Preview
You Might Also Like

Chemistry: Precipitation Reactions 
Chemistry: Electron Shielding 
Chemistry: Understanding Electron Spin 
Chemistry: FirstOrder Reactions 
Chemistry: AcidStrong Base Reactions 
Chemistry: Organic Polymers 
Chemistry: Rates of Disintegration Reactions 
Chemistry: Reviewing OxidationReduction Reactions 
Chemistry: Elimination Reactions 
Chemistry: Acids and Conjugate Base Reactions 
College Algebra: Solving for x in Log Equations 
College Algebra: Finding Log Function Values 
College Algebra: Exponential to Log Functions 
College Algebra: Using Exponent Properties 
College Algebra: Finding the Inverse of a Function 
College Algebra: Graphing Polynomial Functions 
College Algebra: Polynomial Zeros & Multiplicities 
College Algebra: PiecewiseDefined Functions 
College Algebra: Decoding the Circle Formula 
College Algebra: Rationalizing Denominators

Chemistry: Acids and Conjugate Base Reactions 
Chemistry: Elimination Reactions 
Chemistry: Reviewing OxidationReduction Reactions 
Chemistry: Rates of Disintegration Reactions 
Chemistry: Organic Polymers 
Chemistry: AcidStrong Base Reactions 
Chemistry: FirstOrder Reactions 
Chemistry: Understanding Electron Spin 
Chemistry: Electron Shielding 
Chemistry: Precipitation Reactions
About this Lesson
 Type: Video Tutorial
 Length: 8:17
 Media: Video/mp4
 Use: Watch Online & Download
 Access Period: Unrestricted
 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, 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.
About this Author
 Thinkwell
 2174 lessons
 Joined:
11/13/2008
Founded in 1997, Thinkwell has succeeded in creating "nextgeneration" textbooks that help students learn and teachers teach. Capitalizing on the power of new technology, Thinkwell products prepare students more effectively for their coursework than any printed textbook can. Thinkwell has assembled a group of talented industry professionals who have shaped the company into the leading provider of technologybased textbooks. For more information about Thinkwell, please visit www.thinkwell.com or visit Thinkwell's Video Lesson Store at http://thinkwell.mindbites.com/.
Thinkwell lessons feature a starstudded cast of outstanding university professors: Edward Burger (PreAlgebra through...
More..Recent Reviews
This lesson has not been reviewed.
Please purchase the lesson to review.
This lesson has not been reviewed.
Please purchase the lesson to review.
Suppose we have a reaction that's secondorder in [A]. I showed you this plot earlier and I said that this was for a reaction that was firstorder in A and it turns out, for a reaction that's secondorder 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 secondorder 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 firstorder 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 secondorder reaction, and again (this is assuming that the reaction is secondorder) 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 secondorder 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 righthand side, we're going to integrate from zero to T. On the lefthand 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 firstorder reaction; it's now +k and the b is a different b from what it was for a firstorder reaction.
But here's the point: if the reaction is secondorder, 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 secondorder.
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 firstorder in HI." If it were firstorder 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 secondorder in HI. Well, if it's secondorder 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 firstorder 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 firstorder 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 secondorder 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 secondorder.
Now, another property of secondorder reactions is that we can calculate a halflife, just like we calculated a halflife of firstorder reaction. So we're still going to define the halflife to be when [A] at the halflife time is equal to exactly onehalf 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 firstorder reaction except now we're using the equation for secondorder 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 halflife, 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 halflife for secondorder reaction in the way that they did for a firstorder 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 secondorder in that reactant. If the natural log of the concentration as a function of time fits, then that's a firstorder 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
SecondOrder Reactions Page [1 of 2]
Get it Now and Start Learning
Embed this video on your site
Copy and paste the following snippet:
Link to this page
Copy and paste the following snippet: