(I) HCl+NaOH→NaCl+H2O ∆H1=−57.3KJmol−1 (II) CH3COOH+NaOH→CH3COONa+H2O ∆H2=−55.3KJmol−1 Reaction (I) can be written as (III) NaCl+H2O→HCl+NaOH ∆H3=57.3KJmol−1 By adding (II) and (III) CH3COOH+NaCl→CH3COONa+HCl∆Hr ∆Hr=∆H3+∆H2=57.3−55.3 =2kJmol−1