If heat of dissociation of `CHCl_2COOH` is 0.7kcal /mole , the , `DeltaH` for the reaction `CHCl_2COOH+KOHrarrCHCl_2COOH " is " +H_2O`

Options

A. `-13 kcal`
B. `+13kcal`
C. `-14.4kcal`
D. `-13.7 kcal`

Answer

To solve the problem, we need to determine the enthalpy change (ΔH) for the reaction involving the weak acid CHCl₂COOH and the strong base KOH.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Heat of dissociation of CHCl₂COOH = 0.7 kcal/mole.
- The reaction is:
\[ \text{CHCl}_2\text{COOH} + \text{KOH} \rightarrow \text{CHCl}_2\text{COO}^- + \text{H}_2\text{O} \]

2. **Understand the Nature of the Reactants**:
- CHCl₂COOH is a weak acid.
- KOH is a strong base.

3. **Use the Known Heat of Reaction for Strong Acid and Weak Base**:
- When a weak acid reacts with a strong base, the heat evolved is generally around -57.7 kJ/mole.
- Converting this to kcal:
\[ -57.7 \text{ kJ/mole} \times \frac{1 \text{ kcal}}{4.184 \text{ kJ}} \approx -13.8 \text{ kcal/mole} \]

4. **Calculate the Total ΔH for the Reaction**:
- The ΔH for the reaction can be calculated using the formula:
\[ \Delta H = \text{Heat of reaction} + \text{Heat of dissociation} \]
- Substituting the values:
\[ \Delta H = -13.8 \text{ kcal/mole} + 0.7 \text{ kcal/mole} = -13.1 \text{ kcal/mole} \]

5. **Final Result**:
- The ΔH for the reaction CHCl₂COOH + KOH → CHCl₂COO⁻ + H₂O is approximately:
\[ \Delta H \approx -13 \text{ kcal/mole} \]

### Conclusion:
Thus, the final answer is:
\[ \Delta H = -13 \text{ kcal/mole} \]