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JEE Main & Advanced Sample Paper JEE Main Sample Paper-43

  • question_answer
    A reactant  forms two products
    \[A\xrightarrow{{{k}_{1}}}B\] Activation energy \[{{E}_{{{a}_{1}}}}\]
    \[A\xrightarrow{{{k}_{2}}}C\] Activation energy \[E{{\,}_{{{a}_{2}}}}\]
    If \[{{E}_{{{a}_{2}}}}=2{{E}_{{{a}_{1}}}},\] then \[{{k}_{1}}\] and \[{{k}_{2}}\] will be related as

    A)  \[{{k}_{2}}={{k}_{1}}{{e}^{-{{E}_{{{a}_{1}}}}/RT}}\]    

    B)  \[{{k}_{2}}={{k}_{1}}{{e}^{-{{E}_{{{a}_{2}}}}/RT}}\]

    C) \[{{k}_{1}}={{k}_{2}}{{e}^{-{{E}_{{{a}_{1}}}}/RT}}\]     

    D)  \[{{k}_{1}}=2{{k}_{2}}{{e}^{{{E}_{{{a}_{2}}}}/RT}}\]

    Correct Answer: A

    Solution :

      \[\theta ={{45}^{o}}\] \[n'=\frac{v}{v-{{v}_{s}}}\times n\] \[\Rightarrow \] (Since, \[\frac{n'}{n}=\frac{v}{v-{{v}_{s}}}\])


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