Heat of dissociation of acetic acidis 0.30 kcal`mol^(-1)` . Hence,enthalpy change when 1 mol of`Ca(OH)_(2)` is completely neutralized by acetic acid would be

Options

A. `-13.4 kcal`
B. `-27.1 kcal`
C. `-26.8 kcal`
D. `-27.4 kcal`

Explanation

To solve the problem, we need to determine the enthalpy change when 1 mole of calcium hydroxide (Ca(OH)₂) is completely neutralized by acetic acid (CH₃COOH). We are given the heat of dissociation of acetic acid as 0.30 kcal/mol.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The neutralization reaction between acetic acid and calcium hydroxide can be represented as:
\[ \text{Ca(OH)}_2 + 2 \text{CH}_3\text{COOH} \rightarrow \text{Ca(CH}_3\text{COO)}_2 + 2 \text{H}_2\text{O} \]
Here, 1 mole of Ca(OH)₂ reacts with 2 moles of acetic acid.

2. **Determine the Number of Moles of H⁺ Ions**:
Each mole of acetic acid can provide 1 mole of H⁺ ions. Since we need to neutralize 2 moles of OH⁻ ions from 1 mole of Ca(OH)₂, we will need 2 moles of acetic acid, which will provide 2 moles of H⁺ ions.

3. **Calculate the Heat of Dissociation for Acetic Acid**:
The heat of dissociation of acetic acid is given as 0.30 kcal/mol. Since we need 2 moles of acetic acid:
\[ \text{Total heat of dissociation} = 2 \times 0.30 \text{ kcal} = 0.60 \text{ kcal} \]

4. **Calculate the Total Heat Produced**:
The enthalpy change for the neutralization reaction can be calculated by considering the heat produced during the reaction and subtracting the heat absorbed during the dissociation of acetic acid.
- The heat produced when 1 mole of Ca(OH)₂ is neutralized by acetic acid is approximately 27.4 kcal (as per the video).
- Therefore, the heat produced after accounting for the dissociation of acetic acid is:
\[ \text{Heat produced} = 27.4 \text{ kcal} - 0.60 \text{ kcal} = 26.8 \text{ kcal} \]

5. **Determine the Enthalpy Change**:
Since heat is released during the reaction, the enthalpy change (ΔH) will be negative:
\[ \Delta H = -26.8 \text{ kcal} \]

### Final Answer:
The enthalpy change when 1 mole of Ca(OH)₂ is completely neutralized by acetic acid is:
\[ \Delta H = -26.8 \text{ kcal} \]