• # question_answer The energies of activation for forward and reverse reactions for $\frac{1}{4}$ are $\frac{1}{2}$and$5\,\Omega$respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by$(1-e)$ The enthalpy change of the reaction $({{A}_{2}}-{{B}_{2}}\,\to 2\,\,AB)$ in the presence of catalyst will be (in$(1-{{e}^{-1}})$)       AIEEE  Solved  Paper-2007 A)  300                       B)         120                       C)         280                       D)         -20

$Ge$ ${{E}_{a\,(forward)}}=180\,kJ\,mo{{l}^{-1}}$ $Ge$ In the presence of catalyst, $v=E\times B/{{B}^{2}}$ $v=B\times E/{{B}^{2}}$ $v=E\times B/{{E}^{2}}$ $v=B\times E/{{E}^{2}}$