JEE Main & Advanced JEE Main Paper (Held On 11-Jan-2019 Morning)

  • question_answer
    For the chemical reaction \[xY,\]the standard reaction Gibbs energy depends on temperature T (in K) as \[{{\Delta }_{r}}{{G}^{o}}(in\,kJ\,mo{{l}^{-1}})=120-\frac{3}{8}T.\] The major component of the reaction mixture at T is [JEE Main Online Paper (Held On 11-Jan-2019 Morning]

    A) X if T= 315 K                           

    B) Y if T= 280 K

    C) X if T= 350 K               

    D)   Y if T= 300 K

    Correct Answer: A

    Solution :

    \[{{\Delta }_{r}}{{G}^{o}}={{\Delta }_{r}}{{H}^{o}}-T\Delta {{S}^{o}}\] At equilibrium,\[\Delta {{G}^{o}}=0\] \[{{\Delta }_{r}}{{H}^{o}}=T\Delta {{S}^{o}}\] So,\[T=\frac{{{\Delta }_{r}}{{H}^{o}}}{\Delta {{S}^{o}}}=\frac{120}{3}\times 8\] \[\therefore \]\[T=320K\] Thus, if temperature is less than 320 K reaction will be non-spontaneous means X will be the major component while if temperature is more than \[\Delta G=-ve\]and reaction will be spontaneous and Y will be the major component.


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