# Enthalpy of neutralization

The enthalpy of neutralization (ΔHn) is the change in enthalpy that occurs when one equivalent of an acid and one equivalent of 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 it is called the standard enthalpy of neutralization (ΔHno).

The heat (Q) released during a reaction is

${\displaystyle Q=mc_{p}\Delta T}$

where m is the mass of the solution, cp is the specific heat capacity of the solution, and ∆T is the temperature change observed during the reaction. From this, the standard enthalpy change (∆H) is obtained by division with the amount of substance (in moles) involved.

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

When a strong acid, HA reacts with a strong base, B(OH), the reaction that occurs is

${\displaystyle {\ce {{H+}+OH^{-}->H2O}}}$

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

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. ${\displaystyle {\ce {{HCN}+{NaOH}->{NaCN}+{H2O};\ \Delta H=-12kJ/mol}}}$ at 25C

The heat of this reaction is equal to (–12 + 57.3) = 45.3 kJ/mol at 25C.[1]

## References

1. ^ "Enthalpy of Neutralization" (PDF). Community College of Rhode Island. Retrieved 24 February 2014.