

To solve the question regarding the heat of neutralization of a strong dibasic acid in dilute solution by NaOH, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Components**:
- We have a strong dibasic acid (which can donate two protons, H⁺) and NaOH (a strong base that provides OH⁻ ions).

2. **Determine the n-factor**:
- The n-factor of NaOH is 1 because it can donate one OH⁻ ion.
- The n-factor of a dibasic acid is 2 because it can donate two H⁺ ions.

3. **Understanding Neutralization**:
- In the neutralization reaction, one equivalent of the dibasic acid will react with two equivalents of NaOH since it can donate two protons.
- The reaction can be represented as:
\[ \text{Dibasic Acid} + 2 \text{NaOH} \rightarrow \text{Salt} + 2 \text{H}_2\text{O} \]

4. **Heat of Formation of Water**:
- The heat of formation of water (ΔH) during the neutralization reaction is approximately -13.7 kJ/mol. This value represents the energy released when one mole of water is formed from the reaction of H⁺ and OH⁻.

5. **Calculate Heat of Neutralization**:
- Since one mole of dibasic acid reacts with two moles of NaOH, the heat of neutralization for the reaction is still -13.7 kJ, as the heat of neutralization is defined per mole of water formed.

6. **Final Answer**:
- Therefore, the heat of neutralization of a strong dibasic acid in dilute solution by NaOH is approximately -13.7 kJ per equivalent.

### Conclusion:
The correct answer is -13.7 kJ per equivalent.