For an endothermic reaction, where $\mathrm{\Delta H}$ represent the enthalpy of reaction in $\mathrm{kJ} {\mathrm{mol}}^{-1}$, the minimum value for energy of activation (for forward reaction) will be

We know that, $\mathrm{\Delta H}={\mathrm{E}}_{\mathrm{f}}-{\mathrm{E}}_{\mathrm{b}}$

where, ${\mathrm{E}}_{\mathrm{b}}$ and ${\mathrm{E}}_{\mathrm{f}}=$ activation energy for backward and forward respectively.

Thus,${\mathrm{E}}_{\mathrm{f}}=\mathrm{\Delta H}+{\mathrm{E}}_{\mathrm{b}}$, means ${\mathrm{E}}_{\mathrm{f}}>\mathrm{\Delta H}$.

 

$\mathrm{\Delta H}=$heat of reaction (endothermic)

${\mathrm{E}}_{\mathrm{a}}=$energy of activation

Thus, ${\mathrm{E}}_{\mathrm{a}}>\mathrm{\Delta H}$

Heat is absorbed in an endothermic reaction. Thus, for an endothermic reaction, where $\mathrm{\Delta H}$ represent the enthalpy of reaction in $\mathrm{kJ} {\mathrm{mol}}^{-1}$, the minimum value for energy of activation (for forward reaction) will be more than $\mathrm{\Delta H}$.