[mzen]

need help with this problem, its confusing to me,can anybody explain plz

"Suppose a reaction has an activation energy of 100 kJ/mol. A catalyst lowers the
activation energy to 80 kJ/mol. Assuming that the preexponential factor, A, in the
Arrhenius equation remains the same and that the reaction proceeds at room temperature,
how much faster will the reaction be in the presence of the catalyst?"

thank you

[LQ43]

using the Arrhenius equation, everything is the same except that with the catalyst the activation energy Ea is 80% (or 0.8 ) of the uncatalyzed reaction

I would find the ratio of the rate constant (uncatalyzed) vs rate constant (catalyzed) using Ea = 1 for uncatalyed ; and Ea = 0.8 catalyzed

[Kryolith]

Does that help?