JEE Main & Advanced Chemistry Chemical Kinetics / रासायनिक बलगतिकी Question Bank Collision theory, Energy of activation and Arrhenius equation

  • question_answer
    Activation energy is given by the formula             [DCE 1999]

    A)                 \[\log \frac{{{K}_{2}}}{{{K}_{1}}}=\frac{{{E}_{a}}}{2.303R}\left[ \frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}{{T}_{2}}} \right]\]

    B)                 \[\log \frac{{{K}_{1}}}{{{K}_{2}}}=-\frac{{{E}_{a}}}{2.303R}\left[ \frac{{{T}_{2}}-{{T}_{1}}}{{{T}_{1}}{{T}_{2}}} \right]\]

    C)                 \[\log \frac{{{K}_{1}}}{{{K}_{2}}}=-\frac{{{E}_{a}}}{2.303R}\left[ \frac{{{T}_{1}}-{{T}_{2}}}{{{T}_{1}}{{T}_{2}}} \right]\]              

    D)                 None of these

    Correct Answer: A

    Solution :

                    It is modified form of Arrhenius equation.


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