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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    \[\Delta H\]and\[\Delta S\]for a reaction are\[+30.558\text{ }kJ\] \[mo{{l}^{-1}}\]and\[0.066\,kJ\,{{K}^{-1}}mo{{l}^{-1}}\]at 1 aim pressure. The temperature at which free energy change is equal to zero and the nature of the reaction below this temperature are:

    A) 483 K, spontaneous

    B) 443 K, non-spontaneous

    C) 443 K, spontaneous

    D) 463 K, non-spontaneous

    E) 463 K, spontaneous

    Correct Answer: D

    Solution :

    \[\Delta G=\Delta H-T\Delta S\] Where, \[\Delta H=30.558\,kJ{{K}^{-1}}\,mo{{l}^{-1}}\] \[\Delta S=0.066\,kJ{{K}^{-1}}\,mo{{l}^{-1}}\] \[\Delta G=0\] \[0=\Delta H-T\Delta S\] \[\Delta H=T\Delta S\] \[30.558=T\times 0.066\] \[T=463\,K\] Reaction is non-spontaneous reason being\[\Delta G=0\].


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