The enthalpy of neutralisation of `NH_(4)OH` and `CH_(3)COOH` is -10.5 kcal `mol^(-1)` and enthalpy of neutralisation of `CH_(3)COOH` with strong base is -12.5 kcal `mol^(-1)`. The enthalpy of ionisation of `NH_(4)OH` will be

Options

A. 4.0 kcal `mol^(-1)`
B. 3.0 kcal `mol^(-1)`
C. 2.0 kcal `mol^(-1)`
D. 3.2 kcal `mol^(-1)`

Answer

To find the enthalpy of ionization of ammonium hydroxide (NH₄OH), we can use the enthalpy values given for the neutralization reactions. Here's the step-by-step solution:

### Step 1: Write the reactions
1. **Neutralization of NH₄OH with acetic acid (CH₃COOH)**:
\[ \text{NH}_4\text{OH} + \text{CH}_3\text{COOH} \rightarrow \text{NH}_4\text{CH}_3\text{COO} + \text{H}_2\text{O} \]
The enthalpy change for this reaction is given as:
\[ \Delta H_1 = -10.5 \text{ kcal mol}^{-1} \]

2. **Neutralization of CH₃COOH with a strong base (like NaOH)**:
\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]
The enthalpy change for this reaction is given as:
\[ \Delta H_2 = -12.5 \text{ kcal mol}^{-1} \]

### Step 2: Relate the enthalpy of ionization to the neutralization reactions
The enthalpy of ionization of NH₄OH can be related to the enthalpy changes of the two neutralization reactions:
\[ \Delta H_i = \Delta H_1 - \Delta H_2 \]
where:
- \(\Delta H_i\) is the enthalpy of ionization of NH₄OH,
- \(\Delta H_1\) is the enthalpy of neutralization of NH₄OH with acetic acid,
- \(\Delta H_2\) is the enthalpy of neutralization of acetic acid with a strong base.

### Step 3: Substitute the values
Now substitute the values into the equation:
\[ \Delta H_i = (-10.5 \text{ kcal mol}^{-1}) - (-12.5 \text{ kcal mol}^{-1}) \]
This simplifies to:
\[ \Delta H_i = -10.5 + 12.5 \]

### Step 4: Calculate the result
Now, perform the calculation:
\[ \Delta H_i = 2.0 \text{ kcal mol}^{-1} \]

### Final Answer
The enthalpy of ionization of NH₄OH is:
\[ \Delta H_i = 2.0 \text{ kcal mol}^{-1} \]