Find the enthalpy of dissociation of `H_2C_2O_4` from the given data: Enthalpy of neutralisation of strong acid and strong base is `-13.7 kcal mol^(-1)` and that of oxalic acid by a strong base is `-25 kcal mol^(-1)`

Options

A. `-11.3 kcal mol^(-1)`
B. `2.4 kcal mol^(-1)`
C. `1.2 kcal mol^(-1)`
D. `11.7 kcal mol^(-1)`

Answer

To find the enthalpy of dissociation of oxalic acid (H₂C₂O₄), we can use the given data about the enthalpy of neutralization of a strong acid with a strong base and the enthalpy of neutralization of oxalic acid with a strong base.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Enthalpy of neutralization of a strong acid with a strong base (ΔH_neutralization, strong acid) = -13.7 kcal/mol
- Enthalpy of neutralization of oxalic acid (H₂C₂O₄) with a strong base (ΔH_neutralization, oxalic acid) = -25 kcal/mol

2. **Understand the Reaction:**
- When a strong acid reacts with a strong base, the enthalpy change is the heat released during the neutralization.
- For oxalic acid, which is a weak acid, the neutralization with a strong base involves the dissociation of the acid.

3. **Write the Equation:**
- The enthalpy of neutralization of a weak acid (like oxalic acid) can be expressed as:
\[ \Delta H_{\text{neutralization, weak acid}} = \Delta H_{\text{neutralization, strong acid}} + \Delta H_{\text{dissociation}} \]
- Here, we denote the basicity of oxalic acid (H₂C₂O₄) as 2 because it can donate two protons (H⁺ ions).

4. **Substituting the Values:**
- Using the equation:
\[ -25 \text{ kcal/mol} = -13.7 \text{ kcal/mol} + \Delta H_{\text{dissociation}} \]

5. **Rearranging to Find ΔH_dissociation:**
- Rearranging the equation gives:
\[ \Delta H_{\text{dissociation}} = -25 \text{ kcal/mol} + 13.7 \text{ kcal/mol} \]
- Calculating this gives:
\[ \Delta H_{\text{dissociation}} = -25 + 13.7 = -11.3 \text{ kcal/mol} \]

6. **Considering the Basicity:**
- Since oxalic acid has a basicity of 2, we need to account for this in our final calculation. The enthalpy of dissociation per mole of H⁺ released is:
\[ \Delta H_{\text{dissociation}} = \frac{-11.3 \text{ kcal/mol}}{2} = -5.65 \text{ kcal/mol} \]

### Final Answer:
The enthalpy of dissociation of H₂C₂O₄ is approximately **-5.65 kcal/mol**.