The standard enthalpies of formation of ions in aqueous solutions are obtained by arbitrarily assigning a value of zero to H + ions; that is, ΔH° [H f ⁺ (aq) ] = 0. (a) For the following reaction HCl (g) ⟶H₂O H⁺ (aq) + Cl⁻ (aq) ΔH° = - 74.9 kJ/mol calculate ΔH° for the Cl f ⁻ ions. (b) Given that ΔH° for OH f - ions is - 229.6 kJ/mol, calculate the enthalpy of neutralization when 1 mole of a strong monoprotic acid (such as HCl) is titrated by 1 mole of a strong base (such as KOH) at 25°C.

A) - 74.9 kJ B) - 304.5 kJ

Explanation:

Hess's Law says that the ΔH for one reaction is the sums of the ΔH of all the steps to that reaction.

A) As ΔH [H+] = 0, we can say that all the ΔH for the reaction

HCl (g) - -> H + (aq) + Cl - (aq)

Is from the fornation of the Cl - ions, so ΔH [Cl-] = - 74.9 kJ

B)

HCl (g) - -> H + (aq) + Cl - (aq) ΔH = - 74.9 kJ/mol

KOH (aq) - -> K + (aq) + OH - (aq) ΔH = - 229.6 kJ/mol

KOH (aq) + HCl (g) - -> H2O (l) + K + (aq) + Cl - (aq) ΔH = - 74.9 + (-229.6)

ΔH = - 304,5 kJ/mol

The enthalpy of neutralization for 1 mole is - 304,5 kJ