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5. Boltzmann statistic
Basel, 2008
Summary
Supplementary
1.
Introduction
2.
The most probable configuration
3.
Boltzmann relation
4.
Statistical thermodynamics
5.
Applications of Boltzmann relation
material:
German version for this chapter
(Prof. Huber lecture from 2007).
Web
tutorial:
Datenauswertung
Statistik
References:
1.
P. Atkins, P. Atkins, J. de Paula,
“Atkins‘ Physical Chemistry”, Oxford
Univ. Press, Oxford, 8th ed., 2006,
Chapter 16.
2.
Tinoco, K. Sauer, J.C. Wang, J.D.
Puglisi “Physical Chemistry, Principles
and
applications
in
biological
sciences”, Prentice-Hall, New Jersey,
4th ed. 2002, Chapter 11
und
1. Introduction
Almost all chemical properties can be understood by considering the manner in
which the molecules are occupying the energy levels. The total energy of a
system formed by N particles/molecules, E is shared between the
particles/molecules due to their collisions, which:
- redistribute the energy between the molecules
- redistribute the energy between their mode of movement (rotation, vibration,
etc).
Population of the state: each state of the system is characterised by a number
of molecules, Ni with an energy Ei
Aim: to calculate the populations of states for any type of molecules, in any
mode of movement, at any temperature.
1.1 Population of states: characteristics
Characteristics of the populations of states:
- Remain almost constant, even if the identity of the molecules in each state
may change at every collision.
- The molecules
interactions)
are
independent
(we
neglect
the
intermolecular
- Principle of a priori probabilities: all possibilities for the distribution of
energy are equally probable.
Vibration states with Ei are equally
populated as the rotational ones, with Ei.
Total energy of the system is:
E = ∑ Ei
(5.1)
i
Population of states depend only on one parameter: temperature !
1.2 Instantaneous configurations
At any instant the system contains: N0 molecules with E0, N1 with E1, N2
with E2 ...
E0 – zero-point energy (reference
energy, by convention) : E0 = 0
Ei – energy of each molecule
Instantaneous configuration of the system: a set of populations N0, N1, N2, ...
Nk, specified as {N0, N1, ...}, and which has E0, E1, ... energies.
A system of molecules has a
very large number of instantaneous
configurations, which fluctuate with time due to the populations change:
{N, 0, 0, 0, ...}, {N-1, 1, 0, 0, ...}, {N-1, 0, 1, 0, ...}, {N-2, 2, 0, 0, ...}, {N-2, 0, 2, 0, ...},
All molecules in
groundstate
One molecule
is excited
Two molecules
are excited
Weight of the configuration
A system free to switch between groundstate and an excited state will show
properties characteristic almost exclusive to the second configuration, as it is
a more likely state (when the number of molecules, N is high).
A general configuration {N0, N1, ...} can be achieved in W different ways.
Weight of configuration, W: the way in which a configuration can be
achieved.
W = N! /(N0!N1!…Nr!)
(5.2)
Example: Calculate the weight of a configuration in which 20 molecules are
distributed in the arrangement: 0,1,5,0,8,0,3,2,0,1.
W =
20!
= 4 .19 × 10 10
0!1!5!0!8!0!3!2!0!1!
2. The most probable configuration
The most probable configuration is the one which has such a high weight hat
the system will be always found in it and with properties characteristic for this
configuration.
Find the dominating configuration:
(5.3)
W = maximum > dW = 0
Conditions:
- Total energy criterion: take into account only configurations which
correspond to a constant total energy of the system, E:
E = ∑ N i Ei
(5.4)
i
-Total number criterion: total number of molecules is fixed, N.
N = ∑ Ni
i
(5.5)
2.1 Find the most probable configuration
To find a criterion for which W is maximum, is simpler to use lnW and find its
maximum.
ln W = ln
N!
= ln N !−(ln N 0 !+ ln N1!+ K ln N r !) = ln N !−∑ ln N i !
N 0 ! N1!K N r !
i
We simplify the factorials, using Sterling‘s approximation:
ln(n!) = n ln n - n
(5.6)
(5.7)
The approximate expresion of the weight of the configuration is:
ln W = ( N ln N − N ) − ∑ ( N i ln N i − N i ) = N ln N − ∑ N i ln N i
i
i
When a configuration changes so that all Ni
⎞dN
d (ln W ) = ∑ ⎛⎜ δ ln W
⎟ i
N
δ
i
⎝
⎠
i
(5.8)
(5.9)
Ni+dNi, lnW
lnW +d(lnW).
Find the most probable configuration
At a maximum: d(lnW) = 0
(5.10)
and 5.4 and 5.5 are subject to constraints:
∑ E dN
i
i
=0
(5.11)
i
∑ dN
i
=0
(5.12)
i
We will use Lagrange
method of undetermined
multipliers (α, β) to 5.9.
(5.13)
⎞dN + α dN − β E dN
d (ln W ) = ∑ ⎛⎜ δ ln W
⎟ i
∑
∑
i
i
i
δ
N
i
⎠
⎝
i
i
i
⎧
⎞ + α − βE ⎫dN
= ∑ ⎨⎛⎜ δ ln W
⎟
i⎬
i
δ
N
i
⎝
⎠
⎭
i ⎩
As d(Ni) are treated as independent, in order to satisfy 5.10 it is necessary that
for each i:
⎛ δ ln W
⎞ + α − βE = 0
⎜
i
δN i ⎟⎠
⎝
(5.14)
Then Ni have
probable value!
their
most
Find the most probable configuration
Since N is a constant, the diferentiation with respect to Ni gives (using 5.8):
⎞⎫
⎞ ≈ − ⎧δ ⎛ N j ln N j
⎛ δ ln W
⎟⎬
⎜
⎨ ⎜
δN i ⎟⎠ ∑
δ
N
j
⎝
⎠⎭
j ⎩ ⎝
(5.15)
By differentiating the product of the denominator in 5.15 we obtain :
⎛ δ ln W
⎞ = −{ln N + 1} (5.16)
⎜
i
δN i ⎟⎠
⎝
Equation 5.14 becomes:
− (ln N i + 1) + α − β Ei = 0
Because when:
⎛ δN j
i=j
⎜
(5.17)
⎞
⎞ ⎛ δN j
=1
=⎜
⎟
⎟
δN i ⎠ ⎝
δN j ⎠
⎝
⎞ (5.18)
⎛ δ ln N J
⎞ = ⎛ 1 ⎞⎛ δN j
⎜
⎟
⎜
⎜
⎟
δN i ⎠ ⎝ N j ⎠⎝
δN i ⎟⎠
⎝
i≠j
(5.20)
⎛ δN j
⎞
⎜
⎟=0
δ
N
i
⎠
⎝
The most probable population of the state of
energy, Ei is:
(5.19)
N i = eα −1e − βEi
(5.21)
2.2 Boltzmann distribution
Final step: to evaluate constants α and β:
- We use total number criterion 5.5 >
N = ∑ N i = ∑ eα −1e − βEi
i
- We introduce 5.22 in 5.21 and obtain the
number of molecules in a state (Boltzmann
distribution:
(5.22)
i
Ne − βEi
Ni =
− β Ei
e
∑
(5.23)
i
The constant β can be obtained from the mean energy of one molecule, by
taking into account the translation movement of the molecule,
(Ei = pxi2 / 2m): (5.24)
<E> = ΣNi Ei / N = Σ Ei
1
E = 1 =
2 β 2kT
(5.26)
e- βEi
/Σe-
βEi
(5.25)
Ni =
Ne
− Ei
∑e
i
(5.27)
kT
− Ei
kT
2.3 Molecular partition function
The molecular partition function, q represents the summ in the denominator
of the Boltzmann expression for the most probable population (5.27):
q = ∑e
(5.28)
− β Ei
i
The sum is over the
states of an individual
molecule
If several states, gi have the same energy,Ei, the expression for the
molecular partition function is:
gi – multiplicity of the states
q = ∑ g i e − β Ei
(5.29)
i
This function contains all the thermodynamic information about a system of
independent particles/molecules at thermal equilibrium.
3. Boltzmann relation
The distribution of particles energy as function of temperature in a system
formed by a high number of particles:
at T = 0 all particles have the groundstate energy
at T > 0 there are particles with a higher energy than the groundstate-one.
Boltzmann relation for the number of particles(molecules) in a state, En as
function of the number of particles in the groundstate:
−En kT (5.30)
Nn =Noe
N0 = number of particles
with E0 = 0
Nn = number of particles
with En
k = Boltzmann constant
(5.31)
En << kT : Nn ≈ No
(5.32)
En >> kT : Nn ≈ 0
En = kT : N = N0/e
(5.33)
Total number of molecules
Total number of molecules of a system, N, in thermal equilibrium is:
∞
∞
N = N0 + N1 + N2 +…= ∑Ni = ∑Noe−Ei
i= 0
i= 0
∞
kT
= No ∑e−Ei
kT
(5.34)
i= 0
Using Boltzmann relation 5.30 we obtain the ratio of molecules which are
in the state with the energy En:
N n
=
N
e
− E
i= 0
When gn =1
kT
(5.35)
∞
∑
n
e
− E
i
kT
3.1 Proportion of isomers in a system
Which is the probability of each isomer
to be found in the system?
Using 5.30 and taking into account gi we
obtain:
n(gauche) g(gauche) − ∆E
(5.36)
=
⋅ e kT
n(anti)
g(anti)
4
Boltzmann factor ≡ e
∆E
kT
= e
−
0
0
T = 300K
R = kT
∆E
60
120
180
240
Br
Br
300
Br
ϕ 360
F
F
∆ E molar
RT
Example: ∆ E molar = 5 kJ / mol
1-Br-2-F-ethane
2
gi- number of states with the same
energy, Ei (states multiplicity)
−
V(ϕ)
F
≤ 1
gauche
5000
n(gauche) 2 − 8.314⋅
300
= e
= 0.269
1
n(anti)
ngauche + nanti = 100%
anti
gauche
21% gauche
79% anti
If the temperature is increasing, exp(-∆E/kT) is decreasing and thus more
„gauche“ isomers are present !
4. Statistical thermodynamics
THERMODYNAMICS
Chemical & physical properties
(frozen point, melting point, etc)
STATISTICAL
THERMODYNAMICS
QUANTUM
CHEMISTRY
GLOBAL
SIMULATIONS
Micro-domain: atoms, molecules,
bonds, intermolecular interactions...
Calculations
of
the
fundamental
constants (c, Planck constant), definitions
for fundamental properties (nuclear
mass)
4.1 Population distribution-rotation levels
The distribution of molecules with different energy values (discrete energy
levels, equidistant), as function of the population of the groundstate, n0 :
(5.37)
∆E
n i gi − kTi
= e
n 0 g0
ni =
(5.38)
∆E
− i
kT
n0
⋅ gi ⋅ e
g0
ϕ (E ) =
ni
=
n tot
∞
n0
ni
∑n
=
j
j= 0
Where:
n0
g 0 g ie
j= 0
∆E i
∞
g0
∑g
j
e
−
kT
∆E
j
=
ni
n tot
g ie
kT
:
(5.39)
−
∆E i
kT
q
(5.40)
q = ∑ gie
∆E
− i
kT
n3
n2
j= 0
∞
ntot = ∑ n j
−
of
n0 – number of
molecules with E0 = 0
gi- number of states with the same energy, Ei
Relative occupancy of every energy level,
ni – number
molecules with Ei
n1
(5.41)
no
∆E3
∆E2
∆E1
i
Example: vibration energy levels of a diatomic molecule in the harmonic
approximation, or rotation energy levels
5. Applications of Boltzmann relation
> Variation of the pressure as function of high (Barometer formula):
P = P0 e
Mgh
−
RT
(5.42)
g - gravitation acceleration
P – pressure at high h
P0 – pressure at the see level
> Temperature dependence of the reaction rates (Arrhenius law):
k = k 0e
−
∆E A
RT
(5.43)
k - rate constant of the reaction
Ea- activation energy
k0 – pre-exponential factor
Applications of Boltzmann relation
> Fraction of molecules in a gas with velocity components vx in the domain vx
to vx+dvx : one-dimension Maxwell-Boltzmann distribution of molecules
speeds (at a temperature and energy):
ϕ (v x ) =
M
e
2πRT
−
Mv x2
2 RT
(5.44)
> Fraction of molecules in a gas in the velocity range: vx to vx+dvx vy to vy+dvy
and vz to vz+dvz : three-dimension Maxwell-Boltzmann distribution of
molecules speeds (at a temperature and energy):
3
2
Mv 2
−
2
2 RT
⎛ M ⎞
ϕ (v ) = 4 π ⎜
⎟ v e
⎝ 2πRT ⎠
M – molecular mass
(5.45)
T- temperature
R = kT
Applications of Boltzmann relation
> The populations of the energy levels of two isomers (i-Butane and nButane) as function of the temperature:
i - Butan
n - Butan
Low Temperature
> Microscopic temperature definition:
T = −
∆Ei
n
k ln i
n0
i - Butan
n - Butan
High Temperature
(5.46)
5.1 Occupancy of the energy levels
High temperatures:
kT = 2∆E
(5.47)
Ei
3%
8%
5%
11%
9%
14%
14%
17%
24%
19%
39%
15%
- 5%
< 5%
gi = 1
gi = i
At high temperatures the energy levels with higher multiplicity of the states
have a higher occupancy degree, than in the case of energy levels with
multiplicity 1.
When the ∆E is increasing, the occupancy of the higher energy levels is
decreasing.
Occupancy of the energy levels
kT = ∆E
Normal temperatures:
Ei
gi = 1
(5.48)
gi = i
3%
8%
9%
16%
- 5%
23% 29%
< 5%
63% 40%
Low temperatures:
kT = 0.5∆E
(5.49)
Ei
gi = 1
1%
4%
12%
20%
86%
75%
- 5%
gi = i
< 5%
Populations of the rotational energy levels
Example: The relative populations of the rotational energy levels of Co2
• Only states with even J values
are occupied
• The full line shows the
smoothed, averaged population
of levels.
Relative populations of the rotational
energy levels
To understand and learn
- Was ist der Boltzmannfaktor?
-Wann wird im Exponent k, wann R verwendet? Warum?
-Wie hängen k und R zusammen?
-Was kommt in der Boltzmannformel zusätzlich zum Boltzmannfaktor vor?
-Befinden sich bei hoher oder tiefer Temperatur mehr Teilchen in oberen Zuständen?
-Befinden sich bei kleinen oder grossen Quantenabständen mehr Teilchen in oberen Zuständen?
-Was ist die Zustandssumme? Als was kann sie bei der Boltzmann-Verteilung betrachtet werden?
-Wie sieht die Normierungsbedingung bei einer kontinuierlichen bzw. diskreten Verteilung aus?
-Nennen Sie mindestens 3 Anwendungen der Boltzmann-Formel!
-Welche Energie finden Sie im Exponenten der Barometerformel?
-Welche Energie finden Sie im Exponenten der Arrheniusgleichung?
-Welche Energie finden Sie im Exponenten der Maxwell-Boltzmann-Verteilung?
-Worauf beruht der zusätzliche Faktor 4πv2 in der 3- gegenüber der 1-dimensionalen MB-Verteilung?
-Diskutieren Sie die Temperaturabhängigkeit des Gleichgewichts n-Butan i-Butan!
-Warum sind im i-Butan die Niveaux weiter auseinander als im n-Butan?
-Wie ist die mikroskopische Temperatur definiert?
-Wie sind die rotatorische, vibratorische und elektronische Temperatur definiert?
-Was bedeutet es, wenn diese nicht gleich sind?
-Nennen Sie ein Beispiel, wo negative Temperaturen hergestellt werden können! Verletzt das den nullten Hauptsatz der Thermodynamik?
-Worauf beruht die Boltzmann-Verteilung statistisch gesehen?