JEE Main & Advanced Chemistry Chemical Kinetics / रासायनिक बलगतिकी Sample Paper Topic Test - Chemical Kinetics

  • question_answer
    Figure shows a graph in \[{{\log }_{10}}\,k\,vs\,\frac{1}{T}\] where, k is rate constant and T is temperature. The straight line BC has slope, \[\tan \,\theta \,=-\frac{1}{2.303}\] and an intercept of 5 on y-axis. Thus, \[{{E}_{a}},\] the energy of activation, is         

    A) 4.606 cal                      

    B) \[\frac{0.2}{2.303}cal\]

    C) 2 cal

    D) None of these

    Correct Answer: C

    Solution :

    Rate constant, \[k=A{{e}^{-{{E}_{a}}/RT}}\]
    In \[k=\frac{-{{E}_{a}}}{RT}+\ln \,A\]
    \[2.303{{\log }_{10}}k=\frac{-{{E}_{a}}}{RT}+2.303{{\log }_{10}}A'\]
    \[{{\log }_{10}}k=\frac{-{{E}_{a}}}{2.303R}\cdot \frac{1}{T}+{{\log }_{10}}A'\]
    Now, \[\frac{-{{E}_{a}}}{2.303R}=\tan \theta \,=-\frac{1}{2.303}\]
    \[\therefore \]      \[{{E}_{a}}=R=2\,cal\]

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