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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?

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The lower the energy of the intermediate, the faster the transition from the reactants to the products, at a given temperature.

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  • $\begingroup$ So this means the statement is true? And could you please give me a source/citation for this? $\endgroup$ – Mathrix Apr 15 at 19:51
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    $\begingroup$ The rate constant k = A*e^(-Ea/RT) this is the Arrhenius equation, where Ea is the activation energy. Consider that the temperature is kept constant. This means that the only variable is the activation energy Ea, and as we decrease the activation Ea, the rate constant k becomes higher. $\endgroup$ – Aleksejus Pacalovas Apr 15 at 19:57
  • $\begingroup$ It is not really related to this question but does the amount of catalyst used affect activation energy? From its relationsheep with rate, I found this on web: Increasing the amount of catalyst used will not increase the rates of reaction beyond a certain point. $\endgroup$ – Mathrix Apr 15 at 20:08
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    $\begingroup$ The catalyst provides an "alternative route" for the reaction, this route has a lower activation energy, so the amount of catalyst doesn't affect activation energy of that "alternative route". I would expect there to be some asymptotic relationship where when you add more catalyst, the rate of the reaction keep increasing, but more slowly, since each molecule of the reactant would have a molecule of catalyst nearby, so yes I'd expect there to be a limit. $\endgroup$ – Aleksejus Pacalovas Apr 15 at 20:22
  • $\begingroup$ 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. $\endgroup$ – Poutnik Apr 15 at 20:33

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