Lecture 19. Observational Chemical Kinetics
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Calcium hydroxideChemistryAnomalie <Medizin>KupplungsreaktionSurface finishingProcess (computing)Paste (rheology)Lecture/Conference
00:43
Haber processOctane ratingIronHydrogenStickstoffatomAmmoniaHaber processMedical historyLecture/Conference
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Haber processBiosynthesisLibrary (computing)Chemical reactionPressureReducing agentLeft-wing politicsDistickstofftetroxidSurface scienceWursthülleLecture/Conference
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Gene expressionPondMan pageProlineMethylmalonyl-CoA mutaseKipp's apparatusChemistryCaffeineBohriumGoldAnomalie <Medizin>Detection limitGene expressionChemical reactionPressureSystemic therapyReaktionskinetikDistickstofftetroxidBase (chemistry)ÜbergangszustandStuffingOctane ratingMolekulardynamikChemistryVerdampfungswärmeStoffmengenanteilQuantum chemistryGesundheitsstörungSet (abstract data type)HomöostaseAmount of substanceHuman body temperatureAction potentialSunscreenGleichgewichtskonstanteGolgi apparatusPhosphorusLeft-wing politicsOrganische ChemieWursthülleStockfishKlinisches ExperimentAreaWeinfehlerLecture/ConferenceComputer animation
11:37
French friesHydrolyseMan pageSpeciesChemistryCobaltoxideHydrogenEnolClenbuterolWine tasting descriptorsVolumetric flow rateDigital elevation modelMortality rateChemical reactionStoichiometryElementarreaktionReaktionsgleichungController (control theory)CobaltoxideHerzfrequenzvariabilitätSunscreenReaction mechanismSense DistrictEnzymkinetikOctane ratingElectronic cigaretteStuffingNobeliumAmount of substanceHydroxyl radicalHydrogenSpeciesCalculus (medicine)BromineElectronegativityCollisionSetzen <Verfahrenstechnik>Hydro TasmaniaWaterFunctional groupChemistryMoleculeExplosionAreaWhitewaterComputer animation
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Man pageMortality rateAlpha particleChemical reactionBeta sheetSense DistrictConcentrateSunscreenStoichiometryReaction rate constantChemical reactorMolar volumeProcess (computing)HerzfrequenzvariabilitätPotenz <Homöopathie>Setzen <Verfahrenstechnik>Octane ratingSpeciesGene expressionCubic crystal systemPressureElectronegativityChemistrySpawn (biology)Wasserwelle <Haarbehandlung>Nuclear fissionWaterThermoformingPressureWalkingElementarreaktionWursthülleSulfurBody weightComputer animation
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Mortality rateAcetaldehydeKnotBohriumTrihalomethaneMan pageChemical reactionOctane ratingInitiation (chemistry)StoichiometryWine tasting descriptorsLactitolMolecularityRepeated sequence (DNA)ConcentrateMolar volumeSystemic therapyElementarreaktionAcetaldehydeProcess (computing)Reaction rate constantWursthülleDeathAlpha particleFunctional groupPeriodateLevomethadonHydroxybuttersäure <gamma->Beta sheetReaction mechanismDecompositionGene expressionEnzymkinetikMethanisierungIce frontCobaltoxideSetzen <Verfahrenstechnik>Electronic cigaretteHydrogenStuffingGleichgewichtskonstanteDiallyl disulfideBody weightTidal raceAreaComposite materialAcidStorage tankCheminformaticsAgeingWaterSet (abstract data type)Chemical reactorOrganische ChemieOrigin of replicationCaveSpawn (biology)TeaDiagram
Transcript: English(auto-generated)
00:06
Today is the day that we're going to finish thermodynamics. We'll finish it in just a couple of minutes. I'm very pleased.
00:20
I'm talking about the Chatelier on Monday. And we talked about the hyperbolic process. And I said, this is going to be the extra credit problem on midterm 2.
00:40
So I wrote it down here so that we're clear on what it is. And once you think, I just threw it out there. Extra credit problem on midterm 2, the Haber-Bosch process is among those that does not conform to the prediction of the Chatelier's principle. Explain why, very briefly and specifically, I'm going to want a chemically specific reason
01:01
why the Chatelier's principle is not obeyed by the Haber-Bosch principle. It has something to do with the iron catalyst. What is it about the iron catalyst is that means that you can't just increase the partial pressure of hydrogen and get the rate of production
01:21
of ammonia to increase. That's what I mean by doesn't follow the Le Chatelier's principle. You can't 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,
01:43
though, isn't it? Le Chatelier'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.
02:02
It'll be worth either 5 or 10 points, depending on how much we can get right, or 0, of course, if what you're saying is going to be done. Don't say the truth. So that's the question. There's going to be a very clear case about that.
02:22
Le Chatelier's principle says, with an increase in the total reaction surface position equilibrium should shift to the left because of this shift in total reduction total reaction pressure. If I increase the pressure because there's two moles of NO2 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
02:42
to the left. So I went through and showed that we can make that 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.
03:00
But here's the reaction that we wrote down on Monday. We can go through and calculate the position of equilibrium. We've seen these expressions. Calculate the mole fractions of NO2 and N2O4. The mole fraction of N2O4 is just the number of moles of N2O4 divided by the total moles, N2O4 plus NO2.
03:26
So if we just plug in this and this for the number of moles, Y, here for the NO2, that's an example. Here's the number of moles. Here's the total number of moles, that plus that.
03:41
OK? So I get this expression. If I write the values for N2O4, I get this expression right here. OK, so now I can write the expression for the equilibrium constants in terms of pressure. Sometimes we call that K sub P. That just means the equilibrium constant written down in terms of pressure.
04:02
OK? And so it's just NO2 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 when we're done canceling all these P's.
04:21
Yes, that's the pressure corresponding to the standard state, 1 bar. And then we can write this in terms of the total pressure, by using the mole fraction. Because the mole fraction times the total pressure is just the partial pressure. So I can write the partial pressure of NO2
04:42
in terms of the mole fraction, same thing for N2O4. And then I can plug in our expressions to the mole fraction that I derived two slides ago. Here's the expression for the mole fraction of NO2. Here's the expression for the mole fraction of N2O4. And then if I do a little algebra and simplify that, I get this equation right here.
05:01
And then if I solve for X, I get this equation right here. But this is indeed very sad, this line. Because what it tells us is that if we increase the total pressure by increasing the pressure of either NO2 or by reducing the volume
05:22
of the reaction vessel, either way, if that P goes up, this quotient's going to get smaller, isn't it? If I increase the size of the denominator, this quotient's going to get smaller. X is going to get smaller just to the left.
05:44
As this gets smaller, X gets smaller. X is the progress of the reaction. Of course, the progress would also be true. Now, you don't have to be able to derive this. But it is rather satisfying to see
06:02
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.
06:21
That's the same value for X that we got on Monday. It means we probably did it correctly. OK. So that's the effect of pressure. You can usually tell just by inspection what effect pressure will have on a gas phase equilibrium.
06:43
Right? 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't Hoff equation. And I don't want to spend a lot of time talking about it or going through what it means.
07:01
But basically, if a reaction is exothermic, in other words, delta H for the reaction is less than 0, right? The reaction will show d log K over dt of less than 0, of course, according to this reaction. In other words, K will get smaller with increase in T.
07:23
In other words, the reaction will shift to the left in 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't Hoff equation
07:41
that talks about the temperature dependence of chemical reactions. It's very important. We can spend 20 minutes talking about it right now. But I don't want to. OK? So I'll ask you something about this equation on Friday with the reaction to this, changing the temperature
08:06
to this, what's going to happen to the reaction rate or the reaction equilibrium rate. OK? We are done with thermodynamics. We are done with statistical mechanics, all right?
08:25
In chapter 17, what we covered was we talked about section 17.1 that gives energy minimum, the reaction gives energy, and how to calculate those things. We talked about chemical potentials. That's section 17.2.
08:42
There was a section on the statistical description of the equilibrium constant that is very interesting. But we didn't talk about it. I'd read it. We did talk about the effect of pressure today and the effect of temperature. We sort of talked about it.
09:01
I showed you the van't Hoff equation. And I asked you to read about it. And that's section 17.5. Right? This is everything we did in chapter 17. There's a lot more stuff focusing on electric chemistry, which is a subject near and dear to my own heart. It's what I do. But we don't have time to talk about it.
09:24
OK? 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.
09:43
All right. 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 law defined and derived? What is the integrated rate law? Why do we need it? How does the reaction law reduce the integrated law?
10:01
Blah, blah, blah, blah, blah. It's all background on chemical kinetics. But chapter 19 is very important. Now, where are we in the scheme of things? We're called badly off track if we come. This is where we hope to be. This was our scheme on day one. We were brought out by the Bush and Hale.
10:21
Now, here we are today. OK? We're right in the middle of week seven, here we come. And we're not three lectures into chemical kinetics as shown on this scheme right here. We are, in fact.
10:41
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 course this summer
11:01
because everyone now recognizes that you can't do all of this in one quarter. You can't have two quarters of quantum mechanics, this microscopy, and then have one quarter of statistical mechanics, thermodynamics, chemical kinetics, and chemical dynamics. It can't be done. So this doesn't really affect you.
11:21
We're going to try and do this to improve some justice. All right? 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 course on this. You'd enjoy it. But we're going to do it all in about three weeks, two, three.
11:47
That's how much time we're going to have. OK. So on the midterm, midterm's a week from Friday now. It's really going to focus on the stuff since midterm one. So the stuff at the end of mainly the thermodynamic stuff,
12:04
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
12:21
going to be on midterm two as well. OK. 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
12:42
when thermodynamics and statistical mechanics worked out. The first Nobel Prize was in 1901. And the whole slew of guys just missed earning one of those because their work wasn't appreciated until after they died.
13:03
You can't win a Nobel Prize after you die. And so this is earlier than that. The first kinetic measurements were made 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
13:22
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 variables that we have control over? All right? So we need a lot of notation.
13:41
And we want to be very clear about this so that there isn't 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.
14:00
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
14:20
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. It just describes the stoichiometric relationship
14:41
between reactions. So here's a 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, all right? That would be something having to do with the mechanism. But this reaction is not conveying
15:01
any mechanistic information. Something totally different might have happened. Stoichiometric reaction between 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
15:22
between an H2 and an O2. It might be kind of surprising to hear that, 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, a mechanism.
15:48
Hydrogen dissociates to get two hydrogen atoms. Oxygen dissociates to get two oxygen atoms. And then hydrogen atom, that collides with an O2, to get us an all intractable and an oxygen atom,
16:00
and so on and so forth. OK, 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 in the sense that it
16:22
can't be made any more primitive than what it is. All right? An OH radical actually collided with an H2. That's what actually happened, and we got H2 and hydrogen atom out of it.
16:41
OK? So there are elementary reactions that comprise a chemical mechanism, and then there's a stoichiometric reaction that is just somehow the sum of all this produced water. You with me? It's confusing. These are elementary reactions, reaction mechanisms that comprise an elementary reaction, blah, blah, blah,
17:02
blah. For example, this elementary reaction hydro PS. 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.
17:21
There are explosions that can occur because of this complex mechanism. We'll talk about it later. So an elementary reaction is one in which the indicated products are formed directly from the reactants. Notice, as in this case, that the elementary reactant
17:41
molecule with the oxygen atom actually collided with H2 and we got OH radical and hydrogen atom. OK, so we've already discussed this thing called the extent of the reaction with the Greek symbol, the name I don't know how to pronounce.
18:02
Here's the technical definition of what it is. I don't think we actually broke down the technical definitions 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
18:21
that applies in 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
18:43
and it's in units of moles. OK, so this is not as confusing as it looks. Here's an example. H2 plus Br2 is 2HBr. Now, let's say that the change in HBr is plus 0.002 moles.
19:01
All right, 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 and so it's positive by definition.
19:23
OK, so I get 0.001 moles. That's the extent of the reaction. Moles of pressure, concentration, usually moles. So now I can do the same thing for H2. I can equally equalize the extent of the reaction
19:40
in terms of the change in H2 of minus 0.001 moles. All right, 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. All right, so I get minus over minus is plus
20:01
and those two numbers happen to be the same or I did the calculation wrong. OK, so we can calculate the extent of the reaction if we just know how many moles of something was generated I can then deduce what the extent of the reaction is. OK, now in terms of the extent of the reaction,
20:23
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 is a function of time. Yes, that's the volume.
20:42
Like the extent of the reaction itself, the rate of the reaction does not depend on the other 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 of any reactant and product.
21:00
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,
21:23
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.
21:42
I can define the reaction rate as minus 1 over A. A is the stoichiometric coefficient on A. It's negative times DAT, where that A in brackets there, that's the concentration of A, or it's partial pressure.
22:07
So 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.
22:23
The rate is 1 over the stoichiometric coefficient times the concentration DT. OK, we're glad to do this kind of fast. You guys getting all this?
22:45
Now, this is a change in concentration with time. All right, the units are typically molar per second or ATM per second.
23:00
Right? 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. OK, this factor of 1 over volume is gone because we're talking about concentrations here. We had 1 over volume here, all right, momentarily,
23:22
until we converted that volume over this extent of the reaction into a concentration, and we're never going to go back. OK, a rate law relates the concentration reactions as a product. OK, so this is a rate law.
23:42
That's the reaction rate. OK, that's why it's called a rate law. All right, the rate law predicts the reaction rate. Makes sense. Where K, alpha and beta are independent of the concentration, right? That's a rate constant.
24:01
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
24:23
that you're writing the rate law for is an elementary reaction, and you know that. If you know the reaction's 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,
24:41
you can't assume that that's true. Instead, alpha and beta are change of 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
25:03
of the chemical species in this reaction. So 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 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,
25:21
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, units that depend on the,
25:42
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.
26:01
All right? The rate constant tells us what the overall reaction order is. All right? 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,
26:24
I'm telling you it's a first order reaction. All right? 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. Okay? For example, if alpha is one and beta is zero,
26:42
then the reaction is this because if beta is zero, that's just one, right? The reaction is first order in A and first order overall because one plus zero is one. Okay? So what's K going to be? You can always figure it out. You take the rate.
27:02
You take, 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. So if you want to know what the units of the rate constant are, solve for it in the rate law,
27:24
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. Okay? So if you see units of per second, you know immediately it's a first order reaction.
27:42
No one has to tell you that. If on the other hand, alpha is one and beta is one, then you've got this rate law and you've got an overall second order reaction, right? 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 and so I've got molar squared
28:00
and so now the units are per molar per second, right? If I see those units, I know immediately that I've got a second order reaction, right? The units are telling me what the overall reaction order of the reaction is, R, was.
28:25
So stoichiometric reactions, the rate law can be deduced from 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.
28:42
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?
29:00
This is just expressing the overall reaction that occurs as a consequence of about 20 steps, all right? And so it's tempting to say, oh, the rate of the reaction is just K times concentration of hydrogen squared times concentration of oxygen. No. That is untrue.
29:21
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, all right?
29:42
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.
30:02
Right? 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.
30:22
You do have to know what they are. All right? 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,
30:43
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 HI decomposition. In this case, if the products, so HI decomposition would mean the reaction is running in the backward direction, right?
31:03
In that case, these are products of the decomposition, right? So if we remove them, then the rate of the back reaction would be this, because this is an elementary reaction
31:21
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 not going to talk about reversible reactions, the kinetics of reversible reactions very much for a lecture or two. We'll get to it. Right? But I'm just pointing out that
31:41
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 in this. 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
32:02
if these guys are removed or if they diffuse away very quickly, and no back reaction can happen. The stoichiometric reaction for acid aldehyde decomposition, this is acid aldehyde. It decomposes to get methane and CO.
32:23
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
32:42
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? All right? Because there's a one here. That would be the exponent.
33:02
All right? And the rate constant would be this. 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. All right? But that's not an elementary reaction right there. Tell that immediately.
33:21
We did not know this was stoichiometric. 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?
33:41
What are its units? And all I have to do is solve. Here's the rate. Here's the concentration. Concentration's at 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.
34:01
Okay? 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.
34:27
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. All right? But the rate constant, the little K does.
34:41
All right? And the units contain information. Now, how do we experimentally determine these rate laws? This is important. We have some methods that we can use for doing that
35:01
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, all right,
35:20
you want to know what the stoichiometric coefficient is for A, but you don't know what it is. All right? You want to figure out what alpha, beta, and gamma are. All right? The way that you do that is you isolate reactant A by making all of the other reactants enormous compared to A.
35:40
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. All right? B and C are not rate limiting. You've flooded the system with B and C.
36:03
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.
36:20
So 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
36:41
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. All right? And essentially I've turned this into a pseudo alpha order reaction.
37:06
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.
37:22
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.
37:40
All right? I have to use tiny concentrations of A, but I can vary it. All right? And as I do that, here's the lowest concentration of A. Here's the initial rate that I measure. All right? Here's a higher concentration of A. Here's the initial rate I measured there. That's that red line. And here's a higher concentration. All of these concentrations of A here are tiny compared to B and C.
38:01
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
38:20
of the reaction with respect to reagent A. All right? It's a lot of work. All right? 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. All right? So I can take the reaction apart piece by piece
38:51
and figure out what the molecularity is of every reactant. I'm looking for a word that I can't think of.
39:02
Okay. 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. All right? 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.
39:22
All right? And the reaction better not change. All right? 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, all right, they're big enough. Reactions only depending on the concentration of A. Now I can actually run my experiment.
39:43
Okay? 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. All right? That's the method of initial rates. Yes. Yes. Method two, drive an integrated rate law for the reaction.
40:09
Right? An integrated rate law. All right? 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.
40:21
All right? 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. All right? If we integrate our differential rate law, we'll get an integrated rate law, and the integrated rate law explicitly predicts
40:40
what A does as a function of time. Take this for example. All right? The rate is minus DADT. That equals K times A. All right? Very simple reaction. We can integrate both sides. All right? I'm going to move the minus sign over the right-hand side, integrate from the initial concentration of A
41:01
to some final concentration. All right? I'm going to just integrate this DT, and so when I'm done doing that, I'm going to have minus KT, right? 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
41:22
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. All right? 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.
41:42
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. All right? 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,
42:01
a third order reaction, a half order reaction. All right? 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,
42:22
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.
42:44
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,
43:02
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. It's one button in EGOR to do that, and then you get the rate constant from that fit. You don't need to turn it into a straight line,
43:20
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. I remember that very well, sadly.
43:40
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.
44:01
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, 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,
44:20
but the half-life is the time needed for half of a reactant and chemical reaction to be depleted, half of the reactant to be depleted. So here's the integrated rate law for a first-order reaction, which we just derived. All right, the half-life just says,
44:40
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.
45:01
Okay, so if I solve for T to the one-half, it's just log 2 over K. That's dead simple. All right, now, note that in this case of a first-order reaction, the half-life is independent to the initial concentration of A zero. That's the hallmark of a first-order reaction. You start with .2.
45:21
You measure the first half-life. That's how long it takes to get to .1. Now, you pretend .1 is your starting point. You measure how long it takes to get to .5, or rather .05. All right, same amount of time. All right, you measure the third half-life. That's the time it takes to get to .025. That's the same amount of time.
45:40
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 as a function of time. You could just fit this green curve, this equation right here, and you're done.
46:02
All right, 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
46:24
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
46:41
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.
47:01
In other words, if something funny happens, 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.
47:22
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,
47:43
this rate law really holds up. So this is superior. One type of second order reaction. Okay, so I think we'll just stop right here.
48:07
Okay, so this stuff that we talked about today could be on the quiz. Let me just say that. Thank you.