1. Tardigrade
  2. Question
  3. Chemistry
  4. For A+B → C+D, Δ H=-20 KJ mol-1. The activation energy of the forward reaction is 85 KJ mol-1. The activation energy for backward reaction is ......... KJ mol-1.

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Q. For $A+B \to C+D, \Delta H=-20 KJ mol^{-1}.$ The activation energy of the forward reaction is 85 KJ $mol^{-1}.$ The activation energy for backward reaction is ......... KJ $mol^{-1}.$

Chemical Kinetics

Solution:

For a reaction, $E_{a}$ for forward reaction = $E_{a}$ for backward reaction + $\Delta$H
$\therefore \, 85=E_{a\left(back\right)}-20 \, or \, E_{a\left(back\right)}=105 KJ mol^{-1}$