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JEE Main & Advanced JEE Main Solved Paper-2017

  • question_answer
    Two reactions\[{{R}_{1}}\]and\[{{R}_{2}}\]have identical preexponential factors. Activation energy of \[{{R}_{1}}\]exceeds that of \[{{R}_{2}}\]by \[10\,kJ\,mo{{l}^{-1}}.\] If \[{{k}_{1}}\]and \[{{k}_{2}}\]are rate constants for reactions \[{{R}_{1}}\]and \[{{R}_{2}}\]respectively at 300 K, then ln \[({{k}_{2}}/{{k}_{1}})\] is equal to:- \[(R=8.314\,J\,mo{{l}^{-1}}{{K}^{-1}})\     JEE Main Solved Paper-2017

    A)  8                            

    B)  12

    C)  6                            

    D)  4

    Correct Answer: D

    Solution :

     From arrhenius equation, \[K=A.{{e}^{\frac{-Ea}{RT}}}\] So, \[{{K}_{1}}=A.{{e}^{-{{E}_{{{a}_{1}}}}/RT}}\] ?.. \[{{K}_{2}}=A.{{e}^{-{{E}_{{{a}_{2}}}}/RT}}\]        ?.. So, equation \[(2)(1)\Rightarrow \frac{{{K}_{2}}}{{{K}_{1}}}={{e}^{\frac{({{E}_{{{a}_{1}}}}-{{E}_{{{a}_{2}}}})}{RT}}}\] (As pre-exponential factors of both reactions is same) \[\ln \left( \frac{{{K}_{2}}}{{{K}_{1}}} \right)=\frac{{{E}_{{{a}_{1}}}}-{{E}_{{{a}_{2}}}}}{RT}=\frac{10,000}{8.314\times 3000}=4\]


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