 So today is the day that we're going to finish our dynamic. We'll finish it in just a couple of minutes talking about the shot of the A on Monday. And we talked about the paper box process that this is going to be the extra credit problem on midterm two. So I wrote it down here so we're clear on what it is and once you think I just do it out. You're actually going to talk about midterm two. The paper box process is a long note that does not conform to the prediction of the shot of the A's principle. Explain why very briefly and specifically. I'm going to want to chemically specific reason why the shot of the A's principle is not abated by the paper box process. It has something to do with the iron catalyst. What is it about the iron catalyst? That means that you can't just increase the partial pressure of hydrogen and get the rate of production of ammonia to increase. That's what I mean by it doesn't follow the LaChiopla A's principle. You can simply increase the partial pressure of hydrogen or the partial pressure of nitrogen by itself and accelerate the rate of ammonia production. That's what you would expect looking at this reaction though, isn't it? The shot of the A's principle says, hey, if I increase this then people are going to shift to the right. Think about that. Research it on the web. Research it in the library. It'll be worth either five or ten points depending on how much you get right or zero, of course. What you're saying is what we're not saying. Don't say that to everybody. That's what's going to be a very, very case by that. So a shot of these principles says, with an increase in the whole reaction pressure, this should be looking to shift to the left because this shift in subreduction is a whole reaction pressure. I increase the pressure because there's two moles of amortune here and one mole of N2O4 here. One mole of gas versus two moles of gas. If I increase the pressure, I should drive the reaction to the left. So I went through and showed that we can make that as a quantitative. We can make a quantitative prediction about the total pressure. I guess I got carried away. We don't necessarily have to be able to do this, but here's the reaction that we wrote down on Monday. We can go through and calculate the position of each word. We've seen these expressions. Calculate the mole fraction, the amortune N2O4. The mole fraction of N2O4 is just the number of moles of N2O4 if I drive the total moles N2O4 plus N2O2. So if we just plug in this and this for the number of moles, like here for the N2O4, as an example, here's the number of moles. Here's the total number of moles, that plus that. If I write the amortune N2O4, I get this expression right here. So now I write the expression for the equilibrium constant in terms of pressure. Sometimes we call that K sub P. That just means the equilibrium constant written out in terms of pressure. And so it's just N2 squared over N2O4 squared. Of course, always normalized by the standard pressure. And so we're going to end up with an extra factor of P0 in the denominator, where we're going to cancel all these P's. Yes, that's the pressure corresponding to the standard state on bar. And then we can write this in terms of the total pressure by using the mole fraction. Because the mole fraction has the total pressure, so it's the partial pressure. So I can write the partial pressure of N2 in terms of the mole fraction, same thing for N2O4. And then I can plug in our expression to the mole fraction that I derived two slides ago. Here's the expression for the mole fraction of N2O4. Here's the expression for the mole fraction of N2O4. And then if I do an algebra and simplify that, I get this equation right here. And then if I solve for x, I get this equation right here. But this is indeed very satisfying. Because what it tells us is that if we increase the total pressure by increasing the pressure of either N02 or by reducing the volume of the reaction to the vessel, either way, right, if that P goes up, this portion is going to get smaller, isn't it? If I increase the size of the denominator, this portion is going to get smaller. x is going to get smaller. Does it shift to the left? If this gets smaller, x gets smaller. x is the progress of the reaction. And of course the counter is going to also be true. Now you don't have to be able to derive this, but it is rather satisfying to see that you can do that if you want to. It wasn't too difficult. On Monday we actually calculated the position of equilibrium using these sets of conditions. And we did that by solving the quadratic formula. On Monday, this equation can be used to arrive at exactly the same conclusion. That's the same value for x that we got on Monday. That means we probably did it. Correctly. Okay, so that's the effect of pressure. You can usually tell just by inspection what effect pressure will have on a gas-based equilibrium. If you apply pressure, the system tries to minimize the number of moles of gas. Very simple. In the case of temperature, what matters is delta H. We have this thing called the Van Pocke equation. And I don't want to spend all of my time talking about it or going through what it means. Basically, if a reaction is exothermic, in other words, delta H for the reaction is less than zero. The reaction will show d log k over dt of less than zero, of course, according to this reaction. In other words, k will get smaller with increase in t. In other words, the reaction will shift to the left, fading reactants, when t is increased. What I want you to do is go and read about this in chapter 17. There's a short section on the Van Pocke equation that talks about the temperature dependence on chemical reactions. It's very important. We could spend 20 minutes talking about it right now, but I don't want to. Okay, so I'll ask you something about this equation on Friday with changing the reaction to this, changing the temperature to this. What's going to happen to the reaction rate? Or the repeated reactant equilibrium rate. Okay? We are done with thermodynamics. We are done with statistical mechanics. In chapter 17, what we covered was, we talked about section 17.1, which gives energy minimum and the reaction gives energy and how to calculate those things. We talked about chemical potentials, that's section 17.2. There was a section on the statistical description of the equilibrium constant that is very interesting, but we didn't talk about it. I read it. We didn't talk about the effect of pressure today and the effect of temperature. We sort of talked about it. I showed you the Van Pocke equation. And I asked you to read about it in that section 17.5. This is everything we did in chapter 17. There's a lot more sped off focusing on electric chemistry, which is a subject mirrored dear to my own heart. It's what I do, but we don't have time to talk about it. Okay, so this is everything that we talked about in chapter 17. And this is the last stuff that we'll talk about that pertains to the thermodynamics. Now, we're also going to skip chapter 18. We don't have time for it. It's sort of a transitional chapter anyway. We're going to jump straight to chapter 19, which is about chemical kinetics. Now, we are, in fact, yes, the basics. How are reaction rates measured? How is the reaction a lot of fine and dry? Where's an integrated rate law? Why do we need it? How does the reaction mechanism reduce the thermodynamic rate of a lot, a lot, a lot, a lot, a lot? It's all background on chemical kinetics. But chapter 19 is very important. Now, where are we in the scheme of things, Steve? How badly off-track have we come? This is where we hope to be. This was our scheme on day one. We were bright, light, we were push and pale. Now, here we are today. We're right in the middle of week seven, and we're not three lectures into chemical kinetics. I'm showing off this scheme right here. We are, in fact, we're just starting chemical kinetics here. So, I regret to tell you what's happened to reaction dynamics. I don't think we're going to have a lot of time to talk about it. We're going to completely revamp this force this summer because everyone now recognizes that you can't do all of this in one quarter. You can't have two cores of quantum mechanics, it's microscopy, and then have one quarter of the statistical mechanics, thermodynamics, chemical kinetics, and chemical dynamics. They can't be done. So, this doesn't really affect you. We're going to try and do this problem here with some justice. We try to do this subject right here, here's some justice. We spend a lot more time on it. We're not going to have as much time to spend on this. There should be a whole 10-week course on this. You'd enjoy it. But we're going to do it all in about three weeks. Two, three, that's how much time we have left. Okay, so on the midterm, midterm is a week from Friday now. It's really going to focus on the stuff since midterm one, so the stuff at the end, mainly the thermodynamics stuff, I don't think there's any stat mech at all on midterm two. All right, and the stuff that we're going to do in this lecture on Friday and on Monday, mainly. All right, so there's going to be three lectures on kinetics that are going to be on midterm two as well. Okay, so I'll be telling you more about that. So this is the stuff that we're going to talk about. This guy, we're sort of backtracking. We're going back to the early 1800s. When thermodynamics and statistical mechanics worked out, the first Nobel Prize was in 1901. The whole slew of guys just missed earning one of those because their work wasn't appreciated until after they died. You can't win an Nobel Prize after you died. And so this is earlier than that. The first kinetic measurements remain sort of in the early 1800s. So what we're talking about now is the rate of reactions. Kinetics is all about how fast the reaction occurs and what factors control the rate of the reaction, how we measure it, how do we talk about it, how do we accelerate it, how do we decelerate it, how does the rate depend on variables that we have controlled. All right, so we need a lot of notation and we want to be very clear about this so that there's not any confusion because it's intrinsically confusing. Notation always is. So there's something called a stoichiometric reaction. Now that may not sound like a technical term, right? All reactions, stoichiometric, yes they are, right? In one sense, but we're going to use the terminology stoichiometric reaction to mean something special. We're going to mean that the reaction that we're writing down is balanced, but it doesn't necessarily convey any mechanistic information. A stoichiometric reaction is a balanced chemical reaction that does not convey any mechanistic information. So you can't tell by looking at a stoichiometric reaction anything about the reaction mechanism. That's what it means. Okay? Stoichiometric... just describes the stoichiometric relationship in reactions to the clock. So here's the stoichiometric reaction, A moles of A, B moles of B, you get the idea. This does not mean that A moles of A collided with B moles of B at any point, right? That would be something having to do with the mechanism, but this reaction is not conveying any mechanistic information. Something totally different might have happened. Stoichiometric reaction to an oxygen and hydrogen is this. Yes, all right? But the mechanism has nothing to do with this. And when hydrogen and oxygen react to make H2O, there's virtually never a collision between an H2 and an O2. It might be kind of surprising to hear that. Here's the stoichiometric reaction, but when this reaction actually occurs, there's virtually never a collision between an H2 and an O2. What happens instead is this. A mechanism. Hydrogen dissociates to get two hydrogen atoms. Oxygen dissociates to get two oxygen atoms. And then hydrogen atom, that collides with an O2 to give us an O2 and an oxygen atom, and so on and so forth. These are elementary reactions. An elementary reaction is a balanced chemical reaction, just like a stoichiometric reaction, but it does convey mechanistic information, right? It is as stripped down. It's elementary to the sense that it can't be made any more primitive than what it is. An OH radical actually collides with an H2. That's what actually happened and we got H2 in hydrogen atom out of it. So there are elementary reactions that comprise the chemical mechanism and then there's a stoichiometric reaction that is just somehow the sum of all this produced water. With me? It's confusing. Elementary reactions comprise elementary reactions, blah, blah, blah, blah. For example, this elementary reaction hydrogen ES and we can classify elementary reactions as different types. We will get to this later, don't worry about it now. It's very complex how hydrogen and oxygen interact. There are explosions that can occur because of this complex mechanism. We'll talk about it later. So an elementary reaction is one of which the indicated products are formed directly from the reaction. As in this case, the elementary reaction of hydrogen atom collided with H2 and we got OH radical in the hydrogen atom. So we've already discussed this thing called the extent of the reaction with a great simple name I don't know how to pronounce. Here's the technical definition of what it is. If you scroll down the technical definition earlier we need a more precise and general definition for the extent of the reaction. This is the initial moles of some species J. The number of moles is a function of time of species J and this is the stoichiometric coefficient that applies to species J. It's positive for products and negative for reactants. The stoichiometric coefficient for reactants is negative. I'm sure that's something we haven't said earlier. Stoichiometric coefficient for reactants is negative. And it's in units of moles. So this is not as confusing as this looks. Here's an example. H2 plus Br2 is 2HBr. Now let's say that the change in HBr is plus 0.002 moles. That's the change in the number of moles of HBr as this reaction is occurring. That's all. So the stoichiometric coefficient the extent of the reaction is 0.002 moles divided by plus 2 because that's a 2 and it's a stoichiometric coefficient for a product so it's positive by definition. So I get 0.001 moles that's the extent of moles concentration. Usually moles. So now I can do the same thing for H2. I can equally equal the extent of reaction in terms of the change in H2 of minus 0.001 moles. Then I have that minus 0.001 moles divided by minus 1 because the stoichiometric coefficient here is minus 1. It's minus 1 because it's a reactant. So I get minus over minus is plus and those two numbers have to be the same or I do the calculation wrong. So we can calculate the extent of the reaction. We just know how many moles something was generated and then be used for the extent of the reaction. Okay. Now in terms of the extent of the reaction the reaction rate is given by this equation right here. What is this? This is the volume of the reaction vessel and this is just the extent of the reaction the change in the extent of the reaction of the function of time. Yes, that's the volume. Like the extent of the reaction itself the rate of the reaction does not depend on the species used to define it. For the products the way it's defined it has to be the same for the reactants of products. You can define the reaction rate in any of these reactants you should get the same number. There's only one reaction rate. Since this is in moles and this is the volume we have a concentration here. The concentration can be units of pressure or it can be units of concentration. Moles per cubic centimeter. Moles per liter. So usually we're not looking at an expression that looks like this. Usually we're looking at an expression that looks like this. Here's our generic reaction. I can define the reaction rate as minus 1 over a is negative times where that A in brackets there that's the concentration of A or it's partial pressure. So I can define the rate with regard to A or with regard to B or with regard to C all of these rates have to be equal to one another and they will be if I conform to this definition of what the rate is. The rate is 1 over the stoichiometric coefficient times we'll have to do this kind of fast. You guys getting all this? Now, this is a change in concentration with time or the units are typically molar per second or ATM per second. If this is a partial pressure then it's ATM per second if it's a concentration it's typically molar per second but it could be some other concentration unit. This factor of 1 over volume is gone because we're talking about concentrations here we had 1 over volume here momentarily until we converted that volume over this extent of the reaction into a concentration and we're never going to go back. A rate law relates to the concentration reaction so this is a rate law that's the reaction rate that's why it's called a rate law. The rate law predicts the reaction rate makes sense. Or K, alpha and beta are independent of the concentration. That's a rate constant. It's called a constant because it's independent of concentration. Note also that alpha and beta are not necessarily stoichiometric coefficients for species A and B. See how that's alpha and beta? You might think that that's the stoichiometric coefficient on A and that's the stoichiometric coefficient on B. In general that is not true but yes the reaction that you're writing the rate law for is an elementary reaction and you know that. If you know the reaction is an elementary reaction then that is the stoichiometric coefficient on A and that is the stoichiometric coefficient on B. I'll say more about that. But in general unless you know that you can't assume that that's true. Instead alpha and beta are alternative experiments. So from this rate law we can say the order of this reaction with respect to A is alpha and the order with respect to B is beta. That's what that means. So there's an order with respect to all of the chemical species in this reaction. If I say the order with respect to A is alpha the reaction is first order with respect to A or second order that means alpha is 1 or 2. We commonly speak of reactions that are first order in A, yes that means alpha equals 1 or second order in A, yes that means alpha equals 2. In the total order of the reaction is alpha plus beta that's the overall order is the sum of those stoichiometric coefficients. The sum of those exponents rather. So K is the rate constant's units that depend on. So this is important. We mentioned that there's this thing called the rate constant. One of the confusing things is the rate constant has different units depending on what type of reaction it is. We have to be conversant with all these different units. The rate constant tells us what the overall reaction order is. The overall reaction order is information that's embedded in the units of the rate constant. So if I tell you the rate constant for the reaction was this 26.25 per second I'm telling you it's a first order reaction. You have to know those units are units of first order reaction. Reaction has overall first order. Let's look at this a little more detail. K is the rate constant for example if alpha is 1 and beta is 0 then the reaction is this because of beta is 0 that's just 1. The reaction is first order in A and first order overall because 1 plus 0 is 1. So what's K going to be? You can always figure it out. You take the rate here's your rate law all you do is solve for K. K is rate over A, the concentration of A I know the rate is molar per second A is molar and so that means the units of K have to be per second I can just figure that out. If you want to know what the units of the rate constant are solve for it in the rate law plug in the dimensions of the other variables and just calculate what the dimensions of K have to be. Just do a little dimensional analysis on K. If you see units of per second you know immediately it's a first order reaction no one has to tell you that. If on the other hand alpha is 1 and beta is 1 then you've got this rate law and you've got an overall second order reaction and if I solve for K here I get rate over A B and the rate is always going to be molar per second and so I've got molar squared and so now the units are per molar per second. If I see those units I know immediately that I've got a second order reaction the units are telling me what the overall reaction order of the reaction is was. So stoichiometric reactions the rate law can be deduced from an inspection yes, yes the rate law cannot that should be bolded and in italics in other words the rate of this reaction is not this. So this is what I said earlier. We mentioned that this is a stoichiometric reaction you might want to just think about this as being a rate constant which it is not there is no rate constant for this arrow because this process never occurs ok this is just expressing the overall reaction that occurs as a consequence of about 20 steps alright and so it's tempting to say all the rate of the reactions is K times concentration of hydrogen squared times concentration of oxygen no it is untrue not even close to being true but for an elementary reaction the rate law can be generated by inspection for example if I look at this reaction right here which I pulled out of the mechanism for that why the reaction rate is given by K times the concentration of hydrogen atoms times the concentration of oxygen alright if that's a single ended arrow these products don't have anything to do with the reaction rate the single ended arrow tells me it's an irreversible reaction and I don't have to think about what those guys are I could just write products here and I have all the information I need to know about the rate right only the reactants appear in the expression often reactions are significantly reversible and both the forward and backward rates are important so if you have the double ended arrow now you've got a reversible reaction you do have to think about the products you do have to know what they are alright in this case if we study the reaction from left to right remove the H i as it's produced that's important if we remove the H i as it's produced then the rate of the reaction will be this in other words if I remove the H i as it's produced I'm basically turning off the reverse reaction I'm not allowing it to happen at all so this would be the rate of the forward reaction but we can also study H i decomposition in this case if the products so H i decomposition would mean the reaction is running in the backward direction alright in that case these are products of the decomposition right so if we remove them the rate of the back reaction would be this alright because this is an elementary reaction that we're talking about here alright so we can write these rate laws directly from inspection that's the main point we're trying to make here and we're we're not going to talk about reversible reactions the kinetics of reversible reactions very much for a lecture or two we'll get to it alright but I'm just pointing out that if it's a elementary reaction we can deduce what these rates are going to look like just by looking at this there's H i there's two in front of the H i that means there's got to be a two in this example that's the rate constant for the decomposition of the H i and if I get rid of these guys there's going to be no back and so that would be the rate of the decomposition if these guys are removed or if they diffuse away very quickly no back reaction can happen the stoichiometric reaction for acid aldehyde decomposition this is acid aldehyde it decomposes to give methane and CO the rate law for this reaction is this now is that an elementary reaction it's not and I can tell that just by looking at the reaction law right the rate law this rate law would never apply to this reaction if this was an elementary reaction if this was an elementary reaction what would the rate law be that would be a one wouldn't it? because there's a one here that would be the exponent and the rate constant would be that so just looking at this rate law I can tell immediately that this acid aldehyde decomposition has a complex mechanism made up of a series of elementary reactions that I don't know what they are but that's not an elementary reaction right there tell that immediately we did not know this was stoichiometric reaction if we did not know this was a stoichiometric reaction we would know once we looked at this we know it's not an elementary reaction just based on what the rate law is but I can ask what are the units of that rate constant right there what are its units and all I have to do is solve here's the rate here's the concentration concentration to the three halves if I work that out I'm going to have molar one half seconds to the minus one those are going to be the units of the rate constant and if I wanted to I could take those units for the rate constant and deduce that this overall reaction order is one and a half right it's a perfect quiz question here's the rate constant tell me what the overall order of the reaction was one two three one point five okay so it's a little confusing because we've been talking about big K the equilibrium constant and the equilibrium constant never has units alright but the rate constant the little K does and the units have contained information now how do we experimentally determine these rate laws this is important we have some methods that we can use for doing that that we need to know about one of the methods is the method of initial rates the idea is very simple if you've got a complex reaction a bunch of reactants A B and C alright you want to know what the stoichiometric coefficient is for A but you don't know what it is alright you want to figure out what alpha beta and gamma are alright the way that you do that is you isolate reactant A by making all of the other reactants enormous compared to A you make A tiny and B and C large if you do that what will happen the rate of the reaction will depend on the amount of A that you've got right because you've got an excess amount of B and C alright B and C are not rate limiting flooded the system with B and C you put a tiny amount of A and now the reaction will only occur at a rate that's dictated by the amount of A and you can interrogate A and learn everything about how the reaction depends on A here's how this works if I make B and C large they're going to become pseudo constant because the amount of the reaction so B and C are large and there's a tiny amount of A A is only going to allow the reaction to occur at a relatively slow rate and the amount of B and C are not going to be significantly perturbed from their total concentrations because their total concentrations are enormous so I can treat them as constants I can define a K prime that's just that K times B to the beta at C to the gamma alright and essentially I've turned this into a pseudo alpha order reaction I've eliminated the dependence of B and C the reaction only depends on A and I want to know what alpha is so if I want to know what alpha is I just take the log of both sides log of the rate equals log of K prime plus alpha times the log of A I measure now the rate for different initial concentrations of A but I change the initial concentration of A while maintaining it tiny compared to B and C I have to use tiny concentrations alright and as I do that here's the lowest concentration of A here's the initial rate that I measure here's the higher concentration of A here's the initial rate I measure there that's that red line and here's the higher concentration all of these concentrations of A here are tiny compared to B and C I want to emphasize that and now I can just plot log of the rate log of the initial rate is a function of log of A and the slope of that is alpha by golly so I have teased out the molecularity of the reaction with respect to reagent A it's a lot of work but now I know what the order of the reaction is with respect to A and then I make B small make A and C big and repeat the experiment so I can take the reaction apart piece by piece and figure out what the molecularity is of every reactant I'm looking for a word that I can't think ok now how do you know whether A is small enough well you know A is small enough when you change you've got big concentrations of B and C alright let's say A is one millimolar and B and C are one molar how do you know if one molar is big enough you make it 1.1 and the reaction better not change you have to actually check to see if the reaction rate depends on B and C by changing their concentrations when they're big to convince yourself they're big enough reactions only depending on the concentration of A now I can actually run my experiment ok so big and small are relative terms they're nebulous you have to in the lab empirically figure out if they're high enough so that A is truly isolated what's the method of initial rates yes yes method 2 drive an integrated rate law for the reaction an integrated rate law we've been talking about rate laws but we haven't said it but we've been talking about differential rate laws DADT now if we integrate that we can determine explicitly what the time dependence of A is for reactions of different molecularities first order, second order, third order and so on if we integrate our differential rate law we'll get an integrated rate law and the integrated rate law explicitly predicts what A does as a function of time take this for example the rate is minus DADT that equals K times A very simple reaction we can integrate both sides I'm going to move the minus sign over the right hand side integrate from the initial concentration of A to some final concentration I'm going to just integrate this DT and so when I'm done doing that I'm going to have minus KT since that's a zero and this is going to turn into log A over A zero and now I've got an equation that predicts the concentration of A as a function of time this equation here does not predict the concentration of A as a function of time it predicts the rate so the integrated rate law tells me explicitly here's what A does as a function of time measure the concentration of A as a function of time and then see if you can fit it with this equation if you can chances are your reaction is first order with respect to A and you can back out the rate constant from the fit now this is laborious because what if it's not first order in A well you've got to fit it to first order reaction a third order reaction a half order reaction it becomes a fitting exercise but eventually you figure it out the nice thing about this process so let's look at data use the integrated rate law here's what the integrated rate law is for the first order reaction and it predicts A gets smaller as the log and the slope should be negative and the slope of this is actually minus K from the slope or you could just take the raw data which is a curve you could fit that directly in the old days every process had to be converted into a straight line so that you could do the fitting because there were no computers and a lot of the textbooks that we now use to study these processes continue the convention of turning every reaction in such a way that you get a straight line it's not really necessary to do that anymore we can fit curve lines trivially anytime we want to it's one button in Igor to do that and then you get the rate constant from that fit you don't need to turn it into a straight line but this was what was done back before 1990 or so because in 1990 that's pretty much when we started to use personal computers it might have been about 87 remember that very well sadly okay so we can use an integrated rate law there's integrated rate laws for all different molecularities of reactions but another way that we can deal with the integrated rate law is to define the half life of the reaction the half life is defined using the integrated rate law so you need the integrated rate law first then you can define the half life and then you measure the half life measuring I mean you basically have to measure the concentration of the reactant or product as a function of time anyway to get the half life so you're basically measuring the same thing but the half life is the time needed for half of a reactant to chemical reaction to be depleted half of the reactant to be completed depleted completed so here's the integrated rate law for a first order reaction which we just derived alright the half life just says hey I want to know how long it takes for a to fall to half of the value of a zero right so if that is half of that it's just half that ratio is just half okay and so this is now my expression for t to the one half that's the half life t to the one half okay so if I solve for t to the one half it's just log 2 over k that's dead simple alright now note that in this case of a first order reaction the half life is independent of the initial concentration of a zero that's the hallmark of a first order reaction you start with 0.2 you measure the first half life that's how long it takes to get to 0.1 now you pretend 0.1 is your starting point you measure how long it takes to get to 0.5 rather 0.05 same amount of time the third half life that's the time it takes to get to 0.025 that's the same amount of time if the half life stays the same it's a first order reaction but notice what you're measuring here you're measuring the concentration of your reactant is a function of time you could just fit this green curve this equation right here and you're done there's no particular advantage to measuring the half life but if for some reason that happened to be a convenient way to do it you can get the order of the reaction that way one of the advantages of both of these approaches where we're using the integrated rate law is that we're ensuring that the molecular molecularity of the reaction is staying the same over a period of time and over a range of reactant concentrations that's not true in the method of initial rates in the method of initial rates we're measuring the initial rate always for different initial concentrations of the reactant that we're isolating and so in principle we're only getting the molecularity right at the beginning of the reaction if something funny happens and that changes which often occurs we're going to miss it we're not going to know what happens to the molecularity later in the reaction but here we can see if something funny happens in other words if something funny happens we're going to see we're going to have a good fit to our curve here and then something is going to go wrong we're going to fall off this curve at some later point in time and that's information that we can use to understand what's going on we're not going to have that information in the method of initial rates see what I'm talking about this is superior here we're ensuring that we have a fit over a wide range of reacting concentrations in a range of times and if we see a fit all the way to the end of our time window we can be confident that over that time window, over that range of reacting concentrations this reaction, this rate law really holds up so this is superior one type of second order reaction I think we'll just stop right here so this stuff that we talked about today could be on the quiz let me just say that