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 NH₄OH, we can use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps of the reaction, regardless of the pathway taken.

### Step-by-Step Solution:

1. **Identify the Reactions**:
- The neutralization of HCl by NaOH can be represented as:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \quad \Delta H = -55.84 \, \text{kJ/mol} \]
- The neutralization of NH₄OH by HCl can be represented as:
\[ \text{NH}_4\text{OH} + \text{HCl} \rightarrow \text{NH}_4\text{Cl} + \text{H}_2\text{O} \quad \Delta H = -51.34 \, \text{kJ/mol} \]

2. **Apply Hess's Law**:
- According to Hess's law, we can express the enthalpy change of the reaction involving NH₄OH in terms of the enthalpy changes of the reactions we have:
\[ \Delta H_{\text{neutralization of NH}_4\text{OH}} = \Delta H_{\text{ionization of NH}_4\text{OH}} + \Delta H_{\text{neutralization of HCl by NaOH}} \]
- Rearranging gives us:
\[ \Delta H_{\text{ionization of NH}_4\text{OH}} = \Delta H_{\text{neutralization of NH}_4\text{OH}} - \Delta H_{\text{neutralization of HCl by NaOH}} \]

3. **Substitute the Values**:
- Now substitute the known values:
\[ \Delta H_{\text{ionization of NH}_4\text{OH}} = -51.34 \, \text{kJ/mol} - (-55.84 \, \text{kJ/mol}) \]
- This simplifies to:
\[ \Delta H_{\text{ionization of NH}_4\text{OH}} = -51.34 + 55.84 \]

4. **Calculate the Result**:
- Performing the calculation:
\[ \Delta H_{\text{ionization of NH}_4\text{OH}} = 4.50 \, \text{kJ/mol} \]

### Final Answer:
The enthalpy of ionization of NH₄OH is **4.50 kJ/mol**.

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