Reaction mechanisms, catalysts, and reaction coordinate diagrams are going to be the topic of this lesson. We'll start off with reaction mechanisms and learn how to take a sequence of steps that composes a mechanism and turn that into a rate law. And then we're going to talk about catalysts and some of their characteristics, and then finish the lesson off looking at reaction coordinate diagrams, highlighting the activation energy and things of that sort. My name is Chad, and welcome to Chad's Prep, where my goal is to take the stress out of learning science.
Now, in addition to high school and college science prep, we also do MCAT, OAT, and DAT tests. You'll find those courses at chadsprep.com. I'll be sure to leave links in the description. Now, this lesson's part of my new general chemistry playlist. I'm releasing several lessons a week throughout the school year.
So if you want to be notified every time I post a new lesson, subscribe to the channel, click the bell notification. All right, so let's have a little fun here. So reaction mechanisms. So reaction mechanisms are a sequence of steps, and these steps always have to be what we call elementary reactions. And you might think of elementary, my dear Watson, from like Sherlock Holmes, and elementary here means simple.
And so in saying elementary reactions or elementary steps, they are steps that cannot be broken down any further than what they are. And so most reactions have more than one step and they're composed of more than one step. And so, and that's actually why you can't usually just look at the coefficients in a balanced reaction and get the rate law because it turns out it's a combination of steps. And what you always, what turns out want to focus on is what we call the slow step or the rate determining step, that slowest step in the whole reaction.
So Let's talk about these elementary reactions. We'll come back to that concept of slow step in a minute. So, but these elementary reactions come in three classes. Unimolecular, bimolecular, termolecular. So, and that just comes from the fact of how many reactant molecules do you have in that step.
If you only have one, it's unimolecular. If you only have two, it's bimolecular. And if you have three, it is termolecular.
Well, what if there's four? Chad, great question. I'm glad you asked that. There's no such thing.
And so it turns out For these elemental reactions, like let's say you've got a bimolecular, what happens is for a bimolecular elemental reaction to take place, the two reactant molecules need to collide. And so it turns out for a term molecular, you'd actually have to have three reactants all colliding and converging on the same space at the same time. So and that turns out is a very rare event, and so these term molecular reactions are not very common. But like to get more molecules than even three all colli- converging on the same point in space at exactly the same moment is almost essentially impossible.
It's never going to happen. And so we don't even have any classes past term molecular, and the term molecular themselves are pretty rare. So let's say I told you that, you know, on my way to work today, I saw two cars collide. What are the odds? Well, I mean, they're not terrible.
It's not really the most outlandish story. But what if I told you that eight cars coming from eight different directions all converged on the same place and collided at exactly the same moment? Well, we like look around and look for cameras. Is this a movie? Like what's going on?
That would have to be rigged. It's never going to happen spontaneously, right? All on its own without it being rigged. And same thing with molecules. You're never going to have, it turns out, more than three molecules colliding.
And that's why there aren't anything more than termolecular elementary steps. All right. Now a mechanism is composed of one or more of these elementary steps.
And sometimes, you know, an overall balanced reaction is simply an elementary reaction. It's just that single step. And that's how it works. And life is good when that's the case, but most of them are going to be a combination of multiple elementary steps to make it. So take a look here.
Here I've got a mechanism, and this is a two-step reaction. And we've got some questions to answer in this case. And the first off is just if these are the two steps that compose the overall reaction, well then what is the overall reaction? And so what we're going to do is we're just going to add up these two steps together. We're going to put all the reactants on one side and all the products on one side, then anything that shows up on both sides essentially or effectively cancels out of the reaction.
So in this case, we're gonna have H2 plus PT plus H2PT plus C2H4 goes to H2PT plus C2H6, and then finally plus PT. And if we look across both sides, we see that in this case, first PT, just plain old PT by itself shows up. on both sides.
And then we've got this H2PT complex here, this hydrogen platinum complex, that's also showing up on both sides. And so the overall reaction is now what we have left. And so in this case we're going to have H2 plus C2H4 going C2H6.
And so this is one of the questions you might get is what, you know, given this mechanism, what's the overall reaction? Well, there it is. There's our overall reaction.
So the other question you might have to answer here, the one I'm going to ask you here, is identify catalysts and intermediates. So and it turns out that anything that cancels out, so when you add the steps together, anything that cancels out on your way to the overall reaction is either going to be a catalyst or an intermediate. And so let's take a look at this and let's look at that word intermediate.
An intermediate, and I'll write that out, So intermediate literally means in the middle. And so an intermediate is only present in the middle of a reaction. You don't start with it, and you don't end up with it.
You make it in an early step, but you use it up, you consume it before you get to the end of the reaction. And so it's not a reaction that you start with, but it's not a product that you end up with either. And so typically with only two steps it's easier to identify, but it has to be produced in the first step, but then used up in the second step before the reaction's over. And again, it's not gonna show up in this overall reactant.
So we can see that our intermediate in this case is this H2PT complex made in the first step, so used up in the second. Okay, now the other thing we've gotta identify is a catalyst. All right, and so in this case, a catalyst is the exact opposite of how we identified an intermediate.
An intermediate was a product first. and a reactant second. Well, a catalyst, it turns out, the way we're going to identify it is that's going to be a reactant before it's a product.
And so a catalyst is actually a reactant in the reaction. However, we're going to find out a characteristic of catalysts here soon is that they are not consumed in the reaction. So if it's going to get used up early on, it's going to have to get regenerated, reproduced before the reaction's over.
And so in this case, it's both a reactant in an earlier step, but a product in a later step. That way, again, it doesn't show up in the overall reaction here in this case. canceled out. And so in this case, that catalyst is simply the platinum. So in one way, I like identifying catalysts is you can say that a catalyst is a reactant first and a product second.
And so you can say crap to remember that. So a catalyst is a reactant first and a product second. If that helps you, great. And if it doesn't, well, say lovey.
All right, so from any sort of mechanism now, you should be able to get the overall reaction. You should be able to identify both intermediates and catalysts as well. Cool, let's make this a little harder now.
So we're going to put another mechanism up here and actually figure out the rate law based on this mechanism. Let's take a look. All right, so this is the mechanism we're going to look at here.
It's got two steps. We're first going to have O3 break apart into a molecule of O2 and then a single O atom here. So, and then we're going to have that O atom combined with another molecule of O3, which is ozone.
So to form two O2 molecules. And if you add up these two steps together, you'll find out that the O's are going to cancel. And in this case, it's a product first and a reactant second.
So notice it's not a reactant first and a product second, but a product first and a reactant second. You don't start with it. You don't end up with it.
It's only present in the middle of the reaction. That's an intermediate. And again, the intermediate, again, doesn't show up in the overall reaction. If you add these two steps together. Here's your overall reaction.
You have two ozone molecules, O3, producing three oxygen molecules, diatomic oxygen, in this case O2. Okay, so there's our overall reaction. We've identified an intermediate.
A couple of things you might look at is you might look at the first step and say, oh, there's just one reactant molecule that is unimolecular. But in the second step, we've got two reactant molecules, and so this would be bimolecular. And then neither of the steps is term molecular. Again, those are pretty rare.
All right, so when you're given a mechanism, first thing you should know... is that every step in a mechanism is an elementary step. And so it turns out when you've got an elementary step, you can just look at the coefficients of the reactants and figure out the rate law.
So, and it turns out, again, you can only do this for elementary reactions. So back in the last lesson, we talked about overall balanced reactions. We said, you just can't look at the coefficients and assume those are gonna be the orders in the rate law.
Well, here's the caveat, and I pointed towards it in the last lesson. If you know a reaction is an elementary reaction, well then yes you can. The coefficients will be the orders in the rate law.
So how do you know if a reaction is an elementary reaction? Well you've really got two main ways. One if the question states in this elementary reaction, if it's just plain told to you that it's an elementary reaction, well then you know. So or if it's one of the steps in a mechanism, mechanisms are always composed of just elementary reactions so it would have to be an elementary reaction. So In this case, this is a mechanism, a sequence of steps.
Both of these have to be elementary reactions. And we could figure out both of their rate laws as a result. And so like if we take a look at this first one, so rate law here would be K times concentration of O3.
And you know what? I'm going to give myself a little more room on that overall reaction. Well, you've seen it now. Don't need it anymore anyways. So the second one here would be...
Rate equals k times concentration of O. times concentration of O3. That would be the rate law for the second step.
This would be the rate law for the first step. Again, it's only reactants that typically show up. And however many show up, that's going to be the order.
So for O, there was just one of them. And that's why it's just to the first order, first power. Same thing with O3 here.
And same thing. Oh, I lost. I forgot to erase the extra two. Same thing for O3 here, just one of them. Had there been two, then we would have an exponent of two, and it would have been second order, so on and so forth.
Now, here's the deal. I can write rate laws for both steps, but they don't all... always matter.
And so it turns out that what we call the slow step is the rate determining step. And you can kind of think of it this way. So let's say me, you and your best friend are going to form an Olympic relay team because we're like Olympic athletes.
So, and I'm going to go around and run my quarter mile and, you know, and I'm in amazing shape for a 45 year old. So I do it in 60 seconds. Uh-huh.
And you're in even more phenomenal shape. So you do yours in 57 seconds. And then your best friend, just one tick off is at 58 seconds.
And so we need an anchor though. And so we just don't know who to ask. So we ask my four-year-old son. And we get my four-year-old son and, you know, he starts running around the track and he's like, oh, squirrel.
So and then he's running around the track. He's like, I just want to play army man. Or, you know, it takes him, it turns out, three hours to get around the track by the time he finally makes it. Okay. So in this case, how long is our overall relay time?
Well, my son. My four-year-old son is at three hours, and each of us is really close to a minute, so it's like three hours and three minutes, which for all practical purposes is just three hours. Now, if me, you, and your best friend all work on our relay times and shave off a couple seconds, does it really matter? No, because my son is really what's limiting this race, that crazy kid, and so what we really need is for him to improve his time, because effectively his time is pretty much the rate of the whole reaction, because it's so much slower. than all the other steps.
And for a lot of reactions, that's kind of how it works. You have one step that's much slower than all the other steps. And so that slow steps rate is the overall rate. And so we call the slow step the rate determining step. And so being given a sequence of steps, so you might need to be told if I want you to get the overall rate law for this reaction, I have to tell you which one's slow.
And so in this case, there's really two options. You're either going to have the slow step as the first step, or you're going to have the slow step as the second step. And so... Again, you really, I mean, in principle, you could have the slow step being a third step or fourth step in some more complex case. It's just nothing you're likely to ever encounter in your general chemistry course.
So the two scenarios you're likely to see are the slow step is the first step and the slow step is the second step. So let's go with option number one first. And let's just say that the first step is the slow step, which would make the second step the fast step.
So, and when your first step is the slow step, that is the best possible news. If you saw such a question on the test where they gave you a mechanism and they labeled the slow and fast steps and you were supposed to figure out the rate law, if the first step is the slow step, you should go up and hug your professor real tight and say, thank you, thank you, because this is the easier of the two scenarios. Because it turns out that everything after the slow step is not going to affect the rate at all.
Everything after that slow step. And so in this case, if the slow step is the first step, well then everything after it, i.e. the only other step, is not going to affect anything. And so it turns out in this case that the rate law of that slow step is the entire rate law.
So in this case, this rate equals k times O3 is not just the rate law of the first step, it would be the rate law of the entire overall reaction. So again, we had that overall reaction we had written earlier where it was 2 O3 going to 3 O2. And it turns out if, again, this was the sequence of steps and the first one was slow, the rate law would just be simply K times O3.
Life would be great. Okay. However, if instead the first step is your fast step, or sometimes they label it rapid equilibrium, and your second one is the slow step, now instead of getting up in the middle of that exam and hugging your professor profusely, you should get up and scowl at them and give them a... dirty look because they've just done you wrong because this is a much worse situation. This is a much more challenging problem now.
It is not simply just going to be the first steps rate law and it's also unfortunately not simply going to be the second steps rate law either. It's going to be more complicated. So let's take a look at the way this works here.
So your slow step is still the rate determining step and we're going to start with the slow steps rate law. That's where we start. So I'm going to write that out here. Rate equals k times, whoop, let's try and get that right, times O, times O3.
But that's not actually good enough here, because it turns out the first step affects the slow step. Because if you look, we've got an intermediate in our slow step, the O. And where does that O come from?
Well, it's produced in the first step, and that's why the first step affects the second step. Because how quickly the first step can produce the O is going to affect how quickly the second step can go. And so that's why we can't ignore the first step in this case.
Again, if there had been a third step, we'd get to ignore that because everything after the slow step doesn't affect the rate. All right. So, but this isn't going to be good enough since both steps affect the rate and both have rate constants. Well, these aren't the same k value.
They're two different rate constants. And so to distinguish them, we often go k1 and k2 for the rate constant in step one and the rate constant in step two. And so I better put a two right here in this case and get a little more specific. I didn't worry about that before because it was only the first steps rate law that showed up. And so we only had to worry about one K value.
Well, we're going to have to worry about both of them. In fact, one more, it turns out, on top of those two as well. So, but this is where we're going to start. This is our rate law. So, and that's where we go.
Now, let's say that some company hired you to figure out the rate law of this reaction. They want to know, you know, if we produce too much O2, well, that's explosive. And if we produce it too fast, our power, you know, our plant might explode and things of this sort. So.
How do we limit its production? How do we keep this reaction from going too fast? And you come back and be like, well, I figured out here's your rate law.
And so if you keep the O3 concentration low, that'll slow down the rate. And if you keep the O concentration low, that'll slow down the rate. And they just look at you and they say, okay, well, we can limit how much O3 we put in that reactor. So we can keep that low, but there's no O in our reaction. We start with O3, we get O2 out.
What are you talking about this O concentration? What is that? Well, again, that was an intermediate in the reaction.
It's not a reactant in the overall reaction. It's not a product in the overall reaction. It's not even in the overall reaction. And so by telling them to limit its concentration, they're like, well, we never put it in the reactor to begin with, and we don't get it back out.
What are you talking about? And you get fired. And so this is the way it works. I mean, this is not wrong.
It's just premature because technically, an intermediate should not show up in a properly written overall rate law. And so that's got to go. This O here.
That's not acceptable in this rate law. The K2 and the O3, no problem there. K2 is a rate constant. O3 is a reactant.
And reactants and products, no problem. They can show up in rate laws. But intermediates should not show up in a proper rate law.
And so something, some equivalent expression is going to have to be found for that concentration of that intermediate. And we've got a couple of different ways we can come about getting this. And so it turns out this first step is going to reach.
a rapid equilibrium. And what happens when you get a rapid equilibrium is that it's going in the forward direction and in the reverse direction at exactly the same rate. And so we can talk about this first reaction both in the forward and reverse reactions.
And it turns out the rate constants for those are different. So sometimes if all you have is one reaction, we might call the forward rate constant Kf and the reverse rate constant Kr. But if it's in the context of a mechanism like this, what they will typically do...
is give you a k1 and a k-1. So k1 meaning the forward rate constant for step one, and k-1 meaning the reverse rate constant for step one. Now, even though we write k1 and k-1, don't think that these have some special mathematical relation like, you know, one's positive and one's negative, like one's positive 0.01 and one's negative 0.01. Not true. Don't think that they're inverses, like one again is 0.01 and the other one's 1 over 0.01.
That's not true either. There's not some mathematical relationship implied by these labels. They're just labels. One means step one forward direction, and minus one means step one reverse direction. Okay, now if we say that this first step reaches equilibrium here, what that means, again, is that the forward rate is going to equal the reverse rate.
And we can write that as the forward rate law is going to equal the reverse rate law. And so the forward rate law has K1 in it, and then just the concentration of... Let's write that down over here. In fact, I still need this room back. So we're going to have K1 times concentration of O3.
There's our forward rate law for step one. And then if we looked in the reverse direction, these would now be the reactants that would show up in the rate law, and the rate constant would be K minus one. And so we'd have K minus one times concentration of O2. times concentration of O. And so here's the forward rate law, here's the reverse rate law, and again, the idea is that this first reaction's essentially going to finish before this second reaction ever really gets going.
And we're going to learn in the next chapter that a reaction doesn't really finish, like in some of the limiting reagent problems we did back in chapter 3. What often happens, what most typically happens, is they reach this intermediate state of equilibrium, where you still have some reactants around, you still have some products around, And the concentrations are no longer changing, not because the reaction is stopped, but because it's proceeding in both directions at exactly the same rate. And this will make more sense after the next chapter, which is often typically on this same exam with this material here in chapter 14. All right, so what this means then is that if the forward rate and reverse rates are equal, well then we can say that their rate laws are equal. And that's convenient because again, our overall rate law is the rate law from this slow step. We have to start there, but you can't have this intermediate in there. So the K2 is good to go, the O3 is good to go, but we need to find some sort of equivalent expression for our intermediate.
Well, we just wrote an equation that has that intermediate's concentration in it from the first step reaching some sort of rapid equilibrium, because it's the fast one. And so what we're going to do is we're going to rearrange this equality here to solve for the concentration of O, and then we're going to substitute it right back into our slow steps rate law. So again, we're always going to start with the slow steps rate law, but if it's the second step instead of the first, then you're not done.
And you've got to identify what your intermediate is, and there always will be one. And then you've got to go back to the first step and set the forward rate law equal to the reverse rate law, and then solve for that intermediate concentration. And so in this case, it's going to equal K1 times concentration of O3 all over, and we'll bring these and divide by them on the other side. So K minus one.
times O2. And since this all equals the concentration of O, that's what we're going to substitute back in to our slow steps rate law for that intermediates concentration. So if we go back and write this here, got a K2, and now I'm going to put a big parentheses here, we've got K1 times O3 all over K minus 1. times O2. So there's that whole expression we're substituting in for O, and then we still had this O3 concentration.
But then we can combine some terms here because we got an O3 times an O3, which is O3 squared. And so in this case, it is customary, it doesn't have to happen this way, but it's customary to put all the constants together at the beginning. And so in this case, we'll have a K1 times K2 over K minus one, you'll commonly see it that way.
And then times the concentration of... O3 squared all over the concentration of O2. Alright, and that is our overall rate law. And every once in a while, you'll see all these k's combined to just give something like a k prime or something like that. They might write it simply as just k prime.
Because what is a constant times a constant divided by a constant? Well, it's just some other constant. So...
It's been my experience that most of the time you're more likely to see this combination of rate constants, but it just could be written as a single k prime value, since a constant times a constant over a constant is just another constant. All right. Now, it is second order with respect to O3. So in here, you might be like, well, then it's first order with respect to O2.
That's not technically true. This is actually the negative one order. Notice another way we could have written this.
Could have written it like that, right? To the minus one power for O2. Dividing by O2 is the same thing as saying O2 to the minus one power.
So that's a minus one power, that is super rare. And notice that's also interesting because O2 is a product, not a reactant, and products don't routinely show up in rate laws. It's not that they can't, it's just they don't often do, but in some complex mechanisms like this one, they do.
And so in this case, being second order for O3 and negative first order for O2, we would say the reaction, the overall reaction order, just add these together, Well, two plus a negative one is just one. And so the overall order of this reaction is one, and we arrive at that in a very funky way. So having a negative order is not common, but it does show up, especially when your second step is the slow step in your mechanism and not the first. So once again, if your first step's the slow step, its rate law is the entire reaction's rate law. But if your slow step is the second step, you've got a fair amount of work ahead of you.
But just know that, first, that... You know, the slow steps rate law as initially written with an intermediate in it, you're probably not done. It's probably not going to be an acceptable answer.
So, and you've got a little more work to go to get a better expression. And notice this expression has a product in it, has reactant in it, does not have intermediates in it whatsoever. So now we've got to talk about catalysts and reaction coordinate diagrams.
We're going to start with catalysts, but we're going to allude to one of these reaction coordinate diagrams along the way. Now, a catalyst, simply put, is something that's going to speed up a chemical reaction. And the most common catalysts in your life are the ones inside your own body. And it turns out most of the chemical reactions going on inside your body are actually catalyzed by a variety of different enzymes.
And enzymes are biological catalysts. And most of these are... proteins, almost all of them in fact are proteins, and so these biological catalysts are protein catalysts, these are enzymes.
So, but we have a lot of, you know, non-biological catalysts we use in everyday kind of inorganic chemistry as well, and so first thing you should know is that a catalyst is going to speed up a reaction. Well, the question is then, well, how do they speed up reaction? Well, they speed up a reaction by lowering the activation energy of a reaction, so it turns out that a reaction is going to need some sort of energy even to get going. And we'll kind of take a look at this here.
And so if we take a look at this diagram here, so we've got energy plotted on the y-axis and what we call the reaction coordinate, and you'll commonly see Rxn as my abbreviation for reaction, but the reaction coordinate, or sometimes called the reaction progress, on the x-axis. And it just kind of follows the reaction as it goes from reactant to product in terms of energy. And so in this case, it turns out it's not just a straight line.
It turns out there's going to be a hill. you're going to have to get over to go from reactant to product. And the amount of energy it takes you to go from where you start on the reactant to the top of that hill is called the activation energy, abbreviated E sub A. Cool.
And so it turns out, you know, molecules, you know, typical molecules in the air, let's say, are all moving at some different speeds. And there's a certain average speed associated with that particular temperature. And as you raise the temperature, the overall average goes up, but you still have a big distribution of speeds.
And so at some particular temperature, you're going to have some molecules that have enough energy to get over this hill. And at higher temperatures, you'd have a greater percentage of molecules that have enough energy to get over that hill or that energy barrier, that activation energy. Well, what a catalyst actually does is it lowers the activation energy barrier. You get a different path to get from reactant to product that ends up with a lower activation energy.
And so a catalyst, first thing you should know is it speeds up a chemical reaction. How does it do that? By lowering the activation energy. Well, how does it lower the activation energy? Well, it turns out it's going to take a totally different pathway or mechanism for this reaction.
So it's gonna be the same overall reaction, but it's going to take a different pathway, a different mechanism. And so in this case, it turns out we're going to find out that one hill from reacting to product implies one step. But you might have, you know, two hills, it might go two humps as you go across, and that would be corresponding to two steps.
Well, your uncatalyzed reaction and your catalyzed reaction, they don't even have to have the same number of steps. But the thing that will always be true, again, is that a catalyst will speed up a reaction by lowering the activation energy by providing an alternate pathway or mechanism for the reaction to occur. All right.
So you should also know that a catalyst is not consumed in a reaction. And we saw this back in we talked about mechanisms and that that was why a catalyst did not show up in the overall reaction. It was a reactant early on and got used up, but then it got spit back out at the end. It got regenerated at the end so that it fell out of the overall reaction.
So a catalyst is not going to get consumed. So it turns out one catalyst can catalyze the same reaction over and over and over and over again. And so you don't actually have to add a lot of catalyst to get a whole lot of catalysis going on in your beaker or in your cells in your body or things of this sort.
So finally, you should also realize that a catalyst is not going to shift the equilibrium, and that's going to make more sense in the context of next chapter, but you're not going to get any more product than you'd otherwise get. You're just going to get the same amount of product you'd normally get, but you're going to get to that point faster. So that's kind of the deal.
Again, this idea of shifting the equilibrium, again, will make more sense when we get into the next chapter on equilibrium. Okay, so that's our catalyst, and all those parts of the definition of catalyst are important. So again, a catalyst speeds up a reaction.
By lowering the activation energy, by providing an alternate pathway or mechanism for the reaction to occur, catalysts are not consumed and they do not shift the equilibrium towards reactants or products at all. You're going to get the same amount of products as you would normally get, you're just going to get to that point faster. Okay, so we'll take a little more look at these reaction coordinate diagrams from here.
And so we've already kind of identified some things, but let's identify a few more and let's go big here. So again, you're going to have energy on the y-axis and you're going to have this reaction coordinate on the x-axis. And so first off here, I'm going to put the reactant at this energy level here and then product up here. And if your product is higher in energy than your reactant, this is the hallmark of an exothermic reaction. I'm sorry, this is the hallmark of an endothermic reaction.
So we've got to go uphill in energy. We need to absorb heat to get up there. And it turns out this difference in energy... That is what corresponds to delta H.
And so if you have to go uphill to get from reactant to product overall, delta H is positive, reaction is endothermic. However, if we reverse this, if the reactant had been higher energy than the product, then we would be releasing energy overall, and delta H would have been negative, and it would have been exothermic. So from these reaction chord diagrams, you should be able to identify both endothermic and exothermic reactions. All right, once again, let's put...
So our hill there, and again from where you start to the top of the hill, that is your activation energy. And so it turns out a couple other things we can identify here. So it turns out you got reactant here, you got product here, well at the top of the hill is kind of your commitment point.
Once you get there you have a chance to either fall back down to becoming going back and turning back into a reactant, or going forward and turning into a And we call this the transition state, which we abbreviate with this double dagger signal. And so that is our transition state. And sometimes that is actually referred to as the activated complex.
And sometimes, you know, some bonds need to be weakened or broken, you know, on the way to forming this transition state and stuff like this. Now, one thing you should know though, is that you cannot isolate this transition state. You can't just like freeze a reaction and hope to find what this transition state looks like. It is not something that's isolatable because we're going to compare that to intermediates in a little bit, which often are something you can actually isolate in the middle of the reaction if you freeze it.
So, but transition states you can't. So we've got activation energy here, we've got the transition state and the activated complex. And so it turns out the higher your activation energy, the slower the reaction is going to go, because you're going to have fewer molecules that have enough energy at whatever temperature you're at. to get over that activation energy barrier and form product.
Okay, so a couple other things we've got to identify here, and we're gonna draw one more of these lovelies so we can actually see them. So turns out again for every hill you have a step, so this would be an example of a one step reaction. Alright, so here I've got a second reaction here, and you'll notice what's different here is there is actually two hills, and that would be indicative of having two steps in the reaction. So this would be an example of an elementary reaction, and this was definitely not an elementary reaction because it can be broken up into two steps, not just one. And so when you've got more than one step here, then you're going to have more than one transition state.
But you're also going to have this local minimum in the middle, and that is an intermediate. And so in step one, your reactant turns into the intermediate, and in step two, your intermediate turns into the product. And that's the way this is going to work. And so you always have one fewer intermediates than steps.
So two-step reaction is going to have one intermediate. A three-step reaction would have two intermediates, so on and so forth. All right, now we've got two transition states here.
So however, you can also identify what is the slow step. So and if we look here from the first step, we start here and have to get all the way up to the top of the hill there. There's the activation energy for step one.
For step two though, we're starting at this energy level here of the intermediate, and we only have to get up to here, and it's a much lower activation energy. And it turns out, whichever step has the higher activation energy is the slower step. And so in this case, this is much higher activation energy in step one than in step two. And so we'd call step one in this diagram here for whatever reaction this typifies, this would be the slow step or the rate determining step in the reaction. So that's kind of how that works.
And these are the things you need to identify on these reaction coordinate diagrams. So in the second one, just as it was in the first, your product is higher than your reactant in energy. And it doesn't matter the pathway at all. So from where you start, from reactant to product, so that difference is going to be delta H.
And again, if your product is higher than your reactant, then delta H is going to be positive. It's an endothermic reaction. Had the product been lower energy and the reactant higher energy, well, then it would have been an exothermic reaction and delta H would have come out. Negative. All right, one last little thing I want to address here.
And so turns out with these reaction coordinate diagrams You can look at them both forward and backward. You can look at the forward reaction You can look at the reverse reaction. And so if we look at this going forward we saw that it was an Endothermic reaction Delta H is positive.
Well, if we look at it going backwards Well, then it's going to be an exothermic reaction If you look at it in reverse and then going backwards from where you start energetically to the top of the hill That's our new activation energy for the reverse reaction. Maybe I'll even abbreviate that REV for reverse. So it turns out we can often give you a couple of different things to have you calc... there's a relationship you can establish here between the activation energy in the forward direction, the activation energy in the reverse direction, and delta H. And so if I told you this activation energy in the forward direction was 50 kilojoules...
per mole, and I told you that this activation energy in the reverse direction was 30 kilojoules per mole, then the question I could ask you is what is the value of delta H? And so notice if your forward reactions activation energy is the higher one, that implies that your reactant is lower energy than your product. So here the 50 kilojoules per mole was higher than the 30 kilojoules per mole, and so that would be the hallmark of an endothermic reaction.
And in this case, you can see that, you know, delta H and the reverse have to total up to the forward activation energy here. And you can see that delta H itself here needs to be positive 20 kilojoules. And so it turns out there's a lovely formula here. You can kind of derive from that, that delta H is just simply equal to the activation energy of the forward reaction minus the activation energy of the reverse reaction.
And so in this case, that'd be 50 minus 30, and we got 20. Had this been an... exothermic reaction, then when you perform this calculation it would have come out to be a negative number. But again for an endothermic it'll come out positive just as we've done. Cool, that concludes this lesson. Now if you found this lesson helpful hit that thumbs up button to support the channel.
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