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JEE Main & Advanced AIEEE Solved Paper-2008

  • question_answer
    Standard entropy of \[{{X}_{2}},{{Y}_{2}}\] and \[X{{Y}_{3}}\] are 60, 40 and \[50\,J{{K}^{-1}}mo{{l}^{-1}}\], respectively. For the reaction, \[\frac{1}{2}{{X}_{2}}+\frac{3}{2}{{Y}_{2}}\xrightarrow{{}}X{{Y}_{3}}\], \[\Delta H=-30\,kJ\], to be at equilibrium, the temperature will be       AIEEE  Solved  Paper-2007

    A) 750 K

    B)                        1000 K  

    C)        1250 K  

    D)        500 K

    Correct Answer: A

    Solution :

                    \[\frac{1}{2}{{X}_{2}}+\frac{3}{2}{{Y}_{2}}\xrightarrow{{}}X{{Y}_{3}}\,;\,\Delta S=50\]                     \[-\left[ \left( 60\times \frac{1}{2} \right)+\left( 40\times \frac{3}{2} \right) \right]=-40\,J{{K}^{-1}}\,;\,\Delta H=-30\,kJ\]                \[;\,\Delta G=\Delta H-T\Delta S\] At equilibrium, \[\Delta G=0\,\,;\,\,\therefore \Delta H=T\Delta S\,\,;\,\,T=\frac{\Delta H}{\Delta S}\]\[=\frac{-30000J}{-40\,J{{K}^{-1}}}=750K\].


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