Enthalpy of neutralization of HCl by NaOH is `-55.84` kJ/mol and by `NH_(4)OH` is `51.34` kJ/mol. The enthalpy of ionization of `NH_(4)OH` is :

Options

A. 107.18 kJ/mol
B. 4.5 kJ/mol
C. `-4.5` kJ/mol
D. None of these

Explanation

To find the enthalpy of ionization of \( NH_4OH \), we can use the relationship between the enthalpy of neutralization, the enthalpy of ionization, and the enthalpy of the reaction.

### Step-by-Step Solution:

1. **Understand the Formula**:
The enthalpy of neutralization (\( \Delta H_{neutralization} \)) can be expressed as:
\[ \Delta H_{neutralization} = \Delta H_{ionization} + \Delta H_{reaction} \]

2. **Identify Given Values**:
- For \( HCl \) neutralized by \( NaOH \):
\[ \Delta H_{neutralization} = -55.84 \, \text{kJ/mol} \]
- For \( NH_4OH \) neutralized by \( OH^- \):
\[ \Delta H_{neutralization} = -51.34 \, \text{kJ/mol} \]

3. **Determine the Reaction Enthalpy**:
The enthalpy of reaction (\( \Delta H_{reaction} \)) for the neutralization of \( NH_4OH \) can be derived from the enthalpy of neutralization of \( HCl \) and \( NaOH \):
\[ \Delta H_{reaction} = -55.84 \, \text{kJ/mol} \]

4. **Set Up the Equation**:
For \( NH_4OH \):
\[ -51.34 = \Delta H_{ionization} + (-55.84) \]

5. **Rearranging the Equation**:
To find \( \Delta H_{ionization} \):
\[ \Delta H_{ionization} = -51.34 + 55.84 \]

6. **Calculate \( \Delta H_{ionization} \)**:
\[ \Delta H_{ionization} = 4.50 \, \text{kJ/mol} \]

### Final Answer:
The enthalpy of ionization of \( NH_4OH \) is \( 4.50 \, \text{kJ/mol} \).

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