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Heat of neutralization `(DeltaH)` of `NH_(4)OH` and `HF` are `-51.5` and `-68.6 kJ` respectively. Calculate their heat of dissociation? (i) `HCl (aq)+NaOH(aq) rarr NaCl (aq)+H_(2)O, " "DeltaH= -57.3 kJ` (ii) `HCl (aq)+underset("(weak base)")(NH_(4)OH (aq)) rarr NH_(4)Cl (aq)+H_(2)O, " "DeltaH= -51.5 kJ` |
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Answer» `:.` The heat of dissociation of `NH_(4)OH`, `DeltaH=-51.5-(-57.3)=5.8 kJ` Similarly we have `HF(aq) +NaOH(aq) rarr NaF(aq)+H_(2)O, DeltaH=-68.6 kJ` `:.` The heat of diffociation of `HF`, `DeltaH =-68.6-(-57.3)=-11.3 kJ` |
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