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 find the enthalpy change (ΔH) for the reaction:

\[ H_2C_2O_4 \rightarrow C_2O_4^{2-} + 2H^+ \]

given that the heat of neutralization of oxalic acid with NaOH is -106.7 kJ/mol, we can follow these steps:

### Step 1: Understand the Neutralization Reaction
The neutralization reaction between oxalic acid (H₂C₂O₄) and sodium hydroxide (NaOH) can be represented as:

\[ H_2C_2O_4 + NaOH \rightarrow C_2O_4^{2-} + H_2O + Na^+ \]

In this reaction, the hydroxide ions (OH⁻) from NaOH react with the protons (H⁺) from oxalic acid to form water (H₂O).

### Step 2: Write the Enthalpy Change for the Neutralization
The heat of neutralization for the reaction is given as:

\[ \Delta H_{neutralization} = -106.7 \, \text{kJ/mol} \]

This value represents the energy change when one mole of oxalic acid reacts with NaOH to produce one mole of water and the oxalate ion.

### Step 3: Consider the Formation of Water
From the neutralization reaction, we know that:

\[ H^+ + OH^- \rightarrow H_2O \]

The enthalpy change (ΔH) for the formation of water from H⁺ and OH⁻ is:

\[ \Delta H_{formation} = -57.1 \, \text{kJ/mol} \]

### Step 4: Set Up the Equations
We can set up the following equations:

1. For the neutralization reaction:
\[ H_2C_2O_4 + NaOH \rightarrow C_2O_4^{2-} + H_2O \]
\[ \Delta H_2 = -106.7 \, \text{kJ/mol} \]

2. For the formation of water:
\[ H^+ + OH^- \rightarrow H_2O \]
\[ \Delta H_1 = -57.1 \, \text{kJ/mol} \]

### Step 5: Relate the Reactions
To find the ΔH for the reaction:

\[ H_2C_2O_4 \rightarrow C_2O_4^{2-} + 2H^+ \]

We can manipulate the equations. We can express the desired reaction in terms of the neutralization and water formation reactions.

### Step 6: Calculate the Enthalpy Change
Using Hess's law, we can combine the enthalpy changes:

\[ \Delta H_3 = \Delta H_2 - 2 \times \Delta H_1 \]

Substituting the values:

\[ \Delta H_3 = -106.7 \, \text{kJ/mol} - 2 \times (-57.1 \, \text{kJ/mol}) \]

Calculating this gives:

\[ \Delta H_3 = -106.7 + 114.2 = 7.5 \, \text{kJ/mol} \]

### Final Answer
Thus, the enthalpy change (ΔH) for the reaction:

\[ H_2C_2O_4 \rightarrow C_2O_4^{2-} + 2H^+ \]

is:

\[ \Delta H = 7.5 \, \text{kJ/mol} \]