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JEE Main & Advanced Sample Paper JEE Main Sample Paper-15

  • question_answer
    For photochemical dimidiation of \[{{C}_{12}}{{H}_{14}}\]to obtain \[{{C}_{24}}{{H}_{24}},\] the proposed rate law is \[\frac{d[{{C}_{24}}{{H}_{28}}]}{dt}=\frac{{{K}_{1}}{{K}_{3}}{{I}_{abs}}[{{C}_{12}}{{H}_{14}}]}{{{K}_{2}}+{{K}_{3}}[{{C}_{12}}{{H}_{14}}]}\] \[{{I}_{abs}}=\] Intensity of absorbed light. What will be the value of rate of formation of \[{{C}_{24}}{{H}_{28}}\] if the reaction follows zero order.

    A) \[{{K}_{1}}\]                                      

    B) \[{{K}_{1}}{{I}_{abs}}\]

    C) \[\frac{{{K}_{1}}{{K}_{3}}{{I}_{abs}}}{{{K}_{2}}}\]                             

    D) \[\frac{{{K}_{1}}{{I}_{abs}}}{{{K}_{2}}}\]

    Correct Answer: B

    Solution :

    \[\frac{d[{{C}_{24}}{{H}_{28}}]}{dt}=\frac{{{K}_{1}}{{K}_{3}}{{I}_{abs}}[{{C}_{12}}{{H}_{14}}]}{{{K}_{2}}+{{K}_{3}}[{{C}_{12}}{{H}_{14}}]}=\frac{{{K}_{1}}{{K}_{3}}{{[I]}_{absorb}}}{{{K}_{2}}+{{K}_{3}}}={{K}_{1}}{{[I]}_{absorb}}\]\[{{C}_{12}}{{H}_{14}}\frac{{{K}_{1}}{{K}_{3}}{{[I]}_{absorb}}}{{{K}_{2}}+{{K}_{3}}}={{K}_{1}}{{[I]}_{absorb}}{{C}_{12}}{{H}_{14}}*\] \[{{C}_{12}}{{H}_{14}}^{*}+{{C}_{12}}{{H}_{14}}\xrightarrow[{}]{{{K}_{3}}}{{C}_{24}}{{H}_{28}}\]\[[B]\]as\[[{{C}_{2}}{{H}_{14}}]\]in excess.


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