2. Solved Papers
  3. Manipal Engineering

Manipal Engineering Manipal Engineering Solved Paper-2012

  • question_answer
    One mole of ice is converted into water at\[273\text{ }K\]. The entropies of\[{{H}_{2}}O(s)\]and\[{{H}_{2}}O(l)\]are\[38.20\]and\[60.01\,\,J\,\,mo{{l}^{-1}}{{K}^{-1}}\]respectively. The enthalpy change for the conversion is

    A) \[3\,\,kJ\,\,mo{{l}^{-1}}\]                            

    B) \[4\,\,kJ\,\,mo{{l}^{-1}}\]

    C) \[5\,\,kJ\,\,mo{{l}^{-1}}\]            

    D)        \[6\,\,kJ\,\,mo{{l}^{-1}}\]

    Correct Answer: D

    Solution :

    \[\because \]\[\Delta G=\Delta \Eta -\Tau \Delta S\] At equilibrium,\[\Delta G=0\] \[\therefore \]       \[\Delta H=T\Delta S\]            \[\Delta H=273\times (60.01-38.20)\]                     \[=5954.13\,\,J\,\,mo{{l}^{-1}}=5.954\,\,kJ\,\,mo{{l}^{-1}}\]                     \[\approx 6\,\,kJ\,\,mo{{l}^{-1}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner