# Does lower activation energy definitely mean higher rate?

Does lower activation energy definitely mean higher rate? For example the decomposition of $$\ce{H2O2}$$. Can we say it has the highest rate with the catalyst making the reaction's activation energy the lowest?

• The rate constant would be rather $k=A1.\exp{-E1/kT} + A2.\exp{-E2/kT}$. A1 is frequency of collisions leading to possible reactions without catalyst, A2 with catalyst. E1,E2 are then the respective activation energies. With more if catalyst, A1 gets lower, A2 gets bigger. Apr 15, 2020 at 20:33