Heat of neutralisation of oxalic acid is -106.7 KJ `mol^(-1)` using NaOH hence `Delta H` of : `H_(2)C_(2)O_(4)rarr C_(2)O_(4)^(2-)+2H^(+)` is :-

Options

A. 5.88 KJ
B. `-5.88` KJ
C. `-13.7` K cal
D. 7.5 KJ

Explanation

To solve the problem, we need to determine the enthalpy change (ΔH) for the reaction of oxalic acid (H₂C₂O₄) dissociating into oxalate ions (C₂O₄²⁻) and protons (H⁺) in an acidic medium. We are given the heat of neutralization of oxalic acid with sodium hydroxide (NaOH) as -106.7 kJ/mol.

### Step-by-Step Solution:

1. **Identify the Reactions:**
- The heat of neutralization reaction can be represented as:
\[ H_2C_2O_4 + 2NaOH \rightarrow C_2O_4^{2-} + 2H_2O + 2Na^+ \]
- The heat of neutralization (ΔH) for this reaction is given as -106.7 kJ/mol.

2. **Write the Formation of Water:**
- The formation of water from hydroxide ions (OH⁻) and protons (H⁺) can be represented as:
\[ H^+ + OH^- \rightarrow H_2O \]
- The enthalpy change for this reaction (ΔH₁) is approximately -57.1 kJ/mol.

3. **Set Up the Equations:**
- Let’s denote the enthalpy change for the dissociation of oxalic acid as ΔH₂:
\[ H_2C_2O_4 \rightarrow C_2O_4^{2-} + 2H^+ \]
- The enthalpy change for this reaction is what we want to find.

4. **Relate the Enthalpy Changes:**
- The overall reaction can be divided into two parts:
- The neutralization reaction (ΔH₂ = -106.7 kJ/mol)
- The formation of water (ΔH₁ = -57.1 kJ/mol)
- Since we need to find ΔH for the dissociation of oxalic acid, we can express it as:
\[ ΔH_{\text{reaction}} = ΔH_2 + 2ΔH_1 \]
- Rearranging gives:
\[ ΔH_2 = ΔH_{\text{reaction}} - 2ΔH_1 \]

5. **Substitute Values:**
- Substitute the known values into the equation:
\[ ΔH_2 = -106.7 \, \text{kJ/mol} - 2(-57.1 \, \text{kJ/mol}) \]
- Calculate:
\[ ΔH_2 = -106.7 + 114.2 = 7.5 \, \text{kJ/mol} \]

6. **Conclusion:**
- The enthalpy change (ΔH) for the dissociation of oxalic acid into oxalate ions and protons is:
\[ ΔH = 7.5 \, \text{kJ/mol} \]

### Final Answer:
The ΔH for the reaction \( H_2C_2O_4 \rightarrow C_2O_4^{2-} + 2H^+ \) is **7.5 kJ/mol**.