We’re being asked to determine the activation energy for the catalyzed reverse reaction.
We’re given the following information:
ΔH = –75 kJ: Recall that the change in enthalpy (ΔH) is the difference in energy between products and reactants. ΔH for the reaction is negative, which means we have an exothermic reaction. This signifies that the reactants are higher in energy than the products.
A certain reaction has a ΔH = –75 kJ and an activation energy of 40 kJ. A catalyst is found that lowers the activation energy of the forward reaction by 15 kJ. What is the activation energy of the reverse reaction in the presence of this same catalyst?
(A) 25 kJ
(B) 60 kJ
(C) 90 kJ
(D) 100 kJ
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