Transcript Chem 125 Lecture 10 9/26/07 Preliminary

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
Or maybe this is the observed transition at 257.3?
Maybe this is the observed transition at 257.3?
C+O
DHf C
light
energy
graphite O
X*’+ Y
X* + Y
X +Y
H-H 104.2
kcal/mole (DHf H = 52.1)
_
DHf H 2 __ 0
O=O 119.2
kcal/mole (DHf O = 59.6)
_
DHf O
DHf O 2 __ 0
graphite O2
CO 257.3 kcal/mole (DHf C = 171.3)
CO
X-Y
DHf C=O = -26.4
257.3
257.3
spectroscopic
value precise,
141?
but uncertain
125?
Atom Energy from Equilibrium K
K=
-DE/kT
e
=
-(3/4)DE
10
kcal/mole
@ Room Temp
measure K to find DE
=
4
-(3/4)
10
=
-127
10
!
< 1080 atoms in universe (est)
at 10 x room temperature (~3000K)
=
-(3/40)
10
=
-13
10
Need to Plot ln(tiny Pressure of C Atoms) vs. (1/T)
at VERY high T
[Catom]
" [C
"
]
graphite
 e
-DHf C / RT
Pressure of Catom  PC = b e
-DHf C / RT
ln( PC ) = ln( b ) - DHf C / RT
(-DHf C / R ) is the slope of ln( PC ) vs. (1 / T)
Graphite Liner
Cn
gas
ChupkaInghram
Tiny Hole
(lets a little gas escape for
sampling while maintaining
Oven
gas-graphite equilibrium)
(1955)
Tungsten Filament
Tantalum Can
(mp 3293K!)
(electrons boil off to bombard
and heat tantalum can)
C3+
C2+
Detected +
C
Separately
1
Electron
Beam
Magnetic Field of
“Mass Spectrometer”
C + Ion Beam
n
Cn Beam
Cn
gas
Tantalum
Shielding
keeps
highest heat
inside
ChupkaInghram
Oven
(1955)
Optical
Pyrometer
measures oven
Temp by color
through hole in
shielding and
quartz window
Heat of Atomization
of Graphite
(DHf of Carbon Atom)
DH fC / RT
PC  be
C3
C1
ln PC   ln b 
C
2
2450 K
2150 K
DH fC
RT
APPENDIX I
DHf
HEATS OF FORMATION
William Chupka 1923-2007
From Streitwieser, Heathcock, & Kosower
Bond Strengths in CH4
Heat of Atomization of CH4 = 397.5 kcal mol-1
(from heat of combustion, etc.)
Average Bond Energy = 397.5 / 4 = 99.4 kcal mol-1
Bond Dissociation
(from spectroscopy, etc.)
Energy
CH3 -H
104.99 ± 0.03
CH2-H
110.4 ± 0.2
CH-H101.3 ± 0.3
C-H
Sum
80.9 ± 0.2
No individual bond
actually equals the
“average” C-H bond.
(because of changes in
hybridization, etc.)
from B. Ellison
397.5 ± 0.6(good experiments!)
& his friends
ve Bond Energies
“2nd C-O bond” 90-93 kcal/mole!
(Carbonyl group pretty stable)
“3rd bond” 54 kcal/mole
“2nd C-C bond” 63 kcal/mole
Can one sum these average bond energies
to get useful "Heats of Atomization”
for other molecules?
From Streitwieser, Heathcock, & Kosower
DHAtomization by Additivity of
Average Bond Energies?
Seems Pretty Impressive!
Ave. Bond Energy (kcal/mole)
83 99 146 86 111
C-C C-H C=C C-O O-H
S Bond
Energies
DH
Error
atomization kcal/mole
-4.3
Error
%
Ethene
0
4
1
0
0
542
537.7
-0.8
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
a-Glucose
5
7
0
7
5
2265
2248.9 -16.1
-0.7
How accurate must you be to be useful?
Kcalc = 10-(3/4)(DHcalc)
Kcalc = 10-(3/4)(DHtrue + DHerror)
Kcalc = Ktrue  10-(3/4)(DHerror)
kcal error not % error
determines K error factor
To keep error less than 10
need <1.3 kcal/mole error!
Ketone
"Enol"
O
O
ve Bond Energies
H
H
C
C
C
H
H
H H
H H
H
C
C
C
H
H
H
C=O 179
C-O
86
Bonds that change C-C 83
C=C 146
(the others cancel
C-H
99
O-H 111
in the difference)
sum 361
sum 343
Can one sum bond
energies
to get
-(3/4)
18
-13.5
Kcalc = 10
= 10
accurate"Heats of Atomization"?
Kobs = 10-7 = 10-(3/4) 9.3
Ketone
Why is Enol
9 kcal/mole
"Too" Stable?
C(sp2)-H
stronger than
C(sp3)-H
(don’t actually cancel)
"Enol"
••
O
O
H
H
C
C
C
H
H
H H
H H
H
C
C
C
H
H
H
C=O 179
C-O
86
Intramolecular
C-C
83
C=C 146
+O H
HOMO-LUMO
C-H
O-H 111
Mixing 99
sum 361
sum C 343
"Resonance
H
H
Stabilization"
Kcalc = 10-(3/4) 18 =C10-13.5 C
H
K = 10-7 = 10-(3/4)H9.3 H
obs
“Constitutional Energy” from
bond additivity needs correction
*
for effects such as:
• Resonance
••
HO C
H H
H
vs.
H
C
C
sp
sp
H
H
(HOMO/LUMO)
• Hybridization
• Strain
CH2
2
3
* Polite name
for error
Energy determines what can
happen (equilibrium)
-DE/kT
e
K=
-(3/4)DE
= 10
kcal/mole
@ room Temp
and how fast (kinetics)
‡
13
-DE
/kT
10 e
k (/sec) =
13-(3/4)DE
= 10
‡
kcal/mole
@ room Temp
What's so great
about low energy?
Statistics
Gibbs 1902
1902
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
Considered the
ofimplications
Thermodynamics
and
of random
Probability
distribution Calculation
of energy.
regarding the laws of
Thermal Equilibrium
(1877)
Ludwig Boltzmann
1844 - 1906
S = k ln W
Random Distribution of 3 “Bits” of Energy
among 4 “Containers”
How many
“complexions”
have N bits
in the first
container?
N
3
2
#
1
3
Random Distribution of 3 “Bits” of Energy
among 4 “Containers”
How many
“complexions”
have N bits
in the first
container?
N
3
2
1
#
1
3
6
Random Distribution of 3 “Bits” of Energy
among 4 “Containers”
How many
“complexions”
have N bits
in the first
container?
N
3
2
1
0
#
1
3
6 10
30 in 20
30 bits of energy
in 20 molecules
3 bits of energy
in 4 “molecules”
N
3
2
1
0
#
1
3
6 10
Eave = 1/2 kT
Boltzmann Constant
1.987 cal/moleK
(N)
(E)
(Note: temperature is average energy)
-E/kT
e
E
If all “complexions” for a given Etotal are equally likely,
Boltzmann
limit
shifting energy
to any oneshowed
degree ofExponential
freedom of any one
molecule isfor
disfavored.
reducing the energy
available
lots of By
infinitesimal
energy
bits
elsewhere, this reduces the number of relevant complexions.
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
Syrup
If disorder
is in the eye of
the beholder, how
can it measure a
fundamental
property?
Top
View
Ink
line
The rotated state only seemed to be disordered.
“A disordered arrangement”
seems to be an oxymoron.
The situation favored at equilibrium has
particles that have diffused every whichaway.
“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.
Entropy is Counting in Disguise.
Free Energy & 1.377 entropy units
K = e-DG/RT
e-(DH - TDS)/RT
e-DH /RT e TDS/RT
-DH
/RT
DS/R
e
e
e-DH /RT e R ln 2/R
e-DH /RT e ln 2 Y
e-DH /RT x 2
1.377 e.u. (R ln 2)
is a common DS.
e.g.
Conclusions:
difference in
1.377 e.u. just means
entropy
a factor between
of two.
gauche
and
K depends
on anti
T
butane
because
of
X
X DS.
DH, not
Y
X
G (and S) sometimes
obscure what is
fundamentally simple.Y
Anti
Gauche Gauche
End of Lecture 36
Dec. 8, 2008
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J. M. McBride, Chem 125. License: Creative Commons BY-NC-SA 3.0