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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2011

  • question_answer
    The activation energy of a reaction at a given temperature is found to be \[2.303RT\text{ }J\text{ }mo{{l}^{-1}}\]. The ratio of rate constant to the Arrhenius factor is

    A)  \[0.01\]            

    B)  \[0.1\]

    C)  \[0.02\]

    D)  \[0.001\]

    Correct Answer: C

    Solution :

    \[k=A{{e}^{-{{E}_{a}}/RT}}\] \[\frac{k}{A}={{e}^{-{{E}_{a}}/RT}}\] In \[\left( \frac{k}{A} \right)=\frac{-{{E}_{a}}}{RT}\] \[\log \left( \frac{k}{A} \right)=\frac{-{{E}_{a}}}{2.303RT}\] \[\log \left( \frac{k}{A} \right)=\frac{-2.303RT}{2.303RT}\] \[\log \left( \frac{k}{A} \right)=-1\] \[\therefore \] \[\frac{k}{A}=0.1\]


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