MGIMS WARDHA MGIMS WARDHA Solved Paper-2013

  • question_answer
    Each\[N{{H}_{3}}\]molecule has six other\[N{{H}_{3}}\]molecules as nearest neighbours in solid state; AH of sublimation of \[N{{H}_{3}}\] molecule at the m. p is\[30.8\text{ }kJ\text{ }mo{{l}^{-1}}\]and in the absence of H-bonding estimated\[\Delta H\]of sublimation is\[14.4\text{ }kJ\text{ }mo{{l}^{-1}}\]. Hence, strength of H-bond in solid\[N{{H}_{3}}\]is

    A)  \[5.47\text{ }kJ\text{ }mo{{l}^{-1}}\]                     

    B)  \[\text{10}\text{.93 }kJ\text{ }mo{{l}^{-1}}\]

    C)  \[\text{16}\text{.40 }kJ\text{ }mo{{l}^{-1}}\]                     

    D)  \[\text{-16}\text{.4 }kJ\text{ }mo{{l}^{-1}}\]

    Correct Answer: A

    Solution :

                    Total strength of all H-bonds \[=30.8-14.4\] \[=16.4kJ\,mo{{l}^{-1}}\] There are six nearest neighbours, but each hydrogen bond involves 2 molecules. Hence, effective neighbours = 3 Hence, strength of H-bond \[=\frac{16.4}{3}=5.47\,kJ\,mo{{l}^{-1}}\]


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