G. H. CHEN Department of Chemistry University of Hong Kong Intermediate Physical Chemistry

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Intermediate Physical Chemistry Contents: Distribution of Molecular States Perfect Gas Diatomic Molecular Gas Fundamental Relations

Distribution of Molecular States  Configurations and Weights  Boltzmann Distribution, and Physical Meanings of   Molecular Partition Function and its Interpretation  The Internal Energy and the Entropy  Independent Molecular and their Partition Function

Perfect Gas  Partition Function  Energy  Heat Capacity  Pressure and the gas law

Diatomic Molecular Gas  Factorization of Partition Function  Rotational Partition Function  Vibrational Partition Function  Electronic Partition Function  Mean Energy and Heat Capacity

Fundamental Relations  Helmholtz Energy  Pressure  Enthalpy  Gibbs Energy

Statistical Mechanics provides the link between the microscopic properties of matter and its bulk properties. Statistical Mechanics:

THE DISTRIBUTION OF MOLECULAR STATES Consider a system composed of N molecules, and its total energy E is a constant. These molecules are independent, i.e. no interactions exist among the molecules. Countless collisions occur. It is hopeless to keep track positions, momenta, and internal energies of all molecules.

Population of a state For example, for 1000 coins, Population for the head is about 500, and that for the tail is about 500. THE DISTRIBUTION OF MOLECULAR STATES The average number of molecules occupying a state. Denote the energy of the state  i If we have only 10 coins, what is the population of head and tail, and why?

Principle of equal a priori probabilities: All possibilities for the distribution of energy are equally probable provided the number of molecules and the total energy are kept the same. That is, we assume that vibrational states of a certain energy, for instance, are as likely to be populated as rotational states of the same energy. THE DISTRIBUTION OF MOLECULAR STATES

For instance, four molecules in a three-level system: the following two conformations have the same probability l-l  l  l   l THE DISTRIBUTION OF MOLECULAR STATES

Configurations and Weights Imagine that there are total N molecules among which n 0 molecules with energy  0, n 1 with energy  1, n 2 with energy  2, and so on, where  0 <  1 <  2 <.... are the energies of different states. The specific distribution of molecules is called configuration of the system, denoted as { n 0, n 1, n 2,......} THE DISTRIBUTION OF MOLECULAR STATES

For instance, a system with 17 molecules, and each molecule has four states. The above configuration is thus, { 4, 6, 4, 3 } THE DISTRIBUTION OF MOLECULAR STATES

{N, 0, 0,......} corresponds that every molecule is in the ground state, there is only one way to achieve this configuration; {N-2, 2, 0,......} corresponds that two molecule is in the first excited state, and the rest in the ground state, and can be achieved in N(N-1)/2 ways. A configuration { n 0, n 1, n 2,......} can be achieved in W different ways, where W is called the weight of the configuration. And W can be evaluated as follows, W = N! / (n 0 ! n 1 ! n 2 !...) THE DISTRIBUTION OF MOLECULAR STATES

Justification 1.N! different ways to arrange N molecules; 2.n i ! arrangements of n i molecules with energy  i correspond to the same configuration; THE DISTRIBUTION OF MOLECULAR STATES

1.Calculate the number of ways of distributing 3 objects a, b and c into two boxes with the arrangement {1, 2}. Answer: | a | b c |, | b | c a |, | c | a b |. Therefore, there are three ways 3! / 1! 2! Example: THE DISTRIBUTION OF MOLECULAR STATES | a | c b |,| b | a c |,| c | b a | To eliminate overcounting of these configurations

2.Calculate the number of ways of distributing 20 objects into six boxes with the arrangement {1, 0, 3, 5, 10, 1}. Answer: 20! / 1! 0! 3! 5! 10! 1! = note: 0! = 1 Example: THE DISTRIBUTION OF MOLECULAR STATES

Stirlings Approximation: When x is large, ln x!  x ln x - x xln x! x ln x - xln A THE DISTRIBUTION OF MOLECULAR STATES Note: A = (2  ) 1/2 (x+1/2) x e -x

Therefore, ln W  ( N ln N - N ) -  ( n i ln n i - n i ) = N ln N -  n i ln n i xln x! x ln x - xln A Stirlings approximation(cont’d): THE DISTRIBUTION OF MOLECULAR STATES

The Dominating Configuration Imagine that N molecules distribute among two states. {N, 0}, {N-1, 1},..., {N-k, k},..., {1, N-1}, {0, N} are possible configurations, and their weights are 1, N,..., N! / (N-k)! k!,..., N, 1, respectively. For instance, N=8, the weight distribution is then

The Dominating Configuration

The Dominating Configuration

The Dominating Configuration

When N is even, the weight is maximum at k = N/2, i.e. W k=N/2 = N! / [N/2)!] 2. When N is odd, the maximum is at k = N/2  1 As N increases, the maximum becomes sharper! The weight for k = N/4 is W k=N/4 = N! / [(N/4)! (3N/4)!] The Dominating Configuration

| N | 4 | 8 | 16 | 32 | 256 | 6.0 x |R(N) | 1.5 | 2.5 | 7.1 | 57.1 | 3.5 x | 2.6 x 10 3e+22 Therefore, for a macroscopic molecular system ( N ~ ), there are dominating configurations so that the system is almost always found in the dominating configurations, i.e. Equilibrium The ratio of the two weights is equal to R(N)  W k=N/2 / W k=N/4 =(N/4)! (3N/4)! / [(N/2)!] 2 The Dominating Configuration If the system has more states, would we reach the same conclusion as above? Why?

To find the most important configuration, we vary { n i } to seek the maximum value of W. But how? E.g. One-Dimensional Function: F(x) = x 2 dF/dx = 0 The Dominating Configuration Enough?

Two-Dimensional Case: for instance, finding the minimum point of the surface of a half water melon F(x,y).  F/  x = 0,  F/  y = 0. Multi-Dimensional Function: F(x 1, x 2, …, x n )  F/  x i = 0, i = 1,2,…,n To find the maximum value of W or lnW,  lnW /  n i = 0,i=1,2,3,... The Dominating Configuration

1. The total energy is a constant, i.e.  n i  i = E = constant 2. The total number of molecules is conserved, i.e.  n i = N = constant How to maximize W or lnW under these constraints? Two constraints for the system

Let’s investigate the water melon’s surface: cutting the watermelon how to find the minimum or maximum of F(x, y) under a constraint x = a ? L = F(x, y) - x  L /  x = 0  L /  y = 0 x = a The method of Lagrange Multipliers

Generally, to minimize or maximize a function F(x 1, x 2, …, x n ) under constraints, C 1 (x 1, x 2, …, x n ) = Constant 1 C 2 (x 1, x 2, …, x n ) = Constant 2. C m (x 1, x 2, …, x n ) = Constant m L = F(x 1, x 2, …, x n ) -  i i C i (x 1, x 2, …, x n )  L/  x i = 0,i=1,2,..., n The Dominating ConfigurationThe method of Lagrange Multipliers

JUSTIFICATION dL = dF -  i i dC i under the constraints, dC i = 0, thus dF = 0 i.e., F is at its maximum or minimum. The Dominating ConfigurationThe method of Lagrange Multipliers

The method of Lagrangian Multiplier Procedure Construct a new function L, L = lnW +   i n i -   i n i  i Finding the maximum of L by varying { n i },  and  is equivalent to finding the maximum of W under the two constraints, i.e.,  L/  n i =  lnW/  n i +  -  i = 0

Since ln W  ( N ln N - N ) -  i ( n i ln n i - n i ) = N ln N -  i n i ln n i  lnW/  n i =  (N ln N)/  n i -  (n i ln n i )/  n i = - ln (n i /N) Therefore, ln (n i / N) +  -  i = 0 n i / N = exp(  -  i ) The method of Lagrangian Multiplier

The Boltzmann Distribution n i / N = P i = exp (  -  i ) Interpretation of Boltzmann Distribution Meaning of  : ensure total probability is ONE 1 =  i n i / N =  i exp(  -  i ) exp(  ) = 1 /  i exp(-  i )  = - ln [  i exp(-  i )]  > 0, more molecules occupying the low energy states.

Therefore, E = N = 3N/2 , where is the average kinetic energy of a molecule. Therefore, = = 3/2 . On the other hand, according to the Maxwell distribution of speed, the average kinetic energy of a molecule at an equilibrium, The Boltzmann Distribution

This is the physical meaning of , the reciprocal temperature. 1 /  = kT where k is the Boltzmann constant. Thus, /2 = 3kT/2 (This is actually the definition of the temperature) The Boltzmann Distribution

Starting from the principle of equal a priori probabilities, we evaluate the probabilities of different configurations by simple counting, and find that the dominating configuration whose population obeys the Boltzmann distribution which relates the macroscopic observables to the microscopic molecular properties, and is capable of explaining the equilibrium properties of all materials. The Boltzmann Distribution Summary

Two friends A and B are drinking beer in a pub one night. Out of boredom, the two start to play “fifteen-twenty”. A mutual friend C steps in, and the three play together. Of course, this time one hand is used by each. A is quite smart, and studies chemistry in HKU. He figures out a winning strategy. Before long, B and C are quite drunk while A is still pretty much sober. Now, what is A’s winning strategy? When A plays with B and try the same strategy, he finds that the strategy is not as successful. Why? Example 1: The Boltzmann Distribution

Example 2: Consider a molecular whose ground state energy is eV, the first excited state energy -9.5 eV, the second excited state energy -1.0 eV, and etc. Calculate the probability of finding the molecule in its first excited state T = 300, 1000, and 5000 K. The Boltzmann Distribution

The summation is over all possible states (not the energy levels). If the energy level is g i -fold degenerate, then the molecular partition function can be rewritten as The Molecular Partition Function The Boltzmann distribution can be written as p i = exp(-  i ) / q where p i is the probability of a molecule being found in a state i with energy  i. q is called the molecular partition function, q =  i exp(-  i ) q =  i g i exp(-  i )

As T  0, q  g 0, i.e. at T = 0, the partition function is equal to the degeneracy of the ground state. As T  , q  the total number of states. Therefore, the molecular partition function gives an indication of the average number of states that are thermally accessible to a molecule at the temperature of the system. The larger the value of the partition function is, the more the number of thermally accessible states is. The relationship between q and  : exp(  ) = q -1 Interpretation of the partition function

Consider a proton in a magnetic field B. The proton’s spin (S=1/2) has two states: spin parallel to B and spin anti-parallel to B. The energy difference between the two states is  =  p B where  p is proton’s magneton. Calculate the partition function q of the proton. The Boltzmann Distribution Example 3 Example 4 Calculate the partition function for a uniform ladder of energy levels Calculate the proportion of I 2 molecules in their ground, first excited, and second excited vibrational states at 25 o C. The vibrational wavenumber is cm -1. Example 5

Partition function contains all the thermodynamic information! Statistical Thermodynamics The Internal Energy, the Heat Capacity & the Entropy The relation between U and q If we set the ground state energy  0 to zero, E should be interpreted as the relative energy to the internal energy of the system at T = 0, E =  i n i  i =  i Nexp(-  i )  i / q = - Ndlnq/d .

Therefore, the internal energy U may be expressed as The Internal Energy, the Heat Capacity & the Entropy Statistical Thermodynamics U = U(0) + E = U(0) - N (  lnq/  ) V Where, U(0) is the internal energy of the system at T = 0. The above equation provides the energy as a function of various properties of the molecular system (for instance, temperature, volume), and may be used to evaluate the internal energy.

Statistical Thermodynamics The relation between C V and q C v is the constant-volume heat capacity which measures the ability of a system to store energy. It is defined as the rate of internal energy change as the temperature T varies while the volume is kept constant: C v  (  U/  T) V = N(1/kT 2 ) (  2 lnq/  2 ) V Note that, d/dT = (d  /dT) d/d  = -(1/kT 2 ) d/d 

Example 6: Calculate the constant-volume heat capacity of a monatomic gas assuming that the gas is an ideal gas. U = U(0) + 3N / 2  = U(0) + 3NkT / 2 Statistical Thermodynamics where N is the number of atoms, and k is the Boltzmann constant. C v  (  U/  T) V = 3Nk / 2 =3nN A k/2= 3nR/2 where, n is the number of moles, R  N A k is the gas constant, and N A = 6.02 x mol -1 is the Avogadro constant

The relation between S and the partition function q Statistical Thermodynamics The Statistical Entropy According to thermodynamics, entropy S is some measurement of heat q. The change of entropy S is proportional to the heat absorbed by the system: dS = dq / T The above expression is the definition of thermodynamic entropy.

Boltzmann Formula for the entropy Statistical Thermodynamics S = k lnW where, W is the weight of the most probable configuration of the system. Boltzmann Formula (1) indicates that the entropy is a measurement of the weight (i.e. the number of ways to achieve the equilibrium conformation), and thus a measurement of randomness, (2) relates the macroscopic thermodynamic entropy of a system to its distribution of molecules among its microscopic states, (3) can be used to evaluate the entropy from the microscopic properties of a system; and (4) is the definition of the Statistical Entropy.

JUSTIFICATION Statistical Thermodynamics The energy of a molecular system U can be expressed as, U = U(0) +  i n i  i where, U(0) is the internal energy of the system at T=0, n i is the number of molecules which are in the state with its energy equal to  i Now let’s imagine that the system is being heated while the volume V is kept the same. Then the change of U may be written as, dU = dU(0) +  i n i d  i +  i  i dn i =  i  i dn i [dU(0) = 0 because U(0) is a constant; d  i = 0 because  i does not change as the temperature of the system arises.]

According to the First Law of thermodynamics, the change of internal energy U is equal to the heat absorbed (q) and work received (w), i.e., dU = dq + dw dq = TdS (thermodynamic definition of entropy; or more the heat absorbed, the more random the system) dw = -PdV = -Force * distance (as the system shrinks, it receives work from the environment) Statistical Thermodynamics

dU = TdS - PdV = TdS (dV = 0) dS = dU/T = k   i  i dn i = k  i (  lnW/  n i )dn i + k   i dn i = k  i (  lnW/  n i )dn i = k dlnW (  lnW/  n i = - ln (n i /N) = -  +   i ) d(S - lnW) = 0 S = k lnW + constant What is the constant? Statistical Thermodynamics

According to the Third Law of thermodynamics, as T  0, S  0; as T  0, W  1 since usually there is only one ground state, and therefore, constant = 0. Statistical Thermodynamics

Relation between S and the Boltzmann distribution p i Statistical Thermodynamics S = k lnW = k ( N lnN -  i n i lnn i ) = k  i ( n i lnN - n i lnn i ) = - k  i n i ln(n i /N) = - Nk  i (n i /N)ln(n i /N) = - Nk  i p i ln p i since the probability p i = n i /N. The above relation is often used to calculate the entropy of a system from its distribution function.

The relation between S and the partition function q Statistical Thermodynamics According to the Boltzmann distribution, ln p i = -  i - ln q Therefore, S = - Nk  i p i (-  i - ln q) = k   i n i  i + Nk ln q  i p i = E / T + Nk ln q = [U-U(0)] / T + Nk ln q This relation may be used to calculate S from the known entropy q

Independent Molecules Statistical Thermodynamics Consider a system which is composed of N identical molecules. We may generalize the molecular partition function q to the partition function of the system Q Q =  i exp(-  E i ) where E i is the energy of a state i of the system, and summation is over all the states. E i can be expressed as assuming there is no interaction among molecules, E i =  i (1) +  i (2) +  i (3) + … +  i (N) where  i (j) is the energy of molecule j in a molecular state i

The partition function Q Q =  i exp[-  i (1) -  i (2) -  i (3) - … -  i (N)] = {  i exp[-  i (1)]}{  i exp[-  i (2)]} … {  i exp[-  i (N)]} = {  i exp(-  i )} N = q N where q   i exp(-  i ) is the molecular partition function. The second equality is satisfied because the molecules are independent of each other. Statistical Thermodynamics

The relation between U and the partition function Q U = U(0) - (  lnQ/  ) V The relation between S and the partition function Q S = [U-U(0)] / T + k ln Q The above two equations are general because they not only apply to independent molecules but also general interacting systems. Statistical Thermodynamics

Perfect Gas Perfect gas is an idealized gas where an individual molecule is treated as a point mass and no interaction exists among molecules. Real gases may be approximated as perfect gases when the temperature is very high or the pressure is very low. The energy of a molecule i in a perfect gas includes only its kinetic energy, i.e.,  i =  i T q = q T Statistical Thermodynamics i.e., there are only translational contribution to the energy and the partition function.

Translational Partition Function of a molecule q T Although usually a molecule moves in a three-dimensional space, we consider first one-dimensional case. Imagine a molecule of mass m. It is free to move along the x direction between x = 0 and x = X, but confined in the y- and z-direction. We are to calculate its partition function q x. The energy levels are given by the following expression, E n = n 2 h 2 / (8mX 2 )n = 1, 2, … Statistical Thermodynamics

Setting the lowest energy to zero, the relative energies can then be expressed as, q x =  1 dn exp [ -(n 2 -1)  ] =  1 dn exp [ -(n 2 -1)  ] =  0 dn exp [ -n 2  ] = (2  m/h 2  2 ) 1/2 X  n = (n 2 -1)  with  = h 2 / (8mX 2 ) q x =  n exp [ -(n 2 -1)  ]  is very small, then Statistical Thermodynamics

Now consider a molecule of mass m free to move in a container of volume V=XYZ. Its partition function q T may be expressed as q T = q x q y q z = (2  m/h 2  2 ) 1/2 X (2  m/h 2  2 ) 1/2 Y (2  m/h 2  2 ) 1/2 Z = (2  m/h 2  2 ) 3/2 XYZ = (2  m/h 2  2 ) 3/2 V = V/  3 where,  = h(  /2  m) 1/2, the thermal wavelength. The thermal wavelength is small compared with the linear dimension of the container. Noted that q T   as T  . q T  2 x for an O 2 in a vessel of volume 100 cm 3,  = 71 x T=300 K Statistical Thermodynamics

Partition function of a perfect gas, Q = (q T ) N = V N /  3N Energy E = - (  lnQ/  ) V = 3/2 nRT where n is the number of moles, and R is the gas constant Heat Capacity C v = (  E/  T) V = 3/2 nR Statistical Thermodynamics

Fundamental Thermodynamic Relationships Consider an equilibrium system which is consistent of N interacting molecules. These molecules may or may not be the same. Relation between energy and partition function E i is the energy of the state i, p i is its probability. The average energy of the system E is then, E =  i p i E i =  i exp(-  E i ) E i /  i exp(-  E i ) Statistical Thermodynamics

Statistical Thermodynamics Since the partition function of the system Q Q=  i exp(-  E i ) Thus, E = -Q -1 dQ/d  = -dlnQ/d  If we set the molecular ground state energy E 0 to zero, E should be interpreted as the relative energy to the internal energy of the system at T = 0.

Therefore, the internal energy U may be expressed as U = U(0) + E = U(0) - (  lnQ/  ) V Where, U(0) is the internal energy of the system at T = 0, i.e., the ground state energy. The above equation is the relation between U and q. It provides the energy as a function of various properties of chemical systems (for instance, temperature, volume), and may be used to evaluate the internal energy Statistical Thermodynamics It applies to noninteracting, interacting, distinguishable and indistinguishable molecular systems!

Relation between the entropy S and the partition function Q S = [U-U(0)] / T + k lnQ Helmholtz energy The Helmholtz free energy A  U - TS. At constant temperature and volume, a chemical system changes spontaneously to the states of lower Helmholtz free energy, i.e., dA  0, if possible. Therefore, the Helmholtz free energy can be employed to assess whether a chemical reaction may occur spontaneously. A system at constant temperature and volume reaches its equilibrium when A is minimum, i.e., dA=0. The relation between the Helmholtz energy and the partition function may be expressed as, A - A(0) = -kT ln Q Statistical Thermodynamics

Pressure dA = dU - d(TS) = dU - TdS - SdT dU = dq + dw dq = TdS dw = -pdV dA = - pdV - SdT Therefore, pressure may be evaluated by the following expression, p = -(  A/  V) T = kT(  lnQ/  V) T This expression may be used to derive the equation of state for a chemical system. Statistical Thermodynamics

the entropy may also be expressed as S = -(  A/  T) V = klnQ + kT(  lnQ/  T) V = klnQ -(  lnQ/  ) V / T = klnQ + [U-U(0)] / T Statistical Thermodynamics

Consider a perfect gas with N molecules. Its partition function Q is evaluate as Q = (1/N!) (V /  3 ) N the pressure p is then p = kT(  lnQ/  V) T = kT N (  lnV/  V) T = NkT / V pV = NkT = nN A kT = nRT which is the equation of the state for the perfect gas. Statistical Thermodynamics

The enthalpy During a chemical reaction, the change in internal energy is not only equal to the heat absorbed or released. Usually, there is a volume change when the reaction occurs, which leads work performed on or by the surroundings. To quantify the heat involved in the reaction, a thermodynamic function, the enthalpy H, is introduced as follows, H  U + pV Therefore, H - H(0) = -(  lnQ/  ) V + kTV(  lnQ/  V) T Statistical Thermodynamics

The Gibbs energy Usually chemical reactions occur under constant temperature. A new thermodynamic function, the Gibbs energy, is introduced. G  A + pV At constant temperature and pressure, a chemical system changes spontaneously to the states of lower Gibbs energy, i.e., dG  0, if possible. Therefore, the Gibbs free energy can be employed to access whether a chemical reaction may occur spontaneously. A system at constant temperature and pressure reaches its equilibrium when G is minimum, i.e., dG = 0. The relation between the Helmholtz energy and the partition function may be expressed as, G - G(0) = - kT ln Q + kTV(  lnQ/  V) T Statistical Thermodynamics

Example 7 Calculate the translational partition function of an H 2 molecule confined to a 100-cm 3 container at 25 o C Example 8 Calculate the entropy of a collection of N independent harmonic oscillators, and evaluate the molar vibraitional partition function of I 2 at 25 o C. The vibrational wavenumber of I 2 is cm -1 Example 9 What are the relative populations of the states of a two-level system when the temperature is infinite? Example 10 Evaluate the entropy of N two-level systems. What is the entropy when the two states are equally thermally accessible? Statistical Thermodynamics

Example 11 Calculate the ratio of the translational partition functions of D 2 and H 2 at the same temperature and volume. Example 12 A sample consisting of five molecules has a total energy 5 . Each molecule is able to occupy states of energy j  with j = 0, 1, 2, …. (a) Calculate the weight of the configuration in which the molecules share the energy equally. (b) Draw up a table with columns headed by the energy of the states and write beneath then all configurations that are consistent with the total energy. Calculate the weight of each configuration and identify the most probable configuration. Example 14 Given that a typical value of the vibrational partition function of one normal mode is about 1.1, estimate the overall vibrational partition function of a NH 3. Statistical Thermodynamics

Example 15 Consider Stirling’s approximation for lnN! In the derivation of the Boltzmann distribution. What difference would it make if the following improved approximation is used ? x! = (2  ) 1/2 (x+1/2) x e -x Example 16 Consider N molecules in a cube of size a. (a) Assuming molecules are moving with the same speed v along one of three axises (x-, y-, or z-axis). The motion may be along the positive or negative direction of an axis. On the average, how many molecules move parallel to the positive x direction? Statistical Thermodynamics

(b)When a molecule collides with one of six sides of the cube, it is reflected. The reflected molecule has the same speed v but moves in the opposite direction. How many molecules are reflected from the side within a time interval  t ? What is the corresponding momentum change of these molecules? (c) Force equals the rate of momentum change. Calculate the force that one side of the cube experiences. Pressure is simply the force acting on a unit area. What is the pressure of the gas? Assuming the average kinetic energy of a molecule is 3kT/2, derive the equation of state for the system. Statistical Thermodynamics

Diatomic Gas Consider a diatomic gas with N identical molecules. A molecule is made of two atoms A and B. A and B may be the same or different. When A and B are he same, the molecule is a homonuclear diatomic molecule; when A and B are different, the molecule is a heteronuclear diatomic molecule. The mass of a diatomic molecule is M. These molecules are indistinguishable. Thus, the partition function of the gas Q may be expressed in terms of the molecular partition function q, The molecular partition q where,  i is the energy of a molecular state i, β=1/kT, and  ì is the summation over all the molecular states. Statistical Thermodynamics

B A Statistical Thermodynamics

where T denotes translation, R rotation, V vibration, and E the electronic contribution. Translation is decoupled from other modes. The separation of the electronic and vibrational motions is justified by different time scales of electronic and atomic dynamics. The separation of the vibrational and rotational modes is valid to the extent that the molecule can be treated as a rigid rotor. Factorization of Molecular Partition Function The energy of a molecule j is the sum of contributions from its different modes of motion: Statistical Thermodynamics

Statistical Thermodynamics Factorization of Molecular Partition Function

where Statistical Thermodynamics The translational partition function of a molecule  ì sums over all the translational states of a molecule. The electronic partition function of a molecule  ì sums over all the electronic states of a molecule. The rotational partition function of a molecule  ì sums over all the rotational states of a molecule. The vibrational partition function of a molecule  ì sums over all the vibrational states of a molecule.

hv hv hv hv hv n= 0, 1, 2, ……. kT Vibrational Partition Function Two atoms vibrate along an axis connecting the two atoms. The vibrational energy levels: If we set the ground state energy to zero or measure energy from the ground state energy level, the relative energy levels can be expressed as

Then the molecular partition function can be evaluated Consider the high temperature situation where kT >>hv, i.e., Vibrational temperature  v High temperature means that T>>  v    F 2 HCl H 2  v /K v/cm where Therefore, Vibrational Partition Function

where B is the rotational constant. J =0, 1, 2, 3,…  h/8cI  2 c: speed of light I: moment of Inertia  hcB<<1 where g J is the degeneracy of rotational energy level ε J R Usually hcB is much less than kT, Note: kT>>hcB Rotational Partition Function If we may treat a heteronulcear diatomic molecule as a rigid rod, besides its vibration the two atoms rotates. The rotational energy =kT/hcB

For a homonuclear diatomic molecule Generally, the rotational contribution to the molecular partition function, Where  is the symmetry number. Rotational temperature  R  Rotational Partition Function

where, g E = g 0 is the degeneracy of the electronic ground state, and the ground state energy  0 E is set to zero. If there is only one electronic ground state q E = 1, the partition function of a diatomic gas, At room temperature, the molecule is always in its ground state Electronic Partition Function =g 0 =g E

The internal energy of a diatomic gas (with N molecules) q V = kT/hv q R = kT/hcB The rule: at high temperature, the contribution of one degree of freedom to the kinetic energy of a molecule (1/2)kT Mean Energy and Heat Capacity (T>>1)

the constant-volume heat capacity (T>>1) Contribution of a molecular to the heat capacity Translational contribution (1/2) k x 3 = (3/2) k Rotational contribution (1/2) k x 3 = k Vibrational contribution (1/2) k + (1/2) k = k kinetic potential Thus, the total contribution of a molecule to the heat capacity is (7/2) k Mean Energy and Heat Capacity

Translational energy where n = 1, 2, … n is a measurement of the speed of the molecule Quantum Classical for a H in a one-dimonsional box x= 1cm Mean Energy and Heat Capacity

Vibrational energy QuantumClassical A is the amplitude of the vibration Here the vibrational quantum number n is a measurement of the vibrational amplitude. Mean Energy and Heat Capacity

m j =-J,-J+1……. Mean Energy and Heat Capacity Rotational energy QuantumClassical  is the rotational angulor velocity J is a measurement of angular velocity m J is a measurement of the projection of the angular velocity of the z- axis. i.e. a measurement of the rotation’s orientation.

Molecule with N atoms Degree of freedom: Translation: 3 Rotation:3nonlinear 2 linear vibration: 3N – 6nonlinear 3N – 5 linear Diatomic Molecule A-B Symmetry  = Supplementary note 0 if A  B 1 if A = B

Summary Principle of equal a priori probabilities: All possibilities for the distribution of energy are equally probable provided the number of molecules and the total energy are kept the same. A configuration { n 0, n 1, n 2,......} can be achieved in W different ways or the weight of the configuration Dominating Configuration vs Equilibrium The Boltzmann Distribution P i = exp (-  i ) / q W = N! / (n 0 ! n 1 ! n 2 !...)

q =  i exp(-  i ) =  j g j exp(-  j ) Q =  i exp(-  E i ) Partition Function Energy E= N  i p i  i = U - U(0) = - (  lnQ/  ) V Heat Capacity C V = (  E/  T) V = k  2 (  2 lnQ/  2 ) V Entropy S = k lnW = - Nk  i p i ln p i = k lnQ + E / T A= A(0) - kT lnQ Helmholtz energy Summary

H = H(0) - (  lnQ/  ) V + kTV (  lnQ/  V) T Q = q N or (1/N!)q N q = q T q R q V q E Enthalpy Molecular partition function Factorization of Molecular Partition Function Summary

Schr Ö dinger Equation H  E  Research GH CHEN’s Group Theory is reality !

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