5. Boltzmann statistic

Basel, 2008

Summary

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

1. Introduction

2. The most probable configuration

3. Boltzmann relation

4. Statistical thermodynamics

5. Applications of Boltzmann relation

Supplementary material:German version for this chapter (Prof. Huber lecture from 2007).

Web tutorial: Statistik undDatenauswertung

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 are independent (we neglect the intermolecular interactions)

- 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: ∑=i

iEE

Population of states depend only on one parameter: temperature !

(5.1)

1.2 Instantaneous configurationsAt any instant the system contains: N0 molecules with E0, N1 with E1, N2with E2 ...

Instantaneous configuration of the system: a set of populations N0, N1, N2, ...Nk, specified as {N0, N1, ...}, and which has E0, E1, ... energies.

E0 – zero-point energy (reference energy, by convention) : E0 = 0

Ei – energy of each molecule

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 ingroundstate

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!)

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.

101019.4!1!0!2!3!0!8!0!5!1!0

!20×==W

(5.2)

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: W = maximum > dW = 0

Conditions:

- Total energy criterion: take into account only configurations which correspond to a constant total energy of the system, E:

-Total number criterion: total number of molecules is fixed, N.

∑=i

iNN

ii

iENE ∑=

(5.3)

(5.4)

(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.

( ) ∑−=++−==i

irr

NNNNNNNNN

NW !ln!ln!ln!ln!ln!ln!!!

!lnln 1010

KK

We simplify the factorials, using Sterling‘s approximation: ln(n!) = n ln n - n

The approximate expresion of the weight of the configuration is:

( ) ( ) ∑∑ −=−−−=i

iii

iii NNNNNNNNNNW lnlnlnlnln

When a configuration changes so that all Ni Ni+dNi, lnW lnW +d(lnW).

( ) ii i

dNNWWd ∑ ⎟

⎠⎞⎜

⎝⎛= δδ lnln

(5.6)

(5.7)

(5.8)

(5.9)

Find the most probable configuration

At a maximum: d(lnW) = 0

and 5.4 and 5.5 are subject to constraints:

0

0

=

=

∑

∑

ii

ii

i

dN

dNE

(5.10)

(5.11)

We will use Lagrange method of undetermined multipliers (α, β) to 5.9.

( )

ii

ii

ii

ii

iii i

dNENW

dNEdNdNNWWd

∑

∑∑∑

⎭⎬⎫

⎩⎨⎧ −+⎟

⎠⎞⎜

⎝⎛=

−+⎟⎠⎞⎜

⎝⎛=

βαδδ

βαδδ

ln

lnln(5.12)

As d(Ni) are treated as independent, in order to satisfy 5.10 it is necessary that for each i:

0ln =−+⎟⎠⎞⎜

⎝⎛

ii

ENW βαδ

δ

(5.13)

(5.14)

Then Ni have their most probable value!

Find the most probable configuration

Since N is a constant, the diferentiation with respect to Ni gives (using 5.8):

∑⎭⎬⎫

⎩⎨⎧

⎟⎠⎞

⎜⎝⎛−≈⎟

⎠⎞⎜

⎝⎛

j j

jj

i NNN

NW

δδδδ lnln (5.15)

By differentiating the product of the denominator in 5.15 we obtain :

{ }1lnln +−=⎟⎠⎞⎜

⎝⎛

ii

NNW

δδ (5.16) Because when:

i = j

i ≠ j

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞⎜

⎝⎛

=⎟⎠⎞

⎜⎝⎛=⎟

⎠⎞

⎜⎝⎛

i

j

ji

J

j

j

i

j

NN

NNN

NN

NN

δδ

δδ

δδ

δδ

1ln

1

0=⎟⎠⎞

⎜⎝⎛

i

jN

Nδ

δEquation 5.14 becomes:( ) 01ln =−++− ii EN βα

iEi eeN βα −−= 1The most probable population of the state of

energy, Ei is:

(5.17)

(5.18)

(5.19)(5.20)

(5.21)

2.2 Boltzmann distributionFinal step: to evaluate constants α and β:

- We use total number criterion 5.5 >

- We introduce 5.22 in 5.21 and obtain the number of molecules in a state (Boltzmanndistribution:

iE

iii eeNN βα −−∑∑ == 1

(5.22)

∑ −

−

=

i

E

E

i i

i

eNeN β

β

(5.23)

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):

<E> = ΣNi Ei / N = Σ Ei e- βEi / Σ e - βEi (5.25)

(5.24)

kTE

21

21 == β

(5.26) ∑−

−

=

i

kTE

ikTiE

i

e

NeN

(5.27)

2.3 Molecular partition function

∑ −=i

Eieq β

The molecular partition function, q represents the summ in the denominator of the Boltzmann expression for the most probable population (5.27):

This function contains all the thermodynamic information about a system of independent particles/molecules at thermal equilibrium.

(5.28)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:

∑ −=i

Ei

iegq β (5.29)gi – multiplicity of the states

3. Boltzmann relation

En << kT : Nn ≈ No

En >> kT : Nn ≈ 0

En = kT : N = N0/e

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.

Nn =Noe−En kT

Boltzmann relation for the number of particles(molecules) in a state, En as function of the number of particles in the groundstate:

N0 = number of particles with E0 = 0

Nn = number of particles with En

k = Boltzmann constant

(5.30) (5.31)

(5.32)

(5.33)

Total number of molecules

N = N0 + N1 + N2 +…= Ni

i=0

∞

∑ = Noe−Ei kT

i=0

∞

∑ = No e−Ei kT

i=0

∞

∑

N n

N= e − E n kT

e − E i kT

i = 0

∞

∑

Using Boltzmann relation 5.30 we obtain the ratio of molecules which are in the state with the energy En:

Total number of molecules of a system, N, in thermal equilibrium is:

(5.35)

(5.34)

When gn =1

3.1 Proportion of isomers in a system

n gauche( )n anti( )

=g gauche( )

g anti( )⋅e

−∆E

kT

≡ e−

∆ E

kT = e−

∆ E molar

RT ≤ 1

∆ E molar = 5 kJ / mol

Which is the probability of each isomer to be found in the system?

Br

F

Br

F

Br

F

gauche gaucheanti

0 60 120 180 240 300 3600

2

4

ϕ

V(ϕ)

∆E

Using 5.30 and taking into account gi we obtain:

gi- number of states with the same energy, Ei (states multiplicity)

Boltzmann factor

Example:

T = 300Kn gauche( )

n anti( )=

2

1e

−5000

8.314⋅300 = 0.269 21% gauche 79% anti

If the temperature is increasing, exp(-∆E/kT) is decreasing and thus more „gauche“ isomers are present !

(5.36)

1-Br-2-F-ethane

R = kT ngauche + nanti = 100%

4. Statistical thermodynamics

THERMODYNAMICS

STATISTICAL THERMODYNAMICS

QUANTUM CHEMISTRY

GLOBAL SIMULATIONS

Calculations of the fundamental constants (c, Planck constant), definitions for fundamental properties (nuclear mass)

Micro-domain: atoms, molecules, bonds, intermolecular interactions...

Chemical & physical properties (frozen point, melting point, etc)

4.1 Population distribution-rotation levels

ni

n0

=gi

g0

e−

∆Ei

kT

ntot = n j

j= 0

∞

∑

ni =n0

g0

⋅ gi ⋅e−

∆Ei

kT

n i

n tot

ϕ E( ) = n i

n tot

= n i

n j

j = 0

∞

∑=

n 0

g 0g ie

− ∆ E ikT

n 0

g 0g j e

−∆ E j

kT

j = 0

∞

∑= g ie

− ∆ E ikT

q

no

n1

n2

n3

∆E1

∆E2

∆E3

The distribution of molecules with different energy values (discrete energy levels, equidistant), as function of the population of the groundstate, n0 :

gi- number of states with the same energy, Ei

ni – number of molecules with Ei

n0 – number of molecules with E0 = 0

Relative occupancy of every energy level, :

Where: ∑∆

−=

i

kTE

i

i

egq

(5.37) (5.38)

(5.40)

(5.39)

(5.41)

Example: vibration energy levels of a diatomic molecule in the harmonicapproximation, or rotation energy levels

5. Applications of Boltzmann relation

P = P0e−

Mgh

RT

k = k0e−

∆E A

RT

> Variation of the pressure as function of high (Barometer formula):

> Temperature dependence of the reaction rates (Arrhenius law):

g - gravitation acceleration

P – pressure at high h

P0 – pressure at the see level

(5.43)

(5.42)

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 vxto vx+dvx : one-dimension Maxwell-Boltzmann distribution of molecules speeds (at a temperature and energy):

ϕ v x( ) =M

2πRTe

−Mv x

2

2 RT

(5.44)

(5.45)

M – molecular mass

T- temperature

R = kT

ϕ v( ) = 4π M

2πRT

⎛ ⎝ ⎜

⎞ ⎠ ⎟

32

v 2e−

Mv 2

2 RT

> Fraction of molecules in a gas in the velocity range: vx to vx+dvx vy to vy+dvyand vz to vz+dvz : three-dimension Maxwell-Boltzmann distribution of molecules speeds (at a temperature and energy):

Applications of Boltzmann relation

i - Butan n - Butan i - Butan n - Butan

Low Temperature High Temperature

> The populations of the energy levels of two isomers (i-Butane and n-Butane) as function of the temperature:

> Microscopic temperature definition:

0

lnnnk

ETi

i∆−= (5.46)

5.1 Occupancy of the energy levels

High temperatures: kT = 2∆EEi

gi = 1 gi = i

39%

3%

9%

5%

14%

24%

15%

8%

14%

11%

17%

19%

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.

(5.47)

When the ∆E is increasing, the occupancy of the higher energy levels is decreasing.

- 5%

< 5%

Occupancy of the energy levels

Low temperatures: kT = 0.5∆E

Normal temperatures: kT = ∆E

63%

3%

9%

23%

40%

8%

16%

29%

Ei

Ei1%

12%

86%

4%

20%

75%

gi = 1

gi = 1

gi = i

gi = i

(5.48)

(5.49)

- 5%

< 5%

- 5%

< 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?