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JEE Main & Advanced JEE Main Paper (Held On 10-Jan-2019 Morning)

  • question_answer
    A process has \[\Delta H=200\text{ }J\text{ }mo{{l}^{-1}}\] and\[\Delta \,S=40\text{ }J{{K}^{-1}}\,mo{{l}^{-1}}\]. Out of the values given below, choose the minimum temperature above  which  the  process  will be spontaneous: [JEE Main Online Paper (Held On 10-Jan-2019 Morning]

    A) 4 K                              

    B) 20 K  

    C) 5 K      

    D)                  12 K

    Correct Answer: C

    Solution :

    For a process to be spontaneous, \[\Delta \,G<0\] & \[\Delta G=\Delta H-T\Delta S\] at equillibrium \[\Delta G=0\,,\,\,\,\therefore \,\,\Delta H={{T}_{e}}\Delta S\], \[\therefore \,\,\,\,{{T}_{e}}=\frac{\Delta H}{\Delta S}\,\,=\,\,\frac{200}{40}\,\,=\,\,5\,k\] \[\therefore \] If \[T>{{T}_{e}},\text{ }\Delta G<0\] and the process will be spontaneous


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