Halving enthalpy for half the water produced Steps:

• Reaction 1 uses 2 mol NaOH to form 2 mol H₂O with ΔH = -114 kJ mol⁻¹.
• Reaction 2 uses 1 mol NaOH to form 1 mol H₂O.
• Strong acid-strong base neutralizations have constant ΔH ≈ -57 kJ per mol H₂O.
• Divide reaction 1's ΔH by 2: -114 kJ / 2 = -57 kJ mol⁻¹ for reaction 2. Why A is correct:
• Matches standard neutralization enthalpy for 1 mol H⁺ + 1 mol OH⁻ → 1 mol H₂O. Why the others are wrong:
• B: -76 kJ mol⁻¹ lacks basis in stoichiometry or standard values.
• C: -114 kJ mol⁻¹ applies to 2 mol H₂O in reaction 1, not 1 mol.
• D: -228 kJ mol⁻¹ doubles reaction 1 incorrectly for 4 mol H₂O.

Final answer: A