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VMMC VMMC Medical Solved Paper-2014

  • question_answer
    The heat of neutralization of HCl by NaOH is \[-55.9\,kJ\,mo{{l}^{-1}}\]and that of HCN is -12.1 kJ; the energy of dissociation of HCN is

    A) 43.8 kJ                  

    B)  -43.8 kJ

    C) -68 kJ                    

    D) 68 kJ

    Correct Answer: A

    Solution :

    \[{{H}^{+}}+O{{H}^{-}}\xrightarrow[{}]{{}}{{H}_{2}}O;\Delta H=-55.9kJ\]                                                                                                ?(i) \[HCN+O{{H}^{-}}\to {{H}_{2}}O+C{{N}^{-}};\Delta H=-12.1\,kJ\]                                                                                                               ?(ii) On revers Eq. (i) and then adding to Eq. (ii), we get\[HCN\xrightarrow[{}]{{}}{{H}^{+}}+C{{N}^{-}};\Delta H=+43.8kJ\]                    


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