question_answer

Enthalpy of neutralization of \[HCl\] by \[NaOH\] is \[-55.84\,kJ\,mo{{l}^{-1}}\] and by \[N{{H}_{4}}OH\] is \[-51.34\,kJ\,mo{{l}^{-1}}\]. The enthalpy of ionization of \[N{{H}_{4}}OH\] is

A)  \[4.5\,kJ\,mo{{l}^{-1}}\]                              

B)  \[-4.5kJ\,mo{{l}^{-1}}\]

C)  \[1.8\,kJ\,mo{{l}^{-1}}\]                              

D)  \[-1.8\,kJ\,mo{{l}^{-1}}\]

Correct Answer: A

Solution :

\[\Rightarrow \,-51.34=x-55.84\] \[\Delta {{H}_{ionization}}=4.5\,kJ\,mo{{l}^{-1}}\]