3 Molecular systems

In this chapter we return to the equilibrium statistical thermodynamics of amolecular system. The expressions for the Boltzmann distribution (Eqn.1.27)and the partition function (Eqn1.29) are quite general. No assumption wasmade about the size of the system - apart from the fact that the thermal reser-voir should be much larger than the system. This implies that we can also useEqn.1.27 to describe the Boltzmann distribution over energy levels of a singlemolecule and we can use Eqn. 1.29 to describe the partition function of anisolated molecule. This is important, because it will allow us to use our exper-imental or theoretical knowledge of the energy levels of isolated molecules topredict the thermal properties of dilute gases, and later of condensed phases.

We are interested in the calculation of partition functions of the form

q =

all levels∑

i

exp(−βui) (3.1)

Such sums can easily be evaluated in a number of simple cases where we knowthe energy levels from quantum mechanics. Examples (that will be discussedbelow) are the case of a one-dimensional harmonic oscillator, a particle in aone-dimensional box or a rigid, linear rotor. However, in practice we shallalways be interested in systems containing very many particles that are subjectto translation, vibration and rotation (and possibly also to electronic or spinexcitations). It would seem that the calculation of the partition function ofsuch a complex system is totally intractable. In fact, for macroscopic systems ofstrongly interacting particles, this statement is very nearly true. This is what iscommonly called the ”many-body problem”. Computing the partition functionof - say - liquid water is not possible analytically and sophisticated numericalsimulation techniques have been developed that allow us to make progress inthis direction. However, there is an important class of problems where progresscan be made without recourse to numerical simulation, namely the case wherethe total energy of the system can be written as a sum of simple contributions.This situation typically arises when there is virtually no interaction between thedifferent particles in the system (i.e. in the case of an ideal gas). Let us considera simple example. Suppose that we have a system with a total energy E, suchthat

E =

N∑

i=1

ǫi(ni) (3.2)

where ǫi(ni) is the energy of the i-th molecule in a quantum state labeled by thequantum number ni. If all molecules have the same energy levels6, the partition

6We assume that, although the levels of different molecules have the same energy, they are

not physically the same levels. Later, we shall consider what happens if the levels really are

identical.

72

function of this system is given by

Q =∞∑

n1=1

∞∑

n2=1

...∞∑

nN=1

exp(−βN∑

i=1

ǫi(ni))

=

∞∑

n1=1

∞∑

n2=1

...

∞∑

nN=1

N∏

i=1

exp(−βǫi(ni))

=

(

∞∑

n1=1

exp(−βǫ1(n1))

)(

∞∑

n2=1

exp(−βǫ2(n2))

)

...

(

∞∑

nN=1

exp(−βǫN (nN ))

)

= qN (3.3)

where

q ≡∞∑

n=1

exp(−βǫ(n)) (3.4)

In what follows, we shall often make use of this trick. To give a simple example,assume that a molecule has only two levels: a ground state with energy ǫ0 = 0and an excited state with energy ǫ1 = ǫ. In that case,

q = exp(−β × 0) + exp(−βǫ)

= 1 + exp(−βǫ) (3.5)

and hence the total partition function is

Q = (1 + exp(−βǫ))N . (3.6)

3.1 Single-molecule partition function

Quantum mechanics allows us to compute the energy levels of (simple) isolatedmolecules. Once we know these energy levels, we can use Eqn.1.29 to com-pute the molecular partition function. From this quantity, we can compute ahost of important thermodynamical properties of the individual molecules and,as we shall see later, also of macroscopic systems containing many (O(1023))molecules. For instance, we shall be able to compute the internal energy, theheat-capacity, the pressure and the chemical potential. From the latter quantity,we can predict the equilibrium constants of chemical reaction. But, before wedo this, let us first consider the most important classes of molecular partitionfunctions. Recall that the internal energy of a molecule can be due to variouskinds of motion: translation, rotation, vibration and electronic excitation arethe most common. Usually the energy of electronic excitations is very largecompared to kBT . As a consequence, the internal energy of a molecule at roomtemperature is usually due to translation, rotation and vibration. Translationis the motion of the molecule as a whole. Because the intra-molecular motion(rotation and vibration) does not depend on its center-of-mass motion, we canwrite the energy of a molecule as the sum of the translational energy ǫtr and

73

the intra-molecular energy ǫint due to rotation and vibration. It is often a fairapproximation to assume that rotation and vibration are uncoupled. In thatcase, the total intra-molecular energy can be written as

ǫint = ǫrot +

m∑

i=1

ǫivib (3.7)

where ǫrot is the rotational energy and ǫivib is the energy associated with i-thvibration (diatomic molecules have only one vibration, poly-atomic moleculeshave many - we denote this number by m). The total energy of a molecule isthen

ǫtot = ǫtrans + ǫrot +

m∑

i=1

ǫivib (3.8)

Of course, the value of ǫtot depends on the quantum state that the moleculeis in. This quantum state is characterized by a set of quantum numbers kαcharacterizing the translational motion in directions α (α = x, y, z), a quantumnumber J describing the rotational motion and quantum numbers ni describingthe quantum-state associated with the i-th vibration. If we want to computethe molecular partition function, we have to evaluate the sum

qmol =∑

kx,ky,kz,,J,n1,...,nm

exp(−βǫtot) (3.9)

This expression can be simplified appreciably if the internal energy of themolecule can be written as the sum of translational, rotational and vibrationalenergies. Using

exp(−β[ǫtrans + ǫrot +m∑

i=1

ǫivib]) = exp(−βǫtrans) exp(−βǫrot)m∏

i=1

exp(−βǫivib)

(3.10)we can write the molecular partition function as a product of a translational, arotational and m vibrational partition functions

qmol = qtransqrot

m∏

i=1

qivib (3.11)

where

qtrans =∑

kx,ky,kz

exp(−βǫtrans(kx, ky, kz))

qrot =∑

J

exp(−βǫrot(J))

qivib =∑

ni

exp(−βǫvib(ni))

74

Hence, we can compute the partition functions for translation, rotation andvibration separately. The total molecular partition function is then simply theproduct. Actually, the situation is slightly more complex: when computingthe total energy of a molecule, we need to know ǫground, the energy of themolecule in its ground state. ǫground is a measure for the binding energy of themolecule. It will turn out to be important later on (when we compute chemicalequilibria). However, for the time being we ignore this term and simply pretendthat the energy of a molecule in its ground state is zero. Now let us see ifwe can write down expressions for the translational, rotational and vibrationalpartition functions.

3.1.1 Vibration

Let us start with vibration (because it is easiest). If the vibrational motion isharmonic, we know from quantum mechanics that the vibrational levels are non-degenerate and all equally spaced. If the vibration frequency of the molecule isνi then the spacing between two levels is hνi. If we take the potential energyat the bottom of the harmonic potential to be zero, then the energy of the n-thvibrational state is ǫn = (n + 1/2)hνi. The quantity hνi/2 is the zero-pointenergy of the harmonic oscillator with frequency νi. The partition functionassociated with the i-th vibrational mode is

qivib = exp(−βhνi/2)∞∑

n=0

exp(−nβhνi) (3.12)

If we denote exp(−βhνi) by xi, then we see that the vibrational partition func-tion is of the form

qivib = x1/2∞∑

n=0

xni (3.13)

As exp(−hνi/kBT ) = x < 1, we can sum this series and we obtain

qivib =x1/2

1− x=

exp(−βhνi/2)

1− exp(−βhνi)(3.14)

The probability to find this vibration is in its n-th excited state is

p(ni) =exp(−(n+ 1/2)βhνi)

qivib= [1− exp(−βhνi)] exp(−nβhνi) (3.15)

We can also compute the mean vibrational energy. Using Eqn.1.28 we find that

< ǫivib >= −∂ ln qivib∂β

= hνi/2 +hνi exp(−βhνi)

1− exp(−βhνi)= hνi/2 +

hνiexp(βhνi)− 1

(3.16)Note that in the classical limit (h → 0), we find that < ǫivib >= β−1 = kBT , ir-respective of the frequency. By classical limit, we mean that the typical energies

75

involved are much higher than the spacing of the quantum levels. This is thecase for macroscopic objects or for high temperature. In those cases, insteadof explicitly putting in all the values, we might as well take the limit h → 0 orβ → 0.

When exactly can we use this classical or high temperature approximation?To get a feeling, one can introduce the so-called vibrational temperature θv =hν/kB . Above this temperature the typical energy is high enough to make theapproximation β → 0.

qivib ≈kBT

hνi(3.17)

For example, for H2 the vibrational temperature is 6330 K, rather high becausethe atoms are light, so that the frequency is high, and the spacings are far apart.In that case, expression 3.14 for qi must be used at room temperature. In con-trast, for I2 the vibrational temperature θv = 309K, meaning that the spacingsare much closer and even at room temperature many levels are occupied. Inthat case the high temperature limit is reached much earlier.

3.1.2 Translation

For the translational partition function we use the quantum-mechanical resultfor the energy levels of a particle in a box. For a one-dimensional box of lengthL, these energy levels are given by

ǫ(n) =n2h2

8mL2(3.18)

Figure 3: Because of its lowmass the vibrational levels of H2

are much further apart than forI2. For I2 the thermal energy attemperatures > 309K is alreadylarge enough to validate a classi-cal approximation. But not forH2. H2 I2

ǫ

hν = kBθv

76

The lowest energy level is ǫ(1). However, for macroscopic systems the averageenergy is very much large than ǫ(1). Hence, in what follows we use Eqn.3.18 tocompute the energy of level n, rather than ǫ(n)− ǫ(1). We can now write downthe expression for the translational partition function for a one-dimensionalsystem:

qxtrans =

∞∑

n=1

exp(−βn2h2

8mL2) (3.19)

In macroscopic systems, the translational level spacing is typically much smallerthan kBT . As a consequence, we can replace the sum in the expression for qxtransby an integral

qxtrans ≈∫ ∞

1

dn exp(−βn2h2

8mL2) (3.20)

We now change variables. We define x ≡ n√

βh2/8mL2. Then

qxtrans ≈√

8mL2/βh2

∫ ∞

√βh2/8mL2

dx exp(−x2) (3.21)

But, because√

βh2/8mL2 ≪ 1, we can replace the lower limit of this integralby 0. The integral

∫ ∞

0

dx exp(−x2) =1

2

√π (3.22)

and henceqxtrans =

√

2πmL2/βh2 (3.23)

The result for a three-dimensional box follows because the total translationalpartition function is simply

qtrans = qxtransqytransq

ztrans (3.24)

For convenience, we assume that the particle is contained in a cubic box withvolume V = L3. Then

qtrans =(

2πmL2/βh2)3/2

= V(

2πmkBT/h2)3/2 ≡ V/Λ3 (3.25)

where we have defined the so-called “thermal wavelength” Λ = h/ (2πmkBT )1/2

.We will come back to the meaning of this thermal wavelength. Now we can easilycompute the average translational energy of a particle. First we write qtrans as

qtrans = V(

2πm/h2)3/2

β−3/2 (3.26)

Next, we use the relation

< ǫtrans >= −∂ ln qtrans∂β

= −∂ lnβ−3/2

∂β=

3

2

∂ lnβ

∂β

=3

2β=

3

2kBT (3.27)

77

Hence the average translational energy per particle is 32kBT . If we have a system

consisting of N non-interacting particles, we can simply add the translationalenergies of all particles and hence the total translational energy becomes

Etrans =3

2NkBT (3.28)

3.1.3 Rotation

The rotational energy levels of a molecule depend on its symmetry. Here, weonly consider the simplest case: a linear rotor (examples are HCl and HCN ,but not H2O). The energy levels of a linear rotor depend on its moment ofinertia I

EJ = J(J + 1)~2

2I(3.29)

and the degeneracy of an energy level with quantum number J is equal togJ = (2J + 1). The rotational partition function of a linear rotor is then

qrot =

∞∑

J=0

(2J + 1) exp[−βJ(J + 1)~2/2I] (3.30)

This sum can be easily evaluated numerically. But in the “classical” limitβ~2/2I ≪ 1 it is simple to get an analytical result. In that case we can write

qrot ≈∫ ∞

0

dJ (2J + 1) exp[−βJ(J + 1)~2/2I] (3.31)

This integral is easy to evaluate if we make the substitution J2 + J ≡ x

qrot =

∫ ∞

0

dx exp[−β~2

2Ix]

=2I

β~2=

2IkBT

~2(3.32)

The above approximation 3.31 is valid for β~2/2I ≪ 1. This is the casefor high temperature and heavy molecules. To get a feeling for this limit,we again introduce a crossover temperature, called the rotational temperatureθR = ~

2/(2IkB). As an example, for H2 this temperature is 88K and for I2θR = 0.053K. Apparently, for rotations the high temperature limit is alwaysvalid (at least at room temperature)!

As before, we can now compute the average rotational energy per particle.It is

< ǫlinrot >= −∂ ln qlinrot

∂β

= β−1 = kBT (3.33)

78

In the case of non-linear molecules, qrot ∼ T 3/2 in the classical limit. In thatcase

< ǫnon−linrot >=

3

2kBT (3.34)

Note that, in the first paragraph of this section, we mentioned a heteronucleardiatomic such as HCl as an example of a linear molecule, but not the simplehomo-nuclear molecules such as H2, I2 or N2. The reason is that, due to thesymmetry of these molecules, certain rotational states are forbidden. Whichstates are forbidden depends on the nature of the nuclei. For instance, 14N is aboson (its nuclear spin is zero). For a diatomic molecule consisting of bosons,the total wave-function should remain identical if we permute the two nuclei.However, the wave-functions of a linear rotor have the property that they changesign if we rotate the molecule over 180◦ if J is odd. But rotating the moleculeover 180◦ is equivalent to permuting the two nuclei. And for bosons the wavefunction should not change sign when we do that. Hence in that case, all stateswith odd J are excluded. The result is that the rotational partition function isreduced. In the classical limit, this reduction amounts to a factor 2. Similarly,if the molecule contains two identical fermionic nuclei (e.g. H2 - nuclear spin1/2) then the total wavefunction must change sign if we permute the two nuclei.This would suggest that only odd J-states are allowed. However, matters are abit more complex because it makes a difference whether the two nuclear spinsare parallel or anti-parallel. The total wave-function is in this case a productof the rotational wave function (that is changes sign when J is odd) and a spinwavefunction that changes sign when the spins are anti-parallel. As the totalwavefunction should change sign under permutation of the two nuclei, we findthat if the spins are parallel, then J must be odd. And if the spins are anti-parallel then J must be even. In either event, we have to exclude every second Jvalue (either odd or even). In the classical limit, the net result is that the totalrotational partition function is reduced by a factor 2, just as for bosons. Wecall this factor the symmetry number σ ( σ = 2 for the homo-nuclear diatomics).So, the qrot in Eqn. 3.32 becomes

qrot =2IkBT

~2σ(3.35)

For more complex molecules we can compute the symmetry number systemati-cally using group-theoretical arguments.

3.2 Many-particle partition function

3.2.1 Indistinguishability

In the above section, we have given expressions for qmol, the partition function ofisolated molecules in an ideal gas, and for the total energy of the gas. However,we did not yet write down the partition function for an ideal gas of N molecules.It would seem that this is trivial. By analogy to Eqn.3.3 we would expect that

79

the total partition function simply equals qNmol. However, this is notthe case. Infact, we can demonstrate experimentally that

QN 6= qNmol (3.36)

To see this, consider the following experiment. We start with an ideal gas con-taining N particles that are all of one and the same kind - say, argon atoms. Inthat case, qmol is equal to the translational partition function qtrans (Eqn. 3.25).Of these N atoms, we put N/2 in one half of a container with volume V andthe other N/2 particles in the other half. The two halves of the container arekept at a temperature T . They are separated by a removable wall. If Eqn.3.36were valid, then the total Helmholtz free energy of these two systems togetherwould be

F initialtotal = −2kBT ln

(

V

2

(

2πmkBT/h2)3/2

)N/2

= −2kBT ln

(

V

2Λ3

)N/2

= −NkBT ln

(

V

2Λ3

)

(3.37)

Now assume that we take out the wall separating the two sub-systems. Ifthe two halves of the volume had contained different gases, then we wouldsee irreversible mixing, and the free energy would decrease. But, as the twohalves contain identical particles, removing the wall between them has no effectwhatsoever on the total free energy. Yet, if the relation QN = qNmol were true,the we would predict that the new free energy would be

F finaltotal = −kBT ln

(

V/Λ3)N

= −NkBT ln(

V/Λ3)

(3.38)

This would imply that the Helmholtz free energy would change by an amount

F finaltotal − F initial

total = −NkBT ln 2 (3.39)

Clearly, something is wrong with the assumption that, for identical molecules,QN = qNmol. The resolution of this problem lies in the fact that we have beenovercounting the number of distinct quantum states in the system. IfN identicalparticles can occupy the same set of quantum levels, then any permutation ofthe particles over these levels still yields the same overall quantum state. Thatis, we can label the overall quantum state by indicating what levels {i, j, k, l, ...}are occupied. But it is meaningless to say that particle 1 occupies level i andparticle 2 occupies levels j and so on. There is no way in which we coulddistinguish this situation from the one in which particle 2 occupies level i andparticle 1 occupies level j. In short: permutations of identical particles over ashared set of energy levels, do not correspond to distinct quantum states. Thismeans that there is only one way to place N identical particles in N distinctquantum states, rather than N ! ways. Now look at Eqn.3.3, in particular at the

80

expression

Q =

∞∑

n1=1

∞∑

n2=1

...

∞∑

nN=1

exp(−β

N∑

i=1

ǫi(ni)) (3.40)

This expression is valid if all sums over the quantum numbers ni refer to differentlevels. This is, for instance, the case if we look at N harmonic oscillators: thesituation where oscillator 1 is in the ground state and oscillator 2 is in the firstexcited state is not the same as the situation where oscillator 2 is in the groundstate and oscillator 1 is in the first excited state. However, we have two identicalparticles in the same box, then the situation where particle 1 is in the groundstate and particle 2 is in the first excited state is the same as the one wherethe two particles are permuted. For that reason, we should count all termsin the sum in Eqn.3.40 only once. This means that we should divide the sumin Eqn.3.3 by N !. Hence, the partition function of a system of N ideal gasmolecules is

Q =1

N !

∞∑

n1=1

∞∑

n2=1

...∞∑

nN=1

exp(−βN∑

i=1

ǫi(ni))

=qNmol

N !(3.41)

If we now look again at the two volumes with identical gases that were beingcombined, we find that the initial Helmholtz free energy was

F initialtotal = −2kBT ln

[

(

12V/Λ

3)N/2

(N/2)!

]

(3.42)

and

F finaltotal = −kBT ln

[

(

V/Λ3)N

N !

]

(3.43)

The difference between these two free energies is

F finaltotal − F initial

total = −NkBT ln 2 + kBT ln

(

N !

[(N/2)!]2

)

(3.44)

Using the Stirling approximation lnN ! ≈ N lnN −N , we find that

F finaltotal − F initial

total ≈ −NkBT ln 2 +NkBT ln 2 = 0 (3.45)

Hence the assumption that identical particles are truly indistinguishable is com-patible with the experimental observation that the Helmholtz free energy doesnot change when we bring two reservoirs containing identical particles into con-tact.

We should add that the above discussion is not completely honest: we madethe implicit assumption that no two particles would be in exactly the same quan-tum state. For an ideal gas at room temperature, this assumption is quite rea-sonable. This can be seen as follows: recall that the molecular partition function

81

is a measure for the number of quantum state accessible to a molecule. For thetranslational quantum states, this number is equal to

nstates ≈ V(

2πmkBT/h2)3/2

=V

Λ3(3.46)

where we have again used the “thermal wavelength” Λ of a particle. The num-ber of accessible states is equal to the number of volumes Λ3 that fit into themacroscopic volume V . To give a specific example: for argon at room tempera-ture, Λ ≈ 1.7 10−11 m. Hence, the number of accessible quantum states in onecubic meter is 2 · 1032. This is much larger than Avogadro’s number. Hence,the probability that two atoms in a dilute gas will occupy the same level is verysmall indeed. However, Λ increases as the temperature is decreased and at verylow temperatures we should take the possibility into account that two particlesmay occupy the same quantum state. Then it becomes important to distinguishbetween particles that cannot occupy the same quantum state (fermions) andparticles that can (bosons).

In what follows, we shall therefore assume that the partition function of anideal gas of identical molecules is given by

Q =qNmol

N !(3.47)

Now we start to make true on our promise that Statistical Thermodynam-ics allows us to predict the thermodynamical properties of a macroscopic sys-tem solely on basis of our (quantum-mechanical) knowledge of the constituentmolecules. In particular, we are now in a position to compute all thermodynam-ical properties of a dilute molecular gas.

Let us begin with the derivation of the pressure. This derivation is veryimportant because it allows us to justify the identification

β = 1/kBT.

Using the relation

P = β−1

(

∂ lnQ

∂V

)

N,β

(3.48)

we find

P = Nβ−1

(

∂ ln qmol

∂V

)

N,β

(3.49)

But note that of the different terms in qmol = qtransqrot∏m

i=1 qivib, only qtrans

depends on the volume V . Hence

P = Nβ−1

(

∂ ln qtrans∂V

)

N,T

=N

βV(3.50)

If we compare this to the experimental (Boyle/Gay-Lussac) relation

P =nRT

V(3.51)

82

where n is the number of moles of the gas and R is the gas constant, then itfollows that

β =N

nRT(3.52)

The ratio N/n is simply the number of molecules per mole, i.e. Avogadro’snumber L = 6.022 1023. We denote the ratioR/L by kB (Boltzmann’s constant).Then

β =1

kBT(3.53)

where kB has the value 1.3806503 10−23J/K.We can also compute the internal energy of an ideal gas. However, now we

should take translation, rotation and vibration into account. But, as the totalmolecular partition function is a product of these individual terms, the internalenergy

U = −N

(

∂ ln qmol

∂β

)

(3.54)

is the sum of a translational, a rotational and a vibrational part.

3.2.2 Monatomic gas.

Let us first consider a monatomic gas. In that case qmol = qtrans so

Q =qNmol

N !=

(

V/Λ3)N

N !(3.55)

and we get (from Eqn.3.27) that

U =3

2NkBT (3.56)

It then follows immediately that the isochoric heat capacity of an ideal monatomicgas is

cV =

(

∂U

∂T

)

N,V

=3

2NkB (3.57)

Let us next compute the Helmholtz free energy of an ideal monatomic gas

F = −kBT ln

{

V N

N !Λ3N

}

(3.58)

Using the Stirling approximation for lnN !,we get

F = −NkBT ln(V/Λ3) +NkBT lnN −NkBT

= NkBT {ln(ρΛ3)− 1} (3.59)

where ρ ≡ (N/V ) is the number density of the gas. It is then easy to obtain anexpression for the Gibbs free energy G = F + PV .

G = NkBT {ln(ρΛ3)− 1}+NkBT

= NkBT ln(ρΛ3) (3.60)

83

When we recall that, for a one-component system, G = Nµ, where µ is thechemical potential, then we find that

µ = kBT ln(ρΛ3) (3.61)

or, if we use the ideal gas law, in the form P = ρkBT ,

µ = kBT ln(PΛ3/kBT ) (3.62)

Using F = U − TS, we can also get an expression for the entropy S from theexpressions for the Helmholtz free energy F and for the internal energy U . Wethen find the famous Sackur-Tetrode equation for the entropy of a monatomicgas

S = (U − F )/T

= NkB

(

5

2− ln(ρΛ3)

)

(3.63)

3.2.3 Polyatomic gases

When discussing the statistical thermodynamics of ideal polyatomic gases, itis important to bear in mind that the partition function of such a gas can bewritten as

Qtotal =qNtransN !

qNinternal (3.64)

where qinternal is the part of the molecular partition function that contains thecontributions due to rotation, vibration and electronic excitation. Due to thefactorization of the partition function we can write

Ftotal = Ftrans −NkBT ln qinternal (3.65)

Therefore, we can simply add to the translational part of the free energy thecontribution due to the internal degrees of freedom the molecules. This alsoholds for the Gibbs free energy G:

Gtotal = Gtrans −NkBT ln qinternal

= NkBT ln(ρΛ3)−NkBT ln qinternal (3.66)

Of particular interest is the expression for the chemical potential of an ideal gasof polyatomic molecules of species α

µα = kBT ln(ραΛ3α)− kBT ln qαinternal (3.67)

It is useful to cast this expression for the chemical potential in the form that isoften used in thermodynamics

µ(P ) = µ⊖ + kBT ln(P/p⊖) (3.68)

84

where p⊖ is the pressure of the reference state (usually: p⊖= 1 bar). For anideal gas, ρα = Pα/kBT , where Pα is the partial pressure of species α. Hence,we can write Eqn.3.62 as

µα = kBT ln(Λ3αPα/kBT )− kBT ln qαinternal

= kBT ln(Pα/p⊖) + kBT ln(Λ3

αp⊖/kBT )− kBT ln qαinternal (3.69)

and hence

µ⊖ = −kBT ln

(

qαinternalkBT

Λ3αp

⊖

)

(3.70)

One of the quantities that is often measured in experiments is the (isochoric)heat capacity cV .

cV =

(

∂U

∂T

)

N,V

(3.71)

In the late 19th century, when statistical thermodynamics was being developedon basis of classical mechanics, the behavior of the heat capacity of poly-atomicgases was a mystery. It is easy to see why. The total internal energy of an idealgas is equal to the sum of the translational, rotational and vibrational energies.We have computed all these quantities in the classical limit:

Utrans =3

2NkBT

Urot =3

2NkBT (for non-linear molecules)

Uvib = NkBT (for every vibrational mode)

The number of vibrational modes of a non-linear molecule containing n atomsis equal to 3n− 6. Hence, classically, the total energy should be

Uclass = (3n− 3)NkBT (3.72)

It then follows immediately that, in the classical limit,

cclassicalV = (3n− 3)NkB (3.73)

However, this was not at all what was found in experiment. To be more precise:at high temperatures, this expression appeared to become valid, but at low tem-peratures it failed completely. It appeared as if the vibrations - and sometimeseven the rotations - were not contributing to the heat capacity. This problemwas resolved with the advent of quantum mechanics. Above we have discussedthe correct expression for the average rotational and vibrational energy. For avibration with frequency ν, the contribution to the internal energy is

< ǫvib >= −N

(

∂ ln qvib∂β

)

=

= Nhνi/2 +Nhνi

exp(βhνi)− 1(3.74)

85

At low temperatures, this energy approaches a constant value (hνi/2 per os-cillator). Let us write βhνi ≡ x. Then the expression for the heat capacityis

cV = NkBx2 exp(−x)

(1− exp(−x))2(3.75)

At high temperatures (x → 0) cV approaches NkB. But at low temperatures,the heat capacity goes to zero quite steeply. The explanation of this “freezingout” of the vibrational heat capacity was one of the early successes of the mergerof quantum mechanics and statistical thermodynamics.

3.3 Chemical Equilibria in ideal gas mixtures

One of the important results of classical thermodynamics was that it allowed usto write down the conditions for equilibrium. It starts from the condition that,at constant temperature and pressure, the Gibbs free energy G is at a minimumin equilibrium. A variation of the Gibbs free energy can be written as

dG = −SdT + V dP +∑

α

µαdnα (3.76)

where nα denotes the number of moles of species α and µα its chemical potential.At constant pressure and temperature, dP and dT are zero, and hence thecondition for equilibrium is

∑

α

µαdnα = 0 (3.77)

In a chemical reaction, the variations dnα are related through the stoichiometryof the reaction. This is usually denoted by

dnα = ναdξ (3.78)

where ξ measures the progress of the reaction and να is the stoichiometric co-efficient of species α. Hence, the condition for equilibrium is

∑

α

µαναdξ = 0 (3.79)

or, as dξ is arbitrary∑

α

µανα = 0 (3.80)

Using Eqn.3.67 for the (ideal gas) chemical potential of species α

µα = kBT ln(ραΛ3α)− kBT ln qαinternal

= kBT ln(pαkBT

Λ3α)− kBT ln qαinternal (3.81)

we find that∑

α

να

(

ln(pα/p

⊖

kBT/p⊖Λ3α)− ln qαinternal

)

= 0 (3.82)

86

where p⊖ is the pressure of the reference state (e.g. one bar). It then followsthat

∑

α

να ln(pα/p⊖) =

∑

α

να

(

ln(kBT

p⊖Λ3α

) + ln qαinternal

)

(3.83)

But∑

α

να ln(pα/p⊖) = ln(

∏

α

(pα/p⊖)να) = lnK (3.84)

where K is the equilibrium constant of the reaction. Hence we have now deriveda “molecular” expression for the equilibrium constant of a chemical reaction of(ideal) gases

lnK =∑

α

να ln(kBT

p⊖Λ3α

qαinternal) (3.85)

Before we can actually use this expression, we should note one important point.Until now, we have, somewhat arbitrarily chosen the ground state of the moleculeas our zero of energy. However, when we compare the relative stability of dif-ferent molecules, the difference in energy between the molecular ground statesbecome very important. We can account for this part of the internal energy bywriting

qαinternal = exp(−βǫα0 )qα,0internal (3.86)

where qα,0internal is the internal molecular partition function that is obtained whenthe zero of energy coincides with the molecular ground state. When we introducethe notation

∆ǫ0 =∑

α

ναǫα0 (3.87)

we obtain

lnK = −∆ǫ0kBT

+ ln∏

α

(

kBT

p⊖Λ3α

qα,0internal

)να

(3.88)

Remember that we can determine all terms on the right-hand side of this equa-tion either from quantum-mechanical calculations or from spectroscopic mea-surements. Hence, we have now achieved one of the prime objectives of Sta-tistical Thermodynamics: we have obtained an expression that allows us topredict the equilibrium constant of a gas reaction. Take for instance the reac-tion A2 + B2 ⇐⇒ 2AB. For this reaction, the equilibrium constant is givenby

lnK = −∆ǫ0kBT

+ ln

(

qAB,0

internal

Λ3

AB

)2

(

qAA,0

internal

Λ3

AA

)(

qBB,0

internal

Λ3

BB

) (3.89)

Using

Λ3 =(

h2/2πmkBT)3/2

(3.90)

87

qrot =2IkBT

σ~2(3.91)

qivib =1

1− exp(−βhνi)(3.92)

we can work out the individual (translational, rotational and vibrational) con-tributions to the equilibrium constant. Note that the expression for qivib is nowdifferent from Eqn. 3.14 because we have redefined the ground state energy.

3.3.1 Example

To give an example, consider the reaction

H2 +D2 ⇋ 2HD (3.93)

As H and D are isotopes, it is a reasonable approximation to assume thatall three molecules have the same bond-length and that they have the samevibrational force constant, kvib. The molecular ground-states are, however, notthe same, because the zero-point energy (hν/2) of the molecular vibration isdifferent. Hence,

∆ǫ0 =1

2(2hνHD − hνH2

− hνD2) (3.94)

The individual vibration frequencies can be computed using

vHD =1

2π

√

kvib[mH +mD]

mHmD(3.95)

and similar expressions for νH2and νD2

. Let us compute the equilibrium con-stant in the high-temperature limit.

lnK = −∆ǫ0kBT

+ lnΛ3H2

Λ3D2

Λ6HD

+ ln(qrotHD)2

qrotH2qrotD2

+ ln(qvibHD)2

qvibH2qvibD2

(3.96)

Let us compute the equilibrium constant in the high-temperature limit whereqrot = 2IkBT/(σ~

2) and qivib = kT/hνi. The translational contribution is

lnΛ3H2

Λ3D2

Λ6HD

= lnm3

HD

m3/2H2

m3/2D2

= ln33

23/243/2

= ln93/2

23/243/2

=3

2ln

9

8(3.97)

88

The rotational contribution is

ln(qrotHD)2

qrotH2qrotD2

≈ lnI2HD/σ2

HD

(IH2/σH2

)(ID2/σD2

)

= ln(2/3)2

(1/4)(1/2)= ln

32

9(3.98)

where we have used σH2= σD2

= 2 and σHD = 1. And, in addition,

I = µr2e (3.99)

with re(H2) ≈ re(D2) ≈ re(HD), while µH2= 1

2mH , µD2= mH and µHD = 2

3mH . The vibrational contribution is

ln(qvibHD)2

qvibH2qvibD2

≈ lnνH2

νD2

ν2HD

− 1

2ln

9

8(3.100)

and the total equilibrium constant is

lnK = −∆ǫ0kBT

+ ln 4 (3.101)

At very high temperatures, the first term becomes small, and the equilibriumconstant approaches 4. This seems surprising: why should the equilibrium beshifted towards the HD? The reason is that, due to the symmetry of H2 andD2, the number of accessible (rotational states) of these molecules is divided bytwo, and hence the product is divided by four. And, after all, the equilibriumconstant simply measures the ratio of the number of accessible states of theproducts and reactants.

3.4 Interacting systems

It is of course nice to be able to compute the thermodynamical properties ofideal gases but most substances that we know are not in a phase that resemblesthe ideal gas. This means that we should also be able to compute propertiesof systems consisting of interacting molecules. It turns out that this problemis most easily tackled in the classical limit. For convenience, we shall limit thediscussion to simple, spherical particles and we shall ignore the internal degreesof freedom of this particles. This is an excellent approximation for atoms, andit is quite reasonable for simple molecules. We then have to consider only thetranslation partition function. Above, we have already computed this partitionfunction for an ideal gas molecule

qtrans = V(

2πmkBT/h2)3/2

=V

Λ3(3.102)

Now recall that the molecular partition function was nothing but a measure forthe number of quantum states that was accessible to a molecule. It is simply

89

equal to the number of volumes of size Λ3 that are contained in a volume V . Inthis example, we have assumed that the ground-state energy of the molecule waszero. Now suppose that the molecule moves in a constant external potential U .In that case, all energy levels are shifted by an amount U and the translationalpartition function of the molecule becomes

q′trans =V exp(−βU)

Λ3(3.103)

or, in terms of the number of accessible quantum states: nacc = V exp(−βU)/Λ3.We now assume that the total volume of the system can be divided in to smallvolumes ∆V , such that the external potential may vary from one box to thenext, but is constant within a given box. The number of accessible quantumstates in the i-th box is nacc(i) = ∆V exp(−βUi)/Λ

3 and the total number ofaccessible quantum states (i.e. qtrans) is equal to

qtrans =∑

i

∆V exp(−βUi)/Λ3

≈ 1

Λ3

∫

dr exp(−βU(r)) (3.104)

Next consider the case that we have N interacting molecules. Every molecule(denoted by i) is located in a volume ∆V centered around a position ri. ∆Vis so small that the total potential energy due to the interaction between themolecules does not depend on where inside this volume the molecules are lo-cated. Then the number of accessible quantum states for this specific molecularconfiguration is:

nstates(r1, r2, .., rN ) =∆V N

N !Λ3Nexp(−βU(r1, r2, .., rN )) (3.105)

the factor 1/N ! appears again because permuting identical molecules over thedifferent volumes ∆V does not yield a new quantum state. Finally, to computethe total number of accessible quantum states for the entire system, we shouldsum over all possible positions of the volume elements containing the molecules

Qtrans =1

N !Λ3N

∑

r1

∑

r2

...∑

rN

∆V N exp(−βU(r1, r2, .., rN ))

≈ 1

N !Λ3N

∫

dr1

∫

dr2...

∫

drN exp(−βU(r1, r2, .., rN )) (3.106)

This is the (classical) partition function for a system of N interacting molecules(now derived via a different route as in Chapter 1). If we could compute itfor any substance, we would be able to predict all equilibrium properties ofmatter. In practice, we can only compute the partition function of a system ofinteracting particles in a few very special cases (e.g. for a perfect crystal at lowtemperatures). In general, we need to use numerical simulations to computethe thermodynamic properties of normal liquids or solids. The technicalities ofthese simulations do not concern us here. The essential fact is that StatisticalThermodynamics tells us exactly what to compute.

90

3.4.1 Virial coefficients

As an example of a calculation that involves intermolecular interactions, letus consider deviations from the ideal gas law. Experimentally, we know thatthe pressure of a real gas does not satisfy the ideal-gas relation PV/NkBT =1. Rather, we find that, as the number density ρ (=N/V ) is increased, thedeviations from this relation occur

PV

NkBT= 1 +B2ρ+B3ρ

2 + ... (3.107)

where B2, B3 etc. are called the second, third etc. virial coefficients. The virialcoefficients depend on the intermolecular interactions. Here we shall derive anexpression for B2. First, we multiply and divide the translational partitionfunction by V N . This yields

Qtrans =V N

N !Λ3N

∫

dr1V

∫

dr2V

...

∫

drNV

exp(−βU(r1, r2, .., rN ))

=V N

N !Λ3N〈exp(−βU)〉 (3.108)

where the angular brackets denote the average of the Boltzmann factor exp(−βU(r1, r2, .., rN ))over all possible positions {r1, r2, .., rN} within the volume V . At extremely lowdensities, the molecules are almost always too far apart to interact and hencethe average Boltzmann factor is simply equal to one. At higher densities, weshall notice the effect of interactions between molecules. Let us assume that theinteractions between molecules are such that a molecule must be within a cer-tain distance rc in order to experience the potential of the other molecules. Or,phrased in another way, if there are no molecules within a volume vc = (4π/3)r3cof a given molecule, then that molecule does not contribute to the interactionenergy. Let us denote by P0 the probability that no two molecules are within adistance rc. At very low densities, we can write the average of the Boltzmannfactor as

〈exp(−βU)〉 = P0 × 1 + P1 < exp(−βU) >pair (3.109)

where P1 denotes the probability that there is exactly one pair of moleculesat a distance less than rc. Because the density is very low, we can ignore theprobability that there will be more than two molecules at a distance less than rc.In other words, we have either no molecules that are interacting (probability P0)or just one pair (probability P1). Clearly, P0 + P1 = 1 and hence P0 = 1− P1.Now we should still compute the average Boltzmann factor for a pair of moleculesat a distance less than rc. It is

< exp(−βU) >pair=1

vc

∫

vc

dr exp(−βu(r)) (3.110)

91

where u(r) is the potential energy of interaction of a pair of molecules at adistance r. We can now write

〈exp(−βU)〉 = P0 + P11

vc

∫

vc

dr exp(−βu(r))

= 1− P1 + P11

vc

∫

vc

dr exp(−βu(r))

= 1− P1

(

1

vc

∫

vc

dr

)

+ P11

vc

∫

vc

dr exp(−βu(r))

= 1 + P11

vc

∫

vc

dr[ exp(−βu(r)) − 1] (3.111)

As we have assumed that the intermolecular interaction vanishes outside vc (andhence exp(−βu(r))− 1 = 0), we need not limit the integration to the volume vcbut can extend it over all space. Now we should still compute P1, the probabilitythat there is a single pair of (randomly distributed) molecules within the samevolume vc. At low densities, the probability that there is another molecule in avolume around a given molecule is simply equal to ρvc (where ρ is the numberdensity of the molecules). As there are N molecules in the system, and we couldhave taken any of these molecules as our “central” molecules, the probabilityto find a pair is N/2 times larger (the factor 1/2 comes in to avoid doublecounting). Hence, at low densities, P1 = Nρvc/2 and hence

〈exp(−βU)〉 = 1 +Nρ

2

∫

dr[ exp(−βu(r)) − 1] (3.112)

With this result, we can write

Qtrans ≈V N

N !Λ3N

(

1 +Nρ

2

∫

dr[ exp(−βu(r))− 1]

)

(3.113)

The pressure P is given by

P = kBT∂ lnQtrans

∂V

≈ NkBT

V−

ρ2

2

∫

dr[ exp(−βu(r)) − 1]

1 + Nρ2

∫

dr[ exp(−βu(r)) − 1]

≈ ρkBT +ρ2

2

∫

dr[1− exp(−βu(r))] (3.114)

where, in the third line, we have used the fact that, at sufficiently low densities,Nρ2

∫

dr[ exp(−βu(r)) − 1] ≪ 1. If we compare this expression with the virialseries (Eqn.3.107), we find that the second virial coefficient is equal to

B2 =1

2

∫

dr[1− exp(−βu(r))] (3.115)

92

0 0r

σ

r

ǫ

λσ

σuHS(r) uSW (r)

Figure 4: Left: the hard-sphere potential uHS. Right: the square-well potentialuSW

Again, this is a very important result because it shows that a measurementof the second virial coefficient provides information about the intermolecularinteractions. To give a specific example: assume that molecules are hard sphereswith a diameter σ. For r > σ, u(r) = 0 and hence exp(−βu(r)) = 1. For r < σ,u(r) = ∞ and hence exp(−βu(r)) = 0. Therefore,

BHS2 =

1

2

∫ σ

0

4πr2dr=2πσ3

3(3.116)

Of course, real molecules do not only repel each other at short distances, theyalso attract at larger distances. A very simple model potential that exhibitsboth features is the so-called square-well potential. The square-well potential isequal to the hard-sphere potential for r < σ. But for σ < r < λσ (with λ > 1).the square well potential is attractive:

usw(r) = −ǫ (for σ < r < λσ) (3.117)

anduSW (r) = 0 (for r > λσ) (3.118)

We can easily compute the second virial coefficient for this model potential. Itis

BSW2 =

1

2

(

∫ σ

0

4πr2dr − (exp(βǫ)− 1)

∫ λσ

σ

4πr2dr

)

=2πσ3

3

(

1− (exp(βǫ)− 1)(λ3 − 1))

(3.119)

At very high temperatures (β → 0), BSW2 is equal to the hard-sphere second

virial coefficient. However, at low temperatures, the term with exp(βǫ) domi-nates, and B2 becomes large and negative. The point where B2 changes sign

93

is called the Boyle temperature. At that temperature B2 vanishes and the gasbehaves almost ideal. The value of the Boyle temperature follows from theequation:

(

1− (exp(βǫ)− 1)(λ3 − 1))

= 0 (3.120)

which yields

exp(βǫ) =λ3

λ3 − 1(3.121)

orkBT

ǫ=

1

ln[λ3/(λ3 − 1)](3.122)

If we keep ǫ fixed and vary λ, we see that for large λ (long-ranged attraction),kBT/ǫ is large (high Boyle temperature), while for λ → 1, kBT/ǫ → 0 (lowBoyle temperature). If the Boyle temperature is high, at room temperature thesecond virial coefficient B2 is negative. Then it is possible for gas to condenseto liquid. But if the gas does this depends on the the free energy of the system.

It should be stressed that only the lowest few virial coefficients can easilybe computed. Hence, when we are interested in the behavior of dense systems(liquids, solids), the properties of such systems cannot be computed analyt-ically from Eqn.3.106. The standard approach to resolve this problem is touse numerical simulations instead. With the help of computer simulations, itis possible to compute the thermodynamical properties of arbitrarily complex(macro)molecular systems provided that: 1.) we can use a ”classical” descrip-tion (i.e. Eqn.3.106 is valid) and 2.) we have a good knowledge of the potentialenergy function U(r1, r2, .., rN ).

3.5 Chemical equilibrium for interacting systems

In section 3.3 we encountered ideal gas chemical reactions. Now we have theexpressions to describe the general situation of a reaction equilibrium, startingfrom the classical partition function. We start with the condition for a chemicalequilibrium

∑

α

µαναdξ = 0 (3.123)

We will now derive the expression for chemical potential for interacting systemsin the canonical ensemble, via a scheme that is related to a method devised byWidom to compute chemical potentials in computer simulations. For simplicityof notation, we will focus on a one component system and drop the subscript j.We consider N particles in a cubic box of volume V = L3 and temperature T .The (classical) canonical partition function of this system is given by

Q(N, V, T ) =V N

Λ3NN !

∫ 1

0

dsN exp[−βU(sN , L), (3.124)

in which the scaled coordinates sN = rN/L are introduced. In Eq. 3.124, U is awritten as a function of sN and L to indicate that the energy depends on the real

94

rather than on the scaled coordinates. Note that the interaction U(sN , L) doesinclude all the (now classical) internal vibrational and rotational contributions.The expression for the Helmholtz free energy is given by

F (N, V, T ) = −kBT lnQ

= −kBT ln

(

V N

Λ3NN !

)

− kBT ln

(∫

dsN exp[−βU(sN ;L)]

)

= Fid(N, V, T ) + Fex(N, V, T ). (3.125)

The last line expresses the fact that the Helmholtz free energy consists of an idealgas part plus an excess contribution. The definition of the chemical potential,µ = (∂F/∂N)V,T , shows that for sufficiently large N , the chemical potential isgiven by

µ = −kBT ln

(

QN+1

QN

)

. (3.126)

Using Eq. 3.125, we find that

µ = −kBT ln

(

V

Λ3(N + 1)

)

− kBT ln

(∫

dsN+1 exp[−βU(sN+1)]∫

dsN exp[−βU(sN)]

)

= µid +∆µex (3.127)

This shows that the chemical potential has an ideal gas contribution plus anexcess contribution. The ideal gas chemical potential can be evaluated analyt-ically. Note that for polyatomic gasses the excess chemical potential containsalso the contributions from the internal partition function. The excess chemicalpotential can be computed in a simulation via Widom’s insertion method. Tothis end, we separate the potential energy of the N+1 particle system, U(sN+1)into the potential energy of the N particle system, U(sN ), and the interactionenergy of the N + 1th particle with the rest: U(sN+1) = U(sN) + ∆U . Usingthis separation, the excess chemical potential can be written as

∆µex = −kBT ln

∫

dsN+1〈exp(−β∆U)〉N , (3.128)

where 〈. . . 〉 denotes canonical ensemble averaging over the configuration spaceof the N -particle system. This average can be computed by Metropolis MonteCarlo sampling (see Chapter 4). In practice, the procedure is as follows: wecarry out a conventional Monte Carlo simulation on a system of N particles. Atfrequent intervals during the simulations, we randomly generate a coordinatesN+1, uniformly over the unit cube. We then compute exp(−β∆U). By averag-ing this quantity over all trial positions, we obtain the average that appears inEq. 3.128. To the extent that all positions are equivalent, the excess chemicalpotential is the reversible work to insert a particle at a particular position; or, ifthe system is not homogeneous, the excess chemical potential is the average ofthe reversible work to insert a molecule at a given position. The ideal gas parttakes care of the entropy associated with the translational degrees of freedom.

95

3.5.1 Mixture

Consider a system of NA molecules of A and NB molecules of B, which aresolvated in NS solvent molecules. The partition function is given by

Q(NA, NB, NS) = QidAQid

BQidS (3.129)

×∫

dsNAdsNBdsNSe−β(UAA+UBB+UAB+UAS+UBS+USS).

where Qidj = VNj

/(Λ3NjNj !) denotes the ideal gas partition function of themolecule j ∈ {A,B, S}. We now image that the concentrations of A and Bare low, so that we can ignore the interactions between them. The partitionfunction is then given by

Q(NA, NB, NS) = QidAQid

BQidS

×∫

dsNAdsNBdsNSe−β(UAS+UBS+USS). (3.130)

We are now interested in the free energy of this system relative to that of thesolvent only:

e−β∆F =Q(NA, NB, NS)

Q(NS)

= QidAQid

B

∫

dsNAdsNBdsNSe−β(UAS+UBS+USS)

∫

dsNSe−βUSS.

= QidAQid

B

∫

dsNAdsNBdsNSe−β(UAS+UBS+USS)

∫

dsNSe−βUSS

∫

dsNSe−βUSS

∫

dsNSe−βUSS.

= QidAQid

B

∫

dsNAdsNSe−β(UAS+USS)

∫

dsNSe−βUSS

∫

dsNBdsNSe−β(UBS+USS)

∫

dsNSe−βUSS

= QidAQid

B

∫

dsNA〈e−βUAS〉S∫

dsNB〈e−βUBS 〉S (3.131)

The subscript S indicates that the quantities exp(−βUAS) and exp(−βUBS) areaveraged over the ensemble of configurations of solvent molecules. In going fromEq. 3.131 to 3.131, we used that the solvation of the A and B molecules are as-sumed to be independent events. The energy UAS =

∑NA

j=1

∑NS

k=1 uAS(rA,j , rS,k),and similarly for B. By iterating the trick applied in Eq. 3.131, we obtain

e−β∆F = QidAQid

B

NA∏

j=1

∫

dsj〈e−βuAS 〉SNB∏

j=1

∫

dsj〈e−βuBS〉S

= QidAQid

B qNA

cm,AqNB

cm,B, (3.132)

where we have defined the partition function of the center-of-mass

qcm,A = exp(−β∆µex,A). (3.133)

96

By taking the derivative of ∆F with respect to Nα, it can be verified that thechemical potential of species α is given

µα = kBT ln

(

NαΛ3

V

)

+∆µex,α

= kBT ln(

ραΛ3)

+∆µex,α. (3.134)

with ρα = Nα/V is the number density of species α.

3.5.2 Equilibrium constant

The derivation of the equilibrium constant of a chemical reaction follows thatof section 3.3. The chemical equilibrium condition is given by Eq. 3.80:

0 =∑

α

ναµα. (3.135)

Inserting the Eq. 3.134 into the above equation yields

∑

α

να

(

ln

(

ραρ0Λ3α

ρ0

)

+∆µex,α

)

= 0 (3.136)

where ρ0 is a reference density (e.g. 1 mol/liter). We can rewrite this equationas

ln∏

α

(

ραρ0

)να

= −β∆G0 − ln∏

α

(

Λ3αρ0)να

, (3.137)

where we have defined ∆G =∑r

j=1 ∆µex,j. Using the familiar equilibrium con-stant via the law of mass action

Keq ≡∏

α

(

ραρ0

)να

(3.138)

= exp(−β∆G)∏

α

(

ρ0Λ3α

)να. (3.139)

Note the similarity with Eqn 3.85, when realizing that for polyatomic molecularthe excess chemical potentials contains the internal partition function.

To illustrate the chemical equilibrium condition, consider the reaction

A+B ⇄ C (3.140)

The chemical equilibrium condition implies:

µA + µB = µC , (3.141)

ln

(

ρAρBρCρ0

)

= ∆µexC − (∆µex

A +∆µexB ) ≡ ∆G (3.142)

The free-energy difference ∆G is the difference in excess chemical potential, andis independent of the densities of the species A,B and C. Chemical equilibriumis established when the densities are given by Eq. 3.142. If the densities do notobey Eq. 3.142, the system can lower its free energy until Eq. 3.142 is obeyed.

97

3.5.3 Example: Binding of a molecule

Another illustrative example is the binding of a molecule, e.g. the binding ofa gene regulatory protein to a specific site on the DNA to turn on or turn offthe expression of a gene. We first treat this problems in the grand-canonicalensemble. The grand-canonical partition function is

Ξ =

1∑

n=0

exp[−β(n(ǫ − µ)], (3.143)

where ǫ the binding energy, and µ the chemical potential. The variable n = 0, 1indicates whether the binding site is occupied (n = 1), or unoccupied (n = 0).Writing the sum explicitly yields:

Ξ = [1 + exp(−β(ǫ − µ)], (3.144)

from which follows that

βPV = lnΞ = ln(1 + exp(−β(ǫ − µ)). (3.145)

The average occupation number is given by 〈n〉 = ∂ ln Ξ/∂(βµ):

〈n〉 = 1

exp(β(ǫ − µ)) + 1. (3.146)

Note that this has the form of the Fermi distribution.Let us now consider the problem in the canonical ensemble. We considerN B

molecules with solvation free energy ∆µex,B = −kBT ln qcm,B in a volume V thatcan bind to one binding site with solvation free energy ∆µex,O = −kBT ln qcm,O.The canonical partition function for one unoccupied binding site (n = 0) and Nmolecules is given by (see Eq. 3.132):

Q(n = 0, N, V, T ) =(V/Λ3)N

N !qcm,Oq

Ncm,B. (3.147)

The partition function for an occupied binding site

Q(n = 1, N − 1, V, T ) =(V/Λ3)N−1

(N − 1)!qcm,OBq

N−1cm,B (3.148)

where qcm,OB = exp(−β∆µex,OB) is the partition function of the occupied bind-ing site. The probability that the binding site is occupied is then given by

〈n〉 =Q(1, N − 1, V, T )

Q(0, N, V, T ) +Q(1, N − 1, V, T )(3.149)

=1

Q(0, N, V, T )/Q(1, N − 1, V, T ) + 1(3.150)

=1

exp(−β∆G)/(ρBv) + 1(3.151)

=1

KD/ρB + 1, (3.152)

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where ∆G = ∆µex,O +∆µex,B −∆µcm,OB is the binding free energy and KD =exp(−β∆G) is the dissociation constant for the reaction

OB ⇆ O+B (3.153)

We see that ǫ can be identified with −∆G and µ with the kBT ln ρB.Finally, let us consider M binding sites; binding site j has affinity ǫj . We

use the grand-canonical partition function since that is convenient. Note thatthis problem bears strong similarities of that of Langmuir absorption, discussedin chapter 1. The grand-canonical partition function given by

Ξ =

1∑

n1=0

1∑

n2=0

· · ·1∑

nM=0

exp

−β

M∑

j=1

nj(ǫj − µ)

(3.154)

Since the binding events at the different sites are uncorrelated, we can factorisethe partition function:

Ξ =

M∏

j=1

1∑

nj=0

exp(−β(ǫj − µ)nj)

(3.155)

=

M∏

j=1

[1 + exp(−β(ǫj − µ)] (3.156)

The free energy is thus given by

βPV = lnΞ =

M∑

j=1

ln [1 + exp(β(µ− ǫj))] . (3.157)

We can now obtain the binding occupancy of site j via

〈nj〉 =∂ ln Ξ

∂ − βǫj(3.158)

which yields

〈nj〉 =1

exp(β(ǫj − µ)) + 1. (3.159)

Note that this result is identical to that of Eq. 3.146. The reason, is of course,that binding at the different sites occurs independently.

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