How many kcal of heat is evolved by the complete neutralisation of one mole sulphuric acid with `NaOH` ?

Options

A. `13.7 kcal`
B. `27.4 kcal`
C. `6.85 kcal`
D. none of these

Explanation

To solve the problem of how many kilocalories of heat are evolved by the complete neutralization of one mole of sulfuric acid (H₂SO₄) with sodium hydroxide (NaOH), we can follow these steps:

### Step 1: Write the neutralization reaction
The complete neutralization of sulfuric acid with sodium hydroxide can be represented by the following balanced chemical equation:

\[ \text{H}_2\text{SO}_4 + 2 \text{NaOH} \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O} \]

### Step 2: Identify the moles of H⁺ and OH⁻ involved
Sulfuric acid (H₂SO₄) is a diprotic acid, meaning it can donate two protons (H⁺ ions). Therefore, when one mole of H₂SO₄ is neutralized, it produces 2 moles of H⁺ ions:

- From H₂SO₄: 2 moles of H⁺
- From NaOH: 2 moles of OH⁻

### Step 3: Determine the heat evolved per mole of H⁺ and OH⁻
The standard enthalpy change (ΔH) for the neutralization of one mole of H⁺ with one mole of OH⁻ is approximately -13.7 kcal.

### Step 4: Calculate the total heat evolved for the complete neutralization
Since we have 2 moles of H⁺ reacting with 2 moles of OH⁻, the total heat evolved will be:

\[ \Delta H = 2 \times (-13.7 \text{ kcal}) = -27.4 \text{ kcal} \]

### Step 5: State the final answer
Thus, the total heat evolved by the complete neutralization of one mole of sulfuric acid with NaOH is **27.4 kcal**.

### Summary
The heat evolved during the complete neutralization of 1 mole of sulfuric acid with NaOH is **27.4 kcal**.

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