is the Michaelis constant. It represents the apparent dissociation constant of ES to E and S.

Transcription

1 Lecture 35 Chapt 28, Sections 1-4 Bimolecular reactions in the gas phase Anouncements: Exam tomorrow 2:00 is the primary time. vdw 237 I have gotten several suggestions for lecture ideas, thanks and keep them coming. Outline: hard sphere collision energy dependence impact parameter internal energy distribution Review Enzyme catalysis reaction rate linear at low [S] saturates at high [S] (saturating rate called v max ) Michaelis-Menten mechanism excess substrate, so d[es]/dt = 0 initial rate, so [P] = 0 then where is the Michaelis constant. It represents the apparent dissociation constant of ES to E and S. Often find these things with a Lineweaver-Burke plot 1

2 Crossed molecular beam studies allowed researchers to gain tremendous insight with incredible detail into how reactions occur in the gas phase. This was an area of tremendous development in the 70s and 80s and is still very much active today. We will just scratch the surface of a couple of reactions over the next couple days and get you familiar with development of some simple gas-phase reaction theory and some terminology. Hard-Sphere Collision Theory If we have the simple reaction we expect the rate to be given by? As with all theoretical development, we will start with the simplest possible description that every collision between A and B yields a successful chemical reaction. If this is the case, then we should be able to predict the rate of reaction from our kinetic theory of gases. v = collision rate = Z AB = σ AB <u r > ρ A * ρ B * Concentration and number density are the same thing within some unit conversions, so we can see that the reaction constant is just the collision crosssection times the relative velocities. k = σ AB < u r > For instance, in the reaction H 2 + C 2 H 4 C 2 H 6 σ AB = πd 2 AB = π{½(270 pm pm)} 2 = m 2 (the diameters come from way back in table 25.3) and 2

3 Plugging in for k B and T (298 K) gives <u r > = m/s Finally, k = (1000 dm 3 /m 3 )( mol -1 )( m 2 )( m/s) = dm 3 /(mol s) The experimental value is dm 3 /(mol s), so we aren t looking too good with our model off by over 35 orders of magnitude. But, this really is no surprise since we assumed every single collision lead to a reaction. More complex models of reaction cross-section We want to make an improvement, but again we should try to start simple. We really want σ to be about when a reaction occurs, not just when a collision occurs. It makes sense, then for the reaction cross-section to depend on the relative speeds of the molecules. k(u r ) = σ(u r ) u r or This integral we did before in several forms way back in chapter 25 (except for the σ bit). But, now we would are more interested in energy than speed. 3

4 Relative kinetic energy: E r = ½µ u r 2 (Big Equation!!!) This is the equation we will be using for the rest of the chapter. It allows us to test various models for the reaction cross-section against experimental rate constants. Energy threshold for Lets assume that all collisions with relative kinetic energy above a cutoff lead to reactions and all below lead to nothing. So, a simple step function: Evaluating the above integral with this function gives: Notice that we now have an Arrhenius-looking expression with E 0 playing the role of the activation energy. If we go back to the H 2 + C 2 H 4 C 2 H 6 example, we can set this all equal to the experimental k and solve for E 0. We then get E 0 = 223 kj/mol. Experiment is 180 kj/mol. Doesn t look too terrible, certainly much better than our first model, but remember that k depends strongly on E 0, so the T-dependence of the rate would be pretty poor. 4

5 Impact Parameter Let s consider a little bit more complexity. Think of two molecules hitting head on, or just grazing each other. The amount of collision energy available for reaction is very different in these two cases. We can define an Impact Parameter to help describe the difference in these two cases. So, if b (the impact parameter) is big, no collision or glancing collision. What is b for head on collision? b approaches 0. In the Line of Centers model (loc), we assume that only kinetic energy that lies along the line of centers contributes to the reaction. This is a complicated bit of geometry, but the solution is: What is the σ behavior? if E r just equals E 0, we get σ =0. As E r gets really big then σ approaches πd 2 every collision leads to reaction. Plugging this into our trusty k integral equation gives: 5

6 Note that this looks even more like the Arrhenius equation. In fact, in this equation, E a = E 0 + ½k B T and A = u r σ AB e ½ (there is some T dependence in u r, so E a is not as simple as E 0.) We have also done a better job comparing to experiment. However, the shape of our cross-section function does not really compare very well to experimental measurements and we are still getting rate constants that are too high by a couple/few orders of magnitude: Reaction Expmnt A (L mol -1 s -1 ) Calculated A NO +O 3 NO 2 + O ClO Cl 2 + O H 2 + C 2 H 4 C 2 H So, we are still making it too easy for molecules to react. One factor to consider is the orientation of the molecules when they are colliding. This is surely important, but your book doesn t really treat it in detail, so we won t either. It turns out this is still not enough. Internal Energy Distribution One thing we will consider in detail is the way energy is distributed internally in the reactants. Look at how the cross section depends on the total energy of reactants along with the H2+ vibrational state. H He HeH + + H (See Fig 28.4 from McQaurrie and Simon below) 6

7 The total energy includes the translational and vibrational (and rotational) energies of the reactants. - vibrational levels 0-3 have an energy less than E 0, so additional transitional energy is needed to induce reaction. That s why we see a threshold. - vibrational levels > 4 already have more than E 0 energy so they react even with no additional energy. Thus, it isn t just the total energy that matters, it matters how that energy is distributed in the molecule. Lots of vibrational energy means more reactive. 7