Enthalpy of neutralization

In chemistry and thermodynamics, the enthalpy of neutralization (Template:Math) is the change in enthalpy that occurs when one equivalent of an acid and a base undergo a neutralization reaction to form water and a salt. It is a special case of the enthalpy of reaction. It is defined as the energy released with the formation of 1 mole of water. When a reaction is carried out under standard conditions at the temperature of 298 K (25 degrees Celsius) and 1 atm of pressure and one mole of water is formed, the heat released by the reaction is called the standard enthalpy of neutralization (Template:Math).

The heat (Template:Mvar) released during a reaction is

$Q=m{c}_p\mathrm{\Delta }T$

where Template:Mvar is the mass of the solution, Template:Mvar is the specific heat capacity of the solution, and Template:Math is the temperature change observed during the reaction. From this, the standard enthalpy change (Template:Math) is obtained by division with the amount of substance (in moles) involved.

$\mathrm{\Delta }H=-\frac{Q}{n}$

When a strong acid, HA, reacts with a strong base, BOH, the reaction that occurs is

$\mathrm{H}{\phantom{A}}^{+}+\mathrm{O}\mathrm{H}{\phantom{A}}^{-}⟶\mathrm{H}{\phantom{A}}_2\mathrm{O}$

as the acid and the base are fully dissociated and neither the cation Template:Chem2 nor the anion Template:Chem2 are involved in the neutralization reaction. The enthalpy change for this reaction is -57.62 kJ/mol at 25 °C.

For weak acids or bases, the heat of neutralization is pH-dependent. In the absence of any added mineral acid or alkali, some heat is required for complete dissociation. The total heat evolved during neutralization will be smaller.

e.g. $\mathrm{H}\mathrm{C}\mathrm{N}+\mathrm{N}\mathrm{a}\mathrm{O}\mathrm{H}⟶\mathrm{N}\mathrm{a}\mathrm{C}\mathrm{N}+\mathrm{H}{\phantom{A}}_2\mathrm{O};\mathrm{\Delta }H=-12\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$ at 25°C

The heat of ionization for this reaction is equal to (–12 + 57.3) = 45.3 kJ/mol at 25 °C.