vibrations, kinetics, thermodynamics and energy€¦ · sources sources: keith laidler, “chemical...

Vibrations, Kinetics, Thermodynamics and Energy

CHE 538 Lecture 21 Wednesday November 1, 2006

The Relationship between Kinetics and Thermodynamics

The Difference?Chemical Kinetics focuses on rates of processes. Chemical Thermodynamics focuses on the stabilities of constituents of a dynamic system at equilibrium.

Dynamic is the key word in the last sentence.Without exchange between the elements making up the system, we can not be certain that the system has minimized its chemical potential energy.

The SimilarityBoth chemical thermodynamics and kinetics depend on the stabilities of species.The Equilibrium is a ratio of two or more rates of chemical processes.

Sources

Sources:Keith Laidler, “Chemical Kinetics,” 3rd ed. 1987 Harper and RowSteinfeld, J. I.; Francisco, J. S.; Hase, W. L. “Chemical Kinetics and Dynamics” 2nd ed. 1999, Prentice Hall. P. W. Atkins in Physical Chemistry, 3rd ed. p 455.

Constants

h: 6.626 x 10-34 J·s; Planck's constant

k: 1.380 x 10-23 J/K; Boltzmann's constant R/NADon’t confuse this k with the rate constant for reactions, kR, or Kc (Keq) an equibrium constant. I will reserve just plain k for Boltzmann.

R: 8.314 J/(mol·K); molar gas constantcal = 4.184 J

(we are going to talk in kcal)NA: 6.022 x 1023 mol-1; Avogadro's number

Reactions occur along vibrational modes 1.

Think about a vibration in a diatomic molecule.We are marking time on the t axis aboveWhat is happening on the Energy axis?

A BA B

A BA B

A B

1 2 3 4 5t =

A B

0


Remember our discussion regarding the applicability of the parabolic versus the Morse potential energy curves in molecular mechanics.

rrab1 2

34

5

E

0 2 4 6

Ener

gy


From the preceding, there is some distance that atoms A and B can obtain at which there is no longer a bond.

To apply the concept to transition states and kinetic theory we need to think more deeply about motions in molecules. Molecular motion is strongly related to the concept of degrees of freedom.

This concept is very neatly described by P. W. Atkins in Physical Chemistry, 3rd ed. p 455. If you look this up, just hit the index for “degrees of freedom”in whatever edition you can get your hands on. It should also be adequately covered in other physical chemistry text books.

The concept of degrees of freedom

We begin by calculating the number of vibrational modes of an N-atom molecule.The total number of coordinates needed to specify the locations of each atom is 3N.

x, y and z for each atom.

Each atom may change its location by varying one of its coordinates so there are 3N degrees of freedom in the molecule.

Counting Vibrational Modes 1

It is profitable to group the degrees of freedom.

Three coordinates are needed to specify the location of the centre of mass of the molecule, and so three of these degrees of freedom correspond to the translational motion of the molecule.

All other atoms must follow the center of mass so they are dependent on this locus.For example: fix the coordinates of the C atom in methane to (0,0,0).The remaining 3N − 3 degrees of freedom are non-translational modes of the methane molecule.


We wiped these degrees of freedom out by fixing the C atom to 0,0,0!Let’s decrease the motion further by preventing CH4 from tumbling.

You can visualize this by fixing a H-C bond of methane on the x axis.

To fix the H1 atom on the x-axis we limit its y and z values to 0.It needs freedom on the x-axis to vibrate. This leaves us with 3N-3-2 degrees of freedom.

H1 C

H

HHx

2

yIn the case of a non-linear molecule like methane, three coordinates are needed to specify its orientation


But we have left methane still able to tumble about the x axis.

Let’s nail this down by limiting the z coordinate of H2 to 0. We have now frozen methane in space. It can’t translate and it can’t tumble; only vibrations are left.

H1 C

H

HHx

2

y

This leaves 3N − 6 degrees of freedom for relative displacements of the atoms in the molecule.

For methane (3x4)-6 = 6 vibrational modes.


For a linear molecule (H-Cl) we only needed to limit Cl to (0,0,0) and H to (x,0,0).

This leaves 3N − 5 degrees of freedom for relative vibrational displacements of the atoms in the molecule.

H2O is a triatomic (N = 3) non-linear molecule, and has three modes of vibration (and three modes of rotation). 3N-6 = vibrational modes; 3x5−6 = 9.


CO2 is triatomic (N = 3) and linear, and has four modes of vibration (and only two modes of rotation). 3x3−5 = 4 vibrational modes.

If you have “Chemscape Chime by MDL Information Systems” on the computer you are at you can see these modes at the following site.

http://www.chem.purdue.edu/gchelp/vibs/co2.html

Statistical Mechanics 1

In a reaction (eq. 01)

aA + bB + . . . . . . . . . yY + zZ

The equilibrium constant, (eq. 02)E0 = the total energy change (at absolute zero). This is only a value we can approximate. It is not rigorously measurable since 0 K is not attainable. q = partition function.

gi is the degeneracy of the ith state

( ) RTEbB

aA

zZ

yYc eqqqqK /0..../... −⋅=

ii

iqgq ∑≡

Statistical Mechanics 2gi is the degeneracy of the ith state

For the ground state of Butane at a certain temperature:The anti conformer has g = 1. The gauche conformer has g = 2.

(eq. 04)lower energies count more at low temperature.

q is an expression of stability. Ask yourself what happens to the value of q as the bond strength decreases in a diatomic molecule.q decreases

kT

ii

iegq /ε−∑≡

Statistical Mechanics 3qtot = qeqvqrqt. The partition function can be broken down into contributions from

electronic, vibrational, rotational and translational motions.The equilibrium (hence the free energy) for any reaction can be calculated by knowing the partition functions for all the modes of molecular motion starting from the zero point energy! The vibrational components of the partition function will allow a rate theory for the plain vanilla organic reaction.

Thermodynamic Formulation of Conventional Transition State Theory

A stationary state is in equilibrium with a species that does not exist, the transition state.

This is okay for two reasons. (1) On the other side of the transition state there is a product to catch the material.

We can write an equilibrium involving the reactant and the t-state and the product and the t-state in which the t-state cancels out.

(2) We can assume that the t-state is a very unstable intermediate possessing a lifetime far shorter than our ability to detect it. This way we can do the math without having issues with division by zero.

This approximation puts a very low concentration of the transition state in solution. This approach is really the same thing we are doing in item (1) above.

Transition State Theory 1

In a unimolecular chemical reaction, R ----> P, with X¥ = the transition state.

(eq. 05)

(eq. 06)

R XKc

RTERc eqq

RXK /¥

¥¥ 0/

][][ −⋅==


If R is a non-linear, N-atom molecule then is has 3N-6 vibrations.

One of these vibrations is different than the rest since it corresponds to the reaction coordinate.

The transition state has 3N-7 normal vibrations. The vibrational partition function has to reflect this fact.

The partition function for a vibration is 1/(1 − e−hν/kT)ν = frequency of the vibration.The limit of this function has to be calculated as νapproaches 0. Why?


As the force holding the two atoms together increases the ν increases. Likewise as the force decreases νdecreases.

νννhkT

e kTh /1

1lim /0=

− −→

(eq. 07)

Transition State Theory 4The approximation gets better as ν is smaller

0 . 1 - 4 . 0

0

2

4

6

8

1 0

1 2

0 1 2 3 4 5

1 / ( 1 - e ^ - x )

1 / x

0 . 0 1 - 0 . 4

0

2 0

4 0

6 0

8 0

1 0 0

1 2 0

0 0 . 2 0 . 4 0 . 6

1 / ( 1 - e ^ - x )

1 / x

1 . 0 - 4 0

00 . 20 . 40 . 60 . 8

11 . 21 . 41 . 61 . 8

0 2 0 4 0 6 0

1 / ( 1 - e ^ - x )

1 / x

The functions (1-e-1/x)-1 and 1/x above are compared for various values for x (left: the smallest values) You can see that with smaller arguments the functions look very much alike.

Transition State Theory 5Now we can write a 2-part expression for equation 06 that accounts for:

the reaction coordinate in the transition state the ‘stationary state’ portion of the transition state

We will denote the partition function for the stationary state portion of the transition as q¥. Thus,q¥ = kT/hν • q¥. (eq. 08) and (eq. 06) becomes

(eq. 09)

RTERc eqq

RXK /

¥

¥¥ 0/)kT/h(

][][ −⋅== ν

Transition State Theory 6(eq. 10) RTE

R eqqRX /¥

¥ 0/][kT/h)(][ −⋅=ν

ν is the vibrational frequency corresponding to the elongating bond along the reaction coordinate.

Here is the part that will mess with your mind. The left side of (eq. 10) is the rate of the reaction!!! It is a frequency with no return on the reaction coordinate and it is multiplied by the concentration of the transition state. The units certainly are right, M/sec.

Accepting that this entity is indeed a rate of conversion:

][/][kT/h)(][ /¥

0 RkeqqRdtRd

RRTE

R =⋅= −(eq. 11)


(eq. 12)

A portion of (eq. 11) is stationary: (eq. 13)

RTGeq

RTER eKeqq /¥¥/

¥0/ ∆−− ==⋅

RkR

P

R

P

X

Reaction Coordinate 1

int

¥

By putting an imaginary barrier at X¥ we can posit that X¥ is in equilibrium with R.A portion of (eq. 11) is stationary:


The bit in (eq. 10) and (eq. 11) that turns the argument into a kinetic one is the

h)kT/( νThis is the contribution of the reaction coordinate.Mass is leaking over the barrier via bond breakage.

This is not really equilibrium.All the mass that becomes X¥ can not return to R.

Combining (eq. 1) and (eq. 13) gives


Various forms of the same stuff

][][kT/h)(][ /¥

RkRedtRd

RRTG == ∆−

RTGR ek /¥

kT/h)( ∆−= RSRTHR ek // ¥¥

kT/h)( ∆+∆−=

RSRTHR eek // ¥¥

kT/h)( ∆∆−=

/RTH- /RSkT/h)ln(ln ¥¥ ∆∆+=Rk

Chemical Equilibria

Keq

RTGR ek /1

1 kT/h)( ∆−=

RTGR ek /2

2 kT/h)( ∆−=

RTrxnGe /)(∆−

rxnGln ∆=− eqKRTR

P

X

Reaction Coordinate 1

int

¥

vibrations, kinetics, thermodynamics and energy€¦ · sources sources: keith laidler, “chemical...

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