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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?