Английская Википедия:Enthalpy of neutralization
In chemistry and thermodynamics, the enthalpy of neutralization (Шаблон:Math) 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 (Шаблон:Math).
The heat (Шаблон:Mvar) released during a reaction is
- <math> Q = mc_p \Delta T </math>
where Шаблон:Mvar is the mass of the solution, Шаблон:Mvar is the specific heat capacity of the solution, and Шаблон:Math is the temperature change observed during the reaction. From this, the standard enthalpy change (Шаблон:Math) is obtained by division with the amount of substance (in moles) involved.
- <math> \Delta H = - \frac{Q}{n} </math>
When a strong acid, HA, reacts with a strong base, BOH, the reaction that occurs is
- <chem>H+ + OH^- -> H2O</chem>
as the acid and the base are fully dissociated and neither the cation Шаблон:Chem2 nor the anion Шаблон:Chem2 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. <math chem>\ce{HCN + NaOH -> NaCN + H2O};\ \Delta H = -12 \mathrm{kJ/mol}</math> 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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