Lecture 25. Chemical Kinetics Pt. 4.
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TafelweinChemische ReaktionKonzentratOktanzahlGenexpressionStoffwechselwegElektronische ZigaretteTopizitätChemischer ProzessReaktivitätKörpergewichtMemory-Effekt
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BohriumMannoseVancomycinMagnetometerMaskierung <Chemie>OktanzahlGangart <Erzlagerstätte>PhthiseGenexpressionAktionspotenzialKörpergewichtChemische ReaktionMolekülCupcakeKontusionChemische ForschungChemischer ReaktorWasserSchlag <Landwirtschaft>KonzentratPotenz <Homöopathie>Chemischer ProzessReaktionsgleichungTopizitätGezeitenstromFunktionelle GruppeBiochemikerinMemory-EffektPeriodateElektronische ZigaretteFülle <Speise>SenseMonomolekulare ReaktionEnzymStoffwechselwegComputeranimationVorlesung/Konferenz
31:49
MannoseMaischeAltbierPharmakokinetikReaktionsgeschwindigkeitChemische ReaktionOktanzahlGangart <Erzlagerstätte>Memory-EffektGenexpressionSystemische Therapie <Pharmakologie>WursthülleThermoformenTankKonzentratElektronegativitätReaktionsgleichungWasserfallMaskierung <Chemie>KörpergewichtRestriktionsenzymMassenwirkungsgesetzChemischer ReaktorPeriodateGleichgewichtskonstantePotenz <Homöopathie>SenseComputeranimation
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MaischeZeitverschiebungEthylen-Vinylacetat-CopolymereMannoseHomocysteinHydroxybuttersäure <gamma->MagmaMagnetometerVancomycinSetzen <Verfahrenstechnik>OktanzahlWursthülleGenexpressionChemischer ReaktorChemische ReaktionReaktionsgleichungGangart <Erzlagerstätte>Chemischer ProzessWasserscheideMolvolumenPotenz <Homöopathie>KonzentratElektronische ZigaretteWasserGezeitenstromBaseAzokupplungArzneimittelReaktionsgeschwindigkeitAntigenitätGleichgewichtskonstanteSingle electron transferQuellgebietRadioaktiver StoffThermoformenComputeranimation
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Gangart <Erzlagerstätte>Terminations-CodonInlandeisAusgangsgesteinOktanzahlVorlesung/Konferenz
Transkript: Englisch(automatisch erzeugt)
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okay let's go ahead and start does anybody have any questions before I begin? all right, so
00:22
we're fast approaching the end of this topic and so today what we're going to focus on is we're going to look at reaction pathways or the reaction mechanism if you guys remember, we said in kinetics, we have two important goals one, we want to look at how fast reactions proceed
00:43
and secondly, we want to look at reaction pathways and so up to this time, we spend a lot of time looking at how we determine the rate law the rate law is related to how fast the reactions proceed
01:03
and we said if you want to determine the rate law or the rate expression there are two experimental approaches that we can take if what you're measuring in the laboratory is rate versus concentration then we know we can figure out the rate law by taking two equations and
01:23
dividing one by the other all right, so that's if you're measuring rate versus concentration the other alternative is we can measure concentration versus time. Concentration of reactant versus time and if the second approach is where you're measuring concentration versus time
01:41
then what you do is you make three graphs you plot concentration versus time ln concentration versus time and one over concentration versus time and you find the plot that gives you a straight line and from that you can establish the rate law
02:01
okay? So we've looked at that now we want to know what do we want to do with the rate law? Why is it that we go through this process of trying to figure out what the order of the reaction is and try to establish what the rate law is and the reason we do that is because we want to figure out
02:21
the pathway the reaction takes and the word that we use to describe the pathway is called the mechanism. So we want to understand the reaction mechanism so today we're going to look at reaction mechanism and so I've written down some short notes for you and it's on the class website, you already have a copy of it
02:42
you can see that many reactions do not occur in a single step but proceed through a series of steps. All right? Remember when we looked at Hess's law we said that, you know, a reaction can take place in a single step or you can have multiple steps it turns out many of the reactions, the majority of reactions, actually take
03:02
place in multiple steps and if you take an example if you look at the overall reaction shown here, if you look at the overall reaction where NO2 reacts with CO so that gives us the reactants. These are the starting materials that you start with
03:20
and it reacts to give us the products, which are NO and CO2 so when we write a balanced equation in that format where you give the materials you started with and the products that you end up with, this is the overall reaction where you look at the initial state the initial substances that you started with
03:40
and then you look at the final states, which are the final products so that just gives you the overall reaction that took place so the overall balanced equation tells us the reactants and the products and the stoichiometry the relative ratios in which the reactants react to give you products
04:01
it does not tell us anything about the pathway or how the reaction occurs in steps so in reality, if you look at the details of how this reaction takes place in reality what happens is one molecule of NO2 reacts with another molecule of NO2 to give you NO3
04:21
plus NO. So the first step, if you look at that can everybody look at that and tell me what is really happening there? you have two NO2 molecules giving you NO3 so what's happening? One NO2 molecule has given up what? an oxygen atom to the other and so you see an oxygen atom being transferred to the other so that you have NO3
04:41
and the other one has lost an oxygen atom, which is NO that's the first step now in the second step you can see that NO3 now what does that do? it reacts with CO and what is it doing? it's transferring that oxygen to CO to give you CO2 so can you see the difference between the pathway that the reaction takes place
05:02
and the overall reaction. The overall reaction is when you take the two steps and add them up you end up with the overall reaction so very often in thermodynamics, when we look at initial states and final states what we're looking at is the overall reaction we're not looking at the mechanism. We call this a mechanism or the pathway
05:23
but we're looking at what we started with and what you end up with and we do not look at the intimate details of what's actually happening as the reactants, the overall reactants convert to products so the series of steps are called the reaction mechanism so the goal of
05:41
kinetics is to figure out the reaction mechanism the mechanism gives you the intimate details of how exactly the reactants convert to products we're not just looking at what you start with and what you end up with we're looking at a series of steps that lead to how the reactants convert to products
06:02
so do you understand the difference between the overall reaction that we've looked at before which just looks at what is the reactant and what is the ultimate product and a reaction mechanism which looks at the series of events that lead reactants to convert to products
06:21
so the two steps, so we call this a two-step mechanism and this is the first step and this is the second step so to understand a reaction we must know its reaction mechanism and one of the main purposes of studying kinetics is to learn as much as possible
06:40
about the steps involved in a reaction. So when we talk about pathway or the mechanism, that's what it means all right? So each step in a mechanism, so each of these steps in a mechanism is called an elementary step an elementary step is directly caused by collisions of atoms, ions, or
07:02
molecules the rate expression for an overall reaction cannot be derived from the stoichiometry of a balanced equation as we have kept emphasizing and must be determined experimentally. So if you guys remember overall experiment in the lab where we measure all we have at our disposal is what we start with, the initial state, which are the reactants, and
07:24
the final products and so when we look at reactions in the laboratory, what we're looking at is the overall reaction. We're not looking at individual steps so when you look at the overall reaction remember we kept emphasizing that when you look at a reaction
07:40
you cannot figure out what the rate law is going to be based on an overall reaction it turns out if you are looking at a mechanism and each step in a mechanism is called the elementary step and so if you're looking at steps in a mechanism now you can actually look at the coefficient in that mechanism for each step
08:02
and you can write the rate law based on the coefficients. So the order of the reaction actually corresponds to the coefficient is that clear? so that's the difference between the overall reaction you're looking at and individual steps within a mechanism and each elementary step, so we call each one of this an elementary step
08:23
and an elementary step really looks at it at a molecular level so what this means is that when we write an elementary step in the first step it means that two NO2 molecules are colliding. So you have two NO2 molecules colliding the instant they collide with each other, one oxygen is transferred from one NO2 to
08:42
the other and then it breaks apart and you have NO3 and NO now this NO3 now will go collide with a carbon monoxide molecule, and the instant they collide you have an oxygen being transferred. Because remember, they have to come into contact if you're going to have atoms transferring from one molecule to another
09:02
and so an elementary step always describes collisions between molecules. All right? So the first step, you have two molecules colliding you have NO3 and then the NO3 as it bounces around will collide with a CO, so again you have two molecules colliding and you have an oxygen atom being transferred from one
09:22
to the other so because you're looking at, in an elementary step, because we're looking at molecules colliding, we call that a molecularity. All right? and it turns out molecularity can be molecularity refers to one step in a mechanism. So you have a series of steps
09:42
and each step in a molecule is called an elementary step or molecularity. All right? and it turns out elementary steps or molecularity can be of three types. All right? We call it unimolecular and a unimolecular one
10:00
is where you have- this is a step in a mechanism. Remember, this is an elementary step, we're not looking at the overall reaction now we're looking at individual steps and if you have a unimolecular reaction, it means its reactant is only one molecule. So what happens is A goes to products
10:20
so now we're looking at a single step in a mechanism and if one reactant in a single step goes to give you products, we call that unimolecular and so if you want to write the rate for this the rate equals K rate constant K times A
10:40
so this is for that one step all right? So this is one step in a mechanism now if you have two molecules colliding, like the one that we looked at where NO2 and NO2 collide, now you have two molecules colliding so we call that a bimolecular reaction. So if you have in a mechanism if you look at a single step
11:02
and you have two molecules colliding, so you have A plus A giving you products then we call that bimolecular because now two molecules are colliding with each other and if you look at the rate of this, because it's a bimolecular collision
11:20
now the rate corresponds to only in a mechanism, individual steps in a mechanism, the coefficient in the balanced equation gives you the order so here it's a bimolecular reaction now you can either have the same molecules colliding, or you can have two different molecules colliding like this
11:40
so if this happens you can see rate is K it's first order in A and first order in B but overall it's second order. So it's still second order and you can see that the rate law, if you have two molecules colliding you always call it bimolecular
12:00
and a bimolecular reaction will always be overall second order. Yes? So we're going to figure that out. So her question was how do we know whether it's first order or second order without experimental values? It turns out what we measure experimentally is the overall reaction. All right? Today, our goal is to take what we measure experimentally for the overall
12:24
reaction and figure out how you come up with each of these steps. All right? So I'm going to talk about first kind of introducing you to the fundamentals and then the goal is that experimentally measure the overall rate law we're going to use the overall rate law to figure out the individual steps in a mechanism. Okay?
12:43
So we looked at bimolecular reactions. So when we say bimolecular, what does that mean? Two molecules need to collide and so you can only have two reactants there in an elementary step because you have two molecules colliding Okay? Now if you have three molecules colliding, it's called a termolecular reaction
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Okay? So reactions involving the collision of three molecules these are rare because the probability that three molecules will collide simultaneously in the in the right orientation is rare. Okay? So the third problem is that if you have three molecules colliding it's very very rare
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that you have when molecules are bouncing around that three molecules will hit at the same time. All right? and so even though we talk about termolecular reactions, in reality they're not experimentally observed. Okay? In reality, they're not observed because it's very difficult for three molecules to hit at the same time
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but if you did have termolecular reactions, then what you have is this three molecules colliding, giving you products and that would be rate rate constant times A to the power three because you have three molecules colliding or you can have two species. So if you have A plus A plus B
14:03
giving you products then rate would be K times its second order in A and first order in B overall it's third order or you can have three species that are involved, three different molecules colliding
14:20
and that would give you products and here rate would be first order in A first order in B first order in C but overall, third order. All right? So like I said, when we looked at figuring out orders of reactions, I said, you know, the common ones are first order, second order, zero order. All right?
14:44
and so here we've looked at the fact that if you look at individual steps what we commonly observe is either unimolecular, which is first order, or bimolecular, which will give you, just for a single step, it'll give you a second order. Okay? Now let's go back to our mechanism. All right? So
15:02
we said that each individual step in a mechanism you can figure out the rate and if you're looking at each individual step in a mechanism, now you can look at the coefficient of the balanced equation to figure out the order of that reaction. Okay? So now let's go back to our
15:23
mechanism that we looked at and if you look at this mechanism, you can see that this is a bimolecular collision and when these two molecules collide what you have is you have a new species being formed that's not a reactant or a product so if you're just looking at reactants and products
15:43
we won't see this new species that's being formed, and if you look at this new species that's being formed, it's formed in the mechanism and it's consumed in the mechanism. So if you look at NO3, NO3 is something new that is formed in this mechanism, a pathway and then in that itself it's consumed. So if you look at the overall
16:03
reaction if you're just looking at reactants and just looking at products you don't see this new species that's formed and consumed so in a mechanism, if you have a new species that's formed and consumed, and it's not a reactant nor a product, we call that an intermediate. All right?
16:21
So we say NO3 is an intermediate in this mechanism where it's being formed and consumed. All right? So does anybody have any questions up to this point? So we're looking at pathways and the pathways looks at individual steps. Yes, go ahead. We're going to take some examples. All right?
16:42
So I'm going to show you some examples where we're going to look at zero order. So you're not going to see zero order in individual elementary steps but I'm going to show you how that comes into play. In fact, I'm going to take an example in class and show you how the zero order comes into play. Okay? So we've looked at intermediates, and we said intermediates are formed and consumed
17:04
in a reaction, and finally we want to see how does this fit in with the overall rate law that we apply. So if we look at this, we said in this mechanism we're saying that there are two steps step one and step two now remember, each individual step
17:23
will react at a different speed or different rate. All right? And so if you look at this mechanism you can see that the first step is the slow step and the second step is the fast step. All right? So what you have is in a mechanism, you have reactions taking place in sequence, and
17:41
you know, you have step one taking place, then step two, and if you have four steps, step three, step four, you have multiple steps occurring but can you see that overall the slowest step is what's going to determine how fast that reaction proceeds. We call that the bottleneck. All right? It's much like if you have five lanes in a freeway, and let's say each of these
18:03
five lanes is like five steps in a mechanism and if everybody's driving at the speed limit, which is let's say 65 and there's no problems, the traffic flows smoothly but then suddenly what happens is, let's say the lanes are closed and the five lanes are now closed to one lane
18:22
so now what happens? To begin with, everybody's moving at the speed limit but once you hit that bottleneck, now what's going to happen is how fast people proceed beyond that point is going to depend on how fast the cars are moving through that single lane and so that single lane is now the bottleneck that's the slowest step
18:43
likewise, when you have multiple steps and you have one step which is the slowest the overall speed of that reaction is going to be determined by that slowest step because you have multiple steps, everything can go fast but you have one that is really really slow and so overall what we measure experimentally
19:03
is the slowest step. Do you understand that? So the rate law that we measure, or the rate that we measure experimentally is always the slowest step because the slowest step is called the rate limiting step or the rate determining step. That's the bottleneck. That's what's slowing down the
19:21
whole reaction. All right? So when we look at experimentally what we're looking at is experimentally when we measure rate laws we're looking at the overall reaction. We don't see the individual steps all right? So what we see is the overall reaction and the overall rate that we measure experimentally always corresponds to
19:40
the slowest step in the mechanism because that's the bottleneck. So so if I take this two-step mechanism and if I take the first step, which is bimolecular, and if I write the rate law for that the coefficient is two, so for step one rate one would be K1, which is the rate constant for the first step
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times NO2 raised to the power 2 because it's second order is bimolecular. All right? Now step two is this and that rate corresponds to again, remember the rate always depends on rate constant times the reactants. So the reactants here are NO3
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it's first order in NO3, it's first order in CO so the fast step has that. So can you see that what we measure experimentally will correspond to the slowest step and so what happens is that if we carry out the experiment and look at this reaction
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what we measure experimentally would be, this is what we experimentally measure we say rate equals K observed NO2 raised to the power 2. So if we were to carry this out in the laboratory and measure the rate law this is what the rate law would come out to be. All right?
21:00
and this always corresponds to the slowest step. Because remember the slowest step is the rate limiting step and so in a mechanism you have multiple steps and among those multiple steps the overall rate of that reaction is going to depend on the slowest step and the rate law that you see in the slowest step
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is what we measure experimentally because in the experiment what we're measuring is the overall rate, and the overall rate is going to depend on the slowest step. So if you were to go to the laboratory and look at this reaction this overall reaction is what we're looking at
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and if you carry out the experiment we'll find out that even though you have NO2 and CO, the net result is that it's actually zero order in CO remember, your reactants are NO2 and CO so your general form of this rate law would be so if I take this overall reaction
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and can everybody look at this and tell me? I have two reactants, so my overall rate should be K times NO2 raised to the power M times CO raised to the power N but what we measure experimentally is this and what comes out is from this we know rate equals
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K observed NO2, second order, second order CO, zero order, so that equals K observed NO2 so it's second order in NO2. So this is what we observe experiment. This is from experiment
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now this is what we observe experimentally, and you can see that corresponds to the slowest step the slowest step is this step and that is rate equals K1 times NO2 raised to the power 2 that's what we observe as well Do you guys get that? and so what that tells us is that now
23:00
we can explain. So what we start is we measure the overall rate order that comes from experiment and then we come up with a mechanism that's consistent with that and the mechanism has to be where the slowest step corresponds to the overall rate law. Got it? so let's take another example of applying this
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so here's another example that I put so this is the overall reaction we're looking at. So 2NO2 plus F2 gives you 2NO2F so this is the overall, it tells you the reactants that you start with and the ultimate products that you end up with. Okay?
23:41
Now if you go to the laboratory and carry out this experiment and determine the rate law this is what we observe experimentally. And so to show that its experiment, we say the rate constant is what we observe experimentally All right? So what we observe experimentally is rate equals K observed
24:01
its first order in NO2 and first order in F2. So it's bimolecular and that is the overall so now when you come up with a mechanism what does that tell you? That the slowest step has to involve these two molecules colliding with each other so that means that you have to come up with a mechanism
24:21
where the slowest step is where one molecule of NO2 collides with one molecule of F2. All right? and so here is a possible mechanism where you have NO2 colliding with F2 to produce an intermediate NO2F
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and F. All right? And this will be the slowest step. Got it? The next step would be another molecule of NO2 colliding with F to give you NO2F, which is the product. All right? And so the first step is slow and determines the rate and so you can see now
25:00
the mechanism that I came up with is consistent with what I have observed experimentally. Okay? So it turns out that coming up with a mechanism is really a skill. It's an intellectual exercise and is a skill that you develop
25:20
but whenever you come up with a mechanism you have to look at the overall rate law and make sure that you come up with a mechanism where your slowest step corresponds to what you see experimentally. Okay? So now scientists do this all the time. So when you come up with a new reaction and you want to figure out what the pathway is, and so
25:42
remember, you guys have, you know, if you're a biochemist or you're a biology student, you've looked at enzymes and you guys understand, you know, you've learned about the lock and key mechanism, you know, where how enzymes function and that they function in a lock and key sort of way
26:03
All right? That is a mechanism and how do we come up with that mechanism? It's exactly this way. You use kinetics. Kinetics is the only way to figure out the mechanism but it requires a lot of experience. All right? So how does one go about deducing the mechanism for a given reaction? The first thing you do is you go to the lab
26:22
and you measure the experimental rate law so you either measure rate versus concentration or concentration versus time and use the integrated rate law and make three graphs and figure out what the rate law is. All right? So that's done by experiment. Yes? Got it? Okay.
26:49
And so the first step is to experimentally determine the rate law. All right? Now the next step is then using chemical intuition one constructs all the possible mechanisms
27:01
that are consistent with this rate law. So now you have to just brainstorm and you sit there thinking of every conceivable possible way where the slowest step is going to correspond to the overall rate law that you observe. And this is just coming up with many different mechanisms. All right?
27:20
Then what you do is you carry out. So you list out potential mechanisms. You can write to five to six different mechanisms. All right? Now what you do is you further carry out experiments to see whether you can eliminate some of them. One way to do it is you can in situ measure the intermediate. If you can measure the intermediate, remember the intermediate is formed
27:40
and consumed sometimes the intermediate forms so fast and is consumed so fast that you can't experimentally see it but there are other ways where you can it hangs around long enough that you can actually measure and use different techniques to see whether there's an intermediate being formed and consumed. And stuff like that. So you do further experiments to rule out
28:03
some of those All right? In the end what I want you to remember is a mechanism can never be proved absolutely and one can only say it is possibly correct. All right? So you can never 100% prove that there's the mechanism. Because remember, these are individual steps
28:22
and there's no way to see each individual step separately in the laboratory. Do you understand that? So it's purely an intellectual exercise but you can say that in all likelihood, this is the best mechanism that you can come up with to explain that. Okay? Yes, go ahead. Yes.
28:49
So you look at the overall one that you measure and then you try to see which one it would fit, and that has to be the slowest step. Got it?
29:01
And so the way that I'm going to do this is because it requires a lot of skill to come up with mechanisms. And if this is the first time you're seeing this topic, it's virtually impossible for you to come up with a mechanism. All right? So what I will do is that you know, I have put up a worksheet on the class website, and if we have time I'm
29:21
going to work this in class but so here is a worksheet that I've put up and so it's going to get you to figure out the rate law and take you through each step and lastly what we'll do is, what I'll do is instead of asking you to come up with your own mechanism I'll give you like five different mechanisms. All right?
29:43
Of those, four will be incorrect. Only one actually will fit what you see experimentally. All right? And so you can see there's one mechanism there two is a second mechanism, three is a third mechanism and what I will expect you to be able to do is if I give you several different mechanisms I want you to pick the mechanism
30:01
that is consistent with what you see experimentally. All right? I will not expect you to come up with your own mechanism because that, you know, you need a PhD for that. It requires a lot of experience. All right? and and so it's not it's not practical for me to ask you to come up with a creative mechanism
30:22
Does that make sense to everybody? So what I will expect you to be able to do is given four or five different mechanisms only one will actually work and is consistent with the experimental data. Got it? So now you understand why we went through this whole process of trying to figure
30:40
out what the rate law, the overall rate law for a reaction is and we said when we look at the overall reaction, because that's what we have when we start with you cannot look at the balanced equation to figure out what the order of the reaction would be. All right? But once you've figured that out, now when you look at mechanisms,
31:01
now we know that each individual step in a mechanism actually the coefficients in the balanced equation actually gives you the order. So what you do is for each mechanism, you take each step and write the rate law that corresponds to that step and then you know that the slowest step the bottleneck, or the rate-limiting step, or the rate-limiting step
31:22
has to correspond to what you see experimentally. All right? Any questions up to this point? Now what I want to do is I want to kind of relate this to equilibria as well. So we're going to look at kinetics and equilibrium. All right? Because there is a relationship between kinetics and equilibrium
31:42
and so if you look at kinetics and chemical equilibrium and let's say that we're looking at
32:02
consider a elementary step in a reaction mechanism and we're going to just consider one step in a mechanism. Okay? And let's say we're looking at
32:23
NO plus O3 in equilibrium with NO2 plus O2. All right? So up to this point, we've looked at reactions that proceed only in one direction.
32:41
When we deal with equilibria, you can see that equilibria deals with reversible reactions. All right? There's a forward reaction, and because there's a rate constant associated with the forward reaction, we're going to call that Ksub1 and there's a reverse reaction and the rate constant associated with that, we'll call it
33:02
oh, let me not put it that way, let's put it this way. So can I rewrite that, please? So K-1. All right? and that is to distinguish so let's say this is one step in a mechanism and let's say this is the first step so if it's step one, because it's step one I'm going to call the forward
33:20
rate constant K1 and for the reverse direction I'm going to call it Ksub-1 which is the rate constant for the reverse reaction because there's two reactions going on so if I take the forward rate, remember this is an elementary step so I can look at the balanced equation to figure out the rate law so forward rate
33:41
for the forward reaction rate will equal, let's call that rate one, would equal the rate constant for the forward reaction. Remember, the rate constant always depends on the concentration of the reactants and it's first order in NO and first order in O3. Okay?
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Now if I take the reverse reaction and let's call that rate negative one now the rate constant for the reverse reaction will be Ksub-1 now for the reverse reaction, which are the reactants?
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The products are the reactants for the reverse reaction. So the reverse reaction takes place this way and therefore the rate law for the reverse reaction would be NO2 times O2. All right? and we know that at equilibrium rate of the forward reaction
34:42
equals the rate of the reverse reaction. That's what it means when we say a system is at equilibrium, we said the rate of the forward reaction and the rate of the reaction are the same and that's why the concentrations of reactants and products don't change anymore so because this rate one equals rate two, I can say K1
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times NO times O3 equals K-1 NO2 times O2 and if I rearrange this equation, you can see that if I take the concentrations of the products raised to the power of their coefficient divided by the reactants taken to the power of their coefficients
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this really equals the ratio between the rate constants which is actually the equilibrium constant. All right? So if you want to relate the equilibrium constants to rates it's actually a ratio of the rate constants. The equilibrium constant is a constant
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and that is the ratio of the rate constant for K. The equilibrium constant is K, capital K all right? and if you look at that, that's really the ratio between two constants, which is the rate constant for the forward reaction divided by the rate constant for the reverse direction Does that make sense to everybody?
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So now let's take an example of a mechanism where we have an equilibrium involved as well so let's say that we're looking at this reaction. So I'm going to give you one example of a reaction where not only do you have just a regular reaction taking place in a mechanism, but you also have an equilibrium
36:20
involved so let's take this example where the overall reaction is 2 NO plus O2 giving you 2 NO2 and the experimentally
36:45
measured rate law is rate equals K observed. This is the rate constant that we observe experimentally
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is second order in NO2 and first order in O2. All right? So one way to explain this, one possible mechanism that can explain this is in fact if your mechanism just took place in one step
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all right you can see that if the overall reaction is also an elementary step. In other words, it's a one step mechanism Can everybody see that? If this is an example where this reaction took place in one step and therefore the overall reaction is the same as the mechanism itself, which is a single step, you can see it's second order in NO
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and first order in O2 All right? but that's unlikely if you want to come up with an experiment, one possible mechanism that has been suggested for this is this, where you have NO
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plus NO giving you N2O2 all right, where this is K1 this is K-1 and this is, we call it fast equilibrium equilibrium so this is step one and step two of this mechanism will be now this intermediate that's formed
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N2O2 plus O2 equals K2, which is the rate constant for the second reaction giving you 2 NO2 and this is the slow step. All right? So if you add these two together, you can see this intermediate will cancel out
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so what you end up with is 2NO plus O2 gives you 2NO2 so two ways in which you can explain this experimental data is say that this is actually a single step mechanism where it takes place in one step so that the overall reaction
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is actually just one single step in a mechanism, and it's a single step mechanism or you can explain it by a two step mechanism and in this case you have the first step where you have a fast equilibrium followed by a second slow step. Okay? So now we have to derive an expression
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that fits what we see experimentally. This is what we see experimentally so I need to derive an expression that explains this and so we start with a slow step my slow step is this and I know that the rate equals
39:42
K2 times its first order in N2O2 and first order in O2 All right? So the slow step has an intermediate. Can you see that? I took the slow step and I'm looking at the slow step and I'm writing the rate law that corresponds to slow step. Yes, go ahead.
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Is second order? You mean this step where there's one of this and one of this? Yes. The rate law is first order because the coefficient is one
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the balanced equation, the coefficient is one so it's first order. Do you see that? So a single step in a mechanism the coefficient in the balanced equation corresponds to the order with respect and only for a mechanism. Only in an elementary step you can do that. Okay? But can everybody see this is not complete
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because if you look at the rate law for the slow step, there's an intermediate there. Experimentally, can you measure intermediate? No. Experimentally what you measure is the reactants that you start with. Can everybody see that? So that means I have to replace the terms with intermediate the term that represents intermediate
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with terms that represent reactant. Okay? And how can I do that? I can go to the equilibrium constant and now I have this is a reactant, so we're okay but that one but if I go to the first step, remember that is an equilibrium and I know K equals
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N2O2 over NO squared. Can you see that? Because this is an equilibrium this is the law of mass action that tells me the equilibrium constant. Okay? Now if I rearrange this, can you see now I have a term for this intermediate, which is K
41:42
times NO squared. All right? Now I'm going to place this over there. So if I take this and replace it, now you can see rate equals K2 times the constant, the equilibrium constant, times NO squared times O2
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and therefore rate equals K observed can you see this is what I see experimentally as well so I can come up with a two-step mechanism, all right? But now my K observed is actually K2, which is the rate constant for the second reaction
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times the equilibrium constant. Remember it's a constant times constant is going to give me a constant. Did you guys get that? So so both of those are consistent with the rate law so it could be a single-step mechanism where it just takes place in one step and the whole reaction is just one elementary step
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and it's a single-step mechanism or it could be a two-step mechanism but if you if you want to write a mechanism that is two steps, then you see the formation of an intermediate and that intermediate is being consumed but then you can overall derive an expression that also fits what you see experimentally. Can you see that?
43:01
So we can come up with two mechanisms that are consistent with what we see experimentally and so we need to do further experiments. So it turns out it's the second second mechanism is what really happens and the way you can check that is to see the formation of that intermediate. That intermediate N2 or N2 being formed and appearing as the reaction proceeds
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Do you guys see that? So once again I'm giving you the mechanisms. It's very difficult for you to come up with a mechanism on your own All right? However, if I give you a mechanism and if you look at the homework that you are getting the online homework as well as the homework problems that I picked from the textbook
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you'll see in all those cases there are four different mechanisms and you're asked to pick the one that fits what you see experimentally. Okay? Now, in the last couple of minutes, I want to show you how this whole thing is applied Okay? And that's what that worksheet that I have put on the class website, and you can print it out
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I would like you to work this out because it kind of explains the process as you go along. Okay? So let's look at this. So to begin with consider the reaction. So we're looking at the overall reaction and the overall reaction is given at the top and the overall reaction we're looking at
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is SO2 plus O2 let me straighten that out what you're seeing is 2SO2 plus O2 giving 2SO3. So that's the overall reaction that you're looking at. Okay? Now, the following data for kinetics experiment were taken at 25 degrees Celsius.
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So can you see what kind of measurements are we making here? Are we using the integrated rate law, or are we using rates versus concentration? Can you see rates versus concentration? That means that you have to take two equations write the rate law, divide one equation by the other, and then figure out the order with respect to each one. So the first step is
45:02
find the order with respect to each reactant. All right? Now once you've got that, then we've worked problems where you can determine the rate constant determine the value of the rate constant include the units. Next. Calculate the initial rate of the disappearance of O2 in experiment four. So if you can
45:24
figure it out if you figured out rate constant K and you figure out the order with respect to each one of those, you can figure out what that question mark is in line four. Okay? So let's quickly start by looking at A. We may not get through the whole thing,
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we'll complete it next time, but I do want to go through an exercise of how we build up and figure out the mechanism. Okay? So let's start with this. Find the order with respect to each reactant. So we start with all the information that's given to us, and we'll start by looking at the balanced equation. Can everybody write down the overall rate law, the general form? So we don't know exactly what the order is, so use
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Mn and what would the rate law look like? Can everybody see rate should equal a rate constant K what are the reactants for this reaction? we have SO2 let's say raised to the power M
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and O2 raised to the power M. And can everybody see that in this we know the concentration of rate, we know rate, SO2, O2 concentrations what we don't know is K, M, and M. In that table, concentrations are given and their corresponding rates are given.
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So, if we have three unknowns, how many algebraic equations do we need to have to solve for it? We need three equations. And you can see that three sets of data are given. All right? So we'll start with the first one. The first one would be 7.2 times 10 to the negative 5 moles per liter per second
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equals K concentration of SO2 is .1 molar raised to the power M concentration, which is .1 molar raised to the power N. So that's one equation. Second line is 2.9 times 10 to the negative 4
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moles per liter per second equals K now it's .2 molar raised to the power M times .2 molar raised to the power N. That's equation two. If I take the third equation it'll be 6.5 times 10 to the negative 4 moles per liter per second
47:41
equals K times .3 molar raised to the power M times .1 molar raised to the power N the third equation. So I have three equations. Now the ones that I want to pick to start with is the ones where one of the concentrations are the same. So look at those three equations, and which would you pick to start with?
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Which equations would you pick? Equation one and three. Can everybody see that? Because that's the only one that has the same concentrations. All right? Now we always, as a rule of thumb, it's easiest to start with whichever has a bigger number divided by the smaller number. So this is 10 to the negative 4,
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this is 10 to the negative 5, so I'm going to take equation three and divide it by equation one. Okay? So that means I have 6.5 times 10 to the negative 4 moles per liter per second equals K times .30 molar raised to the power M .10 molar
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raised to the power N and I divide by a smaller number, which is 7.2 times 10 to the negative 5 moles per liter per second K times .1 molar raised to the power M times .1 molar raised to the power N. So you can see K and K will cancel out
49:00
.1 molar raised to the power N and .1 molar raised to the power N will cancel out and so if you divide this by this, you end up with 9 equals this is .3 divided by 3 raised to the power M So what do you think M comes out to be? Two. Okay? So now that I have that
49:22
now I have to turn, so I've looked at these two and these two will cancel out, so now I need to use these two to figure that the next one out. Okay? Now we don't have the same concentration, but it doesn't matter because you already figured out what the order is, so we can put the order in there. Okay? So now what I'm going to do is I'm going to figure out, to figure out N,
49:42
I'm going to divide 2 by one. All right? So 2 would be 2.9 times 10 to the negative 4 moles per liter per second equals K times .2 molar raised to the power M times .2 molar raised to the power N Now we figured out that M is 2, so I'm going to put square there.
50:05
Okay? So that would be squared and I'm going to divide by 7.2 times 10 to the negative 5 moles per liter per second equals K times .1 molar squared, because it's second order in SO2, which is M
50:21
and this would be .1 molar in N. Okay? So K and K will cancel out if you look at this this comes out to be 4 equals over here I have 2 raised to the power 2
50:41
times I have 2 raised to the power M Okay? So the number here is 4. So this is 4 already. 4 is taken. So what does N have to come out to be? Zero order. Because anything to the power of zero is one. Can you see that? So N is zero order. So that means my rate law is what?
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My rate law would come out to be so that's the first step. Figure out what the rate law is and the rate law would be rate equals K times SO2 squared, because it's zero order in O2.
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So the experimentally observed rate law is this. All right? Now can I ask you to complete the rest of that sheet because it takes you step by step in all the kinds of problems that I expect you to be able to do and finally see whether you can figure out what the mechanism is. Okay? So we'll stop there for today.