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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2011

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

    A)  0.01                                      

    B)  0.1

    C)  0.02                                      

    D)  0.001

    Correct Answer: B

    Solution :

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


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