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JEE Main & Advanced Sample Paper JEE Main Sample Paper-14

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


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