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MGIMS WARDHA MGIMS WARDHA Solved Paper-2014

  • question_answer
    A certain reaction is at equilibrium at\[82{}^\circ C\] and the enthalpy change for this reaction is 21.3 kJ. The value of\[\Delta S(J{{K}^{-1}}mo{{l}^{-1}})\]for the reaction is

    A)  55.0                      

    B)  60.0   

    C)  68.5                      

    D)  120.0

    Correct Answer: B

    Solution :

                    Gibbs free energy. \[\Delta G=\Delta H-T\Delta S\] But, at equilibrium, \[\Delta G=0\] \[\therefore \]  \[\Delta H=T\Delta S\] or   \[\Delta S=\frac{\Delta H}{T}=\frac{21.3\times {{10}^{3}}}{355}=60\,J{{K}^{-1}}mo{{l}^{-1}}\]


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