DUMET Medical DUMET Medical Solved Paper-2005

  • question_answer
    The enthalpy change \[(\Delta H)\] for the neutralisation of \[M\,\,HCl\] by caustic potash in dilute solution at 298 K is :

    A)  68 kJ          

    B)  65 kJ

    C)  57.3 kJ        

    D)  50 kJ

    Correct Answer: C

    Solution :

    Key Idea: The heat of neutralisation of strong acid by strong base is always constant. (57.3 kJ) It is infact heat of formation of water by \[{{H}^{+}}\] and \[O{{H}^{-}}\] ions.       \[{{K}^{+}}O{{H}^{-}}+{{H}^{+}}C{{l}^{-}}\xrightarrow{{}}{{K}^{+}}C{{l}^{-}}+{{H}_{2}}O\] or \[O{{H}^{-}}+{{H}^{+}}\xrightarrow{{}}{{H}_{2}}O\,\,\,\,\Delta H=57.3\,kJ\] \[\because \,\,KOH\] is strong base and \[HCl\] is strong acid.


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