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Enthalpy of neutralization

From Wikipedia, the free encyclopedia
Change in enthalpy during an acid-base reaction

In chemistry and thermodynamics, the enthalpy of neutralization (ΔnH) 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 °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 (ΔnH).

The heat (Q) released during a reaction is

Q = m c p Δ T {\displaystyle Q=mc_{p}\Delta T} {\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.

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

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

H + + OH H 2 O {\displaystyle {\ce {H+ + OH^- -> H2O}}} {\displaystyle {\mathrm {H} {\vphantom {A}}^{+}{}+{}\mathrm {OH} {\vphantom {A}}^{-}{}\mathrel {\longrightarrow } {}\mathrm {H} {\vphantom {A}}_{\smash[{t}]{2}}\mathrm {O} }}

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. HCN + NaOH NaCN + H 2 O ;   Δ H {\displaystyle {\ce {HCN + NaOH -> NaCN + H2O}};\ \Delta H} {\displaystyle {\mathrm {HCN} {}+{}\mathrm {NaOH} {}\mathrel {\longrightarrow } {}\mathrm {NaCN} {}+{}\mathrm {H} {\vphantom {A}}_{\smash[{t}]{2}}\mathrm {O} };\ \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.[2]

References

[edit ]
  1. ^ a b Clark, Jim (July 2013). "Enthalpy Change of Neutralization". chemguide.co.uk. Retrieved 4 September 2019.
  2. ^ "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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