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−1Thus, 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 ==−57.32kJ