Kramers' Theory of Reaction Kinetics
Kramers’ Theory of Reaction Kinetics
In this lab module, you will learn about why a chemical reaction occurs with a particular rate and how it relates to the chemical reaction barrier. At the lowest level of theory (aka Freshman Chemistry), you learned the rate of a chemical reaction can be calculated using the Arrhenius equation:
* G kBT k Ae (1) * where A is the pre-exponential, G is the activation energy, kB is the Boltzmann constant, and T is the temperature. There are a few problems with this equation, namely what is the pre- exponential A? The first attempt to define it comes from Transition State Theory (TST), which describes two chemical species with enough energy to cross a single energy barrier to a new state (i.e. a chemical reaction has occurred). Transition State Theory also stipulates that the chemicals cannot react “backwards” to reform the original reactants. Using TST, the Arrhenius equation is transformed into:
* G k T k T k B e B h (2) where h is Plank’s constant. While TST is a good improvement over the empirical Arrhenius equation, there are still several significant problems. For one, it doesn’t seem to work well except for a few cases. One obvious shortcoming is the idea that when chemicals have enough energy, they will always react (and never go “backwards”, which really not correct). Here is a way to “fix” the TST result:
* k T G k B e kB T h (3) The function of is very interesting; it is called the transmission coefficient. It is a unit-less percentage that defines how many molecules will react, form a product (and stay that way) if they have enough energy. You may notice that, if I calculate a TST rate which doesn’t match the experiment, I can just calculate a that “fixes” it. This in my opinion is cheating. Scientists began to analyze how to determine from basic physical principals as opposed to empirical observations. The first question to ask, how does the solvent affect the transmission coefficient? Isn’t it sensible that a reaction will occur faster in water than in molasses? What about a reaction in the gas phase, where there is no solvent rather than a buffer gas that occasionally bumps into the reactants? To test your assumptions, we will examine the simplest scenario possible- the reaction of A and B to form C in the gas phase. Shown here are A and B in their potential energy surface (PES), which is defined the function U(x). We will call “x” the reaction coordinate; it doesn’t really represent Cartesian space rather than measure how much A and B are more like reactants or whether they have become products. The PES contains two wells, one for the reactants and one for the product. There is a simple energy barrier from one to the other. Let’s say that a flash of light strikes A and B to impart enough energy for them to cross the barrier to form C as shown here:
Note- there is a small problem: if the molecules have enough energy to cross the barrier from reactant to product, then they have enough energy to react backwards as well:
It seems that molecules A and B are stuck; after all, the law of constant energy means that they will be able to wander over the energy hill forever. Without a way to lose the energy that we gave them, A and B as just going to wander over the reactant and product wells continuously; this makes the transmission coefficient =0. In this case, it is sensible to ask how can anything ever reactant form products? First, examine some unspoken assumptions; such as, is there anything else in the entire universe but A and B? No, while this reaction is occurring in outer space, there are still some gas molecules floating around nearby. In our hypothetical situation, a buffer gas molecule may strike A and B when they happen to be hovering over the product basin. The gas molecule ends up taking energy out of A and B such that they can now drop into the product state to form C.
It seems that having some buffer gas around is “helpful” in terms of this reaction occurring. Under these conditions (called the Kramers Low Viscosity Regime), the transmission coefficient (and thus the overall reaction rate) increase linearly with solvent viscosity.
Now let’s take things to another extreme, A and B are in a very dense, thick liquid. They have enough energy to react, but the other molecules (the solvent or buffer gas) present a physical barrier for the reaction to occur. This process is illustrated here:
The solvent or buffer gas can slow the reaction down in many ways; for one, they can simply get in-between A and B preventing them from physically making contact. They can also slow down the diffusion of A and B such that it takes longer for them to “touch” each other and react. The solvent interacts so strongly with A and B that it may take away all their energy before they can cross the barrier. Furthermore, once A and B do connect, there may be some spatial rearrangements of the atoms that comprise molecules A and B to from C that the solvent molecules are opposing. This case is called the Kramers High Friction Regime; under these circumstances the transmission coefficient (and thus rate of reaction) is inversely proportional to the solvent viscosity.
The sum of these examples illustrates that the rate of a reaction is controlled by a lot more than the activation energy barrier G*. The pre-exponential has a strong effect, which is largely dictated by the nature of the medium that the reaction is occurring in. The local environment can do this by affecting the transmission coefficient . If we can define , then we may finally understand why a chemical reaction appears to occur at the rate that the scientist measures.
Hendrik Kramers studied this theoretical problem in 1940 using Langevin dynamics via the Langevin Equation. The Langevin equation is: 2 x U(x) x m 2 m f (t) t x t (4) where m is mass of the chemicals that are reacting, x represents the reaction coordinate on a potential energy surface defined by U(x), is the solvent viscosity, and f(t) is a force that randomly knocks things around; f(t) is due to the solvent impacting the chemicals (this is where Brownian motion comes from). In the Langevin equation, the presence of a buffer gas or a solvent is represented by the presence of viscosity that slows down anything moving (this is why x it is multiplied by the velocity . If something isn’t moving, then the viscosity of the solvent t doesn’t do anything to it). This viscous “drag” on the reactants is compensated by the random force f(t) that gets them moving again. Kramers performed very sophisticated mathematical operations on the Langevin equation and solved for the transmission coefficient as a function of solvent viscosity. The result was so confusing to researchers that Kramer’s work was ignored for some time (no one else was smart enough to follow the math). Here is what Kramers calculated for the overall rate of a chemical reaction:
1 2 2 * G R 2B m kB T k 1 1e (5) 4 B m where R and B are related to the curvature of the potential energy surface in the vicinity of the reactants and barrier, respectively. We are fortunate that Equation 5 has two solutions in limiting regimes. For very high viscosity (large ), eq. 5 can be recast as:
* G m k T k R B e B (6) 2 Note how the rate goes down with solvent viscosity as stated previously. The rate equation in the limiting regime where is very small is a bit more complicated, but it shows that the rate is linearly proportional to the viscosity.
Reaction Kinetics In this laboratory exercise, you will study the effect of solvent viscosity on a reaction using transient absorption spectroscopy. Specifically, the light-initiated reaction of benzophenone as described here is examined. When benzophenone (C6H5)2CO absorbs a UV/Visible photon, it is excited to its first electronic state and quickly intersystem crosses to the triplet state:
(C6H5)2CO + h → (C6H5)2CO* (S1) → (C6H5)2CO* (T1) (7) In an alcohol solvent, a proton is abstracted from the alcohol to form a protonated ketyl radical:
• • (C6H5)2CO* (T1) + ROH → (C6H5)2COH + ROH (8) In the presence of potassium hydroxide, the protonated ketyl radicals can disassociate:
KD • •- + (C6H5)2COH ↔ (C6H5)2CO + H (9)
Note that the protonated neutral radical and deprotonated anion radical are in equilibrium and KD is an equilibrium constant. Finally, the protonated and deprotonated forms of the ketyl radicals can dimerize, forming the benzopinacol anion: k • •- - (C6H5)2COH + (C6H5)2CO → (C6H5)2(OH)C-C(O )(C6H5)2 (10) where a new bond is formed between the carbon atoms. This is the reaction your measuring!
We will be probing the rate of decay of the deprotonated ketyl radical anion can be deduced from the above equation to be overall a second order process:
(C6H5 )2 CO k (C6H5 )2 COH (C6H5 )2 CO (11) t If we assume that protonated and deprotonated forms of the ketyl radicals rapidly reaches equilibrium with an equilibrium constant of KD, then its equilibrium relationship can be rearranged to yield
• •- + [(C6H5)2COH ] = [(C6H5)2CO ][H ]/ KD (12)
Now substituting (12) into (11) into yields:
2 (C6H5 )2 CO kobs (C6H5 )2 CO (13) t
+ where kobs = k[H ]/KD. Solving the differential equation 10 yields a rate law of the form: 1 1 kobs t (C6H5)2CO (C6H5)2CO t 0 (14)
In this experiment you will monitor the concentration of the deprotonated ketyl radical anion in a series of several alcohol solvents that have different viscosities. The deprotonated ketyl radical absorbs at ~630 nm, allowing its concentration to be monitored by the absorption of the He-Ne laser. From the data of the reaction rate vs. solvent viscosity, you will be able to determine whether your reaction is occurring in the Kramers high or low friction regime. Next, you will change the temperature of one solution to calculate the activation barrier.
III. Experimental Protocol First, have one lab partner start the laser (see FLASH DATA COLLECTION below), while another makes the solutions. All the glassware must be clean prior to this experiment. To clean the sample cell, rinse it thoroughly with water; do not use soap. Drying is not necessary. Add an appropriate amount of benzophenone to a 100 mL volumetric flask to make a 0.005 M solution. Next, fill the solution with the series of alcohol solvents as directed by your TA to the marking on the volumetric and cover with a glass “size 9” cap; mix thoroughly. You will choose three from the following:
Solvent Viscosity (kg/m/s) methanol 0.0005445 ethanol 0.001078 1-propanol 0.00194 butanol 0.002593
Butanol will be used by everyone as that one will be performed at three different temperatures. When you are ready to perform an experiment, add 1 mL of a stock 0.5 M KOH solution (kept in the hood) using a l mL buret; cap the flask and mix well. Do this step in the dark and be sure to protect the solution from visible light at all times. Make sure the laser is really “constant” until the next step. Before you are ready to measure the kinetics, use the syringe to place the liquid solution into the 5 cm flash cell; fill it until it is a little over 3/4 full. Turn on the nitrogen gas tank regulator so that you get a flow from the 1/8” tubing with the canula at the end of it. Get it bubbling fairly significantly, as shown here. Purge for 15 min with N2 and then perform the flash experiment; make sure you get three good kinetic traces at room temperature from each of the solutions (methanol and butanol, and either ethanol or 1- propanol as directed by your TA). In another series of experiments, you will measure the temperature dependent kinetics of the benzopinacol formation to calculate an activation barrier. Prepare two more 0.005 M benzophenone solutions in butanol. Store one under ice in the styrofoam cooler. As you have already done the room temperature one (this is what we recommend), add 1 mL of 0.5 M KOH solution to the “cool” butanol solution, and purge it while still under ice as shown here. Then get three good kinetic traces of it. Next, pour out the solution and measure the temperature with a thermometer. Last, do the “hot one”. Place a large beaker of water on the hot plate and heat to a moderate temperature, 40-50 ºC (this is about a setting of “3”). Next, place the last solution in the photolysis cell, place the photolysis cell in the warm water and do the 15 minutes of purging. This is shown below, but remember to actually do it with the lights out. Next, get three good “flashes,” pour out the solution into a beaker and measure the temperature.
Remember that the “hot” solution is kept free of light; I just had to turn on the lights to take this picture!
FAILURE TO PURGE THROUGHLY WILL RESULT IN NO SIGNAL! and you will most likely have to repeat the exercise.
This experiment uses an electronic flash unit to induce the photodissociation of benzophenone. The concentration of the deprotonated ketyl radical with respect to time is monitored with a photodiode which records the absorption of light at 632.8 nm. A scheme of the experimental apparatus is shown on the next page. The following procedure was written by the inventor to collect the data with the computer attached to the photodiode:
FLASH DATA COLLECTION 1. Turn on the power strip. 2a. Turn on the Lab Station power supply. 2b. Turn on the laser (switch is on the back of the unit) and warm up for ~45 min. 3. Plug in GO-LINK to USB Extension Cable 4. Make sure the instrumentation amplifier is plugged into GO-LI NK 5. Set instrumentation amplifier to 0-1V Scale 6. Launch LOGGER PRO software. 7. Go to EXPERIMENT-REMOVE INTERFACES-GO! LINK 7a. Go to EXPERIMENT-CONNECT INTEFACES- GO!LINK USB 8. Go to EXPERIMENT- SETUP SENSORS-SHOW ALL INTERFACES 9. Under GO!_ Select Box-Choose Sensor- Instrumentation Amplifier 10. Close the box- now a Time vs. Potential graph should appear (~0 mV should be displayed) 11. Go to EXPERIMENT-Data Collection 12. Time Based- Set length to 60 seconds 13. Set sampling rate 200 samples\second 14. DONE 15. Adjust the laser beam into the center of sensor block onto the face of the phototransistor 16. Adjust the potential (on screen) to >750mV but <1000 mV by steering the laser beam into the photodiode. 17. Once the solution is purged, place the cell between the laser and the photodiode (and as close to the photodiode as possible) as shown below. The potential on the screen will drop slightly. Adjust the position of the cell to minimize the loss of signal by twisting it. If the solution has turned cloudy, you need to remake it. 18. Power up the flash lamp 19. Push the collect sample button (green box upper right) 20. After 5 seconds PRESS THE RED FLASH BUTTON 21. After flashing, don’t touch anything. If you get a nice return to the baseline, try to flash it again. Shown here are two examples:
If you can get three flashes in one 60 second time period like on the right, your done! 22. If you have a good data set, save it using File – Export, save the data using either text or .csv format If you haven’t purged the solution enough, you won’t see anything but noise. If too much time passes from aligning the sample to collecting data you may need to repurge the sample with N2. SHUT DOWN 1.EXIT Program 2.Unplug GO-LINK from USB Cable 4.Turn off the laser 5.Turn off the power strip
Last, make sure you properly dispose of your solutions. Some hints: 1. Remember your doing five experiments; three solvents at room temperature, another cold, and another hot. Try to get at least three good kinetic traces from the five above. 2. Keep the solution out of light by wrapping it in foil. 3. Thoroughly purge the solutions. 4. Make sure the He-Ne laser is properly aligned and going through the solution. 5. Keep the flash as close as possible to the cell. 6. Remember to keep the lights off.
IV. Lab Report For your three good runs for the benzophenone solutions, you must convert the time dependent laser intensity I(t) to absorbance A(t) via
A(t) = -log[ I(t) / I(0) ] (15) where I(0) is the initial laser intensity before photolysis. An Excel workbook has been prepared for you where all these calculations are done and is posed on the class web site; furthermore, an instruction set for analysis with MATLAB is also provided (this is what we recommend). First, to get I(0) you want to know where the experiment “began” i.e. when you hit go on the flash lamp. I(0) is thus the average of the signal during this time. On the second sheet of the workbook, all points prior to the flash have been removed, the time is corrected such that the first data point has t=0, and the absorbance is calculated via eq. 15. To determine kobs, we need to examine the inverse of the absorbance, which is calculated on the third sheet of the Excel workbook. The difficulty here is that you don’t want to use all the data because at long times the signal is really poor. In the example workbook, you can see how considering the first 330 data points, 150 data points or first 100 data points after the flash affects the results. You must balance the fact that you need to use as many points as possible but you know you must have positive slopes and intercepts. Also, examine how the line fits the data and the “goodness” of the fit (r2). In the example, the first 100 points were sufficient to get a good fit but including more points than 100 resulted in a poor fit. This is shown in detail on sheet 4 in the spreadsheet. Once you have determined what the optimal data set to use is, you can use the slope -1 -1 to calculate kobs. Since A = ·c·l where l is 5 cm and is 5000 M cm , the slope of the line you fit with Excel is equal to kobs/·l. Now you can calculate kobs and make another linear least + squares fit to that data. Since kobs is equal to k[H ]/KD you can now calculate k with the fact that -10 + -14 KD is 6×10 M and [H ] is 2.0×10 . Last two things- just plot k (y-axis) versus solvent viscosity for the three room temperature solutions to determine whether you are examining a situation in the Kramers’ High or Low Friction Regime. Next, determine the activation barrier from a linear plot of ln(k× butanol viscosity) (the y-axis) vs. 1/T (x-axis). The slope of this line is -G*/R. The temperature dependence to the butanol viscosity can be calculated from the following equation: viscosity = 10.8084×exp[-0.0279566×T(K)] kg/m/s.
As for error analysis, the error of the fit in the intercept and slope from a linear least squares method is a well-known formula. Sheet 5 of the Excel workbook has all those formulas there for you to use. You should show at least one example of the kinetic A(t) trace and the linearization (i.e. 1/A(t)) over the data range you think is best to work with. You should tabulate the kobs from all runs as well, with errors. Overall, we do recommend using Matlab, with a comprehensive tutorial for doing so in the handout section on the web site.
YOUR FINAL REPORT MUST ADDRESS THE FOLLOWING ISSUES
* G kB T kB T k e * h G 1) The TST rate equation: what are the units of the e kB T term?
k B T What are the units of the h term? Why does this make sense? 2) Is your reaction occurring in the low or high friction regime according to Kramer’s Theory? 3) What assumptions are inherent in Kramer’s Theory? How would these assumptions affect the data analysis? 4) Do you believe that the reaction you’re studying (ketyl radical cross-coupling) has a relatively high or low barrier? You might want to compare to some other examples.