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Lecture 35 Chapt 28, Sections 1-4

Bimolecular reactions in the gas phase

Anouncements: Exam tomorrow 2:00 is the primary time. vdW 237I have gotten several suggestions for lecture ideas, thanks and keep them coming.

Outline:

hard sphere collisionenergy dependenceimpact parameterinternal energy distribution

Review

Enzyme catalysis

reaction rate linear at low [S]

saturates at high [S] (saturating rate called vmax)

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

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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 = ZAB = σAB <ur> ρA*ρB

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Concentration and number density are the same thing within some unit

conversions, so we can see that the reaction constant is just the collision cross-

section times the relative velocities.

k = σAB < ur >

For instance, in the reaction H2 + C2H4 Y C2H6

σAB = πd2AB = π{½(270 pm + 430 pm)}2 = 3.85 × 10-19 m2

(the diameters come from way back in table 25.3)

and

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Plugging in for kB and T (298 K) gives <ur> = 1.83 × 103 m/s

Finally, k = (1000 dm3/m3)(6.022×1023 mol-1)(3.85 × 10-19 m2)(1.83 × 103 m/s)

= 4.24 × 1011 dm3/(mol @s)

The experimental value is 3.49 × 10-26 dm3/(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(ur) = σ(ur) +ur,

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.

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Relative kinetic energy: Er = ½µ ur2

(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 F

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 E0 playing

the role of the activation energy.

If we go back to the H2 + C2H4 Y C2H6 example, we can set this all equal to

the experimental k and solve for E0. We then get E0 = 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 E0, so the T-dependence of the rate would

be pretty poor.

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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 Er just equals E0, we get σ =0. As Er gets really

big then σ approaches πd2 – every collision leads to reaction.

Plugging this into our trusty k integral equation gives:

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Note that this looks even more like the Arrhenius equation. In fact, in this

equation, Ea = E0 + ½kBT and A = +ur,σABe½

(there is some T dependence in ur, so Ea is not as simple as E0.)

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 +O3 6 NO2 + O2 7.94 × 108 5.01 × 1010

2ClO 6 Cl2 + O2 6.31 × 107 2.50 × 1010

H2 + C2H4 6 C2H6 1.24 × 106 7.30 × 1011

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.

H2+ + He Y HeH+ + H

(See Fig 28.4 from McQaurrie and Simon below)

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The total energy includes the

translational and vibrational (and

rotational) energies of the

reactants.

- vibrational levels 0-3 have

an energy less than E0, so

additional transitional energy

is needed to induce reaction.

That’s why we see a

threshold.

- vibrational levels > 4

already have more than E0

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.