Consider the reaction $$A \to 2\,B + C,\Delta H = - 15\,kcal.$$      The energy of activation of backward reaction is $$20\,kcal\,mo{l^{ - 1}}.$$    In presence of catalyst the energy of activation of forward reaction is $$3\,kcal\,mo{l^{ - 1}}.$$    At $$400\,K$$  the catalyst causes the rate of the reaction to increase by the number of times equal to

Solution :
$$\eqalign{ & {E_a}\left( f \right) - {E_a}\left( b \right) = \Delta H = - 15\,k\,cal \cr & \Rightarrow {E_a}\left( f \right) = - 15 + 20 = 5\,k\,cal \cr & \frac{{k\left( {catalyst} \right)}}{k} = {e^{\frac{{{E_a} - {E_{a\left( {catalyst} \right)}}}}{{RT}}}} \cr & = {e^{\frac{{\left( {5 - 3} \right) \times {{10}^3}}}{{2 \times 400}}}} = {e^{2.5}} \cr} $$