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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2010

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

    A) \[{{10}^{-1}}\]                                  

    B) \[{{10}^{-2}}\]

    C) \[2\times {{10}^{-3}}\]                  

    D) \[2\times {{10}^{-2}}\]

    Correct Answer: A

    Solution :

    Arrhenius equation is, rate constant, \[k=A{{e}^{-{{E}_{a}}/RT}}\]                                 \[k=A{{e}^{-2.303\,RT/RT}}\]                                 \[\frac{k}{A}={{e}^{-2.303}}\] On solving, we get                                 \[\frac{k}{A}={{10}^{-1}}\]


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