Lecture 23. Chemical Kinetics Pt. 2.
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Chemische ReaktionReaktionskinetikOktanzahlKonzentratElektronische ZigaretteTopizitätEnzymkinetikStoffwechselwegChemische ForschungGezeitenstromSammler <Technik>WasserfallVorlesung/Konferenz
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Hydroxybuttersäure <gamma->MannoseBohriumHomöopathisches ArzneibuchVancomycinSerotonin-Reuptake-HemmerPotenz <Homöopathie>ZeitverschiebungProlinBlitzschlagsyndromChemische ReaktionKonzentratChemische EigenschaftOktanzahlPotenz <Homöopathie>SenseDistickstoffpentoxidGenexpressionChemische ForschungThermoformenGleichgewichtskonstanteReaktionsgeschwindigkeitMassenwirkungsgesetzHomöostaseSetzen <Verfahrenstechnik>ReaktionsgleichungWursthülleGezeitenstromChemischer ReaktorFaserplatteTankWasserKrankengeschichteWildbachChemisches ElementKörpergewichtComputeranimationVorlesung/Konferenz
12:57
MannoseBohriumPolyurethaneHydroxybuttersäure <gamma->Konkrement <Innere Medizin>KonzentratKlinisches ExperimentOktanzahlGenexpressionPotenz <Homöopathie>Chemische ReaktionEnzymkinetikThermoformenTrennverfahrenHydroxideReaktionsgeschwindigkeitMassenwirkungsgesetzGleichgewichtskonstanteChemischer ReaktorSingle electron transferAlterungMemory-EffektWursthülleKörpergewichtReaktionsgleichungWasserKonkrement <Innere Medizin>StoffgesetzWasserwelle <Haarbehandlung>GezeitenstromWasserfallWasserscheideMicroarrayFleischersatzComputeranimation
22:35
MannoseDifferentielle elektrochemische MassenspektrometrieInternationaler FreinameBohriumIonenpumpeMagnetometerAnomalie <Medizin>TrihalomethaneTau-ProteinSekundärionen-MassenspektrometrieAlu-SequenzVancomycinMagmaOktanzahlKonzentratGenexpressionChemische ReaktionSingulettzustandPotenz <Homöopathie>Single electron transferKonkrement <Innere Medizin>ReaktionsgeschwindigkeitWursthülleThermoformenSetzen <Verfahrenstechnik>GleichgewichtskonstanteMassenwirkungsgesetzChemischer ReaktorGezeitenstromEnzymkinetikElektronegativitätKrankengeschichteAntigenitätElektronische ZigaretteCupcakeWasserComputeranimation
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BohriumMannoseHydroxybuttersäure <gamma->VancomycinMischenImpedanzspektroskopieKirChemische ReaktionSauerstoffversorgungKonzentratOktanzahlWasserscheideWursthülleElektronegativitätKrankengeschichtePotenz <Homöopathie>GenexpressionGezeitenstromReaktionsgeschwindigkeitFunktionelle GruppeChemischer ReaktorKörpergewichtWasserSetzen <Verfahrenstechnik>ZellmigrationChemische FormelChemische ForschungSingle electron transferKetaminMolvolumenElektronische ZigaretteComputeranimation
41:51
MannoseLandwirtschaftChemische ReaktionThermoformenTopizitätVSEPR-ModellElektronegativitätKörpergewichtInitiator <Chemie>Memory-EffektKonzentratChemischer ProzessKonkrement <Innere Medizin>StoffgesetzChemischer ReaktorSenseOktanzahlGenexpressionSpezies <Chemie>AlterungGletscherzungeDeltaKochsalzFunktionelle GruppeRadioaktiver StoffChemische ForschungFettabscheiderTellerseparatorComputeranimation
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SenseKonzentratZusatzstoffElektronegativitätInitiator <Chemie>Vorlesung/KonferenzBesprechung/Interview
Transkript: Englisch(automatisch erzeugt)
00:07
okay, can I have everyone's attention? so let's go ahead and start but before I do, I wanted to find out do you guys have any questions?
00:26
we have moved on to looking at chemical kinetics, and if you guys remember last time we talked about the fact that when we study chemical kinetics we have two important goals one is
00:41
we need to figure out how fast reactions proceed and the second goal of kinetics is to figure out the reaction pathway when we say the reaction pathway what we are interested in looking at is the detailed events that take place
01:00
as reactants convert to products. So we're interested in looking at how fast reactions proceed and secondly, we're interested in looking at pathways and we're going to start looking at both of these in detail but to begin with, we're going to focus on how fast reactions proceed. And we said when we talk about how fast
01:22
reactions proceed the term that we use is rate and last time we talked about the fact that you can talk about rate in terms of average rate or you can talk about in terms of instantaneous rate and we said that it is much more useful to us
01:43
to refer to rate in terms of instantaneous rate and from now onwards when we use the term rate what we refer to is the instantaneous rate and then at the end of the class we talked about the fact that most reactions are reversible and if a reaction is reversible
02:02
that means that there is a rate associated with the forward reaction and there's a rate associated with the reverse reaction. So there are two rates forward and reverse and so if you're looking at a reversible reaction the overall net rate that we measure experimentally is actually the forward- the difference between the forward rate
02:22
and the reverse rate. Okay? But to keep things simple from now onwards, we're only going to consider forward rate only, and we're going to pretend as if the forward rate is the only one that exists All right? In reality you have a forward rate and a reverse rate
02:40
but to keep things simple we're just going to focus on the forward rate. Okay? So last time we talked about rate, we talked about the unit for rate and now you understand what it means when we talk about reaction rate. Okay? Now we're going to look at it in a little bit more detail and we're going to look at the relationship between rate
03:03
and concentration. All right? So to keep things easy for you, I've just made some short notes and these notes are available on the class website, and it just kind of highlights the important things that I want you to remember as you study this topic. All right? So today we're going to look at the dependence of rate
03:22
on concentration and this looks at the relationship between the rate of a reaction and the concentration and the concentration that we look at is the concentration of the reactant. All right? So it turns out that the rate of a chemical reaction, and remember we're focusing only on the forward
03:43
reaction, so the rate of a chemical reaction, the forward reaction nearly always depends on the concentration of one or more of the reactants. So if you're looking at a forward reaction then how fast that reaction proceeds will depend on the concentration of the reactants. All right?
04:02
If you have only one reactant, it would depend on that concentration of that one reactant. If you have two reactants or more, then it would depend on the concentrations of all of the reactants. So the first thing to remember is the rate depends on the concentration of the reactant. Okay? And if you want to mathematically describe this relationship
04:24
in the form of an equation, we call that the rate expression or the rate law. So I'm going to take the same reaction, I'm just going to write it down so if we take a reaction where our reactant, we have only one reactant,
04:41
and this reactant goes to products. All right? So if your balanced equation is that one reactant going to products and our reactant is A and that lowercase a is the coefficient in that balanced equation. So the rate of this reaction is going to depend on the concentration of A.
05:01
Remember, the rate always depends on the concentration of the reactant. All right? So if I want to mathematically state this, then I would say that the relationship is described as rate equals equals a constant k. Now remember, I want you to always make sure this is lowercase k, not
05:23
capital K. When you use capital K, where does capital K represent? Equilibrium constant. Okay? So this is lowercase k and remember the rate depends on the concentration of the reactant raised to a power n
05:41
which is a number n. Okay? and the k here is called the rate constant. All right? So the lowercase k is a constant and that is called the rate constant this will be the concentration of A and this n here
06:00
is called the order with respect to that reactant. All right? So this is order with respect to the reactant and this relationship when you write it in this format where you say rate equals a constant times the concentration raised to the power of n
06:24
this relationship is what is called the rate law or rate expression so this is similar to the law of mass action. You guys remember, whenever you write the chemical equilibrium expression where it's the products, concentration of the products
06:43
raised to the power of the coefficient is called the law of mass action here the relationship between rate and concentration of the reactant is always called the rate law. All right? So that's what I've given you in these notes and so the same thing is written here
07:02
so it's always rate equals K, concentration of the reactant raised to the power n, where K is a rate constant and this is a constant n is the order of the reaction with respect to that reactant this is the power to which the concentration is raised. Okay?
07:21
The power n in the rate expression has no direct relationship to the coefficients A in the balanced equation. Okay? So this number n has no relationship to the coefficient. It's a number that we determine experimentally. So there's no way in advance for anyone
07:41
to look at a balanced equation and say what that value of n would come out to be. Does that make sense to everybody? So there's no way that you or I can look at a balanced equation and say what that value n is going to turn out to be. The only way to determine the value of n is to experiment. And later on I'll show you how we figure it out. Okay?
08:04
This value of n can be a whole number or a fractional number. All right? So let's see if n is called the order. Okay? So in some reactions the value of n comes out to be one
08:20
when the value of n comes out to be one, we call it first order. Remember that n means order of the reaction. All right? and therefore when n equals one, we call it first order. So when n equals one the rate law is written as this. Rate equals K
08:41
A raised to the power one. Obviously when something is raised to the power one, you don't need to state it as one. If you leave it blank you know that it is to the power one. Okay? So So if you take this example of this reaction where n2O5 goes to 2nO2 plus half nO2
09:01
it turns out experimentally we know this is first order. So the rate law would be rate equals a constant K, which is the rate constant, times the concentration, there's only one reactant and that one reactant raised to the power one. So when you write the rate law like that, we call that a first order rate law. All right?
09:21
Because the order n is one, and if n comes out to be one, then the rate law is written where n equals one. Now there are other reactions where n can be two. When n is two, the rate law is written as rate equals K A to the power two. And we call that
09:41
reaction second order. All right? And so if you take an example, this is an example of a reaction that has only one reactant and therefore rate equals K the concentration of the reactant raised to the power two and that would be the rate law for a second order reaction. All right? And this is a reaction where
10:01
you have only one reactant and if the order turns out to be two, it will be second order. Now sometimes this could be zero. When n equals zero, the rate law is written as rate equals K A raised to the power zero. All right?
10:21
And you know that if A is raised to the power zero, you know that's going to be one and so the rate will be when n equals zero, the rate law turns out to be this and we call this so rate will equal K because A raised to the power zero turns out to be one and therefore that means rate equals the rate constant. So if the rate equals the rate constant K,
10:46
then we call that a zero order reaction. Does that make sense to everybody? So it could be zero order, if n equals zero, it's called zero order, if n equals one, it's first order, if n equals two, it's second order, and so on and so forth.
11:01
All right? Usually the common ones are zero, one, and two. All right? And lastly, remember this could be a fractional number. So if we take here in this reaction, you can see the rate constant comes to one and a half. All right? So here the rate equals K times the concentration raised to the power
11:23
of a fractional number this could be three halves, or 1.5 and therefore that is the order with respect to that reactant. So the lesson is, to remember is, for you to remember is, that
11:40
the lesson from all of this is that the reaction order is an experimental property that cannot be predicted from the chemical equation. It has to be deduced from experiment. All right? So every reaction that you look at, the relationship between the concentration of the reactant,
12:01
all right, and the rate is called the rate law. And the rate law is always written in the same form. Rate equals a constant K, which is the rate constant, times the concentration of the reactant raised to the power of the N. And that N is called the order of the reaction. All right?
12:20
So what happens if you have more than one reactant? So if you have more than one reactant, and let's take an example of a reaction where you have more than one reactant so for reactions with two or more reactants, the rate law is given by so now let's say we have a hypothetical general form of the equation where you
12:41
have A plus B plus C plus D, you can have any number of reactants going to products then the rate law is written as rate equals once again the rate constant K now we have two reactants, so it would be A raised to the power of a number M B raised to the power of a number N
13:02
and if you have three reactants, it will be C raised to the power of Y, a number Y, all right? and that would describe the rate law. So if you have only two reactants, the rate law would be the concentrations of two reactants, each raised to the power of a number. Okay?
13:21
Where M is the order with respect to A N is the order with respect to B and if you add them up, that gives you the overall order. All right? So let's take an example so if we take this example of this reactant reacting with hydroxide to give you HPO3 2- plus H2, let's say we're looking at this
13:44
reaction now if you were asked to write the rate law, the first thing you would do is look at that equation and figure out how many reactants there are there are two reactants therefore the rate law of the rate expression would be rate of that reaction
14:01
equals a constant K, which is the rate constant times the concentration of each reactant raised to the power of a number so it turns out experimentally we figure out that this is first order and hydroxide is second order. All right? So that means we say that this reaction is first order in H2PO2-
14:25
and second order in hydroxide and the overall order would be 1 plus 2, which is 3. Is that clear to everybody? and that's what we call the rate law. All right? and I want to kind of remind you this is very similar to the law of mass action
14:43
where we when we write down and we say the equilibrium constant K equals the concentrations raised to the power of their coefficients likewise, in kinetics when we look at rate law, or the rate expression it's always the rate this gives the relationship between rate of that reaction
15:02
and the concentrations of reactants. All right? and the rate of the forward direction remember, we're considering the rate in the forward direction the rate always will depend on the concentrations of reactants and that gives the relationship so now we want to know how do you- remember I said we can't figure this out experimentally I mean, sorry, we can't
15:21
look at a balanced equation and just say what this order will turn out to be this order with respect to reactants vary from reaction to reaction. And as I said before, there's no way for us to look at it in advance and say what the order is going to be. And the only way to figure out the order
15:40
is to figure it out experimentally All right? So let me show you how we do this. So it is a mathematical calculation, and essentially what you do So let's take this problem, and it's in the worksheet that I have put up on the class website but let's take a look at how one determines the rate law
16:01
for a reaction and how one determines the order. Okay? So here we're looking at a reaction, and the first thing I want you to look at this, and if you look at the equation the balanced equation that's provided to you. Can everybody see there's only one reactant? All right? So the rate law is going to only depend on the concentration of that reactant. Okay?
16:24
So now in the lab, what you would do is you would do several different trials and at different concentrations of the reactant you can calculate the rate of that reaction. Remember how do we calculate rate? Experimentally, what you do is in the lab
16:42
you would look at concentration versus time, all right? You would make a plot and then you would pick an instantaneous time and you would take the slope at that instant in time, the tangential slope, and from that you can figure out what the rate is. All right? So we talked about that last time. So what we have is
17:01
we're looking at the rates at three different concentrations of the reactant. So here's the first concentration and the corresponding rate here's the second concentration and the corresponding rate and this is the third concentration and the corresponding rate. All right?
17:20
Now we're going to use this data to figure out first the order of the reaction. So the question is determine the order of the reaction and write the rate expression. So we're going to start by writing the general form of the rate law. Okay? So without figuring out what the order is,
17:41
if I'm going to write the general form of the rate law, what would that look like? Rate equals remember, what does the rate law tell you? The rate equals what? The rate constant times the concentration of the reactant raised to the power of a number n that we need to establish.
18:02
So this is the general form of the rate law now we have to figure out what that exact rate law is. Okay? Now if you look at the experimental measurements, can everybody see that we are measuring rate we know the corresponding concentrations but we don't know what the value of this constant is
18:22
and we don't know what the value of this n is, and our ultimate goal is to figure those out in this problem. All right? So we write the general form of the rate law because we know the rate of the reaction depends on the concentration of the reactor. All right? So if you look at this equation, we have two unknowns
18:41
and mathematically, if you have two unknowns, all of you recognize that you need two equations to solve for it. All right? So algebraically if you have two unknowns, we need two equations. So what we're going to do is we're going to write two equations based on two sets of data. So we'll start with the first set of data. We know that the rate equals k times h raised to the power n.
19:04
So if one of my rates is 7.5 times 10 to the negative 4 moles per liter per second that will be the rate equals k times, we know the concentration for that will be .0050
19:22
molar raised to the power of n. So that's one equation. I'm going to go to the second equation, and I'm going to say when the rate is 10 to the negative 3 times 10 to the negative 3 moles per liter per second
19:41
equals the same constant, k, but now my concentration is .010 molar raised to the power n that gives me 2. So now algebraically, I've got to figure out a way to use both of these equations. Because remember, I have two unknowns, I have two equations, now I can solve
20:00
that and figure out what the value of n is, and I can figure out what the value of k is. And you can see that if I divide one equation by the other and usually it's best to keep the math simple, it's always best to take a bigger number and divide the bigger number by a smaller number than vice versa. You should end up with the same answer,
20:22
but it would just keep the math simple. So this is a bigger number so I'm going to say 3 times 10 to the negative 3 moles per liter per second equals k times 0.010 moles per liter raised to the power n
20:41
and I divide that by 7.5 times 10 to the negative 4 moles per liter per second equals k times .0050 moles per liter raised to the power n. All right? Can you see now, k and k will cancel out
21:02
and so what I've got is now an equation with one unknown. All right? and so if you take this and divide this, what I end up with is 4 over here, both of these are raised to the power n so if I take .5 you can see that this comes out to be 2
21:21
and that raised to the power n and so can everybody see I've figured out what n is? n is 2 so we know that this reaction is second order. All right? by taking the data you know, essentially what we're doing is experimentally we're measuring the rate and we're looking at its dependence on concentration. So we're changing the
21:42
concentration of the reactant and by doing this experiment in the lab we can figure out what the order of the reaction is. So now I've figured out what the rate law is. So my rate law would be that rate equals the constant, rate constant
22:03
times hi raised to the power 2 so that answers the first part and this is how in the laboratory we determine what the order of the reaction is. Remember we said we can't figure this out in advance we can't pre-determine or
22:21
predict what this order n would come out to be. The only way to determine it is to carry out the experiment you measure the rates at different concentrations, and you can figure out what the rate law is. So now that we've figured out what the rate law is, we can move on to looking at the
22:41
second part of the question so determine the order of the reaction and write the rate law. So if I ask you on the exam to write the rate law, this is what you need to give. You cannot give this because this is a general form, this is a generic equation and you have to find out the unique rate law that corresponds to this reaction and we've figured out that it is second order
23:02
now if I go to the second part of that problem we're asked to calculate the rate constant and give its unit. So if I go to part B I need to figure out what K is since I've figured out the second- can everybody see, I can just take one of these equations
23:21
now I know that n is 2, and now I have only one unknown, I can solve for K. All right? So I can say I know that rate equals K times HI raised to the power 2 therefore if I take one set of data and I'll take the first set of data which is 7.5
23:41
times 10 to the negative 4 moles per liter per second equals K times .0050 moles per liter squared and therefore I rearrange this equation, I can say K equals
24:01
7.5 times 10 to the negative 4 moles per liter per second divided by .0050 moles per let's say this is squared so it will be moles squared per liter squared. All right?
24:21
and if you were to put the numbers into your calculator and work this out what you would end up with is a 2 sig fig number so it will be 30 now I'm dividing moles per liter per second by moles squared per liter squared okay? So if I take the moles, you can see
24:42
that one of these moles and one of those will cancel out you can see between these two, one negative one, and this will become negative one so now what I end up with is liters per mole per second. All right? and so the unit now turns out to be liters per mole per second
25:03
and can everybody see that the unit for the rate constant will vary depending on the order of that reaction. So there's no one standard unit for the rate constant. It's like the equilibrium constant. Remember we said the equilibrium constant the unit varies depending on the law of mass action likewise, the unit for the rate constant will change
25:23
depending on the order and so we'll say that the unit for the rate constant will vary depending on the rate law or the rate expression. Okay? So you need to figure out the appropriate unit and so that will be the rate constant
25:40
and then if we go on to part C the part C tells us or asks us to calculate the reaction rate for a concentration of HI equal to .02 molar. Okay? So now we need to figure out the rate. Now that we know what the value of K is
26:02
now that we know what the value of N is if we change the concentration, we should be able to figure out the corresponding rate. Okay? So once again we go back to the equation, the rate law which we know is K times HI taken to the power 2
26:21
now we want to figure out the rate, but now we know what the K value is, so that would be 30 times liter per mole per second times the concentration, which is .0020 moles per liter squared
26:41
and so if you put that into your calculator, it comes out to be 1.2 times 10 to the negative 4 and you can see that the unit now works out to be moles per liter per second remember the unit for rate is always concentration per time. So you know that
27:02
your unit has to come out to be the change in concentration, which is moles per liter and change in time, which is per second. So it will be moles per liter per second. All right? And you'll see that the units work out. Got it? So this is how, I don't know, have you guys carried out any experiments in the lab with kinetics? So this is how you figure out
27:22
what the order of the reaction is if you know what the rates are. So we took an example of a problem where we had only one reactant. All right? Now what happens if you have more than one reactant? And so to take an example of how you figure out the order relative to multiple reactants
27:42
let's take an example where once again we're going to use an equation that has two reactants. All right? So if you look at this, now we're looking at the reaction where we have 2NO reacting with O2 to give you 2NO2. All right?
28:03
And so in the laboratory what we're going to be looking at is now now we have two concentrations to look at and so we have two concentrations and we figure out the corresponding rate. All right? And so we have three sets of data and the way we do this is to keep the math simple
28:23
well, the way we kind of- we construct the experiment in a particular way so that we can figure out what the order of the reaction is. And the way we do that is, if you look at this, what we do is we keep one of the concentrations constant. So if you look at NO and look at the first two sets of data, we keep those two concentrations the
28:43
same but only change the other two and we figure out experimentally what the rates are. All right? Then what we do is now for the next set of experiment we keep these two the same and change these two. You see that?
29:00
So what we do is we collect data where if you have multiple reactants we will keep one concentration of the reactants the same, all right? and change the other two. And that allows us to figure out what the order is with respect to that and now the next time we do this, we keep the other two reactants, the concentrations, the same and change the
29:21
other two. So once again, to solve a problem of this type we start by writing the general form of the rate law. Okay? So can you help me write that out? So if I'm going to write the general form of this rate law, I would start by saying
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rate equals what? What do I write down next? The K, the rate constant K, lower case K times my first reactant is what? NO. It's going to be raised to some arbitrary number M now I have another reactant. So my other reactant would be O2
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raised to some arbitrary number N and now ultimately I have to figure out what M is and what N is if I want to know what the rate law is. So now we have three sets of data, so I can write three equations
30:20
so I can say the first equation would be rate is 2.8 times 10 to the negative 6 moles per liter per second equals the constant K times the concentration of NO would be 1 times 10 to the negative 4 molar raised to the power M
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times 1 times 10 to the negative 4 molar raised to the power N that would be my first data. Okay? Secondly, I have the second data, which is 8.4 times 10 to the negative 6 moles per liter per second
31:01
equals K times 1 times 10 to the negative 4 molar raised to the power M times 3 times 10 to the negative 4 molar raised to the power N that would be equation two and the third one would be 3.4
31:21
times 10 to the negative 5 moles per liter per second equals K times 2 times 10 to the negative 4 molar raised to the power M times 3 times 10 to the negative 4 molar raised to the power M. All right? Now I want to go back to this general form of the equation, and can you see, what we
31:43
measure experimentally is rate we know the concentration of N, we know the concentration of NO2, but what we don't know is what M is, we don't know what N is, and we don't know what K is. And remember, if you have three unknowns algebraically, how many equations do you need to solve for it if you want to
32:01
figure out the three unknowns? You need three algebraic equations to solve for three unknowns. And so I've written down the three equations. All right? Now we start by let's take equation two, which is a bigger number than equation one, and I'm going to divide equation two
32:20
by equation one. All right? So if I take these two, you can see what I end up with is 8.4 times 10 to the negative 6 moles per liter per second equals K times 1 times 10 to the negative 4 molar raised to the power M times 3 times 10 to the negative 4 molar
32:40
raised to the power N. And I divide that by equation one, which is 2.8 times 10 to the negative 6 moles per liter per second equals K times 1 times 10 to the negative 4 molar raised to the power M times 1 times 10 to the negative 4
33:00
molar raised to the power M. And when I divide one equation by the other, can you see the two K's will cancel out? Now by design, we're keeping these two concentrations the same and can you see the reason why we do that? If I keep the concentration the same, this whole term will cancel out as well.
33:20
So essentially, I've eliminated one term for that equation now if I divide this by this, I end up with 3 and over here, if I divide this by that, I end up with 3 raised to the power of N. So now, just very simply, I know that N equals 1.
33:44
So I've quickly figured out what the order is with respect to that by dividing one equation by the other and by design, we always keep the concentrations the same so that they cancel out. Okay? So I figured out what N is now if I want to figure out the other one, can you see by design we're keeping these two
34:04
concentrations the same. So now all I have to do is to figure out this, M, I'm going to divide this equation by that. Okay? So now that I know what N is, I need to figure out what M is, and so for that, once again, I always divide
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the bigger number by the smaller number in terms of rate and so I'm going to take this equation and divide this equation by that. So what I end up with is so I'm going to say 3.4
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times 10 to the negative 5 moles per liter per second equals K, all right, times 2 times 10 to the negative 4 molar times 3 times 10 to the negative 4 molar now I've figured out that this N is 1, so I can put
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the one there to remind me that it's first order. Okay? I'm going to divide that by 8.4 times 10 to the negative 6 moles per liter per second times K times now this turns out to be 1 times 10 to the negative 4 molar
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and this would be 3 times 10 to the negative 4 molar raised to the power 1. So K and K would cancel out by design now that we take the one where the other two concentrations are the same so that these terms will cancel out so what I end up with is now this term divided by this term
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and this term divided by this term and both of these raised to the power of m. Okay? So if we divide this by this, what I end up with is 4 and over here I have 2 divided by 1 which is 2 raised to the power of m and therefore I know m
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comes out to be 2. All right? So by dividing, you know, this is just algebra, simple algebra, and by taking advantage of designing the experiment so that we always keep one of the concentrations the same and only varying the other, we can figure out what the order of the reaction is.
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So now we've figured out what the rate law is. What is my rate law? My rate law that I've determined from experiment turns out to be that rate equals a constant K, lowercase k my reactant is NO, and I know it's second order in NO
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and first order in O2. So if you want to describe this rate law, you'd say it's second order with respect to NO it's first order with respect to O2 and what is the overall order for this reaction? 2 plus 1? 3. All right?
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Okay. So ultimately in many of these problems, our ultimate goal is to figure out this expression. What the order is and that gives us the rate law or the rate expression. Okay?
37:20
Now what is the second- so now the second part is determine the rate law, and now they want us to figure out the value of the rate constant. So to figure out the value of the rate constant, you just pick any one of these equations. So let's take the first one where we want to figure out rate constants, so we know rate equals K
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times NO squared times O2. So if I want to know what the rate constant is, this would be the rate divided by the concentration of NO squared times the concentration of O2 which, if you take the first set of data, it would be 2.8
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times 10 to the negative 6 moles per liter per second divided by the concentration of NO which would be 1 times 10 to the negative 4 moles per liter the whole thing squared times the concentration of O2 which is 1 times 10 to the negative 4
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moles per liter and so if you work that out, the answer comes out to be 2 the answer comes out to be, of course we're dividing everything by one, so the answer comes out to be 2.8 times 10 to the power
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6 liter squared per mole squared per second. All right? So what you need to do is look at this, this will be moles this times this will be moles cubed per liter squared, oh, liter per liter cubed
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and you cancel the units out and you can see, once again, that the unit for the rate constant varies depending on the order. So there's no one standard unit for the rate constant, it's going to vary and you can figure out the appropriate unit. Okay?
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So it's very very important because on an exam you're graded not only on your numerical answer, but also on the unit, so you have to make sure that you check to make sure you cancel all the units out and get the appropriate unit for the rate constant. Okay? So any questions up to this point? So we looked at the facts. So the take-home message is what?
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That when we talk about rate remember last time we introduced you to the idea of what it means when we say the rate of a reaction now that we understand what it means when we say rate now we know that the rate actually depends on the concentration of the
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reactant. All right? And the relationship between rate and the concentration of the reactant is given by the rate law. All right? And whenever you refer to a rate law, it's got two parts to it. It's got a rate constant and it's got an order. And so we have to figure that out, and this is done experimentally. There's no way in
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advance for us to predict what the outcome is going to be so if we can't predict this in advance, then you have to deal with experimental data. The only way to figure it out is to use experimental data, and so the type of problem that you will get will say in an experiment this is the data that we obtained
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now from this data figure out what the rate law is and we took two examples of how you figure out the rate law. All right? If you have just one reactant, how do you figure it out? If you have two reactants, how do you figure it out? Okay? Now while this is interesting and fun
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we run into a problem when we deal with this, is experimentally, it's very difficult to measure the rate remember I told you the way you measure the rate of a reaction is to plot concentration versus time and then pick an instantaneous time
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and then you draw a tangential line that goes to that point and figure out the slope of that line. Okay? Now it turns out experimentally, when you actually try to do this in the lab, it's not it's very difficult to do this precisely. All right? You know, getting the right tangential line
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you know, just a slight twist can change it, and so it turns out when you are measuring rate as a function of concentration it's very very difficult to precisely measure the rate. So while the calculations are pretty straightforward executing this in the laboratory becomes difficult. Okay?
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And so we need a slightly different approach. So so it turns out that we're trying to accomplish the same goal, all right? but now we want to take a slightly different approach all right? So instead of looking at, remember we said we looked at the dependence of rate on concentration. All right?
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and what we find is while it helps us understand the topic actually accomplishing, measuring rates in the lab is cumbersome and difficult All right? So we need a different approach that makes it a lot easier for us to accomplish this. All right?
42:40
So what we do is instead of looking at rate versus concentration what we do is now what is much more easier for us to measure is just concentration versus time. All right? So we're going to take advantage of concentration versus time. So to measure the rate of a chemical reaction, we have to determine small changes in concentration
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occurring during a short-term interval sometimes it can be difficult to obtain sufficiently precise experimental data for these small changes because remember, you have to pick a point and then you have to get the tangential slope that runs to that point and sometimes that is difficult to do. Okay? An alternative
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is to fit all the data over a longer time interval with an equation that expresses the concentration of the species directly in terms of time. So what do I mean by that? So remember the way we figure out
43:42
rate is to plot concentration versus time and usually we look at it in terms of reactants. So we look at the concentration of A versus time in seconds and we said it decays like that. So if I wanted to figure out instantaneous, so if I wanted to carry this out,
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I would make measurements at different time intervals like that so I would measure at different times, I would measure the concentration of A so remember, what we measure is the amount that's left unreacted. So at the beginning we have the neutral concentration
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it reacts a little bit, and then we figure out how much of A is left over then we proceed at a different like the first 50 seconds we figure out how much of A is unreacted then in 100 seconds we figure out how much of A is left unreacted and then at 150 seconds we figure out how much of A is left unreacted and you would end up with a plot like that.
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If I wanted to figure out rate if it's at the instant the reaction begins with the initial, I would have to draw a tangential line through that and figure out what it is. That's hard to do. What's more, what's much more easier is to take this exponential decay that you see and come up with a mathematical equation. So we're going to derive a mathematical
45:03
equation that actually describes this whole process. All right? And we're going to derive a mathematical equation that looks at the dependence of concentration versus time and gives us the whole dependence of that entire plot. All right? So we want to derive a mathematical equation
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that describes this plot. So in order to do that we have to look at each one separately. We want to look at a first order reaction and let's say we're going to derive an equation for a first order reaction where you have only one reactant. All right?
45:41
So if I'm going to derive a mathematical expression that describes the whole process and describes the dependence of concentration versus time, I will start with the rate law. And the rate law for this would be rate equals K, all right? How many reactants do I have?
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Only one reactant. And my reactant is A and it is first order, so you know, I put a one there, but I don't need to put a one there because it's a first order reaction. Okay? Now since I want to derive an equation and I'm going to take advantage of integrating the whole process
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I'm going to say that if you write the rate, remember rate is what? Delta concentration of A divided by delta t. Do you guys remember? So instead of using delta I'm going to say we're looking at small incremental changes. So instead of using delta, which is change,
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I'm going to use D or deva which represents small incremental changes and so I'm looking at small incremental changes in A as a function of small incremental changes in time. So I'm representing rate this way and remember, it's a reactant and the reactant is consumed, and so we put a negative sign in front of that
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that will equal K times the concentration of A. Got it? Now I'm going to rearrange this equation so that now I'm going to say that the small incremental change divided by A
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equals negative K dt All right? So I'm going to just put all the concentration terms on one side and the time terms on the other side and now, remember these are incremental changes, and I'm now going to integrate this so that it explains the whole process
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from when it begins, so that so my limits are I start with the initial concentration of A to concentration of A and over here I'm going to integrate time from t equals zero to t and if you don't understand this process, now essentially
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if I integrate this, what I'll end up with is ln A minus ln A zero equals minus Kt. All right? If you don't understand this all I'm interested is that you know what the final outcome is. All right?
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So if you haven't done any integration in your math classes, all I want you to remember is the outcome of this equation turns out to be this or you can also rewrite this and say this is ln A over ln A zero
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equals minus Kt now what this equation represents is after I integrate it, I'm looking at small incremental changes, and when I integrate it, I'm integrating it to the whole limits starting from initial concentration of A until as A reacts
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at A, concentration of A at time t this is what it would look like. Got it? Now if I were to write this, so I would expect you to memorize this because it's a first order reaction and for a first order reaction you need to know this is what it looks like. All right?
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and this as well now if I want to represent this in the form of a graph can you see if I take ln A minus the initial concentration of ln A zero equals minus Kt if I rearrange this, I'll end up with ln A equals minus Kt
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plus ln A zero so I'm just taking this over there and now this is in the form of, if this were the y-axis and K is the slope, it's a negative slope times x-axis, that would be the x-axis, this would be the slope
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plus C which is the intercept all right? So it's in the form of y equals mx plus c and so you can see if I plot this out what that would tell me is that if I plot ln A versus time
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I should end up with a straight line graph, all right? like this where the intercept would be the initial concentration, that's obvious and the slope would equal K and you can see it's a negative slope
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so the final take home message that I want you to remember is if it's a first order reaction, what you do is you measure, you make a graph you look at concentration, sorry, at different time intervals, you figure out the
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concentration of A. So this is A at different time intervals. At 0, this will be the initial concentration of A at different time intervals you measure the concentration of A and now if you take you know, in Excel, you convert A to ln A
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and if you make a plot of ln A versus time and you end up with a straight line with a negative slope, what does that tell you? It's first order does that make sense to everybody? So now instead of measuring rate, what I would do is in the laboratory, I would measure concentration versus time and then take the natural log of the concentration
51:42
and draw a graph and if I end up with a negative slope, a straight line with a negative slope, what does that tell me? It's first order. All right? So that's a lot more easier to accomplish in the lab.