The enthalpy of neutralization of which of the following acid & base is nearly `-13.6 kcal`.

Options

A. HCN and NaOH
B. `CH_(3)COOH` and `NH_(4)OH`
C. HCl and KOH
D. HCl and `NH_(4)OH`

Answer

To determine the enthalpy of neutralization of the given acid and base that is nearly -13.6 kcal, we can follow these steps:

### Step 1: Understand the concept of enthalpy of neutralization
The enthalpy of neutralization is the heat change that occurs when an acid and a base react to form one mole of water. For strong acids and strong bases, this value is typically around -13.6 kcal.

### Step 2: Identify the options provided
Let's assume the options given are:
A) HCl and NaOH
B) CH3COOH and NH4OH
C) HCl and KOH
D) H2SO4 and NaOH

### Step 3: Analyze each option
1. **Option A: HCl and NaOH**
- Both HCl (strong acid) and NaOH (strong base) are strong, and their reaction will yield a typical enthalpy of neutralization close to -13.6 kcal.

2. **Option B: CH3COOH and NH4OH**
- CH3COOH (acetic acid) is a weak acid and NH4OH (ammonium hydroxide) is a weak base. The enthalpy of neutralization for weak acids and bases is generally less negative than that of strong acids and bases.

3. **Option C: HCl and KOH**
- HCl (strong acid) and KOH (strong base) will also yield a typical enthalpy of neutralization close to -13.6 kcal.

4. **Option D: H2SO4 and NaOH**
- H2SO4 (strong acid) and NaOH (strong base) will also yield a typical enthalpy of neutralization close to -13.6 kcal.

### Step 4: Conclusion
Based on the analysis, the enthalpy of neutralization of strong acid and strong base combinations (like HCl and NaOH, HCl and KOH, H2SO4 and NaOH) will be around -13.6 kcal. However, the most straightforward answer is:

**Correct Answer: A) HCl and NaOH or C) HCl and KOH**