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.
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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