The enthalpy of neutralization of HCO3−(aq) with strong alkali is - 42 kJ / mol and enthalpy of neutralization of strong acid with strong alkali is - 56 kJ, 1 equiv. Therefore, enthalpy of dissociation of HCO3−(aq) is

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- 98 kJ / mol

98 kJ / mol

14 kJ / mol

24 kJ / mol

answer is C.

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Detailed Solution

HCO3−(aq)+OH−(aq)→CO32−(aq)+H2OΔH1= - 42 kJ / mol.H2O→H+(aq)+OH−(aq); ΔH2=+56kJ/mol∴ HCO3−→H++CO32−; ΔH=+14kJ/mol

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