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About this Lesson
 Type: Video Tutorial
 Length: 11:06
 Media: Video/mp4
 Use: Watch Online & Download
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 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: Orders of Reaction (3 lessons, $4.95)
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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For a reaction, such as this one here, A goes to B plus C. If we plot [A], at any time, t as a function of t, we're going to get a curve that looks something like this. We know something about this point here that's that the concentration at time zero. The [A][0], remember, is the concentration at time zero. But it's actually sort of difficult to tell that the rate at which [A] is decreasing is first order in A or second order in A. In other words, that how fast this is going away is either equal to k times A concentration of A or k times [A]^2 or k times [A]^3. In other words, it really isn't obvious how the rate law depends on A from a curve that looks like this. Wouldn't it be great if we could keep all the data, but somehow come up with a way to replot the data so that it was more transparent, so that it was more obvious, what was going on, what the rate law looked like?
So in other words, we can get the data that would give this purple curve, but the way that it's plotted here, it's not really transparent what this has to do with the rate law, how we can sort of get the rate law out in a very simple way. And that is the point of this lecture.
Suppose we could come up with some way to replot maybe a function of [A] as a function of time as a function of time. In other words, do something to these data and then plot the thing that we've done as a function of time to get a new kind of function, which is more transparent.
So here's the same reaction, A B + C, where the rate law... Suppose that the rate law is equal to  =k[A]. In other words, this is a firstorder reaction. We've already looked at one example, which is cyclobutane decomposing into two ethylene molecules. That was a firstorder reaction. And another firstorder reaction, which is really sort of important, is that radioactive decay turns out to be firstorder. So, for instance, ^238U or uranium 238 decomposes into thorium 234 plus the helium nucleus. Don't worry about this if you don't really get it right now. You'll see that if and when you study nuclear chemistry.
All right, so lets go back to now this equation:  = k[A]. And write it again here. And now we're going to do some calculus. And if you don't follow this section of the lecture, don't worry about it. I'll give you the result at the end and you can just use that result. But for those of you who are interested in the calculus, what we're going to do is take this expression and divide both sides by [A] and multiply both sides by DT and that's going to get us to this expression here. = kdt. And now we're going to the definite integral of both sides. So we're going to integrate the righthand side from zero up to t and then this is time. And so that means we're going to integrate the lefthand side from the concentration at times zero, which is a zero up to [A] at time t. And when we do that, remember that the integral of is the natural log, and if we plug in the endpoints of our integral, we get an equation that looks like ln[A][t]  ln[A][0 ]= k times t minus t zero. In other words, how much time has gone by? But since t zero is zero, in other words, we start the clock at zero, then that just is kt. And now, let's turn this around by adding natural log of A zero to both sides and that gives us the second blue equation is. And notice that this equation is in the form y = mx + b. The scientists love straight lines, and this is the equation of a straight line. And the reason is, your eye can always tell immediately if something is a straight line. It can't necessarily tell if something is a hyperbola or parabola or all the other shapes of curves that you might actually, but you can always tell a straight line.
So, what are we going to do? Instead of plotting A is a function of time, in other words, instead of plotting concentration of A is a function of time, what we're going to do is plot the ln[A] as a function of time. And what's it going to be? Well it's going to be a straight line instead of a curve, where the slope is going to be k and the intercept is going to be the natural log of [A][0] or the natural log of the concentration at time t = 0. And this, B, you'll recall is the y intercept, and so the y intercept is going to be ln[A][0]. And that looks like this. So remember now, if and only if a reaction is first order will this be true, because remember, this whole derivation, the whole math thing, started with a rate law was firstorder in A.
So, if and only if the reaction is firstorder in A, then if we plot the natural log of A at each of the times t as a function of t in seconds or whatever units of time, this is now considered to be dimensionless. Then, we'll get a straight line; we can draw a line through our points. The slope is k, so we know something about the rate constant, and we already knew, because this is what we put in, we know what it is at time 0.
Okay, so that's really useful and we're going to look at problems like this, but first, another point about firstorder reactions, and that is suppose that we define t[1/2] as the time at which we have half of what we started with. So, in other words, at the time t[1/2], [A] is equal to the initial [A] divided by two. Well, then if we plug in this expression into our equation, our linear equation, it's the ln[A][t 1/2 ]is equal to k time t[1/2] plus the ln[A][0]. And if we rearrange this a little bit, put ln[A][0] on the other side, and then the difference of two natural logs is the natural log of the quotient. And then this fraction here, the A zeros cancel out, and so this thing is equal to the natural log of two is equal to minus the rate constant times the special time t[1/2.] Now, let's reexpress in terms of t[1/2], and we say that t[1/2] is equal to the natural log of two over k of the rate constant. And the natural log of two has the value of .693. What does that say? That says if you have a certain amount of stuff and you wait for the halflife and the halflife is related to one over the rate constant, half of it is going to be gone. And it's independent of how much you started with. So however much you started with, if you wait a certain amount of time t[1/2] half of it will be gone. And it's independent of how much you started with.
And so, for instance, here's the reaction of N[2 ]O[5 ], which happens to be firstorder, and you can see that we started with one molar at time zero, and if we wait however much time it is from here to here, it will be down to halfmoler. And if we wait that equivalent time again, it will down to a quarter. And if we wait that equivalent time again, it will be down to an eighth. And if we wait that equivalent time again, it will be down a sixteenth, a thirtysecond, and so forth.
Yes, nuclear reactions are very often first order. And so, what happens is we can define a halflife for radioactive decay and that tells you how long you have to wait until half of that radioactive nucleus is gone. So half of your sample has decayed into something else. And so that's why it's interesting to talk about a halflife of radioactive molecules and you'll talk about the radioactive atoms, and you'll talk about that more when you're studying nuclear chemistry.
Okay, so we can use the fact that when we have a firstorder reaction, the plot of natural log of the concentration is a function of time as a straight line. Here's an example of a problem. This is the isomerization, rearranging of the nuclei, of methylisonitryal, CH[3]NC, to a acetynitryal, CH[3]CN. So basically, we're just rearranging the atoms. Electrons are staying exactly the same. And if we know that it's a firstorder reaction, we can calculate the firstorder rate constant k, and that halflife t[1/2]. How do we know that it's firstorder? The way we're going to know it's firstorder is we're going to plot the natural log of concentration as a function of time and if it's a straight line, we know it's firstorder. Now in this particular problem, since it goes on in the gas phase, we're going to use concentration in pressure units instead, so we're going to look at the pressure of the methylisonitryal in towards as a function of time. So here is one data set where we measured the pressure of methylisonitryal in towards the function of time.
So, if we plotted the natural log of these numbers versus time, we'd get a straight line, and that tells us that it's firstorder. Now that's expressed in this equation that says the natural log of concentration at time t is equal to minus the rate constant times t plus the natural log of the concentration at the beginning of the reaction. And since it's in the gas phase, we can replace these concentrations with pressure. Think back to gas laws: pressure is proportional to concentration. P is proportional to . So we can replace these concentrations with pressures at a particular time. So this is just equivalent to this. Now we're going to plug in the pressure at time t and we're going to look at time of 15,000 seconds. So we'll put in the pressure at time 15,000 seconds and we have to put in the pressure at the beginning of the reaction. And we have one equation and one unknown. So we can solve for k, and the rate constant for firstorder reaction is inverse seconds. The units are inverse seconds. We solve for k and we get 2.07 x 10^4 inverse seconds. That's the rate constant. Now the rate constant is not particularly transparent, but for firstorder reaction, we can relate it to something called the halflife, and the halflife again is the natural log of 2 divided by the rate constant, and so that equals 3.35 X 10^3 seconds. What does that say? That says if we have methylisonitryal and it's isomerizing at, what was it, 215^°C, then if we wait on the order of 3,000 seconds, half of it will be gone. And if we wait another 3,000 seconds, half again will be gone. So there will only be a quarter gone and so forth. So the halflife is a very transparent way, experimentally, of expressing how fast this reaction is proceeding. It is related inversely to the rate constant and remember this, of everything that I've talked about here today, only holds when we have a reaction that's firstorder. So, it depends on the data obeying this kind of expression right here.
Chemical Kinetics
Orders of Reaction
FirstOrder Reactions Page [2 of 2]
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