For complete neutralization of HCl with NaOH, the heat of neutralization is

Options

A. `+13.70 kJ mol^(-1)`
B. `-13.70 kJ mol^(-1)`
C. `-57.32 kJ mol^(-1)`
D. `-57.32 kJ mol^(-1)`

Explanation

To determine the heat of neutralization for the complete neutralization of hydrochloric acid (HCl) with sodium hydroxide (NaOH), we can follow these steps:

### Step 1: Understand the Reaction
The neutralization reaction between HCl (a strong acid) and NaOH (a strong base) can be represented as:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]

### Step 2: Identify the Type of Reaction
This reaction is an exothermic reaction, meaning it releases heat. The heat of neutralization is defined as the heat change when one equivalent of an acid reacts with one equivalent of a base.

### Step 3: Determine the Heat of Neutralization Value
For strong acids and strong bases, the heat of neutralization is typically around:
- **-13.7 kcal** (calories)
- **-57.6 kJ** (kilojoules)

### Step 4: Convert Units if Necessary
Since the question may ask for the answer in different units, it’s important to note that:
- 1 kcal = 4.184 kJ
Thus, -13.7 kcal can be converted to kJ:
\[ -13.7 \text{ kcal} \times 4.184 \text{ kJ/kcal} = -57.3 \text{ kJ} \]

### Step 5: Final Answer
The heat of neutralization for the complete neutralization of HCl with NaOH is:
- **-57.6 kJ per mole** (or -13.7 kcal).

### Conclusion
Both values are acceptable depending on the unit used. However, the more commonly accepted value in kJ is:
- **-57.3 kJ per mole**.