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JEE Main & Advanced JEE Main Paper (Held On 9 April 2016)

  • question_answer
    For the reaction, \[A(g)+B(g)\to C(g)+D(g),\Delta {{H}^{0}}\]and\[\Delta {{S}^{0}}\]  are, respectively,\[-29.8kJ\,mo{{l}^{-1}}\]and\[-0.100kJ\,{{K}^{-1}}mo{{l}^{-1}}\]at 298 K. The equilibrium constant for the reaction at 298 K is:   JEE Main Online Paper (Held On 09 April 2016)

    A) 1             

    B) 10

    C) 1.0 × 10.10                          

    D) 1.0 × 1010

    Correct Answer: A

    Solution :

                    \[\Delta {{G}^{o}}=\Delta {{H}^{o}}-T.\Delta {{S}^{o}}\] \[=-29.8+298\times (0.1)\] \[=-29.8+29.8\] \[\because \]\[\Delta {{G}^{o}}=0\] apply relation between \[\Delta {{G}^{o}}\And {{K}_{eq}}\] \[\Delta {{G}^{o}}=-RT\ell n{{K}_{eq}}\] \[\therefore \]\[{{K}_{eq}}=1\]                                


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