Enthalpy of neutralization

{{Short description|Change in enthalpy during an acid-base reaction}}

In chemistry and thermodynamics, the enthalpy of neutralization ({{math|Δ{{sub|n}}H}}) 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 {{convert|298|K|C}} 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 ({{math|Δ{{sub|n}}H{{sup|⊖}}}}).

The heat ({{mvar|Q}}) released during a reaction is

:$Q\; =\; mc\_p\; \backslash Delta\; T$

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

:$\backslash Delta\; H\; =\; -\; \backslash frac\{Q\}\{n\}$

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

:H+ + OH^- -> H2O

as the acid and the base are fully dissociated and neither the cation {{chem2|B+}} nor the anion {{chem2|A-}} are involved in the neutralization reaction.{{cite web |url=https://www.chemguide.co.uk/physical/energetics/neutralisation.html |title=Enthalpy Change of Neutralization |last=Clark |first=Jim |date=July 2013 |website=chemguide.co.uk |access-date=4 September 2019}} 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. $\backslash ce\{HCN\; +\; NaOH\; ->\; NaCN\; +\; H2O\};\backslash \; \backslash Delta\; H$ = −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.{{cite web |title=Enthalpy of Neutralization |url=http://faculty.ccri.edu/aahughes/GenChemII/Lab%20Experiments/Enthalpy_of_Neutralization.pdf |publisher=Community College of Rhode Island |accessdate=24 February 2014 |archive-url=https://web.archive.org/web/20161213060740/http://faculty.ccri.edu/aahughes/GenChemII/Lab%20Experiments/Enthalpy_of_Neutralization.pdf |archive-date=13 December 2016 |url-status=dead }}