• # 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

 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, $\frac{-{{E}_{a}}}{2.303R}=\tan \theta \,=-\frac{1}{2.303}$ $\therefore$      ${{E}_{a}}=R=2\,cal$