JEE Main & Advanced Sample Paper JEE Main - Mock Test - 1

  • question_answer
    A process has \[\Delta H=200\text{ }J\text{ }mo{{l}^{-1}}\]and \[\Delta \text{S}=40\text{ }J{{K}^{-1}}\text{ }mo{{l}^{-1}}\]Out of the values given below, the minimum temperature above which the process will be spontaneous:

    A) \[20\text{ }K\]               

    B) \[\text{12 }K\]     

    C) \[\text{5 }K\]                 

    D) \[\text{4 }K\]

    Correct Answer: C

    Solution :

    \[\Delta H=200J\,mo{{l}^{-1}}\]
    \[\Delta S=40J{{K}^{-1}}\,mo{{l}^{-1}}\]
    For spontaneous reaction, \[\Delta G<0\]
    \[\Delta H-T\Delta S<0;\] \[\Delta H<T\Delta S;\] \[\frac{\Delta H}{\Delta S}<T;\] \[\frac{200}{40}<T\]
    \[S<T\]
    So, minimum temperature is 5 K


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