The standard enthalpies of formation of ions in aqueous solutions are obtained by arbitrarily assigning a value of zero to H+ ions; that is, ∆H°f [H+(aq)]=0.
a) For the following reaction
Calculate the ∆H°f for the Cl- iron.
H+(aq) + Cl- ∆H°=-74.9 kJ/mol
b) Given that ∆H°f for OH- ions is -229 kJ/mol, calculate the enthalpy of neutralization when 1 mole of a strong monoprotic acid (such as HCl) is titrated by 1 mol of a strong base (such as KOH) at 25°C.
H°f (H+ (aq)) + ∆H°f (Cl (aq)) - ∆H°f (HCl (g)) = ∆H°rxn
1(0 kJ/mol) + ∆H°f (Cl- (aq) - 1(-92.3 kJ/mol) = -74.9 kJ/mol
92.3 kJ/mol + ∆H°f (Cl- (aq)) = 74.9 kJ/mol
= ∆H°f (Cl- (aq)) = -167 kJ/mol