The heat of neutralisation for strong acid and strong base forming 2 moles of water is

Options

A. `-2 xx57.1 kJ`
B. `-57.1 kJ`
C. `-57.1/2 kJ`
D. Strong acid and strong base will not undergo neutralisation

Answer

To solve the question regarding the heat of neutralization for a strong acid and a strong base forming 2 moles of water, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Heat of Neutralization**:
The heat of neutralization is defined as the heat change when one mole of hydrogen ions (H⁺) reacts with one mole of hydroxide ions (OH⁻) to form one mole of water (H₂O). For strong acids and strong bases, this value is typically constant.

2. **Value of Heat of Neutralization**:
The standard heat of neutralization for the reaction between a strong acid (like HCl) and a strong base (like NaOH) is approximately -57.1 kJ/mol. This means that for every mole of water formed, 57.1 kJ of heat is released.

3. **Calculating for Two Moles of Water**:
If the reaction produces 2 moles of water, we need to consider the heat released for both moles. Since the heat of neutralization is additive, we can calculate the total heat released by multiplying the heat of neutralization for one mole of water by the number of moles formed:
\[ \text{Total Heat} = 2 \times (-57.1 \text{ kJ}) = -114.2 \text{ kJ} \]

4. **Conclusion**:
Therefore, the heat of neutralization for a strong acid and a strong base forming 2 moles of water is -114.2 kJ.

### Final Answer:
The heat of neutralization for strong acid and strong base forming 2 moles of water is -114.2 kJ.

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