Transcript - Lecture 31

OK.

We are starting in on the last unit. And so we are going to be talking about kinetics for the rest of the semester. The last class is a bit of a review of everything we have done, but this is the last topic.

And it is mostly in Chapter 13, but there is a little bit in Chapter 17.

If you notice, on your problem set, the VSEPR is covered, which is in Chapter 3, transition metals in Chapter 16, kinetics in Chapter 13 and some nuclear chemistry in Chapter 17. We were jumping back and forth between several different chapters in this last part of the course.

Rates of chemical reactions and rate loss.

Today is the introductory lecture to rates, and we will introduce you to a lot of terms that you need to know. All right. Chemical reactions. In addition to asking whether the reaction is likely to go as it is written, which is a thermodynamic question, we also want to know how fast that reaction will go.

Kinetics is rates of reaction.

That is what it is. And a kinetic experiment measures the rate at which a concentration of a substance taking place in the chemical reaction changes with time. You are talking about changes in concentration with changes in time.

Let's take a look at a chemical reaction. This is the oxidation of glucose, glucose and oxygen going to carbon dioxide and water.

As professor Ceyer told you in the first part of the course, this is the reason why we eat.

We need glucose for energy. This is what drives our bodies. This is how we make ATP, which is our source of energy in our bodies. This is the reason why we eat. Oxygen is the reason why we what? Breath.

CO2 is the reason why we? Exhale?. And water is the reason why we? Drink.

Yes. You know this reaction quite well, a really important reaction. Professor Ceyer told you about the thermodynamics of this reaction.

She talked about delta G knot. And delta G knot equals what? Delta H minus T delta S. And so you can consider sort of what you get out of this reaction, how spontaneous, how favorable it is considering delta G, which depends on the value of delta H and the delta S.

And, of course, also what temperature the reaction is.

In this case, delta H is a large negative number. What type of reaction is that? Exothermic. Delta S knot is positive, so that is very favorable. You always want to increase entropy. And so delta G is a very large negative number.

This is a thermodynamically very favorable reaction.

If that is true, if you have glucose around, one might expect that when the glucose is sitting in air, in oxygen, that it would be forming CO2 and water. This is a really favorable reaction. I brought some glucose with me today to class.

In here there are some Hershey's Kisses. There is chocolate and there is sugar. This is all sealed up away from oxygen. But according to this thermodynamics, if I open this up, this is a really favorable reaction.

Air will enter the bag and CO2 should come out. I am going to open this up, and we will see what happens. Of course, if you did this on your own, safety glasses might be a good idea.

All right. I don't really see much happening.

All right. Let's observe this reaction over time. What I want you to do, the TAs are going to hand this out, and during the course of the lecture I want you to carefully study your Hershey's Kiss.

[LAUGHTER] And let me know at the end if you observe any CO2 coming off of this. And we will watch one up here at some point. Christine has one up here, which we can take a look at.

All right.

You can help me with this experiment during the course of the class. OK. This brings up a point. It is thermodynamically favorable, yet we are not observing the reaction happening. Why is that?

Well, it is favorable but slow.

Thermodynamics are very important, but that is not the whole story. Rate or kinetics is also a very important part of what determines whether you are going to be observing this reaction happening.

As these are being handed out, let's just consider these two terms for a moment.

This is going back to what Professor Ceyer talked about, and these are actually right out of one of your handouts. She told you about things that are stable or unstable. If you are talking about stable-unstable, you are talking about delta G, you are talking about thermodynamics.

The thermodynamic tendency to decompose.

But that is not the whole story. You also have to consider whether it is labile or non-labile, also referred to as inert. And that refers to the rate. And so that is kinetics. The reaction that we talked about, it might be that glucose is unable.

Thermodynamically it wants to decompose, but it is also fairly non-labile.

The rate at which it happens is very slow. I think that most likely at the end, and we have done this experiment before in class, no one has observed any CO2.

But somehow the chocolate, by the end of the class, seems to have gone away. [LAUGHTER] But yet there is no evidence of the water that has formed so it is a little bit confusing. In any case, these are important terms to know.

Rate is important.

And just as another example in terms of rate being important, someone might say to you that you could make $5 million. Again, rate is important. Will you make that $5 million in the next year or would you have to life to an age of 1562 to make that $5 million? Rate at which things are happening is very important.

In this case, this reaction has to go to some degree.

Because this is how we make energy to live. This is really slow, but your body is using this reaction to get energy. How does that happen? Well, it happens because there are enzymes.

In the body, it must be fast enough. The body uses catalysts, enzymes, proteins that catalyze the reaction. They speed up the reaction.

And that is what allows this reaction to work as your source of energy.

We will consider catalysts and enzymes a little later in this unit, but that is an important point. Let's talk about some of the factors affecting rates of reaction. What are some factors that would affect how fast a reaction would go?

Temperature? Did someone say heat or temperature? What else might affect rates? Concentration.

What about the one that was just up on that slide about how the oxidation of glucose works on the body?

Use of catalyst.

OK, a couple other things. Nature of material and mechanism, which we are going to be talking about. In the next few lectures we are going to be considering concentrations, nature of material, mechanism.

And toward the end we are going to talk about temperature and catalyst.

That is where we are going in this unit. And how they all affect rates of reactions. First we have to introduce you to some terms and talk about measuring the rates of reaction.

Here we have a reaction of NO2 gas, CO, carbon monoxide, NO gas and CO2 carbon dioxide, which we just mentioned before in the last example.

Suppose you were measuring a rate of this particular reaction. You could, for example, plot the change in concentration of NO gas versus time. You are forming NO as the reaction goes.

You are considering the rate of the reaction you are looking at, how much you form in a certain amount of time.

Here might be what the data would look like. As time goes on, you would see an increase. And then it starts to level off. You can consider the average rate, which would be defined as some change in concentration over a particular change in time.

You could express that as the change or delta of the concentration of NO over the change in time, delta time, and you could pick some places and calculate an average rate.

You could draw lines over, two different concentrations at two different times, and calculate what the rate is in that interval. We can do to that average rate, two different concentrations at those two different times.

Change in concentration over change in time.

And that gives you a value. But this average rate then depends on the time interval chosen. It is going to be a different rate here than it might have been down here or up here. That tells you maybe something, but it doesn't necessarily tell you all the things that are going on because it just depends on a particular area.

Instead of considering average rate, you can start talking about instantaneous rate, the rate at a particular time.

And so that is instantaneous rate. And so you really want to know what does the rate say at 150 seconds? At that particular moment in time, what was the rate? Here you are just talking about the limit as that change in time approaches zero.

And that can also be expressed in terms of d concentration of NOdT, so the change in concentration of NO with time at a particular moment. As that change in time approaches zero, the rate becomes the same as the slope of the line tangent to that particular point in time at time t, the one you are interested in.

And so, say we will find this instantaneous rate at 150 seconds here.

We can do that, the rate at that particular time. We are going to draw a line tangent to the curve and figure out the slope of that line.

We can do that, find the slope, get some different points and look at the instantaneous rate. We can calculate the slope of that line here.

And come up with a rate. In this case, 7.7 times 10 to the minus 5.

And this is in molar per second, which is a common unit for rate. OK. That tells us about a particular instantaneous rate at one given time. And an important thing to consider is initial rate.

And so that is the instantaneous rate at time equals zero, the initial rate of the reaction.

That is a little bit about measuring rates. And we will talk more about plotting data when you are trying to get rate constants and other factors. Rate expressions. We can express the rate of the reaction.

When we are measuring a reaction, you can often pick to measure the increase in the amount of products being formed.

You can also consider the decrease in the amount of reactants, so you can measure rate either way looking for disappearance or looking for appearance.

And often decisions about how you are going to measure the rate depend on how easy it is to measure a change in something. There are usually practical considerations that go on here.

Then the rate can be expressed in terms of the disappearance of one of the reactants, here NO2.

The rate is going to depend on how fast that it is disappearing or the disappearance of the second reactant over here and the appearance then of one of the products or the other product. And this, of course, is only true if we assume there are no intermediates.

Because sometimes the rate at which a reactant will disappear will not be the same as the rate at which it will appear because there are lots of steps involved in between.

This comes back to mechanism, which we are going to talk about next week, but this is what is known as rate expressions. Let's look at an example. Here you have a reaction 2HI gas going to H2 gas plus I2 gas.

The rate could be considered in terms of the disappearance of the reactant.

And here we see that for every two HI that disappears, that is decomposed we have one H2 and one I2. If we are going to set up our rate expression, we are going to have to multiply this term by one-half if it is going to be equal to the rate at which H2 appears or the rate at which I2 appears.

The stoichiometry here matters.

If you just looked at the rate of disappearance of HI, it wouldn't be equal to how fast that is forming because two are going away for every one of each of these that is formed. Then, in general, you would write the expression this way.

Minus one over A where A is the stoichiometry of reactant A.

dAdT minus one over B. dBdT plus one over C. dCdT plus one over D. dDdT, that is hard to say. Those are rate expressions. Now we come to rate laws.

A rate law is the relationship between rates and concentrations.

And those are going to be related by a proportionality constant called little K rate constant. Big K was what? Equilibrium constant. And now we have little K, which is the rate constant.

Rate law is going to tell you how the rate depends on the things in the reactions.

There will be a rate constant and there is going to be concentration terms. And it has to be experimentally determined what that rate law is, unless it is an elementary reaction, and we will come to that in a few minutes.

Suppose you have this particular reaction going. Your rate law would have some expression that looks like this.

It would have rate equals the rate constant little K times some concentrations. And here we are raised to the M and to the N, and those are known as the order of the reaction.

And so M is the order of the reaction in A and N is the order of the reaction in B. And little K, of course, is the rate constant. How do you know what terms to put in there?

And how do you know what the order of the reaction is? Well, these are experimentally determined.

The rate law is an experimental result. And you just cannot look at the stoichiometry of the reaction and predict the rate law unless it is an elementary reaction. And so when you are designing mechanisms you are creating steps in mechanisms that are elementary reactions.

And then you can write the rate law based on the stoichiometry.

And so you will be told this is an elementary reaction, and that is fine. That will be a step in a reaction. But, otherwise, unless you know it is an elementary reaction, you cannot just assume the stoichiometry is going to work.

And there will be experimental evidence to indicate the order of the reaction. Rate laws are not just limited to reactants.

Sometimes product terms will appear. You might see term for C or for D, the products of the reactions.

And the order of the reactions can be integers, fractions, negative or positive. Again, the order would be experimentally determined unless it is an elementary reaction. Let's look at various orders of reactions and consider how you would write the rate law and also how you would figure out what was going on with the rate law.

And the way the experiments are done is if you, say, double the concentration of something and then look at the effect of the rate, that is how you are going to figure out experimentally the order of the reaction, what affect it has if you keep everything else constant.

Double one thing, what happens to the rate? Then try doubling the next thing, keeping everything constant, and you would come up with the order of the reaction.

Most of this is blank. There are a couple of things written in your handout, but most of it is blank so let's fill it all in.

We will start up here with M equals one, which is called first order. For something that is first order, the rate would be K, the rate constant, times the concentration of A. And so this is M is the order of the reaction in A.

That would be your rate law.

And so you can put a little one there if you want, but it is implied. That is the rate law. Now let's think about if you double the concentration of A, with this rate law, what would happen to the rate of the reaction? It would double.

All right. Let's look at second order then.

M equals two, second order. This would be written rate constant K times the concentration of A squared, the second order, so M equals two. Now if you double the concentration, what happens to the rate? It quadruples.

If you tripled the concentration, what happens to the rate?

You had nine times. These are the kinds of problems that you will get. Someone will give you, in a problem set or on a test or in real life sometimes, a whole bunch of data, it will talk about doubling or tripling concentrations and then there is the affect on the rate, that is how the rate changed.

Then you can figure out the order of the reaction. Let's go down here now, M equals minus one. I don't think this has a name. I don't know what the name is, if it does have one, but we can write the rate law for it.

The rate equals rate constant times the concentration of A to the minus one.

M up here is minus one. Let's think about what happens if we double the concentration in this case. What would happen to the rate? You would half the rate.

It can also be a fraction. You can have M equals minus one-half, and that is shown here.

You would write that as K times the concentration of A raised to minus one-half. If you double the concentration of A in this case, what would happen to the rate?

It would be 0.7 times 2.0 raised to the minus one-half.

Some of these get a little complicated in seeing the patterns when you are analyzing the data. Some are more straightforward than others, but the ones where it is a negative in a fraction are a little more complicated to observe what pattern is happening.

Going up here we have M equals one-half, and that is called half order. The positive ones here have names. And so that then would be K times A raised to one-half.

And if we double the concentration here, what happens to the rate?

You have 1.4 times the rate.

What do you think M equals zero is called?

Zero order. And what would the rate law be for this one?

Yeah, rate equals K. What would happen if I double the concentration of A? Nothing. No affect on rate.

You can look for this in a list of examples where they are changing the concentration of A and you can look at the affect. And, if there is no affect on rate, then it is zero order. And so the term does not appear in the rate law.

There are some exercises on problem ten on this. It is a good question for a final exam to think about looking at data and coming up with what the rate law for that expression is for that particular reaction based on data.

You think about what is happening, what they are doing, and then what the affect of rate is.

And you can back calculate then what the order of the reaction had to be. The next thing about rate laws is that the overall order of the reaction is simply the sum of the exponents, so the sum of all the individuals orders.

If your rate law was your rate constant, and it is A to the two, so that is what order in A? Second order.

And what order in B? First order. And your overall order is going to be? Three. It would be overall third order, second order in A, first order in B and overall third.

It always breaks my heart on a test when people get this part right and they say what is the overall order of the reaction? And they cannot add the one and the two to get three.

Remember that part. That is like the ??give me points?? on an exam to be able to add these very small numbers and come up with the sum, so remember how to do that for me. Overall, it is just the sum of the individual orders.

Here is the bad news, and that is the units for the rate constant K vary.

The units are going to have to do with what units you have for concentration and rates and what terms are in there. There is no one unit that you can memorize and always know that that is what it is going to be.

You actually have to think about what the units for the rate constant are.

That is a little bit trickier. This one much easier, just add, but the units are a little bit more complicated here. That's all you need to know about rate loss.

It is not too much. All right. Here are some of the problems with these kinds of experiments. Some of the data that you might get, that you are trying to analyze may not be too perfect. Usually, if it is in the textbook, it is pretty perfect data.

But, in general, these things can be tricky to do.

And they can be tricky to get this good data that figure out what the order of the reactions are, because you are often measuring really small changes in concentration of material over brief intervals of time.

That can be technically challenging to do. And so the approach that is used then is to use these integrated rate laws.

And so here you can express concentrations directly as a function of time, and this makes it easier to get at the information that you want.

Let's look at a first order reaction. You know what first order means now. We just have one thing going to another thing, in this particular example. Your first order for A, rate of a reaction could be written.

The rate expression can be written in terms of the disappearance of A over time.

And you know it is first order, you are told it is first order, so you can also write the rate law. The rate law for first order reaction in A would just be the rate constant times the concentration of A, so M is one here.

Now we can derive the first order integrated rate law from this simple expression right here.

What we are going to do is we are going to take the concentration terms and put them on one side of the expression and we are going to take time terms and put them on the other side.

We are going to bring the concentration of A here over to this side of the expression. There it is, one over the concentration of A. We have dT as well.

And so we have moved that. And we have moved K and the negative sign over here.

And we have moved the change in time over to the other side. We have all the concentration terms on one side. We have rate terms and time terms on the other side. And now we can integrate. And we want to consider the change from the initial concentration of A, A to the O for original or the zero.

The original amount that you have of A to the concentration of A you have at time T.

And so we have our concentration expression. And in terms of time, we are looking from zero time to time T. And now we can take this and solve this expression. This is the same one. I am just moving it up to the top of the screen.

And so now we can express this in terms of natural log.

The natural log of the concentration of A at time T minus the natural log of the original concentration of A equals minus KT. And we can reorder that and put a box around it, which means it is going to be important later on.

We can rearrange this in terms of the natural log of the concentration A of time T minus KT plus the natural log of the initial concentration of A.

And so we will come back to this in a few minutes. We can also keep going and rearrange this in other ways. We can rewrite this term instead of by subtracting them rewrite it as this division. Natural log of the concentration of A at time T over the original concentration equals minus KT.

And then we can take the inverse natural log.

And so then on the other side we have E raised to the minus KT. And we can write this in this expression again, box. This is the way that you often see the first order integrated rate law where you are solving for the concentration of A time T is equal to the original concentration of A times E to the minus KT.

How the concentration of A changes depends on the rate constant and the time that has elapsed.

This is often the way that the integrated first order rate law is written. Sometimes you may see a little bit different expressions, but usually this is the final expression that you see. And this is an equation for a straight line.

And that is why it was boxed because it is how data is going to be plotted.

If you are going to use that now and plot data, you can plot data that you collect, so the concentration of A as it is changing. You are measuring how the concentration of A is changing versus time.

And so here is your equation, the natural log of a concentration at time T equals minus KT plus the natural log of the original concentration of A.

You can see that this is the equation for this line.

And so the intercept over here is going to tell you about the initial concentration of A. And what you really want, the great thing that you want is to determine rate constants for a particular reaction? How do I get the rate constant out of this plot?

From the slope, right.

The slope is going to be equal to minus K here. We can solve then, if we plot our data that we measure over time, how it is changing, after this amount of time we have this much more and we can plot the data, we can pull out a rate constant.

This is a very useful way to get information about our particular reaction that we are interested in.

When you are talking about first order equation reactions, you also often talk about half lives, the amount of time it takes for half of the material to reaction.

Half-life is the time that it takes for the original material to be reduced by a half. And its expression is, in terms of T to the half, shown here.

Let's derive an expression for the half-life a first order reaction.

We can take from above, from the integrated first order rate laws this expression, natural log of the concentration of A at a particular time over the concentration of the original material equals minus KT, the rate constant times time.

All right.

For first order half-life, we are talking about half of the material going away. Then we can go in and substitute in half of the original material being gone and solve for that. We can take this.

The A at time, at a particular time, in this case half of it is gone, so we can substitute in the original material divided by two, because what we are interested in is a first half-life.

Half of that original material is gone, so it is the time at time T2 that we are interest in. And now we have put T to the half, so it is not just at any time but at a particular time at this particular condition for first order half-life.

And then we can rearrange this expression here.

And so this just becomes the natural log of a half. The original concentrations will cancel out in this expression and you are left with one over two. Now we can solve for this, the number value, and so you get minus 0.6931.

And then we want to rearrange this expression to solve for the half-life.

If we bring that over, we have half-life equals 0.6931 over K. The one thing you will notice in this expression, there is no term for A anymore, so half-life does not depend on the concentration.

It depends on what instead? On K.

On the rate constant. And rate constant depends on the material in question. Half-life depends on K and K depends on material. We talked about natural of material affecting rate, so here is one example.

Some materials have much longer half lives than other materials.

Just to think about this question then. For the same material, does it take longer for one ton to go to a half ton, or for one gram to go to a half gram? It is the same, right.

It is hard to think about this sometimes, it seems like that should be wrong, but it isn't. All right. One last thing before you go, let's just look at plots. This takes the same amount of time.

Let's consider what it would look like if we plotted then concentration versus time for something that was first order.

And so you would have a first half-life. You would be at half the concentration of the material. The second half-life, after it had gone through another half-life, you would be down to a quarter. And a third half-life you are down to 0.125.

And so that would give rise to a curve like this.

You have this kind of decay curve. And you can see where we are going on this. On Monday we are going to be talking about examples of half-life first order reactions, which are nuclear decay reactions.

We will talk a little about radioactivity, and that is where Chapter 17 comes in.

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