Reaction Rate

[MedPR]

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To measure the rate of a reaction, which of the following CANNOT be observed?

A. rate of appearance of product
B. rate of disappearance of a reactant
C. change in intermediate concentration over time
D. the change in reactant and product concentration as a function of time

TBR says C.

However, if the reactant you are looking at is not part of the rate determining step, you won't be able to accurately determine the reaction rate. For instance, in a two step reaction with a fast first step and slow second step, you could be consuming first step products like mad without ever producing any product.

I understand how C is a correct answer to the question, but B should be correct as well. Right?

 

[milski]

I don't think you can have intermediate in excess - it usually is not stable enough to hang around. You make only as much of it as you can consume in the next step of the reaction.

 

[chiddler]

sure you can measure rate of disappearance. in my biochem lab, we measured rate of disappearance of NADPH as the reaction proceeded. this was done spectrophotometrically so lower absorbance meant less NADPH in solution.

 

[MedPR]

I don't think you can have intermediate in excess - it usually is not stable enough to hang around. You make only as much of it as you can consume in the next step of the reaction.

If the first step is the fast step, and the second step is the slow step, the amount of intermediate will build up won't it?

Either way, I know why C is correct, I just feel like B should be correct as well.

 

[silverice]

I don't think you can have intermediate in excess - it usually is not stable enough to hang around. You make only as much of it as you can consume in the next step of the reaction.

I think you can have intermediate in excess. If the reaction is two steps, Intermediate is basically the product of the first reaction. If the reactant of the second reaction or enzyme is not present then the first reaction would be in equilibrium. What is too unstable to hang around is the transition state. That's the definition I have in mind. It's like in bio when you have a chain of reaction to make a protein and a build up of a particular intermediate usually results in a defective enzyme or a feedback mechanism.

Regarding to the original question that this post asked. If the first intermediate takes a long time to convert then it means that it builds up, then this would be a problem of experimental design. To accurately measure then reaction rate one should know that this is not a elementary reaction aka single step reaction and therefore one would measure not only the disappearance of reactant but also the appearance rate of the product.

 

[silverice]

If the first step is the fast step, and the second step is the slow step, the amount of intermediate will build up won't it?

Either way, I know why C is correct, I just feel like B should be correct as well.

I agree both could be right. But b is right only when we are in a particular situation and that the intermediate is build up enough for one to detect. But usually everything is throwing in a tube including all the enzymes that means that the intermediate would be quickly consume. This is choice c. But reactant always starts at large amount as one would do when designing the experiment, no matter how fast it is disappearing we would still have data allllll the time. As in how representative is it for the appearance of product, we don't know.

 

[MedPR]

I agree both could be right. But b is right only when we are in a particular situation and that the intermediate is build up enough for one to detect. But usually everything is throwing in a tube including all the enzymes that means that the intermediate would be quickly consume. This is choice c. But reactant always starts at large amount as one would do when designing the experiment, no matter how fast it is disappearing we would still have data allllll the time. As in how representative is it for the appearance of product, we don't know.

You would have data all the time about an intermediate as well. You might not be able to isolate the intermediate, but you can certainly detect its existence. In a multi-step reaction you could have 0 of a reactant for a very long time even though the reaction is proceeding. For instance in a 3 step reaction where the first two steps are both slow.

A+B-->C (slow)
C+D-->E (slow)
E+A-->Product (fast)

You could be monitoring the reaction for E and not find any for hours (depending on how slow the first two steps are). Doing this wouldn't help you figure out the reaction rate at all.

I know C is a better answer, but I wanted to share it due to the confusion that might arise about answer B.

 

[silverice]

You would have data all the time about an intermediate as well. You might not be able to isolate the intermediate, but you can certainly detect its existence. In a multi-step reaction you could have 0 of a reactant for a very long time even though the reaction is proceeding. For instance in a 3 step reaction where the first two steps are both slow.

A+B-->C (slow)
C+D-->E (slow)
E+A-->Product (fast)

You could be monitoring the reaction for E and not find any for hours (depending on how slow the first two steps are). Doing this wouldn't help you figure out the reaction rate at all.

I know C is a better answer, but I wanted to share it due to the confusion that might arise about answer B.

Yeah it could be a tricky one. Thx for sharing

 

[milski]

I think you can have intermediate in excess. If the reaction is two steps, Intermediate is basically the product of the first reaction. If the reactant of the second reaction or enzyme is not present then the first reaction would be in equilibrium. What is too unstable to hang around is the transition state. That's the definition I have in mind. It's like in bio when you have a chain of reaction to make a protein and a build up of a particular intermediate usually results in a defective enzyme or a feedback mechanism.

Regarding to the original question that this post asked. If the first intermediate takes a long time to convert then it means that it builds up, then this would be a problem of experimental design. To accurately measure then reaction rate one should know that this is not a elementary reaction aka single step reaction and therefore one would measure not only the disappearance of reactant but also the appearance rate of the product.

It might be time for me to refresh on intermediate vs transition state. If you had asked me a few hours ago, I would have told you that they are the same thing.

If you are talking about a really stable intermediate, are not you really talking about trying to run two reactions simultaneously?

 

[MedPR]

It might be time for me to refresh on intermediate vs transition state. If you had asked me a few hours ago, I would have told you that they are the same thing.

If you are talking about a really stable intermediate, are not you really talking about trying to run two reactions simultaneously?

Hmm.. Even if the intermediate is unstable, your best bet to accumulate an appreciable amount of intermediate is to carry out a two step reaction where the first step is fast, and the second step is slow.

So you would have a small hump, followed by a nadir, followed by a huge hump. In that nadir, intermediates would have nowhere to go (kind of) because they would need some energy to reform reactant, and even more to form product.

I don't know if this is even important, but it was the concept behind a question I did not too long ago. "In which situation would you produce the maximum amount of intermediate" or something to that effect.

 

[silverice]

Hmm.. Even if the intermediate is unstable, your best bet to accumulate an appreciable amount of intermediate is to carry out a two step reaction where the first step is fast, and the second step is slow.

So you would have a small hump, followed by a nadir, followed by a huge hump. In that nadir, intermediates would have nowhere to go (kind of) because they would need some energy to reform reactant, and even more to form product.

I don't know if this is even important, but it was the concept behind a question I did not too long ago. "In which situation would you produce the maximum amount of intermediate" or something to that effect.

New knowledge is always good.

 

[chiddler]

Hmm.. Even if the intermediate is unstable, your best bet to accumulate an appreciable amount of intermediate is to carry out a two step reaction where the first step is fast, and the second step is slow.

So you would have a small hump, followed by a nadir, followed by a huge hump. In that nadir, intermediates would have nowhere to go (kind of) because they would need some energy to reform reactant, and even more to form product.

I don't know if this is even important, but it was the concept behind a question I did not too long ago. "In which situation would you produce the maximum amount of intermediate" or something to that effect.

this is interesting.

 

[MedPR]

Here's the question I was referring to that lead me to the conclusion about intermediate concentration.

It's #76 in TBR Chem chapter IX. It is a discrete.

Which of the following is NOT true about an intermediate[/b]

A. The intermediate is in its highest concentration just after the start of the reaction.
B. Intermediates have finite lifetimes
C. An intermediate is in its highest possible concentration when it is formed after the rate determining step of the reaction.
D. An intermediate is in its highest possible concentration when it is formed before the rate determining step of the reaction.

My thought process: Intermediates of course have very finite lifetimes, so B is out. I wasn't sure about A, so I ignored it at first. Since C and D are opposites, I was pretty sure one of them was the right answer. I thought about it in terms of a reaction coordinate diagram, and if your intermediate is formed after the rate determining step, that means it will probably be consumed by a fast (or at least faster) step than the slowest step, so it won't have time to build up (this is answer C). For answer D, if the intermediate is formed before the slow step occurs, it might be able to accumulate because the slow step won't be consuming it as fast as it is produced. Again, since C and D are opposite answers, I figured one of them had to be the right one.

C is correct.

Here is the explanation.

C is correct. The intermediate is in its highest concentration just after the start of the reaction, because it is formed from the reactant but is not yet used up to form the product. I'm still not really sure that I believe this, because it will depend on exactly how soon after the start we are talking about. Remember that the concentration of intermediate starts at 0, plateaus for a while, then drops back to 0. As the reaction proceeds toward completion, the intermediate is depleted. Choice A is valid. Intermediates are defined as species that have finite lifetime and exist only during the course of the reaction. If the lifetime is too small to be measured, then the species is referred to as a transition state. Choice B is therefore valid. An intermediate builds up when it collects before the rate-determining step. This makes choice D valid and choice C invalid, which means the latter is the best choice.

 

[milski]

Thanks for posting that. I agree that you will build up some amount of intermediate before the slow step. But that amount is limited by its finite life - at most it's going to be lifetime*(rate of creation). You can say that you'll always have enough for the next (slow) step to proceed but you cannot use how much you have to determine the rate of the reaction.