chemistry 125: lecture 36 bond energies, the boltzmann factor, and entropy after discussing the...
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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
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Graphite to C Atom from Spectroscopy
lightenergy
X-Y
X + Y
H-H 104.2 kcal/mole (Hf H = 52.1)
O=O 119.2 kcal/mole (Hf O = 59.6)
CO 257.3 kcal/mole
X* + YMaybe this is the observed transition at 257.3?
141? 257.3
Hf C=O = -26.4
Hf H 02___
Hf O 02___
X*’+ YOr maybe this is the observed transition at 257.3?
125? 257.3
spectroscopic value precise, but uncertain
CO
Hf C
Hf O
graphite O2
C + O
graphite O
(Hf C = 171.3)
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Atom Energy from Equilibrium K
K = e-E/kT = 10-(3/4)E kcal/mole@ Room Temp
= 10-(3/4)= 10-127 !
= 10-(3/40)= 10-13
at 10 x room temperature (~3000K)
measure K to find E
< 1080 atoms in universe (est)
4
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Need to Plot ln(tiny Pressure of C Atoms) vs. (1/T)at VERY high T
" "
Pressure of Catom PC = b e-Hf C / RT
[Catom]
[Cgraphite]-Hf C / RT e
ln( PC ) = ln( b ) - Hf C / RT
(-Hf C / R ) is the slope of ln( PC ) vs. (1 / T)
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Chupka-Inghram
Oven(1955)
Cn
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)
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Chupka-Inghram
Oven(1955)
Cn
gas
Tantalum Shielding
keeps highest heat
inside
Electron Beam
Cn Beam
Cn Ion Beam+
C1+
C2+C3
+
Magnetic Field of“Mass Spectrometer”
Detected Separately
Optical Pyrometer
measures oven Temp by color through hole in shielding and quartz window
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Heat of Atomization of Graphite
(Hf of Carbon Atom)
2450 K 2150 K
€
PC = be−ΔH fC / RT
€
ln PC( ) = ln b( ) −ΔH fC
RTC1
C3
C2
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Hf
From Streitwieser, Heathcock, & Kosower
William Chupka 1923-2007
APPENDIX I
HEATS OF FORMATION
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from B. Ellison & his friends
Average Bond Energy = 397.5 / 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.)
CH3 -H 104.99 ± 0.03
CH2-H 110.4 ± 0.2
CH-H 101.3 ± 0.3
C-H 80.9 ± 0.2
Sum 397.5 ± 0.6
Bond Strengths in CH4
Heat of Atomization of CH4 = 397.5 kcal mol-1
(from heat of combustion, etc.)
(good experiments!)
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ve Bond Energies
Can one sum these average bond energies to get useful "Heats of Atomization”
for other molecules?
“2nd C-C bond” 63 kcal/mole
“3rd bond” 54 kcal/mole
From Streitwieser, Heathcock, & Kosower
“2nd C-O bond” 90-93 kcal/mole!(Carbonyl group pretty stable)
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HAtomization by Additivity of Average Bond Energies?
Ethene 0 4 1 0 0 542 537.7 -4.3 -0.8
Ave. Bond Energy (kcal/mole)
83 99 146 86 111C-C C-H C=C C-O O-H
BondEnergies
Hatomization
Errorkcal/mole
Error%
c-Hexane 6 12 0 0 0 1686 1680.1 -5.9 -0.4
c-Hexanol 6 11 0 1 1 1784 1778.6 -5.4 -0.3
-Glucose 5 7 0 7 5 2265 2248.9 -16.1 -0.7
Seems Pretty Impressive!
How accurate must you be to be useful?
Kcalc = 10-(3/4)(Htrue + Herror)
Kcalc = 10-(3/4)(Hcalc)
Kcalc = Ktrue 10-(3/4)(Herror)
kcal error not % errordetermines K error factor
To keep error less than 10need <1.3 kcal/mole error!
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ve Bond Energies
Can one sum bond energies to getaccurate"Heats of Atomization"?
H C
O
H
CCH H
H
H
H C
O
H
CCH H
H
HKetone "Enol"
C
O
CH
C
O
C
H
C=O 179
C-C 83
C-H 99
sum 361
C-O 86
C=C 146
O-H 111
sum 343
Kcalc = 10-(3/4) 18 = 10-13.5
Kobs = 10-7 = 10-(3/4) 9.3
Bonds that change(the others cancelin the difference)
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H C
O
H
CCH H
H
H
H C
O
H
CCH H
H
HKetone "Enol"
H
Why is Enol9 kcal/mole
"Too" Stable?
O
C=O 179
C-C 83
C-H 99
sum 361
C-O 86
C=C 146
O-H 111
sum 343
Kcalc = 10-(3/4) 18 = 10-13.5
Kobs = 10-7 = 10-(3/4) 9.3
••
C(sp2)-Hstronger than
C(sp3)-H(don’t actually cancel)
IntramolecularHOMO-LUMO
Mixing
H C
O
H
CCH H
H
H+
"ResonanceStabilization"
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“Constitutional Energy” from bond additivity needs correction
for effects such as:
• Resonance (HOMO/LUMO)
• Hybridization
• Strain
CH
HC
H
HHvs.
HO CCH2
H
• •
sp2 sp3
*
* Polite name for error
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Energy determines what can happen (equilibrium)
K = e-E/kT
and how fast (kinetics)
= 10-(3/4)E kcal/mole@ room Temp
k (/sec) = 1013e-E /kT‡
‡= 1013-(3/4)E kcal/mole@ room Temp
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What's so greatabout low energy?
Statistics
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Gibbs 1902
1902
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Exponents &Three Flavors of Statistics
1) The Boltzmann Factor
2) The Entropy Factor
3) The Law of Mass Action
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On the Relationship between the Second Law of Thermodynamics andProbability Calculationregarding the laws ofThermal Equilibrium
(1877)
S = k ln WLudwig Boltzmann
1844 - 1906
Considered the implications of random distribution of energy.
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Random Distribution of 3 “Bits” of Energy among 4 “Containers”
How many“complexions”
have N bitsin the firstcontainer?
3
N
#
3
1
2
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Random Distribution of 3 “Bits” of Energy among 4 “Containers”
How many“complexions”
have N bitsin the firstcontainer?
6
N
#
123
31
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How many“complexions”
have N bitsin the firstcontainer?
0
6
N
#
123
31
Random Distribution of 3 “Bits” of Energy among 4 “Containers”
10
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0
6
N
#
123
31 10
30 bits of energyin 20 molecules
3 bits of energyin 4 “molecules”
30 in 20
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(N)
E
(E)
e-E/kT
Boltzmann showed Exponential limitfor lots of infinitesimal energy bits
Eave = 1/2 kT
If all “complexions” for a given Etotal 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 Constant1.987 cal/moleK
(Note: temperature is average energy)
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Exponents &Three Flavors of Statistics
1) The Boltzmann Factor
2) The Entropy Factor
3) The Law of Mass Action
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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
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Disorder and Entropy
Which is more ordered?
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Disorder,Reversibility,
& Couette Flow
Click for webpage and "Magic" movie
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CouetteFlow 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
Inkline
Syrup
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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
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Free Energy & 1.377 entropy units
K = e-G/RT e-(H - TS)/RT
e-H /RT e TS/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 2
1.377 e.u. (R ln 2)is a common S.
Conclusions:
1.377 e.u. just meansa factor of two.
K depends on Tbecause of H, not S.
G (and S) sometimesobscure what is
fundamentally simple.
e.g. difference in
entropy between gauche and anti
butane
YX
GaucheY
X
Anti
YX
Gauche
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End of Lecture 36Dec. 8, 2008
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