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Chemistry: Using the Arrhenius Equation

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  • Type: Video Tutorial
  • Length: 6:21
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  • Size: 68 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: Collision Model & Arrhenius Equation (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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Another way we can use the Arrhenius equation is to recognize that we can express it at two different temperatures. For instance, we can say that the natural log of the rate constant k[2] at some temperature T[2] avails this relationship that the natural log of the rate constant is equal to the natural log of the frequency factor minus the activation energy over RT[2], so this is at temperature T[2]. Then if we write the equivalent expression for some other temperature, say T[1] then we will have some rate constant k[1] related to T[1], according to this expression. These things are exactly equivalent. It is just that with two different temperatures we would have two different rate constants reflecting the fact that the reactions go at different rates at different temperatures.
Now if we subtract one from the other -- so we subtract this expression from this expression from this expression; that gets us to the first red expression here. The natural log of the rate constant 2 minus the natural log of the rate constant 1 is equal to minus the activation energy over R times the difference of minus . Another way to express that is that because the difference of two natural logs is the natural log of the quotient; we can express it this way instead. These two have sort of divided out this frequency factor, because very often what we are interested in is not being able to calculate the absolute rate constant, but rather the relative rate constant. Rate constant, if you know it at one temperature, you can calculate it at a second temperature.
Let's look at a couple of problems that use this equation or this equation. Remember these are exactly equivalent in order to determine an unknown. Here is the reaction of 2HI H[2] + I[2] and the rate constant is determined to be 2.5510^-8 M^-1 s^-1 at 650 k. And the activation energy has been measured to be 168 kJ/mol. So, what is the rate constant in 700 Kelvin? Again, what we've got is an expression in which we have one equation and one unknown. If we plug in for rate constant 1... Let's go back. We're going to use the top expression here. We're going to plug in for this one, this one, and this one, and solve for that. The unknown is the rate constant k[2] at temperature T[2]. We know what the rate constant is at T[1.] We know the activation energy, and remember the activation energy was given to us in the problem in kilo joules per mole, but in order to get the units to cancel out, we have to express it in joules per mole. So we introduce it 10^3, then minus . You can work out the algebra here or the arithmetic and you find out that the natural log of k[2,] is equal to E^-15.3, and you get that k[2] is equal to 2.35 times 10^-7. Now, this number is somewhat larger than that number 10^-8. And that makes sense because we are asking, "If we run the reaction at a higher temperature, does the reaction go faster?" The answer is yes. It goes faster by about a factor of 10 in increasing the temperature by 50 Kelvins.
Another kind of problem would be if you're given the rates at two different temperatures and you need to calculate the activation energy. Again, you are going to use the same kind of expression. Here is an example. This is a reaction of carbon monoxide and chlorine to form phosgene. Phosgene was used as a warfare agent. It is a gas that causes a lot of lung damage. I think it also damages your eyes. I think it was used during World War II, and I think by the Geneva Convention you're not allowed to use it any more. It also happens to be a useful organic chemistry reagent, and I've used that in my own research, but we are very careful with it.
So, the reaction at 250C is 1.5 times 10^3 or 1500 times as fast as the same reaction at 150C where you assume that all you are doing is changing the temperature and that the concentrations of the reactants are still the same and that the reaction is only faster because the rate constant is faster at the higher temperature. We can say that K[2], the rate constant at temperature 2, is 1.5 10^3k[1], the rate constant at the first temperature, and then remember that in order to use our expression, these temperatures are in absolute. These are Kelvin temperatures. We have to make sure that we put in Kelvin temperatures.
In this particular problem the temperatures were given in degrees Celsius so we have to convert these temperatures in degrees Celsius into Kelvins or absolute temperature. You do that by adding 273, since we only have three significant figures here. We need these three pieces of information and we are going to plug them into the expression. Now we are going to use the second expression. We have the natural log of = minus activation energy over R, times minus . We don't know specifically what k[2] is, but we know that it's 1.5 times 10^-3 times as large as k[1], so we can substitute for k[2],1.5 times 10^-3 times k[1]. Then we can take the natural log, and the k[1]'s conveniently cancel out, so we're taking the natural log of 1.5 times 10^-3. We have equal minus the activation energy over R equals the 1 over the 2 absolute temperatures, and we have one equation and one unknown so we can reduce this down and we get that the activation energy is 135 kJ/mol.
We can use this Arrhenius expression to do all kinds of things. There are basically five unknowns, and if we have any four of them then we can uniquely determine the fifth one. It is just a matter of plugging in and making sure you don't do any silly arithmetic mistakes.
Chemical Kinetics
Temperature and Rates
Using the Arrhenius Equation Page [1 of 2]

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