Heat of neutralization between HCl and NaOH is `-13.7 kcal "equiv"^(–1)`. Heat of neutralization of `H_2C_2O_4` (oxalic acid) with NaOH is`-26 kcal mol^(-1)`. Hence, heat of dissociation of `H_(2)C_(2)O_(4)` as `H_2C_2O_(4) rarr 2H^(+) + C_(2)O_(4)^(2-)`, is :

Options

A. `12.3 kcal "mole"^(-1)`
B. `1.4 kcals "mole"^(-1)`
C. `-39.7 kcals "mole"^(-1)`
D. `-12.3 kcals "mole"^(-1)`

Explanation

To find the heat of dissociation of oxalic acid (H₂C₂O₄) into its ions, we will use the given heats of neutralization for both HCl with NaOH and H₂C₂O₄ with NaOH.

### Step-by-Step Solution:

1. **Understanding the Reactions**:
- The neutralization reaction between HCl (a strong acid) and NaOH (a strong base) can be represented as:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
The heat of neutralization for this reaction is given as \(-13.7 \, \text{kcal/equiv}\).

- The neutralization reaction between oxalic acid and NaOH can be represented as:
\[ \text{H}_2\text{C}_2\text{O}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2 \text{H}_2\text{O} \]
The heat of neutralization for this reaction is given as \(-26 \, \text{kcal/mol}\).

2. **Relating Heat of Dissociation and Heat of Neutralization**:
- The dissociation of oxalic acid can be represented as:
\[ \text{H}_2\text{C}_2\text{O}_4 \rightarrow 2 \text{H}^+ + \text{C}_2\text{O}_4^{2-} \]
- For the dissociation of a weak acid, we can relate the heat of dissociation (\( \Delta H_{\text{diss}} \)) to the heat of neutralization (\( \Delta H_{\text{neut}} \)) as follows:
\[ \Delta H_{\text{neut}} = \Delta H_{\text{diss}} + \Delta H_{\text{neutralization of strong acid}} \]

3. **Calculating the Heat of Neutralization**:
- Since the heat of neutralization for one equivalent of a strong acid and a strong base is \(-13.7 \, \text{kcal}\), and since two moles of water are produced in the reaction with oxalic acid, we need to account for this:
\[ \Delta H_{\text{neutralization}} = 2 \times (-13.7 \, \text{kcal}) = -27.4 \, \text{kcal} \]

4. **Setting Up the Equation**:
- Now we can set up the equation for the heat of dissociation:
\[ -26 \, \text{kcal} = \Delta H_{\text{diss}} + (-27.4 \, \text{kcal}) \]

5. **Solving for Heat of Dissociation**:
- Rearranging the equation gives:
\[ \Delta H_{\text{diss}} = -26 \, \text{kcal} + 27.4 \, \text{kcal} \]
\[ \Delta H_{\text{diss}} = 1.4 \, \text{kcal} \]

### Final Answer:
The heat of dissociation of oxalic acid (H₂C₂O₄) is \(1.4 \, \text{kcal/mol}\).