Question

explain heat of nutralisation for strong acid-strong base reaction is constant

Answer 1

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. ... The standard enthalpy change of neutralization for a strong acid and base is -57.62 kJ/mol.

Answer 2

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 (ΔHno). The heat (Q) released during a reaction is Q = m c p Δ T {\displaystyle Q=mc_{p}\Delta T} 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}}} \Delta H = - \frac{Q}{n} The standard enthalpy change of neutralization for a strong acid and base is -57.62 kJ/mol.[1] The standard enthalpy of neutralization[2] for organic acids is slightly less exothermic than that of mineral acids because of the partial ionization of weak organic acids. The bond between the proton and its conjugate base requires energy to be broken, hence the lower measured magnitude of the enthalpy change. For weak acids and bases, the heat of neutralization is different because they are not dissociated completely and during dissociation some heat is absorbed - total heat evolved during neutralization will be smaller. e.g. HCN + NaOH ⟶ NaCN + H 2 O ; Δ H = − 12 kJ / mol {\displaystyle {\ce {{HCN}+{NaOH}->{NaCN}+{H2O};\ \Delta H=-12kJ/mol}}} {\displaystyle {\ce {{HCN}+{NaOH}->{NaCN}+{H2O};\ \Delta H=-12kJ/mol}}} The heat of ionization in this reaction is equal to (–12 + 57.3) kJ/mol = 45.3 kJ/mol.