JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Morning)

  • question_answer
    In the following reaction; \[xA\to yB\]                           [JEE Main Held on 12-4-2019 Morning] \[{{\log }_{10}}\left[ -\frac{d[A]}{dt} \right]={{\log }_{10}}\left[ \frac{d[B]}{dt} \right]+0.3010\] 'A' and 'B' respectively can be :

    A) n-Butane and Iso-butane

    B) \[{{C}_{2}}{{H}_{4}}\]and \[{{C}_{4}}{{H}_{8}}\]

    C) \[{{N}_{2}}{{O}_{4}}\]and\[N{{O}_{2}}\]

    D) \[{{C}_{2}}{{H}_{2}}\]and \[{{C}_{6}}{{H}_{6}}\]

    Correct Answer: B

    Solution :

    \[\log \frac{-d[A]}{dt}=\log \frac{d[B]}{dt}+0.3010\]           \[\frac{-d[A]}{dt}=2\times \frac{d[B]}{dt}\]           \[\frac{1}{2}\times \frac{-d[A]}{dt}=\frac{d[B]}{dt}\]\[2A\xrightarrow[{}]{{}}B\]           \[2{{C}_{2}}{{H}_{4}}\xrightarrow[{}]{{}}C{{ & }_{4}}{{H}_{8}}\]


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