Which of the following reactions will have the value of enthalpy of neutralisation as `-57.1"kJ mol"^(-1)` ?

Options

A. `CH_(3)COOH+NaOH rarr CH_(3)COONa+H_(2)O`
B. `HCl+NaOH rarr NaCl +H_(2)O`
C. `HCl+NH_(4)OHrarr NH_(4)Cl+H_(2)O`
D. `HCOOH+NaOH rarr HCOONa+H_(2)O`

Explanation

To determine which reaction has an enthalpy of neutralization of -57.1 kJ mol⁻¹, we need to identify the reactions involving strong acids and strong bases, as this value is typical for such reactions.

### Step-by-Step Solution:

1. **Understand Enthalpy of Neutralization**: The enthalpy of neutralization is the heat change that occurs when an acid and a base react to form water and a salt. For strong acids and strong bases, this value is typically around -57.1 kJ/mol.

2. **Identify the Options**: We need to evaluate the provided options to see which ones involve strong acids and strong bases. The reactions typically considered are:
- HCl (strong acid) + NaOH (strong base)
- CH₃COOH (weak acid) + NaOH (strong base)
- HCl (strong acid) + NH₄OH (weak base)
- HCOOH (weak acid) + NaOH (strong base)

3. **Classify the Acids and Bases**:
- **HCl**: Strong acid
- **NaOH**: Strong base
- **CH₃COOH**: Weak acid
- **NH₄OH**: Weak base
- **HCOOH**: Weak acid

4. **Evaluate Each Reaction**:
- **CH₃COOH + NaOH**: This reaction involves a weak acid and a strong base, so it will not have the standard enthalpy of neutralization.
- **HCl + NaOH**: This reaction involves a strong acid and a strong base, which typically results in an enthalpy of neutralization around -57.1 kJ/mol.
- **HCl + NH₄OH**: This reaction involves a strong acid and a weak base, so it will not have the standard enthalpy of neutralization.
- **HCOOH + NaOH**: This reaction involves a weak acid and a strong base, so it will not have the standard enthalpy of neutralization.

5. **Conclusion**: The only reaction that fits the criteria for having an enthalpy of neutralization of -57.1 kJ/mol is:
- **HCl + NaOH**

### Final Answer:
The reaction that will have the value of enthalpy of neutralization as -57.1 kJ mol⁻¹ is **HCl + NaOH**.