Intermediate Physical Chemistry Driving force of chemical reactions Boltzmann’s distribution Link between quantum mechanics & thermodynamics

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

In 1929, Dirac declared, “The underlying physical laws necessary for the mathematical theory of...the whole of chemistry are thus completely know, and the difficulty is only that the exact application of these laws leads to equations much too complicated to be soluble.” Beginning of Computational Chemistry Dirac

H  E  Schr Ö dinger Equation Hamiltonian H =   (  h 2 /2m      h 2 /2m e )  i  i 2 +     Z  Z  e   r   i     e 2 /r i   i  j  e 2 /r ij Wavefunction Energy

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 Democracy among microscopic states !!! a priori: as far as one knows

For instance, four molecules in a three-level system: the following two conformations have the same probability l-l  l  l   l a demon

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.

Population of a state: the average number of molecules occupying a state. Denote the energy of the state  i The Question: How to determine the Population ?

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,......}

{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 !...)

2.Calculate the number of ways of distributing 20 different molecules among six different states with the configuration {1, 0, 3, 5, 10, 1}. Answer: 20! / 1! 0! 3! 5! 10! 1! = note: 0! = 1

Stirling’s Approximation: When x is large, ln x!  x ln x - x xln x! x ln x - xln A note, A = (2  ) 1/2 (x+1/2) x e -x ln W  ( N ln N - N ) -  ( n i ln n i - n i ) = N ln N -  n i ln n i

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

When N is even, the weight is maximum at k = N/2, 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 isW k=N/4 = N! / [(N/4)! (3N/4)!] | N | x |R(N) | x x 10 3e+22 The ratio of the two weights R(N)  W k=N/2 / W k=N/4 is equal to (N/4)! (3N/4)! / [(N/2)!] 2

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 Dominating Configuration: Equilibrium Configuration

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

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

1. The total energy is a constant, i.e.  n i  i = E = constant 2. The total number of molecules is conserved,  n i = N = constant How to maximize W or lnW ? How to maximize a function ? To find the maximum value of W or lnW,  lnW /  n i = 0,i=1,2,3,... ???

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

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

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.

This method is called the method of Lagrangian Multiplier, or Method of undetermined multipliers 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 Boltzmann Distribution P i = exp (  -  i ) Interpretation of Boltzmann Distribution 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.

Physical Meaning of  Imagine a perfect gas of N molecules. Its total energy E =  i n i  i =  i N exp(-  i )  i /  i exp(-  i ) = -Nq -1 dq /d  = -N dlnq / d  where q =  i exp(-  i ), n i and  i are the population and energy of a state i, respectively. It will be shown later that q = V /  3 and  = h (  /2  m) 1/2 [  is called the thermal wavelength] dlnq/d  = -3dln  /d  = -3/2 

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 reciprocal temperature 1 /  = kT where k is the Boltzmann constant. Physical Meaning of 

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:

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.

Answer The energies of second, third and higher excited states are much higher than those of ground and first excited states, for instance, at T = 5,000 K, the probability of finding the first excited state is, p 2 = exp(  -  2 ) with  2 = -9.5 eV And, the probability of finding the second excited state is, p 3 = exp(  -  3 ) with  3 = -1.0 eV

the ratio of probabilities between second & first excited states is, exp[-( ) x 11600/5000] = exp(-20) [ 1 eV = 1.60 x J; k = 1.38 x J K -1 ; 1 eV  k K -1 ] i.e., compared to the first excited state, the chance that the molecule is in second excited state is exceedingly slim. Thus, we consider only the ground and first excited states, a two-state problem.

The probability of finding the molecule in the first excited state is p = exp(-  2 ) / [exp(-  1 ) + exp(-  2 )] where,  1 = eV, and  2 = -9.5 eV (1) T = 300 K

(2) T = 1000 K (3) T = 5000 K

Molecular Partition Function

July 1998GunaHua

July 1998GunaHua

Hatch of an egg

Fundmental Thermodynamic Relationships

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?

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.

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.

Factorization of Molecular Partition Function The energy of a molecule j is the sum of contributions from its different modes of motion: 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.

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

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 hv hv hv hv hv n= 0, 1, 2, ……. kT 

Then the molecular partition function can be evaluated Therefore, 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

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 where B is the rotational constant. J =0, 1, 2, 3,… where g J is the degeneracy of rotational energy level ε J R Usually hcB is much less than kT, =kT/hcB  h/8cI  2 c: speed of light I: moment of Inertia  hcB<<1 Note: kT>>hcB

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

Electronic Partition Function =g 0 =g E 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

Mean Energy and Heat Capacity The internal energy of a diatomic gas (with N molecules) (T>>1) Contribution of a molecular to the total energy Translational contribution (1/2)kT x 3 = (3/2)kT Rotational contribution (1/2)kT x 2 = kT Vibrational contribution (1/2)kT + (1/2)kT = kT kinetic potential the total contribution is (7/2)kT 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

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

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 W = N! / (n 0 ! n 1 ! n 2 !...) Dominating Configuration vs Equilibrium The Boltzmann Distribution P i = exp (-  i ) / q

Partition Function q =  i exp(-  i ) =  j g j exp(-  j ) Q =  i exp(-  E i ) Energy E= N  i p i  i = U - U(0) = - (  lnQ/  ) V Entropy S = k lnW = - Nk  i p i ln p i = k lnQ + E / T A= A(0) - kT lnQ H = H(0) - (  lnQ/  ) V + kTV (  lnQ/  V) T Q = q N or (1/N!)q N