The heat released in neutralisation of HCI and NaOH is 13.7 kcal/mol, the heat released on neutralisation of NaOH with `CH_(3)COOH` is 3.7 kcal/mol. The `Delta H^(@)` of ionsiation of `CH_(3)COOH` is

Options

A. 10.2 kcal
B. 10 kcal
C. 3.7 kcal
D. 9.5 kcal

Answer

To find the ΔH° (enthalpy change) of ionization of acetic acid (CH₃COOH), we can use the information provided about the heat released during the neutralization reactions.

### Step-by-step Solution:

1. **Understand the Heat of Neutralization**:
- The heat released during the neutralization of HCl with NaOH is given as 13.7 kcal/mol.
- The heat released during the neutralization of NaOH with acetic acid (CH₃COOH) is given as 3.7 kcal/mol.

2. **Identify the Components**:
- The total heat released in the neutralization of a strong acid (HCl) with a strong base (NaOH) is considered to be the maximum heat of neutralization, which is 13.7 kcal/mol.
- The heat released when NaOH neutralizes acetic acid is less (3.7 kcal/mol) because acetic acid is a weak acid and does not fully ionize in solution.

3. **Calculate the ΔH° of Ionization**:
- The difference in heat released between the neutralization of HCl and NaOH and the neutralization of NaOH with acetic acid represents the energy required for the ionization of acetic acid.
- Therefore, we can calculate the ΔH° of ionization of acetic acid as follows:
\[ \Delta H^\circ_{\text{ionization}} = \Delta H^\circ_{\text{neutralization (HCl + NaOH)}} - \Delta H^\circ_{\text{neutralization (NaOH + CH₃COOH)}} \]
\[ \Delta H^\circ_{\text{ionization}} = 13.7 \, \text{kcal/mol} - 3.7 \, \text{kcal/mol} \]
\[ \Delta H^\circ_{\text{ionization}} = 10 \, \text{kcal/mol} \]

4. **Conclusion**:
- The ΔH° of ionization of acetic acid (CH₃COOH) is 10 kcal/mol.

### Final Answer:
The ΔH° of ionization of acetic acid is **10 kcal/mol**.