Enthalpy of neutralization of is the heat change taking place when 1 gram equivalent of an acid (or base) is neutralized by 1 gram equivalent of base (or acid) in a dilute solution.Heat of neutralization of strong acid-strong base is always constant, i.e. ΔH = -57.1 kj/mole.HCl (aq) + NaOH (aq){tex}\\to{/tex} NaCl (aq) + {tex}H_2O{/tex} (l)…ΔH = -57.1 kj/moleEnthalpy of neutralization of any strong acid (like HCl, HNO3, H2SO4) with a strong base (like LiOH, NaOH, KOH) or vice versa is always the same i.e. 57.1 kj/mol. This is because strong acids, strong bases and salt that they form are all completely ionized in dilute aqueous solutions.Thus the reaction between any strong acid and strong base for example in the above case may be written as :NaOH (aq) + HCl(aq) {tex}\\to{/tex} NaCl (aq) + {tex}H_2O{/tex} (l)… ΔH = -57.1 kj/molethey will dissociate as :Na(+) (aq) + OH(-) (aq) + H(+) (aq) + Cl(-) (aq) {tex}\\to{/tex} Na(+) (aq) + Cl(-) (aq) + {tex}H_2O{/tex} (l)common ions will cancel out..H(+) (aq) + OH(-) (aq){tex}\\to H_2O(l){/tex}Therefore, neutralization is simply a reaction between H(+) ions given by acids and OH(-) ions given by base to form one mole of {tex}H_2O{/tex}.Since strong acid and strong base completely ionize in aqueous solution, the number of H(+) and OH(-) produced by 1 gram equivalent of strong acid and strong base is always the same, hence enthalpy of neutralization between a strong acid and strong base is always constant i.e. ΔH = -57.1 kj/mole.