Chemistry 125: Lecture 36 Bond Energies, the Boltzmann Factor, and Entropy After discussing the classic determination of the heat of atomization of graphite by Chupka and Inghram, the values of bond dissociation energies, and the utility of average bond energies, the lecture focuses on understanding equilibrium and rate processes through statistical mechanics. The Boltzmann factor favors minimal energy in order to provide the largest number of different arrangements of “bits’ of energy. The slippery concept of disorder is illustrated using Couette flow. Entropy favors “disordered arrangements” because there are more of them than there are of recognizable ordered arrangements. Synchronize when the speaker finishes saying “…know the heat of atomization of graphite.” Synchrony can be adjusted by using the pause(||) and run(>) controls. For copyright notice see final page of this file

Graphite to C Atom from Spectroscopy light energy X-Y X + Y H-H kcal/mole (  H f H = 52.1) O=O kcal/mole (  H f O = 59.6) CO kcal/mole X* + Y Maybe this is the observed transition at ? 141?  H f C=O =  H f H 0 2 _ _ _  H f O 0 2 _ _ _ X*’ + Y Or maybe this is the observed transition at ? 125? spectroscopic value precise, but uncertain CO Hf CHf C Hf OHf O graphite O 2 C + O graphite O (  H f C = 171.3)

Atom Energy from Equilibrium K K = e -  E/kT = 10 -(3/4)  E Room Temp = 10 -(3/4)  = ! = 10 -(3/40)  = at 10 x room temperature (~3000K) measure K to find  E < atoms in universe (est) 4

Need to Plot ln( tiny Pressure of C Atoms ) vs. (1/T) at VERY high T " Pressure of C atom  P C = b e -  H f C / RT [C atom ] [C graphite ] -  H f C / RT  e ln ( P C ) = ln ( b ) -  H f C / RT (-  H f C / R ) is the slope of ln ( P C ) vs. (1 / T)

Chupka- Inghram Oven (1955) C n gas Graphite Liner Tantalum Can (mp 3293K!) Tungsten Filament (electrons boil off to bombard and heat tantalum can) Tiny Hole (lets a little gas escape for sampling while maintaining gas-graphite equilibrium)

Chupka- Inghram Oven (1955) C n gas Tantalum Shielding keeps highest heat inside Electron Beam C n BeamC n Ion Beam + C1C1 + C2C2 + C3C3 + Magnetic Field of “Mass Spectrometer” Detected Separately Optical Pyrometer measures oven Temp by color through hole in shielding and quartz window

Heat of Atomization of Graphite (  H f of Carbon Atom) 2450 K2150 K C1C1 C3C3 C2C2

HfHf From Streitwieser, Heathcock, & Kosower William Chupka APPENDIX I HEATS OF FORMATION

from B. Ellison & his friends  Average Bond Energy = / 4 = 99.4 kcal mol -1 No individual bond actually equals the “average” C-H bond. (because of changes in hybridization, etc.) Bond Dissociation Energy (from spectroscopy, etc.) CH 3 -H ± 0.03 CH 2 -H110.4 ± 0.2 CH-H101.3 ± 0.3 C-H80.9 ± 0.2 Sum ± 0.6 Bond Strengths in CH 4 Heat of Atomization of CH 4 = kcal mol -1 (from heat of combustion, etc.) (good experiments!)

 ve Bond Energies Can one sum these average bond energies to get useful "Heats of Atomization” for other molecules? “2 nd C-C bond” 63 kcal/mole “3 rd bond” 54 kcal/mole From Streitwieser, Heathcock, & Kosower “2 nd C-O bond” kcal/mole! (Carbonyl group pretty stable)

 H Atomization by Additivity of Average Bond Energies? Ethene Ave. Bond Energy (kcal/mole) C-C C-H C=C C-O O-H  Bond Energies  H atomization Error kcal/mole Error % c-Hexane c-Hexanol  -Glucose Seems Pretty Impressive! How accurate must you be to be useful? K calc = 10 -(3/4)(  H true +  H error ) K calc = 10 -(3/4)(  H calc ) K calc = K true  10 -(3/4)(  H error ) kcal error not % error determines K error factor To keep error less than  10 need <1.3 kcal/mole error!

 ve Bond Energies Can one sum bond energies to get accurate"Heats of Atomization"? H C O H C C H H H H H C O H C C H H H H Ketone "Enol" C O C H C O C H C=O179 C-C83 C-H99 sum361 C-O86 C=C146 O-H111 sum343 K calc = 10 -(3/4) 18 = K obs = = 10 -(3/4) 9.3 Bonds that change (the others cancel in the difference)

H C O H C C H H H H H C O H C C H H H H Ketone "Enol" H Why is Enol 9 kcal/mole "Too" Stable? O C=O179 C-C83 C-H99 sum361 C-O86 C=C146 O-H111 sum343 K calc = 10 -(3/4) 18 = K obs = = 10 -(3/4) 9.3 C(sp 2 )-H stronger than C(sp 3 )-H (don’t actually cancel) Intramolecular HOMO-LUMO Mixing H C O H C C H H H H + "Resonance Stabilization"

“Constitutional Energy” from bond additivity needs correction for effects such as: Resonance (HOMO/LUMO) Hybridization Strain C H H C H H H vs. HO C CH 2 H sp 2 sp 3 * * Polite name for error

Energy determines what can happen (equilibrium) K = e -  E/kT and how fast (kinetics) = 10 -(3/4)  E room Temp k (/sec) = e -  E /kT ‡ ‡ = (3/4)  E room Temp

What's so great about low energy? Statistics

Gibbs

Exponents & Three Flavors of Statistics 1) The Boltzmann Factor 2) The Entropy Factor 3) The Law of Mass Action

On the Relationship between the Second Law of Thermodynamics and Probability Calculation regarding the laws of Thermal Equilibrium (1877) S = k ln W Ludwig Boltzmann Considered the implications of random distribution of energy.

Random Distribution of 3 “Bits” of Energy among 4 “Containers” How many “complexions” have N bits in the first container? 3 N#N# 3 1 2

Random Distribution of 3 “Bits” of Energy among 4 “Containers” How many “complexions” have N bits in the first container? 6 N#N#

How many “complexions” have N bits in the first container? 0 6 N#N# Random Distribution of 3 “Bits” of Energy among 4 “Containers” 10

0 6 N#N# bits of energy in 20 molecules 3 bits of energy in 4 “molecules” 30 in 20

(N)(N) E  (E) e -E/kT Boltzmann showed Exponential limit for lots of infinitesimal energy bits E ave = 1/2 kT If all “complexions” for a given E total are equally likely, shifting energy to any one degree of freedom of any one molecule is disfavored. By reducing the energy available elsewhere, this reduces the number of relevant complexions. Boltzmann Constant cal/moleK (Note: temperature is average energy)

Exponents & Three Flavors of Statistics 1) The Boltzmann Factor 2) The Entropy Factor 3) The Law of Mass Action

Disorder and Entropy "It is the change from an ordered arrangement to a disordered arrangement which is the source of the irreversibility.” The Feynman Lectures on Physics, Vol. I, 46-7

Disorder and Entropy Which is more ordered?

Disorder, Reversibility, & Couette Flow Click for webpage and "Magic" movie

Couette Flow If disorder is in the eye of the beholder, how can it measure a fundamental property? The rotated state only seemed to be disordered. Top View Ink line Syrup

Entropy is Counting in Disguise. “A disordered arrangement” seems to be an oxymoron. “A disordered arrangement” is code for a collection of random distributions whose individual structures are not obvious. It is favored at equilibrium, because it includes so many individual distributions. The situation favored at equilibrium has particles that have diffused every whichaway. every

Free Energy & entropy units K = e -  G/RT e -(  H - T  S)/RT e -  H /RT e T  S/RT e -  H /RT e  S/R e -  H /RT e R ln 2/R e -  H /RT e ln 2 e -  H /RT x e.u. (R ln 2) is a common  S. Conclusions: e.u. just means a factor of two. K depends on T because of  H, not  S. G (and S) sometimes obscure what is fundamentally simple. e.g. difference in entropy between gauche and anti butane Gauche Anti Gauche

End of Lecture 36 Dec. 8, 2008 Copyright © J. M. McBride Some rights reserved. Except for cited third-party materials, and those used by visiting speakers, all content is licensed under a Creative Commons License (Attribution-NonCommercial-ShareAlike 3.0).Creative Commons License (Attribution-NonCommercial-ShareAlike 3.0) Use of this content constitutes your acceptance of the noted license and the terms and conditions of use. Materials from Wikimedia Commons are denoted by the symbol. Third party materials may be subject to additional intellectual property notices, information, or restrictions. The following attribution may be used when reusing material that is not identified as third-party content: J. M. McBride, Chem 125. License: Creative Commons BY-NC-SA 3.0