Transcript - Lecture 32

OK.

Today we are going to continue where we left off with kinetics. And so the topics we are going to cover today, we are going to talk about radioactive decay, which is an example of a first order process.

Then we are going to talk about second order integrated rate laws.

Then we are going to something that I love, which is to go back and connect the first topic that I taught in the second half of the semester, chemical equilibrium with kinetics. So we are integrating all the material from this half of the course.

And then we are going to start talking about reaction mechanisms. And, on Wednesday, we will talk more about reaction mechanisms.

And so this will move you closer to being able to do more of the problems on Problem Set 10.

Again, we are going to be in kinetics for the rest of the semester. Other topics we will cover after mechanism, we will talk about temperature, and we will talk about catalysts as well. Most of kinetics is in Chapter 13, but the little bit we are going to do on radioactive decay is in Chapter 17.

Just to review where we were on Friday, we were talking about half-life for a first order process.

First order half-life. Half-life is just the time it takes for half of the original material to go away, and its symbol is t??.

Last time we took the first order integrated rate law, and we were able to solve for an expression from half-life. And all we had to do is say all right, by definition, when half of the original material has gone away then your t is equal to t??.

It is the half-life.

And then the expression for the original material, so that is the concentration of A to the O, the original material drops out. And we find that for a half-life expression, it is independent of a concentration that you started with, so it depends then on the rate constant K.

An example of a first order process is radioactive decay.

Radioactive decay, the rate at which it happens is independent of the number of nuclei that surround the material. That is another way of saying it is independent of the concentration.

This is a first order process. And so we can use the same things we learned on Friday, when we were talking about concentrations of material and apply them to radioactive decay.

So, because this is first order here, it is independent of the surround nuclei, we can apply these first order integrated rate laws that we derived on Friday.

Here were the ones we derived on Friday. We found that the concentration of a material A, at some given time, is equal to its original concentration times E to the minus Kt where k is the rate constant and t is the time.

And we just looked at this again, that the half-life for a first order process is equal to 0.6931 over that rate constant.

Now we can take these equations and just rewrite them in terms of nuclei because we are talking about radioactive decay. Instead of concentration then, we are going to use this term N, which is the number of nuclei.

We can put that in instead.

Now we have a new expression where N, the number of nuclei at some given time is equal to N knot here, the number of original nuclei in the sample, E to the Kt. And k is a rate constant but now it is called the k constant because we are talking about radioactive decay but it is the same principle.

It is just a slightly different word for it. And you have time. So it is the same expression. Instead of concentration we have number of nuclei.

Chemical kinetics is talking about monitoring changes of concentration over time, whereas, nuclear kinetics is talking about measuring decay events.

And so here we would be measuring with a Geiger counter detecting radiation. Over here you are measuring changes of concentration and material.

I just brought a Geiger counter with me. If we turn this on, we can see if we have any radioactivity in the room.

The radiation will cause an ionization and cause a current flow, and the current flow causes clicking. We will see how clean the classroom is over here. Can you hear it?

It seems pretty good. I will check over there later just to be sure that we are free.

My lab uses x-rays to determine structures of enzymes. We need this to check to make sure we have no leaks in our x-ray generator, that we are not exposing anybody who comes by the laboratory. That is how you could measure whether there was any radioactivity and how it would change.

And, if there is a lot of radioactivity, then you start hearing the clicks much faster.

We can also talk about activity of a sample. And activity is abbreviated A here. And so the activity or the decay rate is just the change in the number of nuclei, remember N is the number of nuclei over time.

And so activity would then be equal to the decay constant times the number of nuclei that you have.

And because the activity would be proportional to the number of nuclei, you can also take the expressions that we just looked at.

We just looked at this expression, that the number of nuclei that you have at some given time is equal to the number that you had originally times E to the minus Kt, the decay constant times time. You can also express that in terms of activity.

Here we have the activity of the sample at some given time is equal to the original activity that the sample had, E to the Kt, the decay constant times time.

You could use either one of these, depending on what you are given in a particular problem. In terms of units then, the SI unit for activity is Bq, becquerel.

And this is equal to one radioactive disintegration per second.

And the older unit, which you probably have heard of, is called the curie. And one curie is equal to 3.7 times 10 to the 10th disintegrations per second. Does anybody know which curie this is named after? I heard some here and some here.

There are only two really guesses here. It is actually Pierre Curie.

I always assumed, until I looked this up recently, that it would have been Marie Curie. After all, she was the first woman to receive a Nobel Prize so I figured she could have a unit named after here.

But, in fact, it was her husband that the unit was named after. And she was involved in naming the unit after her husband. And it was named after him because of his untimely death in 1906, I think it was.

He was killed in some kind of road accident. And some people speculate that the effects of radiation were already upon him.

And he wasn't paying so much attention. Maybe it was the absent-minded professor thing.

I don't know. But he ended up having an unfortunate accident and having the unit named after him. He did share with his wife the Nobel Prize before he died, but his wife went on to win a second Nobel Prize after his death.

She got two Nobel Prizes and he got one and a unit named after him. Although, the unit is no longer used.

She probably got the better deal in long-term history. One thing I will mention is that at the time, the reason why it was changed, this is a pretty inconvenient unit.

This is a really big number right here. And, at the time, they were deciding what this unit should be, it was pointed out that this was kind of an inconvenient number. This is way more radiation than a normal sort of worker with radiation would be handling.

But Marie Curie put her foot down, apparently, and said that she didn't want her husband's name associated with a unit that represented an infinitesimally small quantity of material.

As a consequence, this is a large number that was impractical and later substituted. The person that Marie and Pierre Curie shared the Nobel Prize with was Henry Becquerel. Apparently, his wife had no objections.

And so now a much smaller unit is named after him.

Whenever I talk about this, I always ask, it makes me curious whether this is a unit that is named after a woman. As you go through MIT you might want to check with your professors where these units came from.

And if you find one that is named after a woman, please let me know, because I am not aware of any and it is sort of a curious of mine if such a thing exists. The current unit up here and this is the older unit.

OK.

We are not going to do much in Chapter 17, but a little bit on some of the types. This is just a table in your book. I didn't put it in your handout because I don't expect you to really know any of this.

It will be presented in the problem if you need to know it. But just to mention types of radiation, alpha decays and beta decays. Alpha decay, the particle that you are talking about is equivalent to a helium four nucleus.

Beta decay equivalent to an electron.

Sometimes you get a change in mass. Sometimes you get a change only in atomic number. And the half-lives are very different. Again, you don't need to know this, but you will need to look this up for some of the problems that you do, what the half-life is.

And this is right out of your book. I think a few problems require you to go back to this table and look up the half-life for a particular material in question.

I am going to do an example of radioactive decay.

And, to do this, I am going to turn to poetry for this particular example. One of our chemistry grad students, Mala Radhakrishnan is also a poet. And so this is one of her poems, Days of our Half-Lives.

And this is published in a collection of poetry all about chemistry.

She is now working on a biochemistry book, if anybody is interested. This is Chemistry for the Coach Potato. Here poem has to do with uranium-238.

And this is in your handout, this particular charge. Here alpha decays are indicated by this yellow line and the beta decay is in the blue line here.

And so here is uranium-238 at the very top, and this is all the various things that can happen to uranium-238 as it goes through a series of alpha and beta decays.

This is Days of Our Half-Lives. My dearest love, I am writing you, to tell you that I have been through. I have changed my whole identity, but loved I cannot pretend to be. When I was uranium-238, you were on my case to start losing weight.

For five-billion years I'd hope and I'd pray, and, finally, I had an alpha decay. Two protons, two neutrons went right out the door, and now I was thorium-234. But my nucleus was still unfit for your eyes, not positive enough for its large size.

But this time my half-life was not very long 'Cause my will to change was really quite strong.

It took just a month, not even a millennium to beta decay into protactinium. But you still rejected me, right off the bat.

??Protactinium? Who's heard of that??? So a beta decay I did once more to become uranium-234. Myself again, but a new isotope. You still weren't satisfied but I still had hope. Three alpha decays, 'twas hard but I stayed on through thorium again, then radium, then radon.

I thought that I would finally please you.

My mass was a healthy two-twenty-two. But you said, ??Although I like your mass, I don't want to be with a noble gas!?? You had a point, I wasn't reactive. So in order to please you I stayed proactive.

A few days later I found you and said, ??Two more alpha decays and now I am lead!??

You shook your head. You weren't too keen of my mass number of 214. I had a bad experience with that mass before, an unstable acetone walked right out the door.

So in order to change I went away, but all I could do was just beta decay. My hopes and my dreams started to go under because beta decays do not change a mass number.

To business then polonium I hoped and I beckoned, my half-life was 164 microseconds.

And then I alpha decayed and then I was lead with a prize worthy mass of 210. You've got to admit I was getting quite tired and my patience with you had nearly expired. You were more demanding than any I had dated, and much of my energy had already been liberated.

You still weren't happy, but I had a fix.

I really liked the number 206. So I waited for years until the day which began with another beta decay. And then one more. And finally in the end I alphaed to lead 206, my friend. To change any further, I wouldn't be able, no longer active but happily stable.

It took me billions of years to do, but look how I have changed.

And all just for you. And what did you say? I had gotten so old that you'd rather be with a young lass of gold. Well, I give up. We are through, my pumpkin.

Shouldn't all my efforts be counting for something? Well, you won't be able to rule me anymore because I am leaving you, not for one atom but for four. That is right. While you were away diffusing, I met some chlorines that I found quite amusing.

And we are going to form lead Cl4 and you won't be hearing from me anymore.

So over the years I have grown quite wise. I have learned that love is about compromise. You still have half of your half-lives to give. So now you go out there, it is your turn to give.

OK, so there is an example of radioactive decay that brought us all the way up here to all the way down here.

All right. Now let's actually look at an example of how we might work a problem. And if anyone else has any poetry suggestions for other topics in freshman chemistry, feel free to submit those as well.

OK. Now let's just work an example.

We want to find the original activity and the activity after 17 years, which is equivalent to 5.4 times 10 to the 8 seconds of 0.5 grams of plutonium 239. And this is the half-life in years and also in seconds.

We can use this equation. We want to find activity, and original activity.

And we need to know original number of nuclei and the decay constant. But, before we can use that, we have to find the decay constant and the number of nuclei to be able to find the original activity.

First, we can look at the number of original nuclei. And so we were given that there were 0.5 grams. We can convert that with the atomic weight. And here you want to use the atomic weight that is given in the form of that isotope.

You don't need to look this up in the Periodic Table.

You just take that number, which is given to you, and plug it in down here. Then we need to convert with this number, which is what? Avogadro's number. Convert the number of nuclei that are in one mole.

And so then we can get the number of nuclei originally in this sample, in 0.5 grams of this sample. That is 1.3 times 10 to the 21 nuclei.

Then we can find k. How can we find k with what is given here? What do we know that is useful? We know the half-life, right.

We can easily find k from half-life by the equation for a first order half-life process. That is just 0.6931 over the half-life. And the only thing tricky here is you want to make sure that you are going to have all your units come out OK, so either all in years or all in seconds.

If we plug in the second value here, we can get our decay constant k in seconds, 9.1 times 10 to the minus 13 seconds to the minus one.

Now we have a number of nuclei originally present and a decay constant, so we can calculate the original activity. And so we can just plug that in. And so we get, for our units we have nuclei per second.

And one disintegration per second is a becquerel.

Nuclei per second is just a becquerel. That is in Bq. See, that is a more convenient unit than the curie. That is our original value. Now, if we want to know, after 17 years, how much activity is left, we can use this equation down here.

That the activity at some given time is equal to the activity originally times E to the minus Kt where that is the decay constant and time.

We can plug in our original activity that we just calculated, put in our decay constant which we calculated up here, and here we just have to make sure we have the same units.

Seventeen years was converted to number of seconds, and we can put that in and we can solve and we get the same answer. To the significant figures, it is the same.

So no significant difference. And this is one of the problems with materials that are radioactive is that they are around for a very long time.

And so one problem with radioactive material is how are you going to store it? A lot of times the materials we have to store things do not last as long as the material we are trying to store. This is a problem that every once in a while gets talked about in the general media and in political discussions.

But here is an example that these things are around for quite a long time.

All right. I will just mention a use of radioactive decay. A lot of people think of radioactive materials and get sort of frightened by them, but they actually can be very useful and they can save lives and make money.

Here is an example from MIT. This is a picture of Alan Davison who had just retired last year from the Chemistry Department.

And he paid attention to some freshman chemistry stuff and figured out a good way of making a material that could be used for imaging.

And so what he did, he has this technetium. And he figured out how to put it in a coordination complex that had just kind of the right sort of properties that you want for imaging. It wouldn't be in the system very long, it wouldn't be too harmful, but yet it would give you some good scans.

And so he paid attention to metals in this d block area that we have been talking about. And he said how can I put this together in a coordination complex that would have just the right properties? He took one of these guys, in gray are some of the ones that have been used in medicine that are not naturally occurring in the body, and made a coordination complex.

And the ligands he chose were cyanide ligands to get the desired properties.

And this pattern for Cardiolite has made a lot of money, which is really putting it very mildly. The patents, there is money to MIT, to Alan Davison himself, other people involved.

And the Chemistry Department has gotten money for this, too. And over the last ten years, a lot of sort of discretionary funds and scholarships and other things we have been able to do has all been a result of this.

Now the patent just expired on this.

We are looking for other people who pay attention in freshman chemistry to come up with some basic principles, put it together and have a new patent that would allow us to kind of carry on. We liked having all of this extra money around.

So you can be thinking about that. Just sort of basic principles here. And it has also saved a lot of lives. They have moved on and tried to use it now for imaging different kinds of cancers and things like that.

It is really a pretty wonderful material.

All right. Radioactivity can be difficult to store, but it also can be very useful. That is first order processes. Radioactive decay is one of the best examples of a first order process.

Now let's talk about second order. And so what we are going to do now is we are going to derive, as we did for the first order rate laws, some second order integrated rate laws.

Here is second order.

We have a little equation, 2A going to B. And the rate expression, disappearance of A over time. And the rate law would be equal to k, the rate constant times the concentration of A. And the order of the reaction is second.

It is the second order, so you know that that is a two there. And, again, as we talked about last time, what you see up there, you have A to the M.

This indicates the order of the reaction with respect to A.

This is second order. Now, as we did last time, we are going to separate the terms. We are going to take the terms that have to do with concentration and put them on one side and put everything else on the other side, including the time terms.

And so here we have rearranged things.

We have brought A squared over to one side. We have dA over here. We have taken the negative sign and dT and put it on the other side with k. And so we have these separated.

And now we can integrate. Again, as we did last time, we are going to integrate from some initial concentration of A, the original concentration of A to the concentration of A at time t.

And, on the other side, we are going to go from zero time to time t.

And so here is our expression. And now we can take this and solve this expression. There is just the expression again copied to the front of the top of the page. We can solve this. And we get, in brackets, one over the concentration of A to time t minus one over the initial concentration of A, the original concentration of A, all in brackets, with a minus.

And, on the other side, we have minus Kt, the rate constant and the time.

Now we can rearrange this expression. Now we want to arrange it such that the concentration of the material at time t is on one side, we have Kt over here, and one over the original concentration of A on the other side.

And you will see that this can be written in a format for a straight line.

If we plot information about how A changes with time and, again, with chemical kinetics, it is all about how concentration of things change with time versus time, we can get out information that we want for a second order equation.

Here is what a plot would look like. You are plotting. You're collecting your data.

You are collecting changes of concentration of A, so you are measuring the concentration of A at given time points.

And so you plot that versus time. And you should get a straight line. If you don't, something is seriously wrong with something that you are doing. So, you should get a straight line. And so the intercept of the line would be one over the initial concentration, the original concentration of A.

And then the slope of the line is going to give you your rate constant, which is what you usually want to know.

You want to be measuring rate constants for particular reactions. Now let's talk about second order half life.

We will start with this second order integrated rate law. And now half-life, you are going to consider the time it takes for half of your original material to go away.

At this point, when you are talking about second order, your At, your concentration of A at time t is half the concentration of what you had originally here.

We can plug that in. The concentration at the time t that you are interested in is now half of what you had to start with by definition of half-life. Your time now has sort of a new symbol by it. It is t to the half down here.

And now we can rearrange this and solve for this. As we rearrange it, we can move the two up there. And we can bring these concentration terms to the same side of the equation. We can subtract now, which gives us one over the initial concentration of A.

And then, if we rearrange so that we solve for the half-life, we move the rate constant down over here and we can get our expression. What is the biggest difference between this expression and the one for first order half-life? Dependence on concentration, right.

The concentration term is in there.

With first order process, the term for concentration is no longer in the expression. The half-life does not depend on the concentration of the original material, but here the concentration term is in the expression.

Second order half-life depends on the starting concentration of the material. If you were really in a lab doing experiments, you often need to figure out if something is first or second order.

And in the problem set you will have some different things where you are trying to figure out where I will show you how the rate changes with different concentrations.

And often what you do in a laboratory is you plot data and see whether it does fit. I said if you didn't get a straight line when you are plotting something is wrong. In real life, sometimes you would be plotting data.

You can plot data as a first order plot.

Our first order expression was here. The natural log of the concentration of A at a given time is equal to minus Kt plus the natural log of the initial concentration.

Here would be a plot of that. And then for second order we would use the equation that we just derived. Instead of natural log, you are plotting one over the concentration versus time. And if the data really was a second order process, you should get something that looks good over here.

Whereas, over here you should get something that doesn't really look good.

In real life, often what you are doing is seeing how well data fits to some assumption that you have. And you are deciding the orders of reaction, you find that out experimentally. It is usually not given to you.

That is something that you would be doing. And you would be looking at how rates change with time and you would be plotting out data and seeing how well it fits your assumptions.

OK. Now we get to go back and think about what do rates have to do with chemical equilibrium? What is the connection between rates and chemical equilibrium? At equilibrium, one way we can define equilibrium is that the rates of the forward and the reverse reactions are equal.

Now we are thinking about equilibrium in terms of rates.

I will leave that up, I guess. Let's write out our standard reaction. This chalk is really good. No. Here is a good piece of chalk. In our standard reaction, we have A plus B going to C plus D.

And that is how we have been expressing it.

Now we are going to think about rates, so we are going to give the forward reaction a little k1 and the backward reaction, oh, I am going to do k-1. Over the reverse and the backward direction. Now we can express the rate for the forward reaction.

And we can assume here that this is an elementary reaction.

We can write our rate law exactly the way the equation looks. If we do that, the rate is going to have a rate constant, k1. And it will depend on the concentrations of our two reacting materials, so the concentration of A and the concentration of B.

We can write this rate law without doing any of the experimental stuff, if we are told that it is an elementary reaction, a simple reaction.

We can also write the rate for the reverse reaction, so the reaction going in the other direction. And so that rate law will be equal to the rate constant for the reverse direction, which is k-1 times the concentrations of C and D.

That is the rate in the reverse direction.

What do we know about K? What is big K?

Products over reactants. C times D over AB.

That is our equilibrium constant. At equilibrium then, according to this new definition, the rates of the forward reaction will be equal to the rates of the reverse reaction.

At equilibrium k1, the rate of the forward reaction, times A times B is going to be equal to k-1 times concentration of C times the concentration of D.

That is what happens at equilibrium. Rates of the forward reaction equal rates of the reverse reaction.

Then, if we rearrange that expression, we will see that that means that k is going to be equal to k1 over k-1.

Because, if we rearrange this, CD divided by AB is going to be equal to k1 divided by k-1. These are equal. You can express k in terms of the concentrations.

And it is also going to be equal to those two rate constants.

What does this mean in terms of our big equilibrium constants and small equilibrium constants that we have been talking about? In kinetic terms, when we talk about k being greater than one --

Before, when we were talking about the equilibrium being greater than one, we were talking about more products than reactants at equilibrium.

And we were not really talking anything about the rates of the reaction. What is going to be true in terms of these small k's, k1 and k-1, if big K, if the equilibrium constant is greater than one, what is true about those two?

Here k1 then has to be greater than k minus one, or the rate constant for the forward reaction has to be bigger.

Rate constant for forward reaction is bigger.

And big K, equilibrium constant K less than one means that the rate of the reverse reaction, reverse rate constant is bigger. Now you can think about equilibrium constants in a new way.

You can think about them in kinetic terms.

All right. Now we are going to go in and start talking about mechanisms of reactions. And the mechanism is one of the important things in terms of what affects the overall rate.

How many steps does the reaction go in? Are there fast steps? Are there slow steps? How many slow steps? How many fast steps? How long does it really take for the reaction as written to occur?

And so it is unlikely that reactions take place in one step.

That is not too common. Usually they proceed through a series of steps. And each step is called an elementary reaction. I have been mentioning elementary reactions saying that for an elementary reaction you can write a rate law without any data.

You can just write it as is.

That is what you are doing in terms of mechanisms. You will see an overall reaction and then you will try to break it down into steps. And each step is an elementary reaction.

You can write a rate law for each step and then think about whether that reaction mechanism makes sense in terms of the data that is available for the overall reaction. Again, an overall reaction, the order and the rate law cannot be derived from the stoichiometry.

You wouldn't necessarily know if it is first order with respect to one thing, second order with respect to something else, you cannot do that, but for the elementary reaction it occurs exactly as written.

You have broken it down. It is an elementary reaction or a step. You have broken it down so that it is occurring exactly as written, and then you can predict the rate law.

You are sort of breaking down the mechanism into the simplest steps.

Let's look at an example for this. Here is the decomposition of ozone. You have two molecules over here reacting to give you three molecules of oxygen.

A proposed mechanism then has two elementary reactions or two steps.

And here, in the first step, we have one molecule of ozone. Are there questions? It seems kind of noisy. One molecule of ozone going to O, oxygen by itself there, plus O2. This intermediate oxygen species then comes back and reacts with ozone again to give two molecules of O2.

The first reaction is called unimolecular because there is only one thing reacting.

What do you think the next one is called? Right. It is going to be called bimolecular. Let's put that up there. Oh, I guess I should go back to that. All right. Molecularity is the number of reactant molecules that come together to form product.

That is molecularity.

And the second one would be bimolecular. This is another one of those questions that can turn up on the final. It is supposed to be the easy points. I ask what the molecularity is, and you look, there is one thing.

There are two things, it is bimolecular, but people forget what that term means. That is a good one to kind of keep in the back of your head.

That is just a really simple extra few points that you can get on a final to recognize those terms.

All right. We are not going to get to four things reacting. We will go only as much as three things reacting. Unimolecular, one reactant. An example of that would be some kind of decomposition or radioactive decay.

Bimolecular, two things reacting.

Here you would have two things coming together colliding to form a product. And we will talk in the next few lectures about the energies associated with that reaction occurring. And then termolecular.

I give you that that is a little bit of a harder one to remember than the uni and bimolecular, but the termolecular are three things coming together to form products.

As one might imagine, three things all coming together at the exact same time to form a product sort of in one step would be unlikely to happen.

These are fairly rare. Something that is written this way, if you see something that is written as a term molecular reaction, you may predict right away that that is sort of unlikely to happen in one step.

Any of you just trying to get a group of friends together to do something at the same time, knowing that getting three people together at the exact same time to something that you are planning to do is rare that everyone is actually there at the same time to start.

You can remember that that is sort of a rare thing to happen. But if three things did come to react that would be termolecular. All right. Now we can consider the rate laws for each step.

And because these are elementary reactions or steps to the reactions, you can write the rate law exactly from the reaction that is happening exactly as written.

You can use the stoichiometry to consider the rate law. Let's look at the first one then. The rate is going to be equal to this k1 term. The rate constant for step one or k1 times the reactant, which is O3, the concentration of O3.

That is the rate law for the forward direction.

In this case, there is no reverse direction to consider. We only have this rate law for the first step. In the second step then we have the rate equals k2, so that is the rate constant for the second step.

And, in some of your problems, you might not have these in the book. You should use k1 for the first step.

If there is a reverse reaction that should be k-1, k2 for the second step, if there was a reverse reaction it would be k-2, etc.

Again, these do not have reverse steps so we don't have to worry about this yet. You will see more of these on Wednesday on the problem set. Here the rate is k2 times the concentration of O times the concentration of O3.

We can write those exactly as written because those are elementary steps in an overall reaction mechanism.

We could not write the rate law for this, if it has multiple steps, without experimental data to help us do that. Then the individual steps should be able to be added together to get the reaction in question.

And so here, if we add this together, we have two O3s and we have three O2s.

And the O has cancelled out. That is an intermediate in the reaction. O here is a reaction intermediate. And the big trick in writing reaction mechanisms and figuring out rate laws for overall reactions up here is to cancel out intermediates.

You are going to spend the next few days getting ready for the problem set on Friday getting rid of intermediates in reactions.

It doesn't appear in the overall reaction equation and it cannot appear in any of your rate laws that define the particular reaction.

Just, finally, this is the last slide, that reaction mechanisms are a series of steps. Reaction mechanisms can never be proven to be correct. They just can be consistent with data. You are going to be predicting reactions that are consistent with the data on this next problem set.

OK. Thanks.

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