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Chemistry: Determining the Form of a Rate Law


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  • Type: Video Tutorial
  • Length: 9:24
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 100 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 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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Let me remind you what we're trying to do. The balanced reaction is only a recipe for telling us how we make product from reactants. What we are trying to do now is find out how fast this reaction proceeds. And, that has nothing to do with the recipe. What it has to do with is experimentally determining how fast the reaction goes and how it depends on the concentrations of the reactants.
So, for this reaction H[2] + Br[2] 2HBr, we are going to determine a rate law. We can relate these guys to aA moles of A + bB moles of B cC moles of C. In other words, a recipe. What we are looking for is an expression that looks like the rate, which is times the change in concentration of C with respect to time or times the change of concentration of B with respect to time. In general, it is equal to some rate constant times the concentration of A to some power m, and the concentration of B to some power of n.
Remember, m and n don't necessarily have anything to do with A and B. So, for a particular reaction of hydrogen plus bromine to hydrogen bromine, it turns out that the experimentally determined rate law looks like rate, which is equal to the change in concentration of bromine with respect to time, is equal to some rate constant, times the concentration of hydrogen, and this is instantaneous concentration, times the concentration of bromine to the power. So, in this case, m is equal to 1 and n is equal to .
We haven't talked about how you could get something to be , and we'll talk about that later on. The thing I am trying to remind you of is that these exponents don't necessarily have anything to do with these stoichiometric coefficients.
Let's look at the first method for determining what k, m, and n are for a reaction. The reaction I want to look at is the reaction of nitric oxide plus oxygen to nitrogen dioxide. This is a reaction that proceeds in the atmosphere to produce highly oxidized oxides of nitrogen on their way to nitric acid, which is one of the sources of acid rain.
We could plot the concentration of nitrogen dioxide as a function of time and imagine the curve looks something like this. The red curve is our complete data. Imagine that we could measure the initial rate by checking just what the concentration of NO[2] is as a function time. For the first few data points, drawing a straight line through those first few data points, and assuming that the slope of that line is essentially equal to the tangent at zero, the tangent of the red line at zero. And that is what we call determining what the initial rate is for the reaction.
Again, the initial rate exactly is the tangent to the red line at zero. And, we approximate it graphically by just measuring the concentration of nitrogen dioxide as a function of time over a few data points. And, assuming that there is not too much curvature in the data close to zero, so we can just draw a straight line through it. If you think about it, what the slope of the green line represents is the initial rate of the reaction times 2, because we make 2 nitrogen dioxides every time we run this reaction.
So, a typical initial rate data problem is going to look something like this. Here's, again, our balanced reaction. You are going to see a table. The table here represents four separate experiments. Experiment one we will put in some known amount of nitric oxide, which is one of the reactants, and we will put in some known concentration of O[2], which is the other reactant, and we'll measure the initial rate.
What does that mean? Well, that means what we will do is we will put these two things in. We'll collect data that looks like this. And, we'll measure the slope of the green line. Then what we will do is we will repeat the experiment. But, repeat it with slight change in the concentration of one of the reactants leaving the other one the same.
Second experiment, we measure a new initial rate. What it means is when we double the concentration of this guy in our second experiment the data are going to be slightly steeper. In other words, the tangent is going to be steeper. The reaction just goes faster at the beginning if we double the concentration of one of the reactants.
Now, we repeat the experiment. This time we double the concentration of the other one. We don't necessarily have to double, but it turns out that it is convenient that we just exactly double it. And, we measure an initial rate again. Then we do it a fourth time. So you'll see tables that look just like this.
Again, what we are trying to determine is the rate law which looks like some rate constant times the concentration of one of the reactants to the m^th power times one of the other reactants to the n^th power. And, m and n don't depend on the stoichiometric coefficients.
We can express the initial rate of experiment one, which you will recall was .048 molar per second, and that is equal to the rate constant times the concentration of nitric oxide to the m^th power, times the concentration of oxygen to the n^th power. And, we can do the same thing for experiment two. The initial rate of experiment two was different. And, the concentrations of nitric oxide are different, but we can express it like this. We could do this for each of the four experiments.
Now, what we are going to do is if the top equation is true and the second equation is true, then we can take a ratio of this to this and it is equal to the ratio of this to this. Right? If this is an equality and this is an equality then this divided by this is going to be equal to this divided by this. Now, that's what we are going to do. We are going to actually do 2 over 1, because it is convenient for the math. But, you could do 1 over 2.
If we do 2 over I, if we look back at what the actual initial rate for 2 was it is .192 molar per second and the initial rate for 1 was .048 molar per second. Here are those two rate law expressions. Since we have exact same k in both the numerator and the denominator, and it is the same for both, that's 1. And, conveniently, we chose two data sets in which the concentration of oxygen was the same for both experiments, because we have the same exponent here this is also 1. So this nasty looking expression here reduces actually to this over this, which is 4, and this over this, which gives this expression. Then, knowing a little algebra, something to the power divided by something else to the power is this over this to that same power. So that is 2^m. And, by inspection, we can solve that m equals 2.
Alternatively what we can do is take the natural log of both sides; that gives us this expression. When you take the natural log of 2^m, that is m times the natural log of 2. And, then divide through by the natural log of 2 to get an expression for m, and m is equal to 2. So, what have we done? We found one of the three things that we are trying to find. We found the exponent m.
Now, we are going to repeat the exact same procedure. Except, instead of 2 over I, we are going to look at 3 over 1. If we look at 3 over 1, conveniently now it's the expression that has m in it that cancels out. And so our rate expression reduces to this over this, which is 2. And, the k's cancel out again, and then it is this over this to the n power. And, this over this to the n power becomes 2 to the n. And so by inspection we can get n is equal to 1.
So now we've determined m and we've determined n. Again, we are determining it not from the stoichiometric equation, not from the balanced reaction, we are doing it from experimental data.
Finally, we can figure out what k is by taking any one of the data sets now. Because, if we take, for instance, the initial rate of experiment one. It is equal to k times the concentration of nitric oxide. And now we know what that exponent is. It is a 2. And the initial concentration of oxygen, and we know what that exponent is. It is a 1, but we usually don't even write it. So we can now have one equation and one unknown. And we can solve it. And we get k is equal to 1.4 times 10^4. And the units on k have to be such that k times this squared, times that are all going to cancel out such that the final units are in molar per second, and that you can convince yourself that the units on k have to be inverse molar squared inverse seconds.
So, the complete rate law looks like rate - and we can say that rate is equal to, I'll remind you, is equal to the rate constant times the concentration of nitric oxide squared, times the concentration of oxygen. And we say that this reaction is first order in oxygen, second order in nitric oxide, and third order overall. Remember you add the exponents to get the overall order of the reaction.
Now, what does this say? It says that if we double the concentration of oxygen, but keeping the concentration of nitric oxide fixed, we are going to double the rate because k doesn't change. Similarly what it says is that if we hold the concentration of oxygen constant, and double the concentration of nitric oxide, we are going to quadruple the rate. If you go back and look at the data that is exactly what you are going to see. So, this expression, again, derived from the experimental data, and you can go back and double check and make sure that it actually all hangs together.
Chemical Kinetics
Reaction Rates
Determining the Form of a Rate Law Page [2 of 2]

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