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

14 days is the half-life of phosphorus-32