1.

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

Answer»

Solution :`:.` 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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