Catalyst

[Adam_Alyafaie_1L]

I know that you want to avoid having an intermediate molecule in your rate law but is it okay to have a catalyst in your rate late or are you suppose to substitute it somehow?

[Chem_Mod]

Yes it okay to have a catalyst in the rate law. If the catalyst is involved in the slow step.

See page 652 in the textbook.

[Caleb Lim 2M]

Is it incorrect if it is not in the rate law?

[Christopher Reed 1H]

If the catalyst is in the rate determining step, then it must be listed in the overall rate law because its presence, or lack of, will affect the rate of reaction.

[Jata_Kavya_3A]

So catalysts can be included in the rate law, but what about intermediates?

[Cinthia_Ramirez_3B]

Intermediates cannot be in the rate law, so we use the pre-equilibrium approach to get rid of the intermediate.

[Alex Uy 2D]

If a catalyst changes the pathway for the energy, does that mean the reaction is doing less work?

[sas_budge_1l]

if a reaction rate increases by a factor of 1000 in the presence of a catalyst at 25 degrees, what does it mean? and how does this change the approach to equations involved in activation energy?

[Chem_Mod]

Using the Arrhenius equation, you can see the relationship between Ea and k. Catalysts offer alternative reaction pathways which have lower energy barriers. As such, the smaller Ea results in a higher k so your reaction rate increases.

[Alex Yee - 4I]

Would we ever in this class have to calculate the rate of a reaction with a limited amount of catalyst and much greater amounts of reactant, where the some of the reaction goes without using the catalyst and some of it goes at the faster speed with the catalyst? If not, is this something done in higher levels of chemistry, or is the rate of the uncatalyzed reaction disregarded?

[GabrielaGutierrez2A]

What kind of question would we be asked in regards to the "Pre-Equilibrium Approach"? Also, what is the purpose of this approach, and how do we know when to use it?

[danae_blodgett_1H]

In the Pre-Equilibrium Approach does the slow step always determine the rate law?

[Hue_Vo_1D]

In the Pre-Equilibrium Approach does the slow step always determine the rate law?

In any reaction systems, the slow step will always determine the rate of reaction, thus the rate law.
The purpose of using Pre-Equilirbrium Approach is to substitute other terms with intermediate, which is not supposed to be in the Rate Law.
Suppose you have two-step reaction and
the rate limiting step is Step 1: Rate=k(1)[R], in this case you don't have intermediates in rate law and Rate=k[R] as itself is reaction rate law.
the rate limiting step is Step 2: Rate=K(2)[R], in which [R] is product of step 1, meaning [R] in step 2 is intermediate. This time, you use Pre-Equil. Approach to get rid of intermediate.

Hope that makes more sense!

[Michael_Phillipi_1I]

In the Pre-Equilibrium Approach does the slow step always determine the rate law?

In any reaction systems, the slow step will always determine the rate of reaction, thus the rate law.
The purpose of using Pre-Equilirbrium Approach is to substitute other terms with intermediate, which is not supposed to be in the Rate Law.
Suppose you have two-step reaction and
the rate limiting step is Step 1: Rate=k(1)[R], in this case you don't have intermediates in rate law and Rate=k[R] as itself is reaction rate law.
the rate limiting step is Step 2: Rate=K(2)[R], in which [R] is product of step 1, meaning [R] in step 2 is intermediate. This time, you use Pre-Equil. Approach to get rid of intermediate.

Hope that makes more sense!

Can you include a catalyst in the rate law?

[Kristina98]

it's okay to have a catalyst in your rate law

[Kira_Maszewski_1B]

In a two step reaction, if there is a initial reactant that shows up as a final product, is it a catalyst? Meaning the reactant was unchanged and unaffected?

[Wenqian_Deng_1L]

Yes, @kira, if we have a reactant that shows up as a final product, we know it is a catalyst because it was not an intermediate that was "used up" in the reaction.