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About this Lesson
 Type: Video Tutorial
 Length: 12:56
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 Download: MP4 (iPod compatible)
 Size: 138 MB
 Posted: 07/14/2009
This lesson is part of the following series:
Chemistry: Full Course (303 lessons, $198.00)
Chemistry: Chemical Kinetics (18 lessons, $25.74)
Chemistry: Reaction Rates (3 lessons, $5.94)
This lesson was selected from a broader, comprehensive course, Chemistry, taught by Professor Harman, Professor Yee, and Professor Sammakia. This course and others are available from Thinkwell, Inc. The full course can be found at http://www.thinkwell.com/student/product/chemistry. The full course covers atoms, molecules and ions, stoichiometry, reactions in aqueous solutions, gases, thermochemistry, Modern Atomic Theory, electron configurations, periodicity, chemical bonding, molecular geometry, bonding theory, oxidationreduction reactions, condensed phases, solution properties, kinetics, acids and bases, organic reactions, thermodynamics, nuclear chemistry, metals, nonmetals, biochemistry, organic chemistry, and more.
Dean Harman is a professor of chemistry at the University of Virginia, where he has been honored with several teaching awards. He heads Harman Research Group, which specializes in the novel organic transformations made possible by electronrich metal centers such as Os(II), RE(I), AND W(0). He holds a Ph.D. from Stanford University.
Gordon Yee is an associate professor of chemistry at Virginia Tech in Blacksburg, VA. He received his Ph.D. from Stanford University and completed postdoctoral work at DuPont. A widely published author, Professor Yee studies moleculebased magnetism.
Tarek Sammakia is a Professor of Chemistry at the University of Colorado at Boulder where he teaches organic chemistry to undergraduate and graduate students. He received his Ph.D. from Yale University and carried out postdoctoral research at Harvard University. He has received several national awards for his work in synthetic and mechanistic organic chemistry.
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Let's continue to look at this reaction of dinitrogen pentoxide decomposing to form nitric oxide and oxygen. Instead of just plotting the concentration of 0[2] with respect to time, let's actually, qualitatively, at least, because I've just drawn curves, qualitatively plot the concentration of each of the reactants or the reactant in each of the products as a function of time. So, I've put that all on this plot. Now we are looking at the concentration of three species all at once. So, I have x in a square bracket where x could be either one of these species. The concentrations are all in molarity units.
We've already talked about how 0[2] changes as a function of time or the concentration of 0[2] changes as a function of time. It grows slowly with a curve that, qualitatively at least, looks something like this. If we think something about how the concentration of nitric oxide changes in the plot as a function of time, for every oxygen molecule we produce, we actually produce four molecules of nitric oxide. So, if we plot nitric oxide concentration as a function of time, the curve is going to be qualitatively the same as the 0[2] concentration change as a function of time. Except, it is going to rise to a much higher level and, at any point, the slope is going to be much larger than the slope is for the oxygen. So, the slope for the nitric oxide, change in concentration of nitric oxide with respect to time, which is the slope of a tangent in any point, is going to be much larger than the slope for 0[2]. We'll see how those two relate to each other in a second.
The other thing that we can plot is, "How does the concentration of dinitrogen pentoxide change as a function of time?" The answer is that it has qualitatively very different behavior. It starts at a high value and it goes to zero, whereas the others start at zero and go to a higher value. In other words, reactants are disappearing and so the concentration of dinitrogen pentoxide ought to go down toward zero.
Now, can we relate these data to algebraic expressions that express, sort of, the shape of these curves? The answer is absolutely. For instance, if we think about how fast nitric oxide is produced, it is produced at a rate that is four times as high as the rate of production of 0[2] because for every 0[2] we make, we make four nitric oxide molecules. That idea is expressed in this algebraic relationship that the change in concentration of nitric oxide with respect to time is equal to four times the change in concentration of 0[2] with respect to time, . So, again, whatever this is, whatever the rate of production of 0[2] is, four times that is going to be the rate of production of NO. And, this 4 comes from the fact that we have a 4 here.
Now, think about the rate of change in concentration of nitric oxide with respect to time. That is this expression here, change in concentration of dinitrogen pentoxide with respect to time, . First of all there is going to be a negative sign. Why is there a negative sign? There is a negative sign because, if you think about it, the slopes of the tangent on the dinitrogen pentoxide line, those are all negative. In other words, things that are disappearing are going to be related to things that are appearing by a negative sign. Also, there is a 2 here because for every dinitrogen pentoxides that we lose, we only make one 0[2]. And so, the change in concentration of nitric oxide with respect to time is going to be equal to 2 times the change in concentration of 0[2] with respect to time, .
Now, if we take this expression here and this expression here and try to relate the two equations. What we notice is that the change in 0[2] concentration with respect to time is equal to of the change in nitric oxide concentration with respect to time and it is also equal to  the change in dinitrogen pentoxide with respect to time. We can take these two equations and relate them to each other. We have now two equations that relates the change in concentration of each of the species with respect to time.
This whole idea can be generalized for an arbitrary reaction Aa + Bb
Cc + Dd. In other words, we have aA moles of A plus bB moles of B going to cC moles of C plus D moles of D. This expression holds. In other words, times the change in concentration of C with respect to time, is equal to times the change in concentration of D with respect to time. Where we have minus signs here and here, in front of A and B because these are reactants and these are products. Remember, these are going away and these are increasing so we have to have a negative sign to express that concept. And, this is what we are going to call the rate of reaction.
What we want to get to is to create something called a rate law. What a rate law is, it's an equation relating rate and concentration. In this case it is going to be concentration of reactants. Let me try to express that idea in something you might be able to get your mind around.
We are at a Junior High School dance, at the dance boys and girls might not be really eager to dance. But, if you think about the number of people who are going to be on the dance floor dancing, it is certainly going to be proportional to the number of boys. Right? If there are no boys at the Junior High School dance there are not going to be any boygirl couples formed who are dancing. Similarly it is going to be proportional to the number of girls, because if you don't have any girls, if it is just all boys, you are not going to have any boygirl couples dancing.
So, what we can say is that the rate of production of couples dancing is proportional to the number of boys and the number of girls. And there is more to this. So it is proportional to, if you have no boys, you are not going to get any couples formed and you have no girls. But, also, as you get more and more couples dancing, there are going to be fewer and fewer single guys and single girls. Right? So the rate of production of couples is going to go down.
In other words, you've got, at the beginning, maybe, after the dance starts, you have lots of couples forming. But, pretty soon there are just a few, well, in my case, I was sort of the only socially inept guy so I would just be standing on the sidelines. That would mean that the number of boys that are still hanging around is going down as the reaction proceeds, as couples are forming and they are getting onto the dance floor.
Let's look at another analogy. You go to a sandwich shop. What pisses you off? The thing that really pisses you off is if you have a line of 20 people waiting to have their sandwich made and you only have one person making sandwiches. The rate of production of happy people with their sandwiches is really low. But, as you add more sandwich makers, more sandwich artists, what happens is you produce more happy people eating sandwiches. In other words, there's a proportionality, which means that the rate of production of product, meaning happy people eating their sandwiches, and one of the reactants is the number of people making the sandwiches.
So, here is a chemical reaction. This is a reaction of cyclobutane, which looks like this, to form ethylene and it forms two molecules of ethylene for every mole of cyclobutane. Incidentally, ethylene is the monomer for polyethylene and it also happens to be the gas that ripens fruit, but we won't go there right now.
Let's look at the instantaneous rate of production of ethylene. It turns out that it is equal to k, which is called the rate constant, times the instantaneous concentration of cyclobutane. So, this is the reactant and this is how fast product is being produced. If we think about the plot of the concentration of ethylene as a function of time, qualitatively it is going to look something like this. This is something we have seen already. Even if you don't believe that this expression is true, we can at least look at the limits at zero and at the starting concentration and see that it sort of makes sense.
So, at the beginning of the reaction the concentration of cyclobutane, which is a reactant, is going to be very high. If the concentration of this is high, then this product is going to be high, k times the concentration is going to be high. So, that says the instantaneous rate is going to be high.
What is the instantaneous rate at the beginning of the reaction? It is the slope of the blue curve at the beginning of the reaction. And, the slope is going to be the largest at the beginning of the reaction. At the end of the reaction, the instantaneous concentration of cyclobutane is going to be zero. In other words, this is going to go to zero. The concentration of cyclobutane is going to go to zero. So, what happens? Well, that means the instantaneous rate has to go to zero. And that is exactly what happens. In other words the slope of the curve flattens out and goes to zero. So, at least, at extremes, time equals zero and a time equals infinity, this expression makes sense. And it just turns out that mathematically the shape of this curve is expressed by an expression that looks like this.
Let's look at the details of this. First of all, I already mentioned that k is called the rate constant. The units on k change depending on everything that comes after the k. But, in particular, in this reaction, we have the concentration of cyclobutane, we imply that there is a 1 here, so we call this a first order reaction, I'll come back to that in a second. The units on change in ethylene concentration with respect to time are in molarity per unit time. So, the units on k, the rate constant, have to be appropriate in order to make this expression on the lefthand side have the same units on the righthand side. So, in this case, k has the units of inverse seconds. Inverse seconds times molarity units gives molarity per second. It could be inverse minutes or inverse hours or inverse years or whatever, but the point is that it is the inverse time for this type of reaction and we'll see that more.
Finally, I wanted to say, on this slide that we say that the reaction is first order on cyclobutane because there is an implied one. More often than not you won't even see the 1 there. And we say it is first order overall, meaning that we add up, actually we'll see what that means in a second on the next slide. For the general reaction, again, a little a times a, little moles of a plus little b moles of b going to little c moles of c plus little d moles of d. There is going to be a rate expression that looks like some rate constant k times the concentration of each of the reactants, which is A and B, to some power m and n. We say, in general, that it's m^th order in A, so if that's a 2, we'd say it is second order, but if that's a 7 we'd say it is seventh order, it turns out that it's never going to be 7. Similarly for B we say that it is n^th order in B. Then we add m and n and say that it is m + n order overall.
I'll do some examples later on that will make this more clear. But, basically, you just add up m and n to get the order overall. Now, m and n are usually small whole numbers like 0, 1, or 2, and they could be fractions, and they could even be negative, and we'll look at examples of that. The important point is that m and n generally do not depend on little a and a little b. In other words, this number here and this number here do not generally depend on this number here and this number here. M and n have to be experimentally determined. And, we'll see lots of different ways that you can experimentally determine that.
So, what have we learned? Well, we at least expressed the general idea of what a rate expression looks like. In subsequent tutorials what we are going to do is show many different ways that you can determine what m and n are. And, that is really at the heart of kinetics.
Chemical Kinetics
Reaction Rates
Rate Laws: How the Reaction Rate Depends on Concentration Page [1 of 3]
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