\( \require{mhchem} \)
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 (ΔHn⊖).
The heat (Q) released during a reaction is
\( Q = mc_{p} \Delta T \)
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.
\( \Delta H = - \frac{Q}{n} \)
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]
References
Clark, Jim (July 2013). "Enthalpy Change of Neutralization". chemguide.co.uk. Retrieved 4 September 2019.
"Enthalpy of Neutralization" (PDF). Community College of Rhode Island. Archived from the original (PDF) on 13 December 2016. Retrieved 24 February 2014.
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