So TBR says
Chad says
which one is right?
Chad says
which one is right?
RDS TBR vs Chad?
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They're both right. Chad is a bit more clear, but TBR means the same thing.
The step with the highest activation energy means the step where the hill is the highest. The hill is the free energy change.
For example.
The first step is the RDS even though the second step technically is at a higher energy transition state. The free energy change between the nadir (intermediate) and the activation energy of the second step is not as high as the free energy change between the reactant and the first transition state.
We say step one has a great activation energy (energy required to climb the hill) even though the energy level of the second transition state (Ea of 2nd step) is higher than the first.
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Actually medpr if I remember correctly the second step in that picture is rate determining.
I believe you are correct. The first step would be the slowest, while the second would actually be the rate determining since it's at a higher energy level. Usually, the slowest is the rate determining, but not in this case. I think...
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The RDS is always the slowest step. In that picture, the slowest step is the first step so it is the RDS. Step 2 will consume the products of step 1 (the intermediate) faster than step 1 can manufacture them. This means that step 2 is faster, and is not the RDS.
It's the same concept as a team is only as strong as its weakest link. Step 1 is the weakest link in terms of rate, so the reaction is only as fast as step 1.
It's the same concept as a team is only as strong as its weakest link. Step 1 is the weakest link in terms of rate, so the reaction is only as fast as step 1.
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No. That doesn't make any sense. Higher NRG level doesn't matter, step 2 proceeding faster than step 1 is physically impossible because it is reacting with the products of step 1; step 1 must be the RDS.
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How can step 2 be the RDS if it is faster than the slowest step?
The overall reaction can only proceed as fast as the slowest step. It doesn't matter what order the steps come in, the slowest step is always the RDS.
Say step 1 was fast and step 2 was slow. Step 1 would produce a ton of intermediates that would accumulate in solution because step 2 was too slow to consume them all. Step 2 is the RDS.
If step 1 is the slow step and step 2 is the fast step, the rate of step 2 depends on the rate of step 1 because step 2 requires the product(s) of step 1 to proceed. If step 2 is waiting on step 1, how can the overall reaction depend on step 2? It can't. Step 1 is the slow step, and again, the slow step is the RDS.
This confusion is the whole point of this thread. The highest energy doesn't matter. It's the highest energy change that matters. Step 1 could require a transition from 1kJ to 50billion kJ and step 2 could require a transition from 40billion kJ to 80billion kJ. Step 1 would still be the RDS even though step 2's transition state is at a higher energy level.
Are you absolutely sure about this? The rate determining step is always the one with the highest energy transition state. I'm not sure where I learned this (may have been Chad, actually) but
Highest Hill - Rate determining Step
and
Largest Energy of Activation - Slowest Step
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RDS is just a fancy phrase for slow step. Just like entropy is just a fancy word for disorder. Think about the above rate diagram. If step 1 produces 1X every 10 seconds, but Step 2 takes that 1X and makes 1Y every 2 seconds, how can step 2 be the RDS if step 2 has to wait on step 1?
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This would only be true if the reactions you are comparing are starting from the same energy level, which is not the case here. The step with the highest activation energy is always going to be the slowest step which is always the RDS, all things being equal.
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The pic is a Sn1 reaction. The RDS is step 1. As MedPR says, you need to look at the change in energy. The initial increase in energy is from breaking a bond, formation of the carbocation. So the RDS in this reaction is the formation of the carbocation, the nuc will not attack unless the carbocation is formed so than RDS is the first hump.
The intermediate formed is lower in energy than the second one because it rehybirdzed to sp2 (from sp3), then the next intermediate is the formation of a new bond and rehybirdzing to sp3.
(I did this chapter a little while ago in TBR Orgo Book 1, pg 211-212 for reference)
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I see what all of you guys are saying, and it makes sense. And 99% of the cases we see aren't like this. But Chad pointed out specifically that it wasn't the case for an energy diagram like the one drawn above...but alright, I guess I see what you guys are saying. I'll go with y'all
Wiki
"The concept of the rate-determining step is very important to the optimization and understanding of many chemical processes such as catalysis and combustion.
In a reaction coordinate, the transition state with the highest energy is the rate-determining step of a given reaction."
So then does that mean the rate law would be that of the second step with the higher energy transition state?
"The concept of the rate-determining step is very important to the optimization and understanding of many chemical processes such as catalysis and combustion.
In a reaction coordinate, the transition state with the highest energy is the rate-determining step of a given reaction."
So then does that mean the rate law would be that of the second step with the higher energy transition state?
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I know everyone is going to start saying "DON'T TRUST WIKI" but I think it's right. So, like I said earlier.
Highest hump - highest energy/ rate determining step
Highest Ea - slowest step
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When you read "highest energy" you should be reading "highest energy change." That should clear this all up.
Just like F=ma is technically net force equals ma.
Just like F=ma is technically net force equals ma.
Simply put, RDS is the slow step=most energy change, in this case the first step in the picture.
EDIT: medpr SO True concerning newton's 2nd law, kinda sad how long it took me to get that.
EDIT: medpr SO True concerning newton's 2nd law, kinda sad how long it took me to get that.
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Not sure how its still alive, lol can we get a close on this thread.This topic has been beat to death and after reading this, if anyone gets this question wrong on the mcat they should go and [insert some horrific fate] 😀
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The F=ma thing took me FOREVER too. It made physics so hard in the beginning.
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I know this thread a few months old but I wanted to add my two cents so it doesn't mislead future SDNers doing a search on this topic. Typicalindian and fas are correct; the second step is the rate determining step. After step 1 occurs, you are left with some intermediate. Now the intermediate has to choose between going left back towards the reactants (which has a smaller energy hill) or going right towards the products (which has a larger energy hill). The intermediate will choose to go towards the reactants since this requires less energy. Therefore, it doesn't matter how quick or slow step 1 is occurring. The intermediates that form will choose to go back to the reactants, which makes step 2 the rate determining step.
