The enthalpy of neutralization is defined as the heat envolved when 1g equivalent of an acid is neutralized by 1g equivalent of a base or vice-versa in dilute solution. This is constant and its value is −57.1‌kJ for neutralization of any strong acid by a strong base since in dilute solution they completely dissociate into ions. H+(aq)+OH+(aq)‌H2O(l) ∆H‌neu ‌‌=−57.1‌kJ‌mol−1 Thus, neutralization involves combination of 1 mole of H+ions with 1 mole of OH−ions to form 1 mole of H2O. Now, it is clear that 1g equivalent (or 1 mole) of any acid on complete dissociation gives 1 mole of H+ ions. But this is not true in case of dibasic or diprotic acid, for example, 1 mole of H2SO4 gives 2 moles of H+ions on complete dissociation. However, 1g equivalent of H2SO4(=0.5‌mol) gives 1 mole of H+ions. ‌H2SO4+2‌NaOH⟶Na2SO4+2H2O ‌∆H‌neu ‌=−114.64‌kJ ‌∴‌ Enthalpy of neutralization ‌=‌