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Chemistry: Second-Order Reactions

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  • Type: Video Tutorial
  • Length: 8:17
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  • Download: MP4 (iPod compatible)
  • Size: 89 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: 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, oxidation-reduction 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 electron-rich 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 molecule-based 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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Suppose we have a reaction that's second-order in [A]. I showed you this plot earlier and I said that this was for a reaction that was first-order in A and it turns out, for a reaction that's second-order in A, qualitatively, the curve is going to be exactly the same. Quantitatively, it's going to be different. But the point is, [A] as a function of time, is still going to drop from whatever it is at the beginning along some curve down towards zero. And, again, this is not transparent. For a second-order reaction, is there some transformation we can do, a mathematical transformation, which allows us to plot these data to get something that is a straight line, just as we did for a first-order reaction? And the answer is, yes we can do that.
So here's a general reaction: A B + C. And if we write down what the rate law looks like for a second-order reaction, and again (this is assuming that the reaction is second-order) then the change in concentration of A with respect to time with a minus sign, because remember, A is a reactant, is equal to some rate constant times [A]^2. This too is what makes it a second-order reaction.
So we can play the same calculus games that we did before. Divide through this time by [A]^2 and multiply both sides by dt and we get to this. And then, integrate both sides of this expression. On the right-hand side, we're going to integrate from zero to T. On the left-hand side we're going to integrate from [A][0] up to [A][t, ]exactly like we did before. And when we do that, the integral of 1/(A^2DA) is -1/A. So this definite integral turns into this difference and the integral of -kdt is just kt. So now, let's just get rid of some minus signs and reexpress this, and we can get to the second red equation. And this, again, is beautiful, because it's in the form y = mx +b. Now it's not natural log of A but rather 1/A and the slope is not -k as it was in the first-order reaction; it's now +k and the b is a different b from what it was for a first-order reaction.
But here's the point: if the reaction is second-order, then if we plot the data as versus t, then the slope is going to be k, the rate constant, and the y intercept is going to be .
All right, so let's look at an example. Let me show you what that graphically looks like. We're saying plot one over [A][t ]versus t and the intercept is going to be and the slope is going to be k. The rate constant is going to be k. And this is only true if the reaction is second-order.
So, let's look at an example. An example is the reaction of two HI's going to H[2 ]+ I[2]. And here is a data set. Let's look at the top line first. What we did is we monitored the concentration of HI as a function of time. Now we're doing it in minutes, but minutes or seconds, the point is it's just a unit of time. And we'd get these things. And I'll show you what these look like in a second, but remember, when we plot [A] as a function of time, we'd simply get a curve. And now, suppose we don't know what the rate of the reaction is, the order of the reaction in HI is. And suppose we thought, "Well, maybe it's first-order in HI." If it were first-order in HI, then if we plot the natural log of HI as a function of time, then this second data set ought to be a straight line and maybe it's not. Maybe it's second-order in HI. Well, if it's second-order in HI, then if we plot as a function of time, then this third data set should be a straight line. Okay?
Now we'll look at the plots. So, again, we plotted all three of those. We plotted the raw data. Raw data is exactly what we measured, HI, as a function of time, and here are the points. So there are the five points that were on the previous slide. And if we now check to see whether or not these points all fall on a line or not, we see that it, even to your eye, looks curvy, but let's go ahead and draw the line in, and it will be really clear that the points don't fall on the line. This is, again, the concentration of HI as a function of time.
Now we'll plot the natural log of the concentration of HI as a function of time. Okay, this is the still the time axis here. And if the reaction is first-order in HI, then these data points should fall on a line. So let's see if they do, and we'll go ahead and draw the line in. And you can see that these data points are a little below the line. In other words, they're still curved to this data set, and so the reaction is not first-order in HI either. But now let's plot , and if we plot , look at that, the data now falls perfectly on a straight line. And what that does is that tells us that the reaction is second-order in HI. Now it turns out that the slope of this line is not exactly k. It's actually 2k for reasons having to do with the fact that this is not exactly A B + C, but rather 2 times A B + C). And so there's a factor of two here in the slope. So in other words, the slope of this line is 2k not k. Don't even worry about it.
It's still true that the intercept here, that's one over A zero or one over the initial concentration of HI, so that's still true. But more importantly, the plot of as a function of time is a straight line. And that's what tells us that the reaction is second-order.
Now, another property of second-order reactions is that we can calculate a half-life, just like we calculated a half-life of first-order reaction. So we're still going to define the half-life to be when [A] at the half-life time is equal to exactly one-half of what it was at the beginning of the reaction. And if we do that, so here's plugging now into the expression that we derived for the straight line, we get at the time t[1/2] is equal to k times the time t[1/2] plus one over the initial concentration. This is exactly what we did when we were looking at a first-order reaction except now we're using the equation for second-order reaction. Well, you can go through the algebra, and remember, we have to substitute this expression in for this quantity here, and if you do that, you get that t one half is equal to one over the rate constant times the initial concentration.
What does this say? This says that the half-life, or how long it takes for half of your material to go away, depends on how much stuff you've started with. And so you have to know precisely how much stuff you started with to know how long it's going to take for half of it to go away. And if you think about it, that's sort of useless. If you have to know exactly how much you've got originally to know how long it's going to take for half of it to go away, that's not a really useful thing, because you have to know something precise about how much you started with. And so people typically don't talk about a half-life for second-order reaction in the way that they did for a first-order reaction, where it didn't depend on how much stuff you originally had.
So let's summarize. If you find out that your data fit on a straight line when you plot one over the concentration as a function of time, then that means that the reaction is second-order in that reactant. If the natural log of the concentration as a function of time fits, then that's a first-order reaction. And both of them, if you plot the concentration as a function of time, so no natural log; no one over anything, but just the concentration, it's going to give a curve but it's not real transparent from the curve exactly what's going on, so we use these mathematical tools to get plots of straight lines that are easier to understand.
Chemical Kinetics
Orders of Reaction
Second-Order Reactions Page [1 of 2]

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