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JEE Main & Advanced Chemistry Chemical Kinetics / रासायनिक बलगतिकी Question Bank Mock Test - Chemical Kinetics

  • question_answer
    A hydrogenation reaction is carried out at 500 K. \[C{{H}_{2}}=C{{H}_{2}}+{{H}_{2}}\xrightarrow[no\,catalyst]{500k}C{{H}_{3}}-C{{H}_{3}}\] Activation energy \[-\,{{E}_{a}}\,KJ\,mo{{l}^{-1}}\] \[C{{H}_{2}}=C{{H}_{2}}+{{H}_{2}}\xrightarrow{pd,400k}C{{H}_{3}}-C{{H}_{3}}\] Activation energy = (\[{{E}_{a}}-20\]) KJ \[mo{{l}^{-1}}\] If rate remains constant, then \[{{E}_{a}}\] is

    A) \[120\text{ }kJmo{{l}^{-1}}\]  

    B) \[100\text{ }kJmo{{l}^{-1}}\]

    C) \[20kJmo{{l}^{-1}}\]   

    D) \[80\text{ }kJmo{{l}^{-1}}\]

    Correct Answer: B

    Solution :

    [b] \[k=A{{e}^{-{{E}_{a}}/RT}}\] \[k=A{{e}^{-{{E}_{a}}/R\times 500}}\] \[k'=A{{e}^{-({{E}_{a}}-20)/R\times 400}}\] Rates are equal, hence \[\therefore A{{e}^{-{{E}_{a}}/R\times 500}}=A{{e}^{-({{E}_{a}}-20)/R\times 400}}\] \[\therefore \frac{{{E}_{a}}}{500}=\frac{{{E}_{a}}-20}{400}\] \[4{{E}_{a}}=5{{E}_{a}}-100\] \[\therefore {{E}_{a}}=100KJmo{{l}^{-1}}\]


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