2. Solved Papers
  3. CET Karnataka Medical

CET Karnataka Medical CET - Karnataka Medical Solved Paper-2010

  • question_answer
    The activation energy for a reaction at the temperature T K was found to be 2.303 RT \[Jmo{{l}^{-1}}\]. 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,  \[k=A{{e}^{-{{E}_{a}}/RT}}\] \[k=A{{e}^{-2.303RT/RT}}\] \[\frac{k}{A}={{e}^{-2.303}}\] On solving, we get \[\frac{k}{A}={{10}^{-1}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner