Heats of neutralisation and water ionisation:\nGiven the heat of neutralisation of strong acid–strong base (HCl + NaOH → NaCl + H2O) is −57.46 kJ per kg-mole of water formed, determine the heat of ionisation of water (H2O → H+ + OH−) in kJ per kg-mole.

Introduction / Context:
The heat of neutralisation for strong acids and bases at infinite dilution is nearly constant (about −57 kJ per mole of water formed) because the net reaction effectively is H+ (aq) + OH− (aq) → H2O (l). The reverse process, water ionisation, requires the same magnitude of energy input but with opposite sign. This relationship connects calorimetry with acid–base thermodynamics.

Given Data / Assumptions:

• Heat of neutralisation (strong acid/strong base): ΔH_neut ≈ −57.46 kJ per kg-mole H2O formed.
• Reaction considered for ionisation: H2O (l) → H+ (aq) + OH− (aq).
• Standard dilute conditions; heats are referenced per mole of water.

Concept / Approach:
Ionisation of water is the reverse of the neutralisation step. By Hess’s law, reversing a reaction changes the sign of its enthalpy change while preserving magnitude. Therefore, the heat of ionisation equals +57.46 kJ per kg-mole (endothermic), the energy required to dissociate water into its ions at the same reference conditions.

Step-by-Step Solution:

Verification / Alternative check:
Textbook values for the heat of neutralisation converge near −57 kJ/mol at infinite dilution; reversing the reaction consistently yields +57 kJ/mol for ionisation, aligning with the concept that creating separated ions from neutral water requires input energy.

Why Other Options Are Wrong:

• −57.46: sign error; that is neutralisation, not ionisation.
• 114.92 and −28.73: double or half the correct magnitude; no basis in the stoichiometry here.
• 0: contradicts the well-established thermochemistry.

Common Pitfalls:
Forgetting to reverse the sign when reversing the reaction; mixing per-mole and per-kg-mole conventions; ignoring dilution assumptions.

Final Answer:
57.46