Lecture 22: Chemical Kinetics

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Just a few last remarks. Last day we were intensively looking at the structure and properties of oxide glasses, and I just have a few remarks left to make about that and then we will move onto kinetics today.

Just a reminder, we saw that oxide glasses typically consist of three types of components, primarily network formers. These are covalent oxides that form oxygen bridges between the metal species. And we can tailor the nature of the network by adding so-called modifiers which are ionic oxides, and they donate oxide ions.

The oxide ions go into the network and they break the oxygen bonds in a process we call scission. And then lastly in certain glasses we add a compound of a type called an intermediate. And these are actually network formers, but they form a different number of bonds than the parent network.

Typically a larger number of bonds. And, in doing so, they create some free volume which enhances certain properties, sometimes the mechanical properties, sometimes the thermal properties. Specifically, thermal shock resistance is enhanced by the addition of network formers.

And, at the end of the lecture, we had started talking about strengthening mechanisms. And the way to strengthen oxide glasses -- Remember, oxide glasses are not too bad in compression but are very weak in tension.

And the way to strengthen them is to pre-stress compressively so that the effective yield stress becomes then the sum of the natural yield stress of the material, plus the compressive stress that you have added to the material.

And you don't have to necessarily do this through the entire material. Typically, we simply pre-stress the surface. Because, let's face it, if something is going to fail, it is going to fail through the application of force causing the crack at the free surface.

If we can prevent the crack from forming at the free surface this can, in many instances, preserve the integrity of the object. We saw the first way of improving the strength of an oxide glass called tempering.

And that exploits differential cooling so that if you look on this trace here, volume versus temperature, we know that if we have surface cooling at a high rate, we will have a higher excess volume than if we have cooling in the bulk of the glass at a lower rate.

And that leads to a differential in volume which ultimately has a compressive stress on the surface. There is another way to achieve the same goal, that is compressive stress on the surface, and that involves a chemical treatment.

And, in particular, what we can do is take the glass -- And I am going to give a representative composition here. Let's say we have a glass that consists of, we will have a network former, silica, and a network modifier will be soda, sodium oxide, and we will throw in an intermediate, alumina.

This is a sodium aluminosilicate glass. Former. Modifier. Intermediate. And we know that the sodium ions are present. The oxide ions have gone into the network and have resulted in the formation of terminal oxygens here.

The sodiums are still free as sodium ions. Let's take a look at the sodium ions at the surface of the glass. Some of the surface sites are going to be occupied by sodium ions. What we can do is immerse this glass into a molten salt at elevated temperature.

Typically, let's look at something like potassium chloride. This melts at 775 centigrade, so this might be sitting at 800 degrees C. And, as a molten salt, it consists of potassium cations and chloride anions.

Take a look at this interface here. On the left I have a high concentration of sodium but no sodium in the molten salt. In the molten salt, I have a high concentration of potassium but no potassium in the glass.

By virtue of the concentration gradient, potassium will want to enter the lattice and displace some of the sodium and some of the sodium will want to leave to enter the salt. Ions are soluble in an ionic medium.

What happens if a potassium ion enters the glass and occupies a site previously held by sodium? Well, we know that the radius of the potassium ion is substantially larger than the radius of the sodium ion.

If potassium enters the glass, it is going to be a force fit. This means force fit. And it is going to lead to the generation of compressive stress. And this is not too different from what happens when you put carbon into iron.

The carbon is larger than the interstitial site in iron. This treatment is called ion exchange. By ion exchange, which is a chemical treatment, we can also strengthen the glass and thereby make it a little more robust for use.

I think that is a good place to stop the treatment of glasses, and we may return a little bit later in the semester. What I want to do now is start a new unit. And the unit I want to talk about today is a unit that would be called, from the domain of physical chemistry, I want to talk about chemical kinetics.

And chemical kinetics is the study of reaction rates and mechanisms. And this is very important for a number of reasons. For that we discuss it in 3.091. Let's talk about three reasons that come to mind.

First of all, productivity. If we go into an industrial setting and we want to improve throughput or yield, we need to understand the rates and mechanisms of reactions. This is linked to competitiveness, it is linked to resource utilization, efficient use of materials and energy if we know how to process things effectively.

The second thing is environment. A number of you have come to me after class and, on different occasions, spoken of your interests in environmental matters. Well, one of the simplest things one can do to minimize adverse environmental impact of chemical processes is to be efficient.

Understanding how reactions proceed is critical in minimizing the adverse environmental impact. Productivity issues. Secondly, environmental issues. And, thirdly, societal issues ultimately. You are here at MIT.

You want to get an analytical education and you want to go out and do something in the world. And one of the things that can happen, if you understand your chemistry and understand the kinetics, is you can keep a plant operative, you can keep people working.

There are societal implications for failure to understand kinetics. When a process becomes obsolete, you go out of business and people go out of work. For these, and many other reasons, we want to study kinetics.

And kinetics takes place over a very wide range. That is why I played the song about time. That is what this is about. It is about time. We have at the one extreme very slow chemical processes, and at the other extreme very fast chemical processes.

Slow, we have things that take place along geological or even cosmic time scales. A very, very slow processes. And, on the other extreme, we have such things as explosions. And explosions are not necessarily destructive.

For example, in the airbag that operates in an automobile, it operates thanks to the fast kinetics of a glass solid reaction. There is no pump that can inflate an airbag at an acceptably high rate to do you any good in an automobile crash.

What happens here is the kinetics of a reaction between two solids. What we have is basically solid plus solid goes to gas. And, as you know, the volume of a gas is thousands of times the volume of a solid.

And, in particular, it is the reaction of sodium azide with potassium nitrate. And one of the products is nitrogen gas in huge quantities. You have your accelerometer which decides that you are decelerating at a rate that is alarmingly high.

It probably not a healthy occasion. A tiny electric current flows through a wire which acts as the trigger for this reaction. Once the reaction goes, you can inflate the airbag in a matter of milliseconds.

What we want to do is understand some of the basics of chemical kinetics. We are going to look at a general reaction and understand how, first of all, to characterize chemical reactions. Let's characterize the rate of a chemical reaction.

I am going to write a. Little a is stoichiometry and big A is some chemical identity. A moles of A plus b moles of B react to form little c moles of C plus little d moles of C. That is sort of a prototypical reaction.

And kinetics accounts for the rate of conversion. That is what we want to know. How quickly are we consuming our reactants? How quickly are we generating our products? That is what we are interested in understanding.

We have, first of all, a simple thing. We have conservation of mass operating, which we have seen from the early part of 3.091, that the rate of consumption of all of the reactants must equal the rate of production of all of the products.

There are no sources or sinks. We have no wormholes here. Everything that goes into the plant has to come out of the plant in some other way. And specifically we can even get into looking at the rate of disappearance of a single reactant.

For example, the stoichiometry would dictate that the rate at which I consume a, this is one over a, this is the pre-multiplier, times the change in the concentration. I apologize for the multiple fonts here.

In this case, little c represents the stoichiometry. In this case, this represents concentration of A. The rate of change of concentration of A, the rate of its disappearance. When it is normalized for mole number would be the same as the rate of appearance of D, so d by dt of the concentration of D.

We have these individual relationships. This is simply conservation of mass. We have that. And now comes the interesting point of kinetic theory. What kinetic theory says is that there is a relationship between the instant rate of consumption of a product and its concentration.

In other words, we essentially say that concentration of a reactant is the driving force for reaction. Let's say in terms of a word equation, the rate of change of concentration of any species, this is the concentration of some species i, is proportional to the instant concentration.

And that makes sense. If you have a lot of a substance, it is going to have a high chemical potential to react. If you have a dilute concentration of a substance, it will have a weak propensity to react.

But it is not a linear relationship. It is some power-based relationship. We have to say raised to some power. Conceptually, that is what is going on. Now we can write this in mathematical formulation.

Let's say we are looking at the rate of consumption, so that is d by dt of some concentration of species i. I am going to say it is proportional to the concentration of species i, it is raised to some power, and this n is now the order of reaction.

You have seen n used so many times in 3.091. It was a quantum number in the Bohr model. It was an index of refraction. It was in the Born repulsion. Today it is functioning as the order of reactions.

N is a very talented variable. It is not typecast in a simple role is what I am saying. Now, how do I take a proportionality and turn it into an equality? I put in a constant of proportionality. And that constant of proportionality is called the specific chemical rate constant.

And these have to be determined by experiment. You cannot look at the stoichiometry of a reaction and infer what the order of reaction is. These must be determined by experiment. Let's look at that first equation and write it in its most general form.

What you could say is that the rate of change of concentration of species A goes as some constant times the concentration of species A to some arbitrary power alpha, but it may be influenced by some other concentration.

To be most general, we say concentration of species B raised to some power times the concentration of species C raised to some power. You might say, well, how can the concentration of species C have any influence on the rate of consumption of species A? Well, what if species C acts in some way as an inhibitor to the reaction? As the concentration of the product goes up, it could actually choke off the reaction.

To be general, we will put all four in here. And we will call these alpha, beta, gamma and delta all partial orders of reaction. To be determined experimentally. And, by the way, they do not have to be integers.

They are not necessarily integer. You can have something that is half-order or one third-order reaction, not necessarily integer values. Let's look at one. Here is a simple reaction. Unfortunately, it is a little bit toxic.

This is carbon monoxide plus chloride which will react to generate phosgene. COCl2 is phosgene. I have worked with this stuff in my research. Actually, it has a fantastic ability to desiccate material.

If you have a solid that has a parts per million residual water, if you expose it to phosgene, phosgene will attack the water, generating carbon dioxide and HCl. It is very good. Unfortunately, it was also used in gas warfare during World War I.

It is now banned by international treaty. On the basis of experimental evidence, we know that the rate of consumption of carbon monoxide in this reaction is related to the concentration of carbon monoxide linearly and the concentration of chlorine to the power of three-halves based on experiment.

We would say that this reaction is first-order in carbon monoxide. Because there is a linear relationship between the rate of consumption of carbon monoxide and the concentration. This is C to the power one, but we don't write the one.

It is of order 1.5 in chlorine. Or, we could say wholly or entirely of order 2.5. That gives you a sense of how the rate of disappearance of one constituent can depend on not just its own concentration but other concentrations.

Clearly, if we are trying to build a plant to make phosgene, we now have a quantitative measure of how to improve the rate of throughput. Suppose somebody says we would like to build a plant that produces so many tons per year, well, you know what your limitations are in terms of concentration so you can figure out what you would have to do in a quantitative predictive manner.

Well, there is another way we can improve the yield. What is another way to improve yield in a reaction? Instead of increasing concentration, we can increase temperature. We know that materials tend to react more quickly with one another at elevated temperature.

Where is the elevated temperature going to fit into this equation? I don't see any temperature term. Temperature must be buried in the rate constant. And that is where we are going to go. We are going to look inside the rate constant.

We will increase yield by increasing temperature. And that was known for many, many years, but it was finally the Swedish chemist Arrhenius who in 1889 announced the relationship that bears his name.

What he found was that the relationship between the specific rate constant and temperature was exponential through an equation of this form. Exponential of the ratio of some energy, which he termed the activation energy, divided by the product of the Boltzmann constant and temperature and a pre-exponential factor which we write as A in honor of Arrhenius.

Or, another way you can write it in this manner, A e to the minus Ea over k Boltzmann T. So there is the relationship. And another way of showing it is to put it on a semi log plot. If you plot the natural logarithm of the rate constant as a function of the reciprocal of the absolute temperature, you will find that, according to this equation, you should get a straight line.

One over T is increasing from left to right so high temperature is on the left side, and this should be linear with a slope of minus Ea, which is the activation energy, divided by the Boltzmann constant.

And just to give you a sense of order of magnitude, the activation energy for typical chemical reactions is on the order of about one electron volt, which is on the order of about 100 kilojoules per mole.

If you take this and choose an arbitrary rate for room temperature, plug in the value of 100 kilojoules per mole, what you can show yourself is that with this activation energy you double the rate for every ten degrees rise above room temperature.

Obviously this breaks down for large excursions in temperature. But in the vicinity of room temperature, if you have an activation energy, you typically have about a one electron volt, you can expect to double for every ten degrees.

What is the physical meaning of activation energy? Well, I thought I would try to show it to you in terms of a mechanical analog. What I am going to do here is show you what we can say is going to be, say, a loudspeaker.

What I have done by putting the x on the loudspeaker at the intersection of the x is we will say that that is the center of mass of the loudspeaker. The loudspeaker is now in a gravitational field.

And we can argue that I could represent this as a point of zero dimension at this height off the table. Now, clearly, if I lay this on its side, the point has fallen. It is now at a lower gravitational potential.

Really, if the box had its way, the box would prefer to lie on its side. How come the box doesn't just fall on its side? Why doesn't the box just lie on its side? See, it is a decrease in energy when I do that.

Did you see the photon? It is absorbed. Now look at it go. Why doesn't the box fall on its side? Well, this is a box at zero Kelvin. Now let's put the box at room temperature. What is happening at room temperature? Well, the box is doing this at room temperature.

Even so, the box won't fall on its side. Why? Because, in order for the box in three space to go from the vertical upright locked position to the supine position, what must the box do? The box must first rise up on one corner.

And only when it rises up on one corner and gets to this cusp does it have the ability to fall freely. What we are looking at, in terms of gravitational potential, is activation. I can now give you the same thing.

I could plot this as extent of reaction on the abscissa and some energy relative to gravitational potential on the ordinate, and I have three positions here. I have the box in the vertical position, I have the box in the horizontal position and, clearly, there is a decrease in energy.

This change is the delta E of the reaction, but the box cannot get to the right position from the left position without first rising up on its side. It must, at some point, rise up on its side. And when it rises up on its side its center of mass has risen, and this energy is the activation energy.

That is the mechanical analog. The delta E reaction, this is the driving force for the reaction in the first place. And Ea, in fact, is the gatekeeper. It determines the rate of reaction because, on the basis of Ea, Ea mediates the rate constant.

So, all other things being equal, temperature gives you the rate constant through Ea. And now we will recall the other temperature effect. You say, gee, how does any of this stuff happen? You know that room temperature energy is typically on the order of about one-fortieth of an electron volt, and I just told you that chemical reactions have activation energies on the order of one electron volt.

According to that, we would have to be up at thousands of degrees centigrade in order to promote any chemical reactions. How do we get chemical reactions at 200, 300 degrees centigrade? What we have going on is the distribution of energies.

And I will go back to that plot we saw earlier. Here is T1. Here is the average energy at T1. And out here is one electron volt, or whatever it happens to be, the activation energy. And so only the fraction of the population that has greater than the activation energy is able to participate in the reaction.

And we have seen before the measure of the area under this curve from Ea out to infinity is given by the relationship Ea, k Boltzmann T, exponential of which. And so it was Boltzmann who first gave us this understanding, then Maxwell gave us the distribution and away we go.

What happens by changing temperature, we have seen before. When we change temperature, the curve moves ever so slightly so that the average energy at T2, where T2 is greater than T1, moves a little bit.

But what happens up here is the area under the curve between Ea and infinity goes up by large leaps and bounds. And that is how, by modest increases in temperature, we are able to get substantial increases in the rate of reaction.

Now let's look at some specific reactions of various common orders. I am going to look at first-order which is the most common, and then I am going to look at second-order. Let's look at first-order reactions.

Here is an example. The example is N2O5 decomposes to give us NO2 plus O2. There is the stoichiometry of the reaction, but on the basis of measurements in the laboratory we can write that the rate of consumption of N2O5 -- The square brackets denote concentration.

It is found to be linear in concentration of N2O5, or N2O5 raised to the first power. This we say is a first-order reaction. And I am getting tired of writing N2O5 bracket, bracket, so I am just going to write c.

C represents all of this. I am compressing the font. I am going to rewrite this equation as minus dc by dt equals kc. **-dc/dt = kc** That is a first-order reaction. Why? Because this is a one.

That is why it is first-order. It is linearly dependent. The rate of change you see is linearly dependent on concentration. I do a little bit of calculus here. I can invert this, and I can then write minus dc over c equals k dt.

And what we are going to is integrate this thing out so that we can get the timeline on this. And so I can put an integration sign here. K is independent. We can do that. And I am going to start at time zero where I have some initial concentration and go to some arbitrary concentration c and arbitrary time t.

What I am going to look at is this. This is some initial concentration versus time. We know all of these things are going to attenuate. I want to know what this curve really, really looks like. And I know some of you are sitting there wincing saying that because your math professors have told you that you cannot make the limits of an integral the same as the value of the integrand.

I don't want to go to math jail, so we will put little tildes over this so that all the math weenies are happy here. But what I could do to really make it pedantic is - let's not do that. Let's use a carrier variable.

I will make this s, which now you don't know what I am talking about, and I will make this c. And that is mathematically beautiful. It does not explain what we are talking about so I am going to go back and put the c with a little tilde over it so everybody now is happy.

This is t with a little tilde. And if we integrate this thing out what do we get? We get that the natural log of c goes as the natural log of the initial concentration c naught minus kt. What I am talking about is the way that I can determine what the order of reaction is is to look at the functional shape of this.

In other words, whether it is first-order, second-order or 2.5-order, they all look like this, the curves. What I really need to do is, what the I is capable of doing is detecting something simple like straight line.

If I am cleaver about it, if I take instead of c versus t, if instead I map this into functional. I map c into some function of c and I map t into some function of t, that is embedded in the physical chemistry of the process so that f of c should be a linear function of g of t, whether it is this or whether it this.

Then I can inspect and say that looks linear. Therefore, the assumptions that underlie those functions must be what is going on. And that is the way we determine order of reaction. One other reaction I want to give you, even though it is not chemistry.

I still want you to understand, this falls under the rubric of general culture, that radioactive decay obeys first-order kinetics. Let's just take note of that. Radioactive decay is first-order. For example, here is a reaction.

U-238 can decompose to give thorium-234 plus helium. And if we plot concentration of uranium as a function of time, starting with some initial value c naught, we will just have this attenuation. But if we map it into the natural log of c versus time, we will get a straight line, and the slope here will give us minus k.

And we know that this is n equals one because, if you see a plot of natural log of concentration as a function of temperature, if it is a straight line, bingo, that tells you that this must be first-order and the slope of that line is minus k.

Now, it turns out that people that work in radioactive decay do not like to express the rate of reaction in terms of the rate constant. They prefer to use a different quantity called the half-life.

The half-life is a little more practical. It is more directly obvious what it is related to. The half-life is basically what you get if you take this equation here and plug in c naught over two. It is the time it takes to consume half of the reagent that is present.

If I start with c naught and I plug in c naught over two and solve for time that is going to give me the half-life. And if you do that you will end up with t to the one-half is equal to the natural log of two, which is 0.693 over k.

The half-life is inversely proportional to the rate constant and mediated by the natural log of two. And it turns out for this reaction, the half-life, t one half for this reaction is 4.5 billion years.

If you spill this into the sandbox at the neighborhood park, it is going to be a while before you can go back there and play. It will take you 4.5 billion years just to get the concentration down to half of what it is, and then 9 billion years will get it down to a quarter, etc.

And the nice thing about half-life, and the only time I would ever ask you to calculate half-life is in connection with first-order reactions. Because only for first-order reactions is half-life independent of concentration.

For every other order of reaction the half-life is a function of concentration in which case it is a useless quantity, except to perhaps professors of chemistry. We will not ask you for this except in the case of first-order.

T one-half independent of concentration for n equals one. You can prove to yourself, for higher order, how the half-life varies with concentration. I said that would look at the other, which is second-order reactions, and here is an example.

It is dimerization of aluminum trichloride. Aluminum trichloride can form the dimer Al2Cl6. This is gas phase. And right off the bat I am going to let c be the concentration of aluminum trichloride.

And it is second-order because the rate of disappearance of aluminum trichloride goes as the square. Now, this is one of those instances where it is really tempting because this reaction has a two in front of it.

And it turns out that the rate of disappearance of aluminum trichloride goes as kc squared. But do not, do not, I am going to say it three times, do not assume that this is the rule that because you saw it once that this is always the case that the stoichiometry gives you the order of reaction.

This is not the general case. It happens in this reaction for reasons that we will go into later, but generally you have to measure the order of reaction. And so if we go through the same analysis we will have minus dc over c squared equals k dt.

We will do the integration. Eventually, we end up with this relationship. The reciprocal of concentration is a linear function of time. That means the functional that we need is given here. The way to determine if something is second-order or not is to take this ordinary time line, concentration versus time, which is always going to look like this.

To the naked eye you are going to say I don't know if that is first-order or second-order, but I can detect a straight line. Instead, if I map this into one over c versus time, I should end up with a straight line starting at one over c naught.

And the slope of that line should be k. When I see that this is a straight line, I go n equals two. And the slope is the rate constant. If this were not second-order, I would not get a straight line.

This allows me to proceed and get the order of reaction. And, in fact, what I am showing you is one of the ways people determine order of reaction. They use the integrated form of the rate equation.

And so this is called the integral method to determine order of reaction. Determination of order of reaction. The first way is integral method. And it is essentially a comparison with the integrated form of the rate law.

Integrated rate law equation. This is really inspection. This is trial and error, but we don't say trial and error because we are from MIT so we use fancier words. We say by inspection. That is how you tell a boss you don't know what you are doing.

We really have to do this by inspection, sir. What I do is I try n equals one. If I get an arbitrary data set, I try n equals one, I try n equals two, and after that I say forget it and I am going to go to the other method.

Let me show you a couple of slides that indicate this. Here is a data set. I pulled this out of some book. This is a reaction. This is time. This is the concentration of some species. And don't worry about that.

We will get to that in a second. Here is your data set. This is all you have. And if you plot these, here is what they look like. This is concentration in moles per unit volume as a function of time.

And there is c naught normalized to one unit, and it is attenuating. It is over 500 seconds, we are down to about 20% of the initial value. I say what is the order of reaction? Well, I cannot tell.

What I am going to do is say now let's take natural log of c versus time. Bingo, straight line. On this basis, I conclude that it is a straight line. And the slope of this line is the rate constant.

But just to show you that it doesn't always work, I said just for grins and chuckles, plot that same data set as one over c. And look what happens. When I plot it as one over c, I don't get a straight line, so it really does discriminate.

It does not give me what I was going to expect in terms of a straight line, except when I have the proper order of reaction. Now let's see another way that we can get at this, and this is called the differential method.

We are not going to use the integrated form of the rate equation. We will use the differential form. And, to compress the notation, I am going to let lower case r be the rate. The rate is lower case r which I am going to let stand for minus dc by dt.

See how I get my fonts more and more compressed? I start with the big square brackets, then I get to this and now I get to this. It is getting tighter and tighter. Now I can write the rate equation as the rate goes as kc to the n.

There is the arbitrary rate equation. And now what I am going to do is take the logarithm of the equation. I can map that into the logarithm of the rate. The log of a product is the sum of the log, so that will be the logarithm of the rate constant plus n times the logarithm of concentration.

Now what I can do is start with concentration as a function of time. We know it starts at some value c naught and attenuates. Now I am going to plot the slope, the rate of consumption, not as a function of time but as a function of concentration.

Here is the initial concentration. This is going to represent the slope r naught, which is equal to minus dc by dt at time t naught when concentration equals c naught. I am going to end up with a data pair r naught and c naught.

And, ultimately, here is what I want to do. I might want to make this plot of log of the rate versus log of the concentration. I have my first data pair, r naught, c naught, and go over here. And then some arbitrary time later, here is t naught, here is t1.

At t1, I take the slope, so this is now slope r1 which is minus dc by dt at t1. And now the concentration is equal to some new value c1. This was c naught. This is c1. Now I am taking the data pairs r1, c1.

I take that and they go over here. Let's do one more because I am having so much fun. This is t2 and here is c2. The concentration has dropped to c2. I take the slope r2, blah, blah, blah, c2. And so I take that and will put it over here.

I am going to end up with a straight line. If you don't get a straight line in a log-log plot you are an idiot. You are working with chemical systems. I mean log-log linearizes everything. What do I have here? I have a straight line, log r versus log c.

The slope is n, the order of reaction. And what is this intercept? The value of the intercept is log k. Superb. I did that with the same data set. What I did is I went back. Here is the rate. I took the rates by taking the delta c over delta t.

I went down the curve, differentiated. Those are my rates. And now here is log r versus r c. Straight line. And it has a slope of one. So? In fact, the least squares gives me 1.018. It allows for a little bit of variation in experimental measurement, and it gives me a straight line.

This is the differential method for determining the rate of reaction. Well, I said we can increase productivity by increasing concentration. We are going to increase productivity by increasing temperature.

But there is another way we can increase productivity. And the way we can do that is by modifying the value of k. I am showing you how to do it by playing with temperature. There is another way to modify the value of k, modify the value of activation energy.

And we do that by the use of a catalyst. Increase rate via catalyst. And what the catalyst does is decreases the activation energy for reaction. And, by decreasing the activation energy, all other things being equal, same concentration, same temperature, therefore, that means it increases k at given temperature.

The way we can view the catalyst as working is the following. If I plot this energy versus extent of reaction. This is the reactants at some energy arbitrarily denoted as shown. And the products have to be at a lower energy.

If the products are not at a lower energy than the reactants, there is no driving force for the reaction in the first place, but we know we have to go up to some activated state in order to get up to the corner of the box, so to speak.

Here we are. This is the energy state of the reactants, this is the energy state of the products, but we have to invest this amount of energy in order to allow for the throughput. This is activation energy.

What a catalyst does is it reduces the level of activation needed by assisting with some kind of a physical process such as adsorbing gas molecules so that they can find each other more readily and react.

This is catalyzed. This is Ea. And now this one here is Ea catalyzed. The catalyst reduces the activation energy. And the result is, if we go over to this curve, n(E) versus E, we still have the same temperature distribution.

We have not changed that. If this is the activation energy as stated for the reaction in the presence of the catalyst, the catalyst reduces the activation energy, Ea catalyzed. Therefore, the area under the curve is much, much greater under the influence of catalysis.

And, while we are in the neighborhood, there are other actors that can increase the activation energy. That is up here. This is Ea under the influence of an inhibitor. And sometimes we want inhibitors.

What is an inhibitor? An inhibitor is going to raise the activation energy. Now the area under the curve is even smaller. Inhibited. For example, a catalyst we want in order to improve productivity such as conversion of pollutants in automotive exhaust.

We want to increase the productivity. But an inhibitor, we want to put in to slow down an undesirable reaction such as to reduce the rate of corrosion in your coolant. It flows through the radiator and engine block.

There are chemicals that are added to inhibit the corrosion processes that are undesirable for the long-term maintenance of the automobile. We see how there is a relationship between the catalysis, inhibition and the energy available to drive a reaction.

Let's see what we have up here for five minutes today. Radiocarbon dating. This is from Willard Libby who got the Nobel Prize in 1960. In the upper atmosphere, under the influence of cosmic rays, neutrons are generated.

And the neutrons react with nitrogen to form carbon 14. And we know that carbon 14 is present in about one part per trillion. This is now part of the carbon cycle down here, all the carbon atoms in all living things.

Because we are constantly exchanging carbon with our surroundings, take any carbon atom out of any one of us and we will have dominantly carbon 12, about 1% carbon 13 and one part per trillion carbon 14.

But when we die, we stop exchanging with the surroundings. In fact, if you were looking for a new euphemism for death, instead of saying "he passed away", you could say "he exited the carbon cycle."

He is no longer playing. So, upon death, we stop exchanging. We start at one part per trillion when we die, and then this reaction takes place. The carbon 14 decays radioactivity to nitrogen plus beta.

It is a beta decay. And so if we measure the ratio of carbon 14 to carbon 12, knowing the half-life is 5730 years, we can figure out how long it has been since we checked out of the carbon cycle. These are some things that have been carbon dated.

There are some things that have been looked at over time. One of the assumptions is that the rate of neutron generation and carbon 14 generation has not changed over, let's say, the last 10,000 to 15,000 years, which is really a blink of an eye by cosmic measurements.

Here are some things that have been dated. The Dead Sea scrolls came in at about 1,917 years back from today, but with uncertainty, plus or minus 200. This is Mount Mazama that erupted in Oregon and made Crater Lake.

It quenched some trees. The charcoal from those trees comes in at 6453 from today. The charcoal from the cave painting in Lascaux in France dates back 15,516 years. That is the kind of stuff that you see.

Here is one that is quite controversial, and we are not strangers to controversy. This is the Shroud of Turin. This is a painting showing what is to be the burial shroud of Christ. It shows the two images that it is supposed to have.

This is how the shroud would have been. It is a linen piece about 14 feet long. This is from a National Geographic Article of about six, seven years ago showing various features on the shroud. This is the shroud and this is the painting, just to give you a sense of what is supposed to be there.

And so in the mid ë80s fragments of that shroud were sent to three different laboratories in three different countries, and the fabric was dated to come from this interval of about 1260 to 1390. This shows the cartoon of cosmic rays generating carbon 14 which goes into flax.

Once we pick the flax and make linen, it is out of the carbon cycle. And these are going out like twink lights. Then you can calculate the various things. This is very hotly contested. Some people are ardent believers.

They believe that the data that were gathered were false because there was a fire in 1532. Here is a painting showing people repairing the shroud using candles. And there are arguments that there is paraffin from the candles that is coating the outer surface of the filaments.

In fact, you are analyzing the carbon from the paraffin that was observed on the filament. It is all about sampling. Very interesting. If you get on the Internet, you will find all kinds of stuff on there about it.

There is a question, how do you do the test? You know the way to do it properly is to take the whole thing, put it in a giant blender so you are not looking at surface effects. This is the problem, how to do nondestructive evaluation of an artifact.

I will see you on Friday.