Enthalpy of neutralization

In chemistry and thermodynamics, the enthalpy of neutralization (ΔH_n) 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 (ΔH_n^⊖).

The heat (Q) released during a reaction is

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

where m is the mass of the solution, c_p 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, BOH, 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.^[1] 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.^[1] 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. ${\displaystyle {\ce {HCN + NaOH -> NaCN + H2O}};\ \Delta H=-12\mathrm {kJ/mol} }$ at 25°C

The heat of ionization for this reaction is equal to (–12 + 57.3) = 45.3 kJ/mol at 25 °C.^[2]

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