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NEET Chemistry Thermodynamics / रासायनिक उष्मागतिकी Question Bank Self Evaluation Test - Thermodynamics

  • question_answer
    The enthalpy of neutralisation of \[N{{H}_{4}}OH\] with \[HCl\] is \[-51.46\text{ }kJ\text{ }mo{{l}^{-1}}\] and the enthalpy of neutralization of \[NaOH\] with \[HCl\] is \[-55.90\text{ }kJ\text{ }mo{{l}^{-1}}\]. The enthalpy of ionisation of \[N{{H}_{4}}OH\] is                  

    A) \[-107.36 kJ\text{ }mo{{l}^{-1}}\]         

    B) \[-4.44 kJ\text{ }mo{{l}^{-1}}\]

    C) \[+107.36\,kJ\,mo{{l}^{-1}}\]  

    D) \[+4.44\text{ }kJ\text{ }mo{{l}^{-1}}\]

    Correct Answer: D

    Solution :

    [d]
    \[\underset{Strong\,acid}{\mathop{HCl}}\,\xrightarrow{{}}\underset{(Complete\,\,ionisation)}{\mathop{{{H}^{+}}}}\,+C{{l}^{-}}\]   ...(i)
    \[N{{H}_{4}}OH\rightleftharpoons \underset{Weak\,base}{\mathop{NH_{4}^{+}}}\,+OH{{~}^{-}}\] \[\Delta H=x\,kJ\text{ }mo{{l}^{-1}}\]        ...(ii)
    \[{{H}^{+}}+O{{H}^{-}}\xrightarrow{{}}{{H}_{2}}O\] \[\Delta H=-55.90\text{ }kJ\text{ }mo{{l}^{-1}}\]              ...(iii)
    (from neutralisation of strong acid and strong base) From equation (i), (ii) and (iii) \[N{{H}_{4}}OH+HCl\xrightarrow{{}}NH_{4}^{-}+C{{l}^{-}}+{{H}_{2}}O\] \[\Delta H=-51.46\text{ }kJ\text{ }mo{{l}^{-1}}\] \[\therefore x+\left( -55.90 \right)=-51.46\] \[x=-51.46+55.90\] \[=4.44\text{ }kJ\text{ }mo{{l}^{-1}}\] \[\therefore \]  Enthalpy of ionisation of \[N{{H}_{4}}OH\]is  \[\Rightarrow 4.44\text{ }kJ\text{ }mo{{l}^{-1}}\]


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