 Today, today that we're going to finish the dynamics, we'll finish it in a couple of minutes. We're talking about the shot of the A on Monday. And we talked about the paper box process. 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 blew it out. We'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 a chemically specific reason why the shot of the A's principle is not obeyed 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 shot of the 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. So 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. That's what you're saying. So that's what's going to be a very, very case about that. So a shot of the A's principle says, if you increase the total reaction pressure, this should be blue because it should go to the left because this shift from low-reduction to a low-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 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 and 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 amortune, as an example, here's the number of moles. Here's the total number of moles, that plus that. If I write this expression, if I write the amortune for 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 V. That just means the equilibrium constant written down 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 speeds. 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. The mole fraction and the total pressure is 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. Everybody saw it, we're actually getting this equation right. 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, if that P goes up, this portion's going to get smaller, isn't it? If I increase the size of the denominator, this portion's going to get smaller, this portion's going to get smaller. This is going to be shifts to the left. This gets smaller, X gets smaller. X is the progress of the reaction. And of course the converse will 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. 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. It 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 phase 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 Gogh equation and I don't want to spend all of my talking about it or going through what it means. But basically, if a reaction is exothermic, another word is delta H for the reaction that's 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 favoring the 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 Gogh equation that talks about the temperature dependence of chemical reactions. It's very important. We could spend 20 minutes talking about it right now. I'll ask you something about this equation on Friday with if the reaction is 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 that gives energy minimum and the reaction gives energy and how to calculate those things. We talked about chemical potential to the 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 did talk about the effect of pressure today and the effect of temperature. We sort of talked about it. I showed you the Van Gogh equation, and I asked you to read about it at that Section 17.5. This is everything we did to 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 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 is an integrated rate of law? Why do we need it? How does a reaction rate go? There's a lot of 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 is our scheme on day one. We were bright-eyed and pushed pale. Now, here we are today. We're right in the middle of week seven, in turn, so. And we're not free ledgers in the chemical kinetics as shown on this scheme right here. 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. So 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, quantum mechanics, psychroscopy, and then have one quarter of the statistical mechanics, thermodynamics, chemical kinetics, and chemical dynamics. It can't be done. So, this doesn't really affect you. We're going to try and do this through some justice. We try to do this subject right here, 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 force on this. You'd enjoy it. But we're going to do it all in about three weeks. One, two, three. How much time is left? Okay, so on the mid-term, mid-term is a week from Friday now. It's really going to focus on the stuff since mid-term one, so the stuff at the end of, mainly the thermodynamic stuff, I don't think there's any stat mech at all on mid-term 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 mid-term 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, you know, we're sort of backtracking. We're going back to the early 1800s when thermodynamics and statistical mechanics worked out, right? The first Nobel Prize was in 1901, right? And the whole slew of guys just missed earning one of those because their work wasn't appreciated until after they died, can't win an Nobel Prize after you died. And so this is earlier than that, right? The first kinetic measurement remains earlier than 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 do 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 the 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 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? Aren't 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, all right? But it doesn't necessarily convey any mechanistic information. A stoichiometric reaction is a balanced chemical reaction that does not convey any mechanistic information. You can't tell by looking at a stoichiometric reaction anything about the reaction mechanism. That's what it means, okay? Stoichiometric reactions are just described the stoichiometric relationship between the reactions to the product. 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 does not convey any mechanistic information. Something totally different might have happened. So, if the reaction between 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 H2 all, there's virtually never a collision between an H2 and an O2. It might be kind of surprising to them. Here's the action, right? 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. Hydrogen dissociates to get two hydrogen atoms. Oxygen dissociates to do two oxygen atoms. And then hydrogen atom, that collides with an O2 to give us an oligarchal linen 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 is the, it's elementary to the sense that it can't be made any more primitive than what it is, all right? An OH radical actually collides with an H2. That's what actually happened and we got H2 in hydrogen atom out of it. Okay? 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. It's pretty much one. With me? It's confusing. These are elementary reactions. The reaction mechanism comprises elementary reactions, blah, 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. So as in this case, that the elementary reaction is oxygen atom actually collided with H2. And we've got always track of the nitrogen atom. Okay. So we've already discussed this thing called the extent of the reaction with a degree of simple. I don't know how to pronounce. Here's the technical definition of what it is. I don't think we actually wrote 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 is 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. Okay. So this is not as confusing as this looks. Here's an example. 2 plus Br2 is 2HBr. And 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. So the stoichiometric coefficient. The extent of the reaction is going to be plus 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. Okay. And so I get 0.001 moles. That's the extent of moles of 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 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. Alright. So I get minus over minus is plus and those two numbers have to be the same or I did the calculation wrong. Okay. So we can calculate the extent of the reaction if we just know how many moles of 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. It's the same for reactants and products. The way it's defined it has to be the same for the reactants and products. You can define the reaction rate in any reactant and product. 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. Okay. 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 one over A. A is a stoichiometric coefficient on A. It's negative times D, A, and T. Where that A in brackets there that's the concentration of A or it's partial pressure. So the rate I can define the rate with regard to A or with regard to B or with regard to C or with regard to D 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 one over the stoichiometric coefficient times D concentration DT. We'll ask them 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 one over volume is gone because we're talking about concentrations here. We had one 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 often has a general flow. 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. Unless 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. Yes. 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. With respect to A is alpha the reaction is first order with respect to A or second order that means alpha is one or two. We commonly speak of reactions that are first order in A, yes that means alpha equals one or second order in A, yes that means alpha equals two. And 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 the, okay so this is important. So 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. And 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 in a little more detail. 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 and you solve for K. K is rate over A the concentration of A I know the rate is molar per second A is molar 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 so 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 concentration of A times concentration of B and the rate is always going to be molar per second 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 are was so stoichiometric reactions of 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 okay this is just expressing the overall reaction that occurs as a consequence of about 20 steps alright and so it's tempting to say oh the rate of the reaction is times the concentration of hydrogen squared times the 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 only the reactants appear in this 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 HI as it's produced that's important if we remove the HI as it's produced then the rate of the reaction will be this in other words if I remove the HI as it's produced I'm basically turning off the reverse reaction I'm not allowing it to happen at all the forward reaction but we can also study HI decomposition in this case if the products so HI decomposition would mean the reaction is running in the backward direction in that case these are products of the decomposition so if we remove them then the rate of the back reaction would be this because this is an elementary reaction that we're talking about here 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 but I'm just pointing out that if it's an elementary reaction we can deduce what these rates are going to look like just by looking at this there's HI there's two in front of the HI that means there's got to be a two that's the rate constant for the decomposition of the HI 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 the rate law this rate law would never apply to this 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 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 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 to the minus one half seconds to the minus one those are going to be the units of the rate constant and if I wanted to the constant and deduce that this overall reaction order is one and a half 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 it's a little confusing because we've been talking about big K the equilibrium constant and the equilibrium constant never has units 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 you want to know what the stoichiometric coefficient is for A but you don't know what it is you want to figure out what alpha beta and gamma are 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 because you've got an excess amount of B and C B and C are not rate limiting you've 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 and I can treat them as constants I can define a K prime that's just that K times B to the beta C to the gamma and essentially I've turned this into a pseudo alpha order reaction I've eliminated the dependence of B and C and now 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 I change the initial concentration of A while maintaining it tiny compared to B and C I have to use tiny concentrations of A but I can vary it and as I do that here's the lowest concentration of A here's the initial rate that I measure here's a higher concentration of A here's the initial rate I measured there here's the red line and here's a 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 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 now how do you know whether A is small enough well you know A is small enough when you've got big concentrations of B and C 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 reactions only depending on the concentration of A now I can actually run my experiment 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 that's the method of initial rates method two 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 all right 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 you can 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 second order reaction a third order reaction a half order reaction it becomes a fitting exercise but eventually you figure it out thinking 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 and so we get the rate constant 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 plotting it 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 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 might have been about 87 remember that very well sadly okay so we can use an integrated rate law 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 it's a lot like measuring you basically have to measure the concentration of the reactant or product as a function of time anyway to get the half-life actually measuring the same thing but the half-life is the time needed for half of a reactant the chemical reaction to be depleted half of the reactant to be depleted depleted 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 so if that is half of that it's just half that ratio is just half and so this is now my expression for t to the one-half that's the half-life t to the one-half so if I solve for t to the one-half it's just log 2 over k that's dead simple 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 for 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 or rather 0.05 same amount of time you measure 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 you're measuring the concentration of your reactant as a function of time you could just fit this green curve this equation right here and you're done so 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 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 which is 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 reactant 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 reactant concentrations this reaction this rate law really holds up so this is superior one type of second order reaction ok so I think we'll just stop right here ok so this stuff that we talked about today could be on the quiz let me just say that