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Kramers' Theory of Reaction Kinetics

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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.
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