# Rate Determining Step for Reaction Coordinate Diagram

Discussion in 'MCAT Study Question Q&A' started by Graffiti, 04.30.11.

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1. ### GraffitiBanned

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Let's say you're looking at a reaction coordinate diagram for an exothermic reaction. The reaction has two transition states, the first transition state having a lower hill than the second transition state. Also, the first transition state has the larger activation energy compared to the second transition state. The first transition state would be the slow step because it has the larger activation energy. The second transition step would be the rate determining step. This confuses me because I always thought that the slow step was the rate determining step. Can anyone explain this?

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3. ### Rabolisk

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While there can be more than one rate-limiting step, usually the slowest step is described as such. Not sure why the second step would be considered the rate-limiting step.

4. ### cheesier 2+ Year Member

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The rate determining step is always the one with the highest energy transition state. Why this is, I'm not really sure. Hopefully somebody can explain it.

Last edited: 05.01.11
5. ### GraffitiBanned

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um no, it was just a hypothetical situation explaining that the rate determining step is the one with the highest hill. i just didnt understand why its rate determining since it doesnt have the largest activation energy.

6. ### Rabolisk

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What is the source?

7. ### cheesier 2+ Year Member

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The question is correct. It doesn't matter what extra details there are. If there are 300 transitions states, the one with the highest transition state energy is the rate determining step, 100% of the time. Energy of activation does not matter.

8. ### cheesier 2+ Year Member

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Yes, this is correct because even though the first step is slower, the second step is the one with the higher energy transition state. I'd be curious to know if there's any real life reaction in which this scenario would actually play out, but the question is right.

Actually, I'm thinking about it and while the question is still right, I think it would be impossible to draw out a reaction coordinate diagram that actually showed a first slow step followed by a faster step with a higher energy transition state. The takeaway is that high transition state = rate determining step = slowest step in a RCD that you could actually draw.

9. ### daftpatton 5+ Year Member

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Graffiti,

I've seen the diagram you are explaining before in my studies that showed such results. What helped me to understand and retain the concept and it's reasoning is to just remember the following:

Highest Hill = Rate determining Step

and

Largest Energy of Activation (Ea) = Slow Step

Of course, the two are usually the same, being the highest hill with highest energy of activation, but remember this: "If the first hill goes downward after it's peak, dips and then goes up again to reach another transition state that will be higher than the second hill, the Energy of Activation for that second step will begin at the lowest point right after the first hill. This is what makes a higher energy transition state possible with a lower Ea. The Ea may actually be higher in the graph for the second hill, but it's length is not as long as the first hill, which required more energy."

So, in attempt to summarize, the first hill (reaction) took a much longer time to occur due to it's larger Ea, but since energy was already invested and did not decrease all the way back to the baseline where it began before starting another hill, it does not require a larger amount of additional energy then the first Ea amount to create a higher hill. The second hill's Ea is essentially being added onto a point of where it is already at. Since the second hill is the highest, it is the rate determining. This is because even though it did not take the longest to occur, it is the highest point in the graph and the completion of the reaction depends on getting to that highest point, since it is the "overall peak of energy".

The opposite of this would be a graph with a high hill, dipping about halfway, followed by a lower hill. Going from right to left, measure the hills' Ea's from where the hills begin to go upward until they reach their peaks. The first will have the largest Ea, which makes it the slow step for requiring the most energy in a single step, as well as the highest hill, which makes it the rate determining step since it is the point in the graph with the highest overall energy.

Let me note, when I originally wrote this, I said that the "highest" Ea = the slow step. And came back and edited it to say the "largest" Ea, only because this seems to create confusion for these types of problems. The largest is the the hill with the length from the starting point of the hill (not necessarily of the entire reaction) to the top of the hill. So the only things you need to pay attention to when asked to determine the rate determining step and slow step are: 1.) The highest hill and 2.) which hill has the largest Ea from the hill's low point, not the beginning of the overall reaction, to high point.

I sincerely hope this helps and apologize for writing you a short novel.

Last edited: 05.07.11
10. ### blackandgold1

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But wouldnt the highest hill also have the largest energy of activation.. since it is furthest from the energy threshold? I always thought the higher the hill, the further away from threshold, so the more activation energy needed? Which would mean the higher the hill, the larger the activation energy, the slower the step = rate determining step? Im confused now.

11. ### Mcat35 2+ Year Member

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thats what he said isn't it?

but anyways check out this link for the SN1 reaction

Since the an SN1 mechanism consists of two elementary steps, there are two transition states in the reaction coordinate diagram. Since the energy of activation for the first step is so much higher than that for the second step, the first step of the an SN1 mechanism is the rate-limiting or rate-determining step.

http://guweb2.gonzaga.edu/faculty/c....cfm?L21resource=reaction-coordinate-diagrams

12. ### Biebs 2+ Year Member

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Let me see if I can clear this up for you.

Take two-step exothermic reaction where the FIRST transition state is HIGHER than the second transition state:

The step with the higher activation energy, is the slow step (The first transition state). Because it has the higher activation energy, it will have the lower rate constant. The rate determining step would be the first step as well. Typically, the slow step is the RDS, and the slow step is the one with higher activation energy.

However, in the case *I think you're describing* where you have a two step exothermic reaction where the first transition state is higher than the second (second hump is higher than the first). The first step has the higher activation energy--this is the slow step. However, the RDS is the SECOND ONE. The RDS is the one that has the overall highest hill in the reaction. This is because the slow step is relatively high up, but the second hill is even harder to process through (higher hill--the reaction could either continue in the forward direction or reverse directions and travel back through the first step). Because the second transition state has the highest hill, it is still determining the overall rate of the reaction because it was higher, it was easier to go back and you still have not made any product. Even though it is not the overall highest activation energy step (slow step), it is still the RDS, because it is the highest hill. This is not usually the case, but just the "rule."

The overall, highest hill is typically the RDS, this is usually the slowest step, but as you saw in the second example--it doesn't have to be.

I hope I didn't make things even more complicated.

Last edited: 05.08.11
13. ### Biebs 2+ Year Member

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Remember the activation energy is the energy barrier that reaction has to cross. It is measured from step to step. (IE your starting point is different depending what Ea you're measuring).

For example, in a two step exothermic reaction where the first transition state is higher than the second--the first activation energy is measured from the reactants to the first transition state.

The second activation energy is measured from the intermediate (NOT reactants) to the second transition state. So, it is very possible that the second step can have a lower activation energy than the first, which is usually the case.

Just make sure your starting points are where they are supposed to be. If you were always starting your measurement for Ea from the reactants, then the second step would always have the highest activation energy--but this is not the case.

14. ### daftpatton 5+ Year Member

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Do not measure from the energy starting point, the easiest way for each "individual step" is to measure it from where each hill begings. The highest hill may actually be a "step with a smaller Ea" right after a "step with a larger Ea". The second step didn't necessarily require all of the energy that the first step did, and vice versa. Just measure from the bottom of each hill to it's top, those will be separate steps and you just compare their lengths. The longest step in length will be the slow step because it requires the most energy as an individual step.

15. ### GraffitiBanned

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Thanks everyone!! This makes more sense now

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