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 Template:Convert and 1 bar 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

<math> Q = mc_p \Delta T </math>

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.

<math> \Delta H = - \frac{Q}{n} </math>

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

<chem>H+ + OH^- -> H2O</chem>

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.<ref name="chemguide">{{#invoke:citation/CS1|citation |CitationClass=web }}</ref> 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.<ref name="chemguide" /> 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. <math chem>\ce{HCN + NaOH -> NaCN + H2O};\ \Delta H</math> = −12 kJ/mol at 25 °C

The heat of ionization for this reaction is equal to (−12 + 57.3) = 45.3 kJ/mol at 25 °C.<ref name="Community College of Rhode Island">{{#invoke:citation/CS1|citation |CitationClass=web }}</ref>