Enthalpy of Neutralization

Formula / Core Reaction:

$$\text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l) \qquad \Delta H^\circ_\text{neut} = -57.3 \text{ kJ eq}^{-1} \text{ or } -13.7 \text{ kcal eq}^{-1}$$

Concept: Enthalpy of neutralization for any strong acid and strong base is constant because both are completely ionized, reducing the net ionic process simply to the combination of $$\text{H}^+$$ and $$\text{OH}^-$$ ions to form liquid water.

Notation / Key Observation:

• $$\Delta H_\text{neut}$$: Enthalpy change per gram-equivalent of water formed ($$-57.3 \text{ kJ eq}^{-1}$$);
• Note: For weak acids or weak bases, $$\vert{}\Delta H_\text{neut}\vert{} < 57.3 \text{ kJ eq}^{-1}$$ due to energy consumed in complete ionization.