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The enthalpy of neutralization for the first proton of H2S is -33.7 kJ mol-1.  Calculate the first...

The enthalpy of neutralization for the first proton of H2S is -33.7 kJ mol-1.  Calculate the first acid ionization energy for H2S (aq).

2.               The enthalpy of neutralization of HS- (aq) is -5.1 kJ mol-1.  Calculate the ΔfH° of S2- (aq).

3.               Calculate the ΔsolnH° of hydrogen chloride.

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Answer #1

For the given reaction,

H2S (aq) + OH- (aq) <==> HS- (aq) + H2O (l)

2. enthalpy of neutralization dHneu = -33.7 kJ/mol

We known,

H+ (aq) + OH- (aq) <==> H2O (l)

enthalpy of this reaction = dH = -285.8 kJ/mol

Inverting the IInd reaction and adding to the first,

H2S (aq) + OH- (aq) <==> HS- (aq) + H2O (l)

H2O (l) <==> H+ (aq) + OH- (aq)

-------------------------------------------------------------

H2S (aq) <==> H+ (aq) + HS- (aq)

This is first ionization energy of H2S (aq)

enthalpy of first ionization energy = -33.7 + 285.8 = 252.1 kJ/mol

3. For the reaction,

HCl (aq) <==> H+ (aq) + Cl- (aq)

dHo(soln) = dH(products) - dH(reactants)

taking values from literature,

dHo(soln) = (0 - 230.0) - (-285.8)

                 = 55.8 kJ/mol

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