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 solve the problem, we need to determine the heat of dissociation of oxalic acid (H₂C₂O₄) based on the given heats of neutralization with NaOH.

### Step-by-step Solution:

1. **Identify the Given Data:**
- Heat of neutralization between HCl and NaOH: \( \Delta H_{neutralization} = -13.7 \, \text{kcal/equiv} \)
- Heat of neutralization of H₂C₂O₄ with NaOH: \( \Delta H_{neutralization} = -26 \, \text{kcal/mol} \)

2. **Understand the Reaction:**
- The reaction of H₂C₂O₄ with NaOH can be represented as:
\[ H_2C_2O_4 + 2 \, NaOH \rightarrow Na_2C_2O_4 + 2 \, H_2O \]
- This indicates that 1 mole of H₂C₂O₄ reacts with 2 moles of NaOH to produce 2 moles of water.

3. **Relate Heat of Neutralization and Heat of Dissociation:**
- The heat of neutralization for the reaction of H₂C₂O₄ with NaOH can be expressed as:
\[ \Delta H_{reaction} = \Delta H_{dissociation} + \Delta H_{neutralization} \]
- Here, \( \Delta H_{dissociation} \) is the heat required for the dissociation of H₂C₂O₄ into its ions, and \( \Delta H_{neutralization} \) is the heat released during the neutralization process.

4. **Calculate the Heat of Neutralization for the Reaction:**
- Since 2 moles of NaOH are used, the heat of neutralization for this reaction would be:
\[ \Delta H_{neutralization} = 2 \times (-13.7 \, \text{kcal/equiv}) = -27.4 \, \text{kcal/mol} \]

5. **Set Up the Equation:**
- Now we can set up the equation using the values we have:
\[ -26 \, \text{kcal/mol} = \Delta H_{dissociation} + (-27.4 \, \text{kcal/mol}) \]

6. **Solve for Heat of Dissociation:**
- Rearranging the equation gives:
\[ \Delta H_{dissociation} = -26 \, \text{kcal/mol} + 27.4 \, \text{kcal/mol} \]
\[ \Delta H_{dissociation} = 1.4 \, \text{kcal/mol} \]

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

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