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JEE Main & Advanced Sample Paper JEE Main Sample Paper-27

  • question_answer
    The value of \[{{\log }_{10}}K\] for a reaction \[AB\] is: (Given: \[{{\Delta }_{r}}H_{298K}^{{}^\circ }=-54.07\,kJ\,mo{{l}^{-1}},\] \[{{\Delta }_{r}}S_{298K}^{{}^\circ }=10J{{K}^{-1}}\,mo{{l}^{-1}}\]and \[R=8,314\text{ }J{{K}^{-1}}\text{ }mo{{l}^{-1}};\]\[2.303\times 8.314\times 298=5705\])

    A)  5                                

    B)  10

    C)  95                               

    D)  100

    Correct Answer: B

    Solution :

    \[\Delta {{G}^{o}}=\Delta {{H}^{o}}-T\Delta {{S}^{o}}\,=-54.07\,\times 1000\,-298\times 10\] \[=-54070\,-\,2980\,=-57050\] \[\Delta {{G}^{o}}-2.303\,RT\,{{\log }_{10}}K\] \[-57050\,=-2.303\,\times 298\,\times 8.314\,{{\log }_{10}}K=-5705\,{{\log }_{10}}K\] \[{{\log }_{10}}\,K=10\]


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