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NEET Sample Paper NEET Sample Test Paper-39

  • question_answer
    For vaporization of water at 1 atmospheric pressure, the values of\[\Delta H\]and\[\Delta S\]are \[40.63\text{ }kJ\text{ }mo{{l}^{-1}}\]\[\text{ }and\text{ }108.8\text{ }JK-1\text{ }mol-1,\] respectively The temperature when Gibbs energy change \[(\Delta G)\] for this transformation will be zero is

    A)  273.4 K     

    B)  393.4 K

    C)  373.4 K     

    D)  293.4 K

    Correct Answer: C

    Solution :

     \[\Delta G=\Delta H-T\Delta S\] \[\Delta G=0,\,\therefore \Delta H=T\Delta S\] \[T=\frac{\Delta \Eta }{\Delta S}=\frac{40.63\times {{10}^{3}}}{108.8}=373.4\,K\]


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