For strong acid strong base neutralisation energy for 1 mole `H_(2)O` formation is `-57.1 kJ`.If `0.25` mole of strong monoprotic acid is reacted with `0.5` mole of strong base then enthalpy of neutralisation is

Options

A. `-(0.25 xx57.1)`
B. `0.5 xx57.1`
C. `57.1`
D. `-(0.5 xx57.1)`

Answer

To solve the problem, we need to calculate the enthalpy of neutralization when 0.25 moles of a strong monoprotic acid react with 0.5 moles of a strong base. The enthalpy change for the formation of 1 mole of water during this neutralization is given as -57.1 kJ.

### Step-by-step solution:

1. **Identify the limiting reagent**:
- We have 0.25 moles of strong acid and 0.5 moles of strong base.
- Since the acid is monoprotic, it will react with the base in a 1:1 molar ratio.
- Therefore, 0.25 moles of acid will react with 0.25 moles of base.

2. **Determine the amount of water produced**:
- The reaction will produce water according to the limiting reagent, which in this case is the strong acid (0.25 moles).
- Thus, the amount of water produced will be 0.25 moles.

3. **Calculate the enthalpy of neutralization**:
- The enthalpy change for the formation of 1 mole of water is -57.1 kJ.
- For 0.25 moles of water, the enthalpy change will be:
\[ \Delta H_{\text{neutralization}} = 0.25 \, \text{moles} \times (-57.1 \, \text{kJ/mole}) = -14.275 \, \text{kJ} \]

4. **Final answer**:
- The enthalpy of neutralization for the reaction is -14.275 kJ.

### Summary:
The enthalpy of neutralization when 0.25 moles of a strong monoprotic acid react with 0.5 moles of a strong base is -14.275 kJ.