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

  • question_answer
    The lattice energy of solid \[NaCl\] is \[180\text{ }kcal\text{ }mo{{l}^{-1}}\] and enthalpy of solution is \[1\text{ }kcal\text{ }mo{{l}^{-1}}\]  If the hydration energies of \[N{{a}^{+}}\] and \[C{{l}^{-}}\] ions are in the ratio 3 : 2, what is the enthalpy of hydration of sodium ion?

    A) \[-107.4\text{ }kcal\text{ }mo{{l}^{-1}}\]

    B) \[107.4\text{ }kcal\text{ }mo{{l}^{-1}}\]

    C) \[71.6\text{ }kcal\text{ }mo{{l}^{-1}}\]            

    D) \[-71.6\text{ }kcal\text{ }mo{{l}^{-1}}\]

    Correct Answer: A

    Solution :

    [a] \[\Delta {{H}_{hyd.}}=\Delta {{H}_{sol.}}-\Delta {{H}_{lattice}}\] \[=1-180=-179\text{ }kcal\,mo{{l}^{-1}}\] Then \[\Delta {{H}_{hyd.}}=(N{{a}^{+}})+\Delta {{H}_{hyd.}}(C{{l}^{-}})=-179\] or \[\Delta {{H}_{hyd.}}=(N{{a}^{+}})+\frac{2}{3}\Delta {{H}_{hyd.}}=-179\] or \[\Delta {{H}_{hyd.}}=(N{{a}^{+}})=-107.4\,kcal\,mo{{l}^{-1}}\]


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