Statistical Thermodynamics in Chemistry and Biology - …Statistical Thermodynamics in Chemistry and Biology Introduction Per-Olof Astrand˚ D3-119 Realfagsbygget, Department of Chemistry,

Statistical Thermodynamics in Chemistry andBiologyIntroduction

Per-Olof Astrand

D3-119 Realfagsbygget, Department of Chemistry,Norwegian University of Science and Technology,

[email protected]

January 6, 2015

Introduction

Course in statistical thermodynamics andthermodynamics

I Lecturer: Per-Olof AstrandTue, 10-12, R3; Thu, 08-10, [email protected]

I Exercises: Mehdi Mahmoodinia, Erlend Grøndahl JansenMo, 8-10, K5; Tu, 14-16, F6The first exercise session: Mon 12.01

I Lectures will this year be in English. Teaching assistants also speakEnglish.

I Last date for teaching activities: Tue 21.04

Per-Olof Astrand Statistical Thermodynamics in Chemistry and Biology January 6, 2015 2 / 11

Introduction

Literature

I Book: Ken A. Dill and Sarina Bromberg,Molecular Driving Forces: Statistical Thermodynamics in Biology,Chemistry, Physics, and Nanoscience, 2nd ed., Garland Science, 2010.

I Some exercises are taken from the book.Solutions will not be handed out. A list of answers is provided.

I Additional exercises (including previous exams) are available withsolutions.

I A basic goal is to learn how to solve exercises.


Introduction

Exam

I Exam date: 27.05, 15.00-19.00I Written exam (code A: it is allowed to bring books and notes)I Learn to use the book and own notes!I No obligatory activities


Introduction

Learning outcome

I The course introduces the basis of statistical thermodynamics withexamples from chemistry and biology

I Introductory courseI General course - knowledge applicable in “all” fieldsI Fundamental course - generic phenomenaI Theoretical course


Introduction

Learning outcomeFrom the study handbook

After the course, the student is expected to be able to:I Explain the basic concepts and principles in statistical thermodynamics.I Use lattice models to study basic phenomena in chemistry and

nanoscience.I Construct new models based on the basic principles in statistical

thermodynamics.


Introduction

Learning methods

I One overview lecture on most chaptersI Some chapters (mathematics) are given only as exercisesI Around 3-5 suggested exercises per chapter (from the book)I Around 2-3 guest lectures?I Focus this year: extend the number of own-developed exercises with

solutions (from old exams)


Introduction

Communication

I Reference group - three personsI It’s learning for information/messages from meI Ask questions! Both on new and old topics.I Ok to ask the teaching assistants on ”old” exercises.I A discussion forum on It’s learning.I How to use video recordings from last year?


Introduction

Some advice

I Everything is connected in the course. We use all the previous materialall the time.

I Thus:I Work continuously: not everything is easy to understand from the start, but it

is the same concepts coming over and over again.I Do not lag behind. Drop some of the recommended exercises on the

previous chapter, and return to them later.I Work together in small groups. Both discussing and explaining the text to

each other, and solving the exercises.


Introduction

Contents

I What is statistical thermodynamics (statistical mechanics)?I Connection between microscopic and macroscopic theory

I What is thermodynamics?I Macroscopic theory

I Nanoscience is at the mesoscopic level. What does that mean?


Introduction

Questions to you

I What is entropy?I What is the chemical potential?I What is the hydrophobic effect?

Hand in written answers to me. One page in total for all three questions.


Statistical Thermodynamics in Chemistry andBiology

Statistical thermodynamics

Per-Olof Astrand


[email protected]

January 5, 2015


Last time: Questions to you

I What is entropy?I What is the chemical potential?I What is the hydrophobic effect?



What is statistical thermodynamics?The microscopic world

I The microscopic world is described by quantum mechanics; theSchrodinger equation:

Hψi = εiψi

where εi is the energy of state i .I For matter, terms like atoms, molecules, and electronic states are key

concepts.

The macroscopic worldI The macroscopic world is described by thermodynamics.I A system is described in terms of pressure, density, temperature, free

energy, etc.


Statistical thermodynamics provides the connection between the microscopicand macroscopic worlds. It gives the possibility to determine thermodynamicsproperties from molecular (quantum mechanical) models.



An example of a statistical mechanical modelI The temperature, T , is a macroscopic property and an experimental

observable that may be written as an average:

T = 〈T 〉 = 〈T (t)〉 = 1τobs

τobs∑τ=1

T (τ)

where we emphasize that it is an average in 〈T 〉 and an average overtime, t in 〈T (t)〉. τobs is the number of observations.

I In particular,

T =2〈K 〉3NkB

where kB is Boltzmann’s constant and the kinetic energy, K , is given byclassical mechanics as a sum over all the particles, N, as

K =N∑

i=1

miv2i

2

and can be calculated at each observation.Per-Olof Astrand Statistical Thermodynamics in Chemistry and Biology January 5, 2015 4 / 4


1. Principles of probability

Per-Olof Astrand


[email protected]

January 5, 2015


Foundations of entropy

Probability and entropy

The reason we have to deal with probability, multiplicity and combinatorics isthat it is the foundation of describing entropy.

Definition of entropy, S

In Chapter 6, Boltzmann’s law:

S = kB ln W

where kB is Boltzmann’s constant. We have a link between the macroscopicquantity, the entropy, S, and the multiplicity, W , of the microscopic degrees offreedom.



What is probability?I Let’s have in total N trials/events/measurements.I The outcome of a trial/event/measurement may be A, B, C, etc. In total

we have t distinct outcomes.I nA, nB, nC , etc. are the number of outcomes for each category, A, B, C,

etc.I The probability, pA, of outcome A is

pA =nA

N

wheret∑

J=1

nJ = N andt∑

J=1

pJ = 1

I We will often use its differential form (regard N as a constant)

dpA =dnA

N;

t∑J=1

dnJ = 0 andt∑

J=1

dpJ = 0



Simple example: two dice

Let’s role two dice. The outcome of each event may be described by a pair ofnumbers, (a,b), each in the range 1 − 6.

I What is the probability of the outcome (6,6)? We have in total 6 × 6 = 36possibilities. The probability is thus 1

36 = 0.0278.I What is the probability of the outcome (5,5) or the outcome (6,6)? It is

simply the sum of the probabilities for each outcome: 236 = 0.0556

I What is the probability of the total score 12? What is the probability ofgetting the total score 11? : The probability of the total score 12 is thesame as in the previous question since (6,6) is the only way to obtain 12.There are two ways to obtain 11, (5,6) and (6,5), so the probabilitybecomes 2

36 .



Simple example: two dicePart 2

I Let’s role the dice twice:I What is the probability that the first and the second outcomes are (6, 6)? It is

simply the product of the probabilities for each outcome: 1362 = 0.00077

I What is the probability that the first or the second outcome is (6, 6)? It iseasier to calculate one minus the probability for not obtaining (6, 6) in thefirst and second outcome: p = 1−

( 3536

)2= 0.0548

I What are the limiting values, N →∞, if we repeat the event N times? Theprobability to get (6, 6) in each event approaches zero for large N. Theprobability to get (6, 6) once approaches one for large N.



Second example: multiplicity

I N identical, non-interacting particles (molecules)I Each particle has discrete energy states, εi , given by the Schrodinger

equation in quantum mechanics: Hψi = εiψi

I The total energy, E , is given as a constraintI We have

I n0 particles with energy ε0 on level 0,I n1 particles with energy ε1 on level 1, etc.

I We get (as a sum over all states)

N =∞∑i=0

ni ; E =∞∑i=0

εini

I Example: E = Nε0 (all particles in the ground state)I For E larger than Nε0: many possibilities



Second example: multiplicityPart 2

Simple question

In how many different ways can we put n0 particles on level 0, n1 particles onlevel 1, etc.?

I Example: N = 12, n0 = 5, n1 = 4, n2 = 2, n3 = 1I We start by chosing particles at level 0:

I First particle: N possibilities,I Second particle: N − 1 possibilities, etc.I Total number of possibilities: N(N − 1)(N − 2)(N − 3)(N − 4)I This is correct if the particles are distinguishableI If indistinguishable: 5! identical combinations.I Total number of possibilities:

N(N − 1)(N − 2)(N − 3)(N − 4)5!



Second example: multiplicityPart 3

I For n0 = 5,N(N − 1)(N − 2)(N − 3)(N − 4)

5!I For n1 = 4,

(N − 5)(N − 6)(N − 7)(N − 8)

4!I For n2 = 2,

(N − 9)(N − 10)

2!I For n3 = 1,

(N − 11)

1!I The total number of combinations, Ω, (the multinominial distribution),

Ω =N!

(n0!)(n1!)(n2!)(n3!)=

N!∏i

ni !

which we also denote the multiplicity, W .Per-Olof Astrand Statistical Thermodynamics in Chemistry and Biology January 5, 2015 8 / 11


Third exampleI Constraint: Total energy, E = 4ε, is constantI Model system: 4 molecules in three energy levels: ε0, ε1, and ε2.I Equidistant energy levels: ε0 = 0, ε1 = ε, ε2 = 2εI Three ways to fulfill the constraints, E = 4ε and N = 4:

I 2ε2 + 2ε0I ε2 + 2ε1 + ε0,I 4ε1,

I We get the multiplicities,

W1 =N!∏

ini !

=4!

2!0!2!= 6

W2 =N!∏

ini !

=4!

1!2!1!= 12

W3 =N!∏

ini !

=4!

0!4!0!= 1

I Which one is most common? Entropy? Which one do we regard as mostunordered?



First example of lattice model

I Regard a surface where molecules A and B are placed on a surface:nA = 20, nB = 20, N = nA + nB.

I Regard two cases: We put a label on each molecule (left), and we onlylabel them by the type of molecule A and B (right).

A1 B5 B10A11A17B20 A3 B6

A20 A7 A18B18 B7 A19B19 B8

B17 A8 A2 B9 A14B16A15 A4

B1 A12B14A10B15 A6 A16B13

A9 B3 B12 B2 A13 A5 B11 B4

A B B A A B A BA A A B B A B BB A A B A B A AB A B A B A A BA B B B A A B B

I Left: W = N! Right: W = N!nA!nB !

I The ”right” case will be used all the time for molecular systems.



Summary

I Discussed probabilityI Introduction to combinatorics to give the multiplicityI Not discussed (but used later): average, variance, etc.I Some questions for thought:

I What happens if we put two objects with different temperature in contact?I What happens if we put a drop of ink in water?I Can we understand this behaviour already now?



2. Extremum principles predict equilibria

Per-Olof Astrand


[email protected]

January 13, 2015


What are extremum principles?

I The driving forces on molecules are of two kinds:I The energy (in covalent bonds, intermolecular interactions) is minimized.I The entropy is maximized.

I Extremum (or variational) principles are obtained by minimizing ormaximizing certain mathematical functions.



States of equilibrium - types of extrema (1)

I Stable equilibrium (at x = x∗):

V (x)− V (x∗) > 0 for all x 6= x∗

dVdx

= 0 andd2Vdx2 > 0 at x = x∗

I Neutral equilibrium (at x = x∗):

dVdx

= 0 for all x



States of equilibrium - types of extrema (2)

I Metastable equilibrium (at x = x∗):

V (x)− V (x∗) > 0 for small |x − x∗|

dVdx

= 0 andd2Vdx2 > 0 at x = x∗

V (x)− V (x∗) < 0 for large |x − x∗|

I Unstable system (at x = x∗):

V (x)− V (x∗) < 0 for all x 6= x∗

dVdx

= 0 andd2Vdx2 < 0 at x = x∗



An extremum principle for maximum multiplicity

Maximizing multiplicity predicts the highest probable outcome:I Model system: 4 coin tossesI Two of the possible outcomes: HHHH and HTHHI Which sequence is most probable? They are equally probable: 1

16I Which composition is most probable? 4 of 16 sequences are composed

by 3 H and 1 T; 1 of 16 sequences are composed by 4 H only

Maximizing multiplicity is equivalent to maximizing entropy.



Predicting heads and tails by a principle of maximummultiplicity

I N coin tosses; n heads; (N − n)tailsI The multiplicity, W , is given as

W (n,N) =N!∏

ini !

=N!

n!(N − n)!

I Illustrated for N = 10, N = 100, N = 1000.

I For large N, sharper and sharper peaks.I Examples: W (50,100) = 1.01× 1029 and W (25,100) = 2.43× 1023

I For large N, we only need to consider W max



Stirling’s approximationStirling’s approximation1 is common in statistical mechanics since factorialsare often difficult to manipulate mathematically. It is valid for large n.

I Simpler approximation:

ln n! ≈ n ln n − n ⇒ n! ≈(n

e

)n

I Better (less common) approximation:

ln n! ≈ 12

ln 2π +

(n +

12

)ln n − n ⇒ n! ≈

√2πn

(ne

)n

Exercise:Test Stirling’s approximation forn = 5,10,50,100,500,1000,10000,105,106, . . ..Hint: Use

ln n! =n∑

m=1

ln m

for large n for the ”exact” calculation1See Appendix B



Results for exercise: Stirling’s approximation

n ln n! n ln n − n 12 ln 2π +

(n + 1

2

)ln n − n

5 4.7875 3.0472 4.770810 15.1044 13.0259 15.096150 148.4778 145.6012 148.4761

100 363.7394 360.5170 363.7385500 2611.3305 2607.3040 2611.3303

1000 5912.1282 5907.7553 5912.128110000 82108.9278 82103.4037 82108.9278

100000 1051299.2219 1051292.5465 1051299.22191000000 12815518.3847 12815510.5580 12815518.3847



Lattice model for particles and “pressure”

I N = 3 spherical particlesI Three different “volumes”: MA = 5, MB = 4, MC = 3I What value of M maximizes the multiplicity?I Or what happens if the volume is suddenly expanded?I Analogous to coin tossing: [occupied, vacant, vacant, ...]I Multiplicity obtained as

W (N,M) =M!

N!(M − N)!

I WA(3,5) = 10, WB(3,4) = 4, WC(3,3) = 1I The particles will be “fully spread out”



Lattice model for diffusion and mixing

I Four black particles and four white particlesI The total volume is fixedI What is the most probable composition?I W = Wleft ·Wright

I WA = 4!2!2!

4!2!2! = 36, WB = 4!

1!3!4!

3!1! = 16, WC = 4!0!4!

4!4!0! = 1

I The driving force for getting the same “concentration” everywhere isnamed chemical potential



Lattice model for polymers

I Two-dimensional lattice model for a polymer with 3 monomers andconnected to a wall.

I All the possible sequences are given in the figure.I Which is the most probable composition, i.e. length of the polymer?



Summary

I We will use extremum principles. The (free) energy is minimized. Theentropy is maximized.

I For large systems, only the maximum multiplicity is of importance. Thenalso Stirling’s approximation may be used.

I The multiplicity of the system gives the entropy of the system, i.e. themultiplicity of the system is to be maximized.

I Considered three model systems: pressure, chemical potential, andpolymer length.



Exercise E2.1

Consider N particles with a dipole moment placed on a quadratic lattice(surface). N is a large number so that the system can be regarded to bemacroscopic. The dipole moment can point in one of four directions in theplane of the surface: →,←, ↑ or ↓. Calculate and compare the multiplicity fortwo ”cases”: a) we first place N identical particles on the lattice, then weassume that each dipole moment can point in any direction; b) we have N/4particles with each dipole moment that are placed on the lattice [Hint: useStirling’s formula, ln N! = N ln N − N, that is valid for large N.]



3. Heat, work and energy

Per-Olof Astrand


[email protected]

January 15, 2015


Conservation laws

Some properties are conserved:I Linear momentum (but not velocity)∑

i

~pi =∑

i

mi~vi

I MassI Energy (but not heat)I Not mentioned here: charge is also conserved. Electroneutrality is

actually a very strong macroscopic condition.I A property that is conserved can only flow from one place to another.



Force and workForceNewton’s second law defines the force in terms of acceleration as

~F = m~a = md2~rdt2

Work: ball on a springI Intrinsic force, f = −kx , from the springI Applied force, fap = −fI The work, dw , on the system by moving the ball

towards right a distance, dx ,

dw = fapdx = −fdx

I The total work is given as

w =

x2∫x1

dw =

x2∫x1

fapdx = −x2∫

x1

fdx = k

x2∫x1

xdx =k2(x2

2 − x21)



Energy

Kinetic energy

The kinetic energy, K , of a particle with a mass, m and a velocity, ~v is given as

K =12

m~v · ~v =12

m(v2

x + v2y + v2

z)

=12

mv2

Potential energy

The potential energy, V , is given by the position only; it is the work an objectcan perform by virtue of its position. Coulomb’s law is an example:

V =q1q2

4πε0R12

Total energy

The total energy, E , is conserved:

E = K + V



The first law of thermodynamics

The internal energy, U, which is the sum of heat, q, and work, w , isconserved:

U = q + w

I Heat and work can be interconvertedI Work can be of various forms: mechanical, electrical, magnetic, etc.I Heat radiation: electromagnetic radiation and heat can be interconverted:

Planck’s law:∆E = hν = ~ω



Atoms and molecules have quantized energiesI For a system of non-interacting particles (such as

an ideal gas), the total microscopic energy, E ,

E =∞∑i=0

niεi

I The (average of the) microscopic energy, E , isthe same as the internal energy, U.

I If U is increased by heating, the energy levels donot change. The population, ni , change.

I In contrast, work changes the energy levels, εi ,not the populations.

I Consider a change of the energy, dE ,

dE =∞∑i=0

(dniεi + nidεi ) = dq + dw = dU



Why does heat flow?

I We have previously demonstrated that a gas expands because themultiplicity, W (V ), increases with the volume, V . That defines pressure.

I We have also demonstrated that particles mix because the multiplicity,W (N), increases as particle segregation decreases. That defineschemical potential.

I These are manifestations of the principle that systems tend to their stateof maximum multiplicity.

I We will now investigate the flow of heat from hot to cold.



How does multiplicity depend on energy?

I Miniaturized model of a material: three particles.I i=0, 1, 2,. . ., energy levels with energy, εi = iI W (U) is an increasing function with U.



Why does energy exchange?

I Two systems, A and B. Eachsystem has 10 particles and 2energy levels, ε0 = 0 and ε1 = 1.

I Given internal energies for each subsystem, UA and UB, the multiplicityW : W (UA,UB) = WA(UA) × WB(UB)

Start system: W (2,4) =10!

2!8!× 10!

4!6!= 45 × 210 = 9450

End system: W (3,3) =10!

3!7!× 10!

3!7!= 120 × 120 = 14400

I How does the heat flow? Do we do any work on the system?



Energy can flow uphill

I Same system, but now consider a system with unequal amount ofparticles in A and B.

I Case 1: NA = 10, NB = 4, UA = 2, UB = 2I The multiplicity becomes

W = WA × WB =10!

2!8!× 4!

2!2!= 45 × 6 = 270

I Case 2: NA = 10, NB = 4, UA = 3, UB = 1. What happens?I The multiplicity becomes

W = WA × WB =10!

3!7!× 4!

1!3!= 120 × 4 = 480

I Energy is not equalized.I We will demonstrate shortly that heat flows so that temperature is

equalized.



Second law of thermodynamics

I The principle of maximum multiplicity is the second law ofthermodynamics.



SummaryI Energy:

I In classical mechanics, the energy is divided into a kinetic energy and apotential energy.

I In quantum mechanics, the particles populate energy levels.I In thermodynamics, the internal energy, U, is divided into work and heat.

I Microscopic interpretation of heat and work:I Heating changes the population of the energy levels in a quantum

mechanical system.I Work changes the energies of the quantum mechanical states, not the

populations of the states.I Connection between the microscopic and macroscopic worlds: The

microscopic total energy, E , is regarded as the internal energy, U, inthermodynamics.

I The flow of heat from a hot body to a cold body can be explained by theprinciple of maximizing the multiplicity.

I The principle of maximum multiplicity is the second law ofthermodynamics.



5. Entropy and the Boltzmann distribution law

Per-Olof Astrand


[email protected]

January 20, 2015


Agenda

We will continue adding piece by piece to the puzzle...In this chapter, we will discuss:

I Entropy: connection between entropy, multiplicity, and probabilityI Boltzmann’s distribution law: a first glance



What is entropy?I Boltzmann’s law:

S = kB ln W

I From a probability perspective, we ”know”:

S = −kB

t∑i=1

pi ln pi

I The multiplicity W (e.g. for placing t types of molecules on a lattice):

ln W = lnN!

n1!n2! . . .nt !

Stirling≈ N ln N − N −

t∑i=1

(ni ln ni − ni)

= −t∑

i=1

ni lnni

N= −N

t∑i=1

pi ln pi

where we essentially usedt∑

i=1ni = N in ”both directions”.



What is entropy?Part 2

I To continue:

ln W = −Nt∑

i=1

pi ln pi ⇒ 1N

ln W = −t∑

i=1

pi ln pi =SN

NkB

where SN is the entropy for N trials. The normal definition of entropy isper trial (or per particle or per mole)

S =SN

N



Example 5.1: Order and disorderSpin a pencil N times: north, east , west , and south

W =N!

nn!ne!nw !ns!

I Which case in the figure gives the maximum multiplicity?I A water molecule has a dipole moment. Which dipole moment has liquid

water? Liquid water has on average a zero dipole moment.



Minimization with constraintsSee chapter 4, pages 66-72

I Assume that we would like to minimize a function f (x , y) with theconstraint g(x , y) = 0.

I The method of Lagrange multipliers: construct a function h(x , y),

h(x , y) = f (x , y)− λg(x , y)

where λ is a Lagrange multiplier.I Minimize h(x , y) with respect to x , y and λ,

∂h(x , y)∂x

= 0

∂h(x , y)∂y

= 0

∂h(x , y)∂λ

= g(x , y) = 0



Maximum entropy: no (physical) constraintsI Still, the probability has to sum to 1:

t∑i=1

pi = 1 ⇒t∑

i=1

dpi = 0

I We seek the distribution (p∗1 ,p

∗2 , . . .p

∗t ) that maximizes the entropy, S,

S(p1,p2, . . . ,pt) = −kB

t∑i=1

pi ln pi

subject to the constraint above.I Note the differential of S:

dS =t∑

i=1

∂S∂pi

dpi



Maximum entropy: no (physical) constraintsPart 2

I Solve by the Lagrange multiplier method:

t∑i=1

(∂S∂pi− α

)dpi = 0

where α is the Lagrange multiplier (including also kB).I Has to be fulfilled for each i :

−1− ln p∗i − α = 0 ⇒ p∗

i = e(−1−α)

I To simplify:

p∗i =

p∗i

1=

p∗i

t∑i=1

p∗i

=e(−1−α)

te(−1−α) =1t

I Conclusion: Even (flat) distribution



Maximum entropy: constant energyI The total energy is constant:

E =t∑

i=1

εini ⇒ EN

=t∑

i=1

piεi

I We seek the distribution that maximizes the entropy with an extraconstraint: the probabilities still have to sum to 1 and the energy is E/Nis constant:

t∑i=1

dpi = 0 ;t∑

i=1

εidpi = 0

I Solve by the Lagrange multiplier method:t∑

i=1

(∂S∂pi− α− βεi

)dpi = 0

I Langrange multipliers give (analogously to the previous example):

−1− ln p∗i − α− βεi = 0 ⇒ p∗

i = e(−1−α−βεi )



Maximum entropy: constant energyPart 2

I Same approach again:

p∗i =

p∗i

1=

p∗i

t∑i=1

p∗i

=e(−1−α)e−βεi

t∑i=1

e(−1−α)e−βεi

=e−βεi

t∑i=1

e−βεi

=e−βεi

q

I The Boltzmann distribution law

p∗i =

e−βεi

q

I q is the partition function

q =t∑

i=1

e−βεi

which is a very central concept in statistical thermodynamics.I The Lagrange multiplier, β, is at this stage not determined.



Summary

I Defined entropy from Boltzmann’s law.I Derived the Boltzmann distribution law for two cases.I The partition function is defined.



6. Thermodynamic driving forces

Per-Olof Astrand


[email protected]

January 21, 2015


Thermodynamics is two laws and a little calculus

Three basic tools of thermodynamics:I The first law for the conservation of energy.I The second law or the maximum entropy principle.I Multivariate calculus

Thermodynamic driving forces: (in principle, already discussed)I pressure - the tendency to exchange volumeI temperature - the tendency to exchange energyI chemical potential - the tendency to exchange matter



Thermodynamic systemsThe type of a thermodynamic system is defined by its boundaries:

I Open system: can exchange energy, volume and matter with thesurroundings (e.g. the earth).

I Closed system: matter cannot cross the boundaries; energy can and thevolume can expand (e.g. the balloon).

I Isolated system: neither energy nor matter can cross the boundaries; thevolume cannot change (e.g. an insulating container)

I Semipermeable membrane: restricts the flow of some particles but notothers (e.g. biological membranes)

I Adiabatic boundary: Allows no heat to flow; thus allows to measure work(e.g. a thermos bottle).

I Phase: is a homogeneous part mechanically separable from the rest ofthe system (e.g. solid, liquid, gaseous, liquid crystalline, etc.)

I Simple system: only a single phase, surface effects can be neglected;idealized model system



Properties are extensive or intensive

Extensive properties:

If a property P can be obtained from a sum of subsystems, P = P1 +P2 +P3 +· · · it is extensive, i.e. dependent on the size of the system.

I Spatial extent: Normally volume, but for surfaces, the area is considered.I Number of particlesI Internal energy: E.g. for non-interacting particles, the energy is just the

sum of single particle energiesI Entropy: Why? Two subsystems, A and B: the maximum multiplicity

W = WA ×WB; S = kB ln W = kB ln WA + kB ln WB = SA + SB

Intensive properties:

Some properties are independent on the size of the system, i.e. temperature,pressure, concentration, and chemical potential.



Fundamental thermodynamic equations predictequilibria

I We have previously studied three model systems for maximizing themultiplicity (entropy):

I S(V ) or W (V ) predicted the expansion of gasesI S(N) or W (N) predicted mixingI S(U) or W (U) predicted heat flow

I Using multivariate calculus, we can study more complex processes. Thefundamental equation for entropy is:

S = S(U,V ,N)

where N indicates several components N1,N2, . . ..I The variables can be changed independently (keeping the other variables

constant), and their values are specified (given). All other properties needto be calculated (measured).

I Unfortunately, thermodynamics was developed as

U = U(S,V ,N)



Some definitionsI A small change in internal energy, dU,

dU =

(∂U∂S

)V ,N

dS +

(∂U∂V

)S,N

dV +M∑

j=1

(∂U∂Nj

)S,V ,Ni 6=Nj

dNj

I Define temperature, T , pressure, p, and chemical potential, µj :

T =

(∂U∂S

)V ,N

; p = −(∂U∂V

)S,N

; µj =

(∂U∂Nj

)S,V ,Ni 6=Nj

leading to

dU = TdS − pdV +m∑

j=1

µjdNj

I Intensive and extensive properties appear in pairs, i.e. ∂U∂x is the

conjugate of x . Thus, temperature is the conjugate to entropy, pressure isthe conjugate to volume, and chemical potential to the number ofparticles.



Different wayI A small change in entropy, dS,

dS =

(∂S∂U

)V ,N

dU +

(∂S∂V

)U,N

dV +M∑

j=1

(∂S∂Nj

)U,V ,Ni 6=Nj

dNj

I Using


j=1

µjdNj

gives

dS =1T

dU +pT

dV −m∑

j=1

µj

TdNj

I so that

1T

=

(∂S∂U

)V ,N

;pT

=

(∂S∂V

)U,N

;µj

T= −

(∂S∂Nj

)U,V ,Ni 6=Nj

I Convenient, since we can maximize S by setting dS = 0.



The ideal gas law from a lattice modelI Use

pT

=

(∂S∂V

)U,N

I For S(V ), use a lattice model (e.g. Example 2.2)

SkB

= ln W (N,M) = lnM!

N!(M − N)!

where we place N molecules in M lattice points.I Obtain S(V ) from S(M) by realizing that M is proportional to the volume,

M = kV , where k = M/V :(∂S∂V

)U,N

=

(∂S∂M

)U,N

∂M∂V

=

(∂S∂M

)U,N

MV

I Use Stirling’s approximation for S,

SkB≈ M ln M − N ln N − (M − N) ln (M − N)



The ideal gas law from a lattice modelPart 2

I We hadSkB≈ M ln M − N ln N − (M − N) ln (M − N)

I Differentiate(∂S∂M

)U,N

= kB

(1 + ln M − ln(M − N)− M − N

M − N

)= −kB ln

(1− N

M

)I Use ln(1− x) ≈ −x − x2/2 + . . .. If N/M 1 (i.e. the density is low

which is reasonable for an ideal gas), we only need to keep the first term:

p = −kBTMV

ln(

1− NM

)≈ −kBT

MV

(−N

M

)=

NkBTV

which is the ideal gas law.I Alternative form: pV = nRT where n is in moles and R = kBNA where NA

is Avogadro’s constant.



Driving forces and equilibria

I Recapitulate the differential form of S,

dS =1T

dU +pT

dV −m∑

j=1

µj

TdNj

I We will demonstrate that

1T,

pT,

µ

T

behave as ”forces”, i.e. they drive the system towards equilibrium.I dS = 0 defines thermal, mechanical and chemical equilibrium.



Temperature drives energy exchange

I Two objects are brought into thermal contact (cannot change volume orexchange particles, i.e. dVA = dVB = dNA = dNB = 0)

I System A: UA, SA, TA; System B: UB, SB, TBI Entropy: S = SA + SB, is maximized: dS = 0I Energy: U = UA + UB, is constant, i.e. dU = 0I Differential for S:

dS = dSA + dSB =

(∂SA

∂UA

)V ,N

dUA +

(∂SB

∂UB

)V ,N

dUB = 0

I Note that we used dVA = dVB = dNA = dNB = 0 (whereas(∂S∂V

)U,N and(

∂S∂N

)U,V are non-zero).

I Differential for U: dU = dUA + dUB = 0 ⇒ dUA = −dUB



Temperature drives energy exchangePart 2

I which gives

dS = dSA + dSB =

[(∂SA

∂UA

)V ,N−(∂SB

∂UB

)V ,N

]dUA = 0

⇒(∂SA

∂UA

)V ,N

=

(∂SB

∂UB

)V ,N

I Using1T

=

(∂S∂U

)V ,N

gives1

TA=

1TB⇒ TA = TB

I The temperature is equalized.



Temperature drives energy exchangePart 3

I In what direction does energy flow?I Look at dS out of equilibrium:

dS =

(1

TA− 1

TB

)dUA

I dS > 0 to drive S towards the maximum, leading to that if TA > TB thendUA < 0.

I In words, heat flow from a hot body to a cold body until the temperature isequal.



Pressure is a force for changing volumeI No energy exchange with the surroundings,

dU = 0.I No heat or mass transfer between the two

subsystems: TA = TB and dNA = dNB = 0.I Volume change:

dV = dVA + dVB = 0 ⇒ dVA = −dVB

I Maximize the entropy, dS = 0

dS =

(∂SA

∂VA

)UA,NA

dVA +

(∂SB

∂VB

)UB ,NB

dVB

+

(∂SA

∂UA

)VA,NA

dUA +

(∂SB

∂UB

)VB ,NB

dUB = 0

I The two last terms vanishes since dUA = −dUB,1T = ∂S

∂U and TA = TB.



Pressure is a force for changing volumePart 2

I We get by using dVA = −dVB

dS =

[(∂SA

∂VA

)UA,NA

−(∂SB

∂VB

)UB ,NB

]dVA = 0

I Using pT = ∂S

∂V the condition becomes:

dS =

(pA

TA− pB

TB

)dVA = 0

I Using TA = TB, the condition is satisfied if

pA = pB

i.e. the pressure is equalized.I Again for dS > 0, fulfilled if pA > pB and dVA > 0, i.e. if pA > pB, VA will

increase until the entropy is maximized and the pressures are equal.



The chemical potential drives particle exchangeI N identical particles, NA on side A and NB on side BI dNA = −dNB since dN = 0.I Temperature and pressure are held constant.I Maximize the entropy leads to

dS =

(∂SA

∂NA

)dNA +

(∂SB

∂NB

)dNB = 0

which gives

dS =

(µB

TB− µA

TA

)dNA = 0

I Condition for chemical equilibrium: µA = µB, i.e. thechemical potential is equalized.

I Again for dS > 0, if µB > µA then dNA > 0, soparticles move from high chemical potential to lowchemical potential.



Summary

I We have introduced state functions, S (U,V ,N) and U (S,V ,N), andmore will come.

I We have discussed conjugate properties, i.e. properties appear in pairs:p − V , T − S, µi − Ni .

I The expression for S (U,V ,N),

dS =1T

dU +pT

dV −m∑

j=1

µj

TdNj

defines temperature, pressure, and chemical potential.I Driving forces arising from maximizing the entropy (multiplicity):

I heat flows so that temperature is equalizedI volume changes so that pressure is equalizedI particles flow so that chemical potential is equalized



7. The logic of thermodynamics

Per-Olof Astrand


[email protected]

January 26, 2015


Some key concepts in classical thermodynamics

I First law of thermodynamicsI Quasi-static processesI Heat capacityI Thermodynamic cycles



The first law interrelates heat, work and energy

I First law of thermodynamics:

dU = δq + δw

I δq and δw are path-dependent properties (not state variables)I dU is a state function,

∆U =

B∫A

dU = UB − UA

so ∆U depends only on the start state and the end state (not on thepath).

I The first law is defined so thatI δq > 0 when heat flows into the system.I δw > 0 when work is done on a system.



Quasi-static processes

I Thermodynamics is about equilibrium, not about rates (see irreversiblethermodynamics, see transport processes in chap. 17).

I Processes are quasi-static if they are performed slowly enough that theirproperties are independent of time and independent of the speed of theprocess.

I For a gas in a chamber where work is carried out by a piston, aquasi-static process is defined as

δw = −pextdV



The heat capacity, CVI Measuring the heat capacity in a bomb calorimeter: fixed volume

δw = −pextdV = 0

leading to for U(S,V ,N) and ignoring terms in N,

dU = δq = TdS

I The heat capacity, CV , is defined as the amount of heat needed to raisethe temperature of the system with 1 K at constant V :

CV =

(δqdT

)V

=

(∂U∂T

)V

= T(∂S∂T

)V

I If we know the temperature dependence of CV :

dU = CV dT ⇒ ∆U =

TB∫TA

Cv (T )dT

I Similar expression at constant pressure in chapter 8: Cp



Heat capacities of gases

I Regard U(V ,T ) for a gas,

dU =

(∂U∂V

)T

dV +

(∂U∂T

)V

dT

I For a gas, the dependence of U on V can be ignored,(∂U∂V

)T≈ 0

since the gases are dilute and rarely interact.I So whether we have constant volume or not,

dU = CV dT (for a gas at low density)



Reversible processesI A process is called reversible if returning the system to its initial

conditions also returns the surroundings to its initial conditions (noconversion of energy that cannot be recaptured).

I We can a relate (for a closed system, dN = 0):

dU = TdS − pdV = δq + δw

In a quasi-static process: δw = −pdV , which leads to

dS =δqrev

T

I This is often referred to as the thermodynamic definition of entropy.I Also since (for dV = dN = 0),

dU = TdS = CV dT ⇒ ∆S =

B∫A

dS =

TB∫TA

CV

TdT



Thermodynamic cycles and fictitious processesI Changes in state variables, like p, V , T , N, U and S, do not depend on

the path, e.g. ∆U = U2 − U1 only depends on the start and end states.I Heat and work are not state variables.I For example, you can introduce a fictitious state A,

∆U = (U2 − UA) + (UA − U1)

which may simplify the problem dramatically.Right figure below: The process is divided into two parts, one withconstant pressure and one with constant volume.

I Similarly, thermodynamic cycles (left figure below) can be utilized since∆U = 0 for the entire process (all the way around).



Four processes to describe an ideal gasI Constant-volume (isochoric) process: (p1,V0,T1)→ (p2,V0,T2)

no volume change gives pdV = 0,

∆U = q =

T2∫T1

Cv (T )dT

I Constant-pressure (isobaric) process: (pext,V1,T1)→ (pext,V2,T2)

w = −V2∫

V1

pextdV = −pext (V2 − V1)

I We have (recall, also when the volume is not constant),

∆U =

T2∫T1

CV dT

so that q = ∆U − w .Per-Olof Astrand Statistical Thermodynamics in Chemistry and Biology January 26, 2015 9 / 15


Four processes to describe an ideal gasPart 2

I Constant-temperature (isothermal) process: (p1,V1,T0)→ (p2,V2,T0)in this case, it is assumed that pint = pext in a quasistatic process (slowenough for the internal pressure to adapt to the external pressure):

w = −V2∫

V1

pextdV = −V2∫

V1

pintdV = −V2∫

V1

NkBTV

dV = −NkBT lnV2

V1

where we in the last steps have assumed an ideal gas.I At constant temperature, dU = CV dT = 0, so that q = −w since

∆U = w + q.



Four processes to describe an ideal gasPart 3

I Adiabatic process: (p1,V1,T1)→ (p2,V2,T2)

I Definition of adiabatic process: δq = 0.I We use

CV dT = dU = −pdV

I For an ideal gas,

CV dT = −NkBTV

dV

I Rearrange and integrate,

T2∫T1

CV

TdT = −

V2∫V1

NkB

VdV ⇒ CV ln

T2

T1= −NkB ln

V2

V1

where we in the last term assumed that CV is independent of thetemperature.



Heat engine: Carnot cycle

I Heat engine: take in heat and perform work



Heat engine: Carnot cyclePart 2

I Two adiabatic and two isothermal steps (at Th and Tc).I dU = 0 for the entire process (thermodynamic cycle), so qtot = −wtot.I For the two adiabatic steps: q = 0.I For the two isotermal steps: q = −w ,

qtot = qh + qc = NkBTh lnVB

VA+ NkBTc ln

VD

VC

I To simplify use the adiabatic steps,

VC

VB=

(Th

Tc

) CVNkB

=VD

VA⇒ VB

VA=

VC

VD

which gives

wtot = −qtot = −NkB (Th − Tc) lnVB

VA



The Carnot cycle is reversible

I Demonstrate by showing,

∆S = ∆SAB + ∆SBC + ∆SCD + ∆SDA = 0

I For the two adiabatic steps:

∆Sadiabatic =qT

= 0

I For the isothermal steps,

∆SAB =qh

Th= NkB ln

VB

VA

and∆SCD =

qc

Tc= NkB ln

VA

VBsince

VB

VA=

VC

VD

I Summing these terms gives 0.



Why do engines waste heat?I Consider the following three steps of the cycle:

I ∆U is a state function,w = qh − qc

I Define the efficiency asη =

wqh

= 1− qc

qhI For a reversible process,

∆Stotal = ∆Sh + ∆Sc =qh

Th− qc

Tc= 0

I Rearrange,qc

qh=

Tc

Th⇒ η = 1− Tc

Th



8. Laboratory conditions and free energies

Per-Olof Astrand


[email protected]

January 29, 2015


Switch from maximum entropy to minimum free energyS(U,V ,N)

I The boundaries are controlled by the internal energy, U, volume, V , andnumber of particles, N.

I Reasonable properties to control experimentally.I Equilibrium principle: entropy is maximized

U(S,V ,N)

I The boundaries are controlled by the entropy, S, volume, V , and numberof particles, N.

I The entropy is difficult(impossible) to to control experimentally.

Introduce new independent variablesI It would be natural to use temperature, pressure, and chemical potential

as independent variables.I More convenient to control experimentally (c.f. biological systems)I New thermodynamics quantities: enthalpy, free energiesI New equilibrium principle: minimum free energy



Helmholtz Free EnergyI Constant temperature, T , volume, V , and number of particles, N.I Heat is transferred to and from a heat bath to keep the temperature

constant.

I One step back: treat the system and the bath as acombined system, Scombined(U,V ,N):

dScombined = dSsystem + dSbath ≥ 0

where . . . ≥ 0 denotes that dS is maximized.I The combined system is isolated:

dUcombined = dUbath + dUsystem = 0

I Use the fundamental equation for dS:

dSbath =1T

dU +pT

dV − µ

TdN =

1T

dUbath

for a bath where V and N are constant.Per-Olof Astrand Statistical Thermodynamics in Chemistry and Biology January 29, 2015 3 / 14


The Helmholtz free energyPart 2

I Combining the last two equations,

dSbath = −dUsystem

Twhich leads to

dSsystem −dUsystem

T≥ 0 ⇒ dUsystem − TdSsystem ≤ 0

I Condition for equilibrium in terms of the system (test tube) alone.I Define a quantity, F (sometimes A), Helmholtz free energy

F = U − TS

I Its differential isdF = dU − TdS − SdT

and at constant temperature, the condition above is obtained for dF ,

dF = dU − TdS

I The Helmholtz free energy is minimized at equilibrium: dF = 0.Per-Olof Astrand Statistical Thermodynamics in Chemistry and Biology January 29, 2015 4 / 14


A model for dimerization

I Two particles, N = 2, V lattice points (constantvolume), and constant temperature.

I Minimize Helmholtz free energy, F (actually,compare Fdimer and Fmonomer).

I “Bond energy” given: U = −ε (ε > 0)I Dimer case: the multiplicity becomes

Wdimer = V − 1

I The Helmholtz free energy:

Fdimer = Udimer − TSdimer = −ε− kBT ln (V − 1)



A model for dimerizationPart 2

I Monomer case: the multiplicity becomes

Wmonomer = Wtotal −Wdimer =V !

2!(V − 2)!− (V − 1) =

(V2− 1)

(V − 1)

I The Helmholtz free energy:

Fmonomer = Umonomer − TSmonomer = −TSmonomer

= −kBT ln((

V2− 1)

(V − 1)

)I Recapitulate the Helmholtz free energy for the dimer:

Fdimer = Udimer − TSdimer = −ε− kBT ln (V − 1)

I Which state (dimer or monomer) dominates? Which has the lowest freeenergy? The temperature governs! The dimer is stable at 0 K, but atsome temperature the monomer state will become stable.



Fundamental equation for the Helmholtz free energy

I Multivariate calculus again for F (T ,V ,N):

dF = d(U − TS) = dU − TdS − SdT

I Use the equation for dU:

dF =

TdS − pdV +m∑

j=1

µjdNj

− TdS − SdT

= −SdT − pdV +m∑

j=1

µjdNj

I Compare U(S,V ,N) and F (T ,V ,N).



Fundamental equation for the Helmholtz free energyPart 2

I We can also write dF as

dF =

(∂F∂T

)V ,N

dT +

(∂F∂V

)T ,N

dV +m∑

j=1

(∂F∂Nj

)T ,V ,Ni 6=j

dNj

and we get the following relations:

S = −(∂F∂T

)V ,N

; p = −(∂F∂V

)T ,N

; µj =

(∂F∂Nj

)T ,V ,Ni 6=j

I We will not go into Legendre transforms in detail. See Appendix F.



The enthalpy, HI The enthalpy H(S,p,N), (c.f U(S,V ,N), F (T ,V ,N))I “Standard procedure”:

H = H(S,p,N) = U + pV

I Differentiate:dH = dU + pdV + Vdp

I Substitute with the expression for dU:

dH =

TdS − pdV +m∑

j=1

µjdNj

+ pdV + Vdp

dH = TdS + Vdp +m∑

j=1

µjdNj

I Why not H = U − pV?



The Gibbs free energy, GI The Gibbs free energy, G(T ,p,N) (c.f H(S,p,N), F (T ,V ,N))I Standard procedure:

G = H − TS

I Differentiate:dG = dH − TdS − SdT

I Substitute with the expression for dH:

dG =

TdS + Vdp +m∑

j=1

µjdNj

− TdS − SdT

dG = −SdT + Vdp +m∑

j=1

µjdNj

I Many other possibilities than S(U,V ,N), U(S,V ,N), F (T ,V ,N) ,H(S,p,N), and G(T ,p,N)?Yes, but most of them are not very useful (see table 8.1 for a helpful list)



Heat capacity, Cp

I We regard H(S,p,N),

dH = d(U + pV ) = dU + pdV + Vdp = δq + δw + pdV + Vdp

I Quasistatic processes: δw = −pdV . Constant pressure: dp = 0. Gives:

dH = δq

I The heat capacity, Cp is defined as,

Cp =

(δqdT

)p

=

(∂H∂T

)p

I Analogous to CV .



ThermochemistryI The enthalpy of a molecule is regarded as a sum of enthalpies for

covalent bonds.I The enthalpy for a chemical reaction is defined as

∆∆Htotal = ∆Hreac −∆Hprod

which is the difference between the formation enthalpy of the reactantsand products.

I Best explained with example 8.6 (burning propane):

CH3CH2CH3 + 5 O2(g)→ 3 CO2(g) + 4 H2O

I Use bond enthalpies (in kJ/mol) from Table 8.3. C-C: 347, C-H: 414,O=O: 499, C=O (in CO2): 799, O-H: 460.

I Results in

∆∆Htotal = (8×460 + 6×799)− (2×347 + 8×414 + 5×499) kJ/mol

= −1974 kJ/mol



Summary

I Introduced Helmholtz free energy, enthalpy and Gibbs free energy.I General procedure for introducing new state functions.I Minimization principle for free energies.I Model system for dimerization (first system where we minimize/compare

free energies).



Exam Dec. 2008 - Exercise 4A lattice model for dimerizationConstruct a lattice model for dimerization on a surface. The surface consistsof quadratic lattice with A = L2 sites. The two identical and indistinguishablemolecules (monomers) are bound to the surface. The molecule-surfaceinteraction energy is ε0, and the dimer binding energy is ε.

a) Give an expression for the Helmholtz free energy of dimerization,

∆F = Fdimer − Fmonomer

b) What is the condition for dimerization to occur? In particular, what is thecondition for ε? Will dimerization be more likely at higher or lowertemperatures?

c) Briefly described, cooperativity is the effect of binding to a surface or anenzyme when already a neighbour molecule is bound. For example, anoxygen molecule binds easier to hemoglobin when already one or severaloxygen molecules are bound. Suggest how ε0 may be modified to includecooperativity in this model?



9. Maxwell relations and mixtures

Per-Olof Astrand


[email protected]

February 2, 2015


Mathematics of partial derivatives

Today’s agendaI Design of fundamental equationsI Maxwell relationsI Homogeneous functions to develop the Gibbs-Duhem relationship

(The Gibbs-Duhem equation is not covered in the 2nd ed. of the book)

Per-Olof Astrand Statistical Thermodynamics in Chemistry and Biology February 2, 2015 2 / 20


How to design a fundamental equationI Each extensive degree of freedom in the fundamental equations is paired

with its conjugate force. We already have: p,V, T ,S, µj ,Nj.I Other examples:

I force, length - f L - for elastic materialsI surface tension, area - γ A - for interfacesI electric potential, charge - ψ Q - for charged particlesI magnetic field, magnetic moment - B I - for magnetic systems

I The fundamental equation for the internal energy becomes

dU = TdS − pdV +∑

j

µjdNj + fdL + γdA + ψdQ + BdI

U is special: all variables are extensiveI More general


j

µjdNj +∑

j

FjdXj

where

Fj =

(∂U∂Xj

)S,V ,N,Xi 6=j



Surface tensionI When is it important? soap films, micelles, cell membranes, etc.I System: γ is fixed and the area can changeI Choice of variables: T , p, N, and γI Begin by U(S,V ,N,A) (Nb! all extensive variables):


j

µjdNj + γdA

I Change to (T ,p,N, γ):

−d(TS) + d(pV )− d(γA) = −TdS − SdT + pdV + Vdp − γdA− Adγ

which gives

d(U − TS + pV − γA) = −SdT + Vdp +∑

j

µjdNj − Adγ = d(G − γA)

I The system will be at equilibrium for (T ,p,N, γ) when the thermodynamicfunction G − γA is at a minimum.



Ex. 9.2. Surface tension againI Instead constant T , p, N, and A.I The area is constant; the shape of the object will change (i.e. a drop on a

surface)I Again, begin by U(S,V ,N,A):


j

µjdNj + γdA

I Change to (T ,p,N,A):

−d(TS) + d(pV ) = −TdS − SdT + pdV + Vdp

which gives

d(U − TS + pV ) = −SdT + Vdp +∑

j

µjdNj + γdA = dG

I In this example we obtain the regular Gibbs free energy as the “minimumprinciple”.

I Need to pay attention to the independent variables, in this case(T ,p,N,A).



Maxwell relations: an exampleI Use Euler’s reciprocal relation, for example(

∂2U∂V∂S

)=

(∂2U∂S∂V

)I The fundamental equation for dU


j

µjdNj

I We thus have

T =

(∂U∂S

)V ,N

; p = −(∂U∂V

)S,N

I Substituting this into Euler’s relation gives:(∂T∂V

)S,N

= −(∂p∂S

)V ,N

which is an example of a Maxwell relation.Per-Olof Astrand Statistical Thermodynamics in Chemistry and Biology February 2, 2015 6 / 20


How to obtain a Maxwell relation

I Suppose we for some reason are interested in obtaining(∂S∂p

)T ,N

from the ideal gas law: pV = NkBT .



How to obtain a Maxwell relationPart 2

1. Identify the independent variables and the corresponding fundamentalequation: (use Table 8.1 or 9.1)

(p,T ,N) and thereby G (p,T ,N)

2. Look at its differential form:

dG = −SdT + Vdp +m∑

j=1

µjdNj

3. Construct cross-derivates:

−(∂S∂p

)T ,N

=

(∂2G∂p∂T

);

(∂V∂T

)p,N

=

(∂2G∂T∂p

)4. Euler’s relation: (

∂S∂p

)T ,N

= −(∂V∂T

)p,N

= −NkB

p

for an ideal gas.Per-Olof Astrand Statistical Thermodynamics in Chemistry and Biology February 2, 2015 8 / 20


Thermodynamics of a rubber bandIs the retraction in a rubber band driven by a change in enthalpy or in entropy?

I U(S,V ,L):dU = TdS − pdV + fdL

I We are interested in enthalpy and entropy. Note that dG = d(H − TS).So we get dG from G(T ,p,L):

dG = −SdT + Vdp + fdL

I The force, f , is given as

f =

(∂G∂L

)T ,p

=

(∂H∂L

)T ,p− T

(∂S∂L

)T ,p

I Since we can measure f , use Maxwell relations:(∂S∂L

)T ,p

= −(∂f∂T

)p,L

i.e. the entropic contribution is obtained from measuring f (T ).



A rubber bandPart 2

I To get the enthalpy contribution, combine thetwo previous equations:(

∂H∂L

)T ,p

= f − T(∂f∂T

)p,L

which can be determined from the sameexperiment.

I Figure 9.1. shows that retraction of rubberincreases with temperature (special property ofrubber).

I The explanation is that the entropy decreaseswith length (also demonstrated in Ex. 2.4 wherewe studied the multiplicity of conformations withchain length).



Measuring expansion

I The thermal expansion coefficient, α, is definedas

α =1V

(∂V∂T

)p

and is the fractional change in volume withtemperature at constant pressure.

I For an ideal gas:

α =p

NkBTNkB

p=

1T

I Upper figure: Specific volume v (volume/mass)of polyethylene.

I Lower figure: α for various liquids (water,benzene, etc.)



Measuring compression

I The isothermal compressibility, κ, is defined as

κ = − 1V

(∂V∂p

)T

and is the fractional change in volume withpressure at constant temperature.

I For an ideal gas,

κ =1p

I Graph: The relative volume, V/V0, ofhexadecane. V0 is the volume extrapolated tozero pressure.



Entropy change with pressureI Consider the entropy change with pressure at constant temperature:

dS =

(∂S∂p

)T ,N

dp

I Combine the Maxwell relation(∂S∂p

)T ,N

= −(∂V∂T

)p,N

with the definition of the thermal expansion coefficient, α,

α =1V

(∂V∂T

)p,N

to get

dS = −(∂V∂T

)p,N

dp = −αVdp

I α can be measured as a function of pressure:

∆S = −∫ p2

p1

α (p) V (p) dp



Partial molar volumesMulticomponent systems have partial molar properties

I What is a molar property? For a system with n moles of a singlecomponent, the molar volume, v = V/n. Similarly, the molar Gibbs freeenergy, g = G/n.

I Multicomponent system in mole: n = n1,n2, . . .nm. The partial molarvolume is defined as,

vj =

(∂V∂nj

)T ,p,ni 6=j

I The change in volume, dV , is thus

dV =m∑

j=1

(∂V∂nj

)T ,p,ni 6=j

dnj =m∑

j=1

vjdnj

I In simple cases, vj is independent of the composition and it can beobtained from the pure substance. In general, this is however not true.For an exercise, see E9.1.



The chemical potentialI The chemical potential is a partial molar free energy.I We have for example


j=1

µjdNj

but we can also write

µj =

(∂U∂Nj

)S,V ,Ni 6=j

=

(∂G∂Nj

)T ,p,Ni 6=j

=

(∂F∂Nj

)T ,V ,Ni 6=j

=

(∂H∂Nj

)S,p,Ni 6=j

I Partial molar quantities are defined specifically for quantities at constantT and p. So only the Gibbs free energy

µj =

(∂G∂Nj

)T ,p,Ni 6=j

=

(∂H∂Nj

)T ,p,Ni 6=j

− T(∂S∂Nj

)T ,p,Ni 6=j

= hj − Tsj

is called a partial molar quantity.



Partial molar properties are linkedI We have from the definition of the partial molar volume:

dV =m∑

j=1

vjdnj

I In general,

V =m∑

j=1

vjnj

thus also in general

dV =m∑

j=1

(vjdnj + njdvj

)which leads to

m∑j=1

njdvj = 0

I It shows that the partial molar volumes are not independent of each other.



The Gibbs-Duhem equationSame procedure again. . .

I Consider U(S,V ,N),

U = TS − pV +m∑

j=1

µjNj

I Differentiate,

dU = TdS + SdT − pdV − Vdp +m∑

j=1

µjdNj +m∑

j=1

Njdµj

I Subtract the fundamental equation for U,


j=1

µjdNj



The Gibbs-Duhem equationPart 2

I This gives the Gibbs-Duhem equation.

m∑j=1

Njdµj = Vdp − SdT

and at constant temperature and pressure,

m∑j=1

Njdµj = 0

I Central for example for understanding phase transitions.



Summary

I Design fundamental thermodynamic functions from a set of independentvariables.

I Maxwell relations - provides a way to obtain unmeasurable propertiesfrom measurable ones. List of Maxwell relations: See Table 9.1

I Multicomponent systems - The Gibbs-Duhem equation (and similarexpressions).



Exam Aug. 2011 - Exercise 1Maxwell relations

a) Describe in a few sentences what a Maxwell relation is. What is thetheoretical (mathematical) foundation used to derive Maxwell relations from afundamental equation? Why are they useful, for example in experimentalwork?

b) For a single-component system where each particle has a charge, q, givethe fundamental equation for the Helmholtz free energy for a system in anexternal electrostatic potential, ψ. Give the Maxwell relations for thisfundamental equation which includes ψ.

c) Show the following relation (where we in this derivation assume thatdq = ZedN):(

∂U∂N

)T ,V

= (µ+ Zeψ)− T(∂µ

∂T

)V ,N− ZeT

(∂ψ

∂T

)V ,N



10. The Boltzmann distribution law

Per-Olof Astrand


[email protected]

February 5, 2015


Probability distributions for atoms and molecules

Today’s agendaI The Boltzmann distribution law (slightly different from chapter 5)I Partition functionsI Thermodynamic properties from partition functions



Typical example



Derivation of the Boltzmann distribution lawI System of N particles with discrete energy levels, Ej , j = 1,2, . . . , tI Aim at calculation of the probability, pj , to be in each state j .I Minimize Helmholtz free energy in the form,

dF = dU − TdS = 0

(previous derivation in chapter 5: maximizing the entropy, dS = 0)I Express the entropy in terms of probabilities:

S = −kB

t∑j=1

pj ln pj

and

dS = −kB

t∑j=1

(1 + ln pj

)dpj



Derivation of the Boltzmann distribution lawPart 2

I We postulate that the macroscopic internal energy U is (the average of)the microscopic energy

U =t∑

j=1

pjEj

dU =t∑

j=1

(Ejdpj + pjdEj

)I From quantum mechanics:

I Heat does not change the energy levels (but work does).I Heat changes the populations (probabilities to be in each level): dpj

I We do not want modify the energy levels, just find the optimum pj(constant V and N):

dU =t∑

j=1

Ejdpj




I We want to minimize F ,

dF = dU − TdS = 0

with the constraint that the probabilities, pj , always sum up 1. Theconstraint is expressed with a Lagrange multiplier, α, as

αt∑

j=1

dpj = 0

I We thus get

dF =t∑

j=1

(Ej + kBT

(1 + ln pj

)+ α

)dpj = 0

I It has to be 0 for each value of j ,

ln pj = −Ej

kBT− α

kBT− 1

I (Note that the book uses the notation p∗j for pj when the condition dF = 0is fulfilled. I use pj everywhere.)




I Exponentiate

pj = e−Ej

kB T e−α

kB T −1

I We need to eliminate α. Rewrite

t∑j=1

pj = 1 as 1 =t∑

j=1

e−Ej

kB T e−α

kB T −1

and divide pj by “1” to get the Boltzmann distribution law

pj =pj

1=

e−Ej

kB T

t∑j=1

e−Ej

kB T

=e−

EjkB T

Q

where Q is the partition function.




I The partition function is thus defined as

Q =t∑

j=1

e−Ej

kB T

and is in principle impossible to obtain for large and complex systems(too many states j).

I We note that the relative probability between two states i and j is

pi

pj=

e−

EikB T

Q

e−

EjkB T

Q

= e−(Ei−Ej)

kB T

I Note: more particles will have low energies and fewer particles will havehigher energies

I Example: Constant total energy: there are many more arrangements(larger multiplicity) when many particles have relatively small energiesthan if a few particles have high energies.



Ex. 10.2. The Maxwell-Boltzmann distribution ofvelocities

I Kinetic gas theory: classical mechanics for the kinetic energy, ε,

ε (v) =12

mv2 =12

m(v2

x + v2y + v2

z)

I According to Boltzmann’s distribution law, the probability p (vx ) is

p (vx ) =e−

ε(vx )kB T

∞∫−∞

e−ε(vx )kB T dvx

=e−

mv2x

2kB T

∞∫−∞

e−mv2

x2kB T dvx

=

(m

2πkBT

) 12

e−mv2

x2kB T

which is called the Maxwell-Boltzmann distribution. We have used (fromAppendix D)

∞∫−∞

e−ax2dx =

√π

a



Ex. 10.2. The Maxwell-Boltzmann distributionPart 2

I The mean square velocity 〈v2〉 may be obtained,

〈v2x 〉 =

∞∫−∞

v2x p (vx ) dvx =

(m

2πkBT

) 12∞∫−∞

v2x e−

mv2x

2kB T dvx

I Again use Appendix D (and∞∫−∞

= 2∞∫0

for this integral):

∞∫0

x2e−ax2dx =

14a

√π

a

I We get

〈v2x 〉 =

kBTm⇒ 1

2m〈v2

x 〉 =12

kBT



Ex. 10.2. The Maxwell-Boltzmann distributionPart 3

I Generalize to three dimensions:I Since v2 = v2

x + v2y + v2

z and for an ideal gas 〈v2〉 = 〈v2x 〉+ 〈v2

y 〉+ 〈v2z 〉

12

m〈v2〉 = 32

kBT

i.e. the average kinetic energy has a contribution of 32 kBT per particle. Note

the close relation between kinetic energy and temperature.I Since the velocity components are independent,

p (v) = p (vx) p (vy ) p (vz) =

(m

2πkBT

) 32

e−mv22kB T



What does the partition function tell us?I It is a sum of Boltzmann factors

e−Ej

kB T

I In many cases, E1 = 0,

Q =t∑

j=1

e−Ej

kB T = 1 + e−E2

kB T + e−E3

kB T + . . .+ e−Et

kB T

I Ej → 0 or T → ∞Ej

kBT→ 0 ⇒ pj →

1t⇒ Q → t

I Ej → ∞ or T → 0

Ej

kBT→ ∞ ⇒ (p1 → 1, pj 6=1 → 0) ⇒ Q → 1



Density of states

I In some cases the energy states are degenerate, i.e. different stateshave the same energy.

I We can then sum over all macrostates, i.e. states with distinct energy,and take the degeneracy into account explicitly.

Q =

nmax∑n=1

W (En) e−En

kB T

where W (En) is the density of states or degeneracy.I Alternatively, the density of states, W (En), is the number of states in the

interval En − δ ≤ En ≤ En + δ, where δ is a small number.



Distinguishable or indistinguishable particlesI Consider first distinguishable particles (A and B):

Ej = εAm + εB

n

qA =a∑

m=1

e−εA

mkB T ; qB =

b∑n=1

e−εB

nkB T

I We would like to express Q in qA and qB,

Q =t∑

j=1

e−Ej

kB T =a∑

m=1

b∑n=1

e−εA

m+εBn

kB T =a∑

m=1

b∑n=1

e−εA

mkB T e−

εBn

kB T

=

(a∑

m=1

e−εA

mkB T

)(b∑

n=1

e−εB

nkB T

)= qAqB

I In general for N particles: Q = qN

I For indistinguishable particles, we cannot see the difference betweenqAqB and qBqA

Q =qN

N!



Thermodynamic properties from partition functionsThe internal energy, U, from the partition function

I The internal energy as an average over states,

U =t∑

j=1

pjEj = Q−1t∑

j=1

Eje−βEj

where β = 1kBT .

I Note that (∂Q∂β

)=

∂

∂β

t∑j=1

e−βEj = −t∑

j=1

Eje−βEj

I U is thus obtained as

U = − 1Q

(∂Q∂β

)= −

(∂ ln Q∂β

)and we have expressed U only in terms of the partition function Q andother macroscopic properties (in this case T ).



The internal energy, U, from the partition functionPart 2

I Since β = 1kBT , (

∂β

∂T

)= − 1

kBT 2

such that

U = kBT 2(∂ ln Q∂T

)I The average particle energy for indistinguishable particles, Q = qN/N!,

〈ε〉 =UN

=kBT 2

N

(∂ ln qN

∂T

)+ 0 = kBT 2

(∂ ln q∂T

)= −

(∂ ln q∂β

)I Where did the N! term go? What is 〈ε〉 for distinguishable particles?



The entropy from the partition functionI The definition of the entropy in terms of probabilities is

S = −kB

t∑j=1

pj ln pj

I Substituting the Boltzmann distribution law,

pj = Q−1e−Ej

kB T

gives

S = −kB

t∑j=1

(Q−1e−

EjkB T

)(ln(

1Q

)−

Ej

kBT

)

= kB ln Qt∑

j=1

pj +1T

t∑j=1

pjEj = kB ln Q +UT

= kB ln Q + kBT(∂ ln Q∂T

)where

t∑j=1

pj = 1 ; U = 〈E〉 = Q−1t∑

j=1

Eje−

EjkB T



More thermodynamic properties from the partitionfunction

Constant (T ,V ,N)

Use regular thermodynamics to get more properties:

Internal energy: U = kBT 2(∂ ln Q∂T

)V ,N

Entropy: S = kB ln Q + UT

Helmholtz free energy: F = U − TS = −kBT ln QChemical potential: µ =

(∂F∂N

)T ,V = −kBT

(∂ ln Q∂N

)T ,V

Pressure: p = −(∂F∂V

)T ,N = kBT

(∂ ln Q∂V

)T ,N



Ex. 10.5. The Schottky two-state model

I System: N distinguishable particles with two energy levels for eachparticle, 0 and ε0 > 0

I Useful for many different problems,I Dimer or polymer lattice models (Ex. 8.1 and Ex. 8.2)I Atoms or molecules excited by electromagnetic radiation.I Behavior of spins in magnetic fields (Ex. 10.6)

but here we keep it general.I Find the average particle energy, 〈ε〉, the heat capacity, CV , the entropy

and the free energy per particle from the partition function.I The particle partititon function, q,

q = 1 + e−βε0



Ex. 10.5. The Schottky two-state modelPart 2

I The energy per particle is given as

〈ε〉 = −1q

(∂q∂β

)=

ε0e−βε0

1 + e−βε0

as shown in the top figure as a function oftemperature. Note that 〈ε〉 → ε0

2 when T →∞.I For the heat capacity, CV , use its definition

CV =(∂U∂T

)V ,N and U = N〈ε〉,

CV = N(∂〈ε〉∂T

)V ,N

= N(∂〈ε〉∂β

)V ,N

(∂β

∂T

)

= − NkBT 2

(∂〈ε〉∂β

)V ,N




I Note that d( u

v

)= vu′−uv ′

v2 , to get(∂〈ε〉∂β

)V ,N

=

(1 + e−βε0

) (−ε2

0e−βε0)−(ε0e−βε0

) (−ε0e−βε0

)(1 + e−βε0 )

2

=−ε2

0e−βε0

(1 + e−βε0 )2

which gives

CV =Nε2

0

kBT 2e−βε0

(1 + e−βε0 )2

which is also shown as a function of the temperature on the previousslide.




I The entropy is given as

S =UT

+ kB ln Q =Nε0e−βε0

T (1 + e−βε0 )+ kBN ln

(1 + e−βε0

)I and Helmholtz free energy as

F = −kBT ln Q = −NkBT ln q = −NkBT ln(1 + e−βε0

)



Ensembles

I In this chapter we have considered systems in which (T ,V ,N) isconstant. This is called the canonical ensemble.

I An ensemble has the following meaning:The collection of all microstates of the system that fulfill the macroscopicconstraints (e.g. (T ,V ,N))

I Examples of ensembles:canonical ensemble (T ,V ,N)

isobaric-isothermal ensemble (T ,p,N)grand canonical ensemble (T ,V , µ) (see chapter 28)microcanonical ensemble (U,V ,N)



Summary

I Derived the Boltzmann distribution law for the canonical ensembleI Defined the corresponding partition function.

I Divided the partition function into particle contributions: Q = qN

N! (forindistinguishable particles).

I Connected thermodynamic properties to the partition function.I Defined ensembles



Exam June 2012 - Exercise 1Boltzmann distribution lawa) How is an ensemble defined in statistical thermodynamics? Discuss brieflythe differences between the microcanonical, canonical andisobaric-isothermal ensembles. Which are the variables, fundamental functionand extremum principle for each ensemble, respectively? Which ensemblesare preferred experimentally (motivate the answer)?

b) For which ensemble is the Boltzmann distribution,

pj =gje−Ej/kBT

Q(1)

derived? What is gj (we have also used the notation Wj ) in eq. (1)? How is Qdefined? What does the magnitude of Q tell us?

c) Assume that we have three molecules, N = 3, and also assume that onlythe two lowest molecular states, ε0 = ε and ε1 = 2ε, can be occupied. What isthe probability for that the total energy, E , is 5ε? The temperature, T = 300 Kand ε = 10 kJ/mol.



Exam June 2012 - Exercise 1Part 2 - to be solved when we introduced K and x

Boltzmann distribution law

d) The equilibrium constant, K ,

K =x1

x2

is investigated. Show that it can be written both in terms of the energydifference, ∆E = E1 − E2, as

K = Ae−∆E/kBT

and in terms of the Helmholtz free energy difference, ∆F = F1 − F2, as

K = e−∆F/kBT

Use eq. (1) as a starting point. Define A in terms of properties used in eq. (1).



11. The statistical mechanics of simple gases and solids

Per-Olof Astrand


[email protected]

February 10, 2015


Partition functions for atoms and molecules

Today’s agendaI Derive expressions for the molecular partition function:

I TranslationI RotationI VibrationI Electronic excitations



Quantum mechanics, summary

I The states of atoms and molecules are described by quantummechanics.

I The energy is quantized and light and matter interact: ∆E = hν.I The fundamental equation is the Schrodinger equation: Hψi = Eiψi .I The Hamiltonian, H, is an operator and consists of a kinetic energy

operator, K , and a potential energy operator, V ,

H = K + V

I The Schrodinger equation can only be solved exactly for a handful ofmodel systems because of the complexity of the potential energy.



Quantum mechanicsThe Schrodinger equation

I Eigenvalue problem:Hψi = Eiψi

where H is the Hamilton operator, ψi is the wavefunction (eigenfunction),and Ei is the energy (eigenvalue).

I The energy is quantized, which is denoted by that a quantum mechanicalparticle is in a state i.

I If several states have the same energy, they are said to be degenerate.



Quantum mechanicsThe Hamilton operator

I The Hamiltonian is the energy operator,

H = K + V

and consists of a kinetic energy operator, K and a potential energyoperator, V .

I For N particles, the kinetic energy is always the sum of the particle kineticenergies,

K =N∑i

p2i

2mi

I The potential energy is unique for each type of system. For a molecule, itis the Coulomb interaction between all the particles:

V =N∑

i,j>i

qiqj

4πε0RIJ



Quantum mechanicsSeparation of variables

I If the Hamiltonian can be separated into a sum of the form,

H (x , y , z) = H1 (x) + H2 (y) + H3 (z)

then we can do a variable separation into separate problems, e.g.,

H1ψ(1)i (x) = E (1)

i ψ(1)i (x); H2ψ

(2)j (y) = E (2)

j ψ(2)j (y); H3ψ

(3)k (z) = E (3)

k ψ(3)k (z)

I The total energy becomes the sum of the different contributions,

Ei,j,k = E (1)i + E (2)

j + E (3)k

and the total wavefunction becomes the product,

ψi,j,k (x , y , z) = ψ(1)i (x)ψ

(2)j (y)ψ

(3)k (z)



Quantum mechanicsThe Born-Oppenheimer approximation

I The molecular wavefunction is separated into two parts, one for theelectrons and one for the nuclei:

Em = Ei + En

I The nuclear part is further divided into translation (of the entire molecule),rotation (of the entire molecule) and vibrations (internal motion in themolecule),

En = Ej + Ek + El

I It takes 3 coordinates (degrees of freedom) to describe the translationalmotion of the molecule, 3 coordinates (degrees of freedoms) to describerotation of the entire molecule. Thus, we have 3N − 6 vibrational degreesof freedoms (linear molecules is an exception).



Model systems in quantum mechanics

I Particle-in-a-box (model for translation)I Harmonic oscillator (model for vibration)I Rigid rotor (model for rotation)



Particle-in-a-boxI One-dimensional problem: V = 0 in 0 ≤ x ≤ L, otherwise V =∞.

εn =(nh)2

8mL2 , n = 1,2,3, . . .

ψn (x) =

√2L

sin(nπx

L

)I Calculate the partition function, qt :

qt =∞∑

n=1

e−εn

kB T =∞∑

n=1

e− n2h2

8mL2kB T

I Define the translational temperature, θt as

θt =h2

8mL2kB⇒ qt =

∞∑n=1

e−n2θt

T



Particle-in-a-boxPart 2

I In most cases θt T , which means that the energy spacings are small.Thus approximate as an integral

qt =

∞∫0

e− h2n2

8mL2kB T dn =

√2πmkBT

h2 L

where we have used a standard integral in Appendix D.I Generalize to three dimensions. Separation of variables gives

εnx ,ny ,nz =h2

8m

(n2

x

a2 +n2

y

b2 +n2

z

c2

)

where a, b and c, are the dimensions of the box.I Exercise: For which combinations of values of a, b, and c, do we get

degenerate states? Two of a, b or c are identical



Particle-in-a-boxPart 3

I The partition function in three dimensions,

qt = qxqy qz =

(2πmkBT

h2

) 32

abc =

(2πmkBT

h2

) 32

V

where V is the volume of the box.I Sometimes we introduce Λ (a characteristic length) as

Λ3 =

(h2

2πmkBT

) 32

such thatqt =

VΛ3



Harmonic oscillator

I The vibrations in a molecule is approximated by aHarmonic oscillator.

I The potential energy as a function of an internalcoordinate (e.g. bond length) is often approximatedwith a Taylor expansion:

V (x) = V (0)+xV (1) (0)+12

x2V (2) (0)+16

x3V (3) (0)+. . .

I V (0) only shifts the zero level; the gradient is 0 atthe minimum; the anharmonicity, V (3) (0) andhigher order terms may be neglected in a firstapproximation:

V (x) =12

x2V (2) (0) =12

ksx2

where ks is the force constant.



Harmonic oscillatorPart 2

I The solutions for the Schrodinger equation:I The energy levels are given as

εn =

(n +

12

)hν , n = 0, 1, 2, . . .

where the frequency ν is

ν =1

2π

√ks

µ

and the reduced mass, µ, is given for a diatomic molecule as

µ =m1m2

m1 + m2

I We have a zero-point vibrational energy: ε0 = 12 hν

I The wave function is given in terms of Hermite polynomialsI Alternative expressions: hν = hc

λ = ~ω.



Harmonic oscillatorPart 3: as it was done in the 1st ed. of the book

I The single particle partition function for vibrations, qv ,

qV =∞∑

n=0

e−β(n+ 12 )hν = e−

βhν2

(1 + e−βhν + e−2βhν + e−3βhν + . . .

)= e−

βhν2(1 + x + x2 + x3 + . . .

)I Use the series expansion (Appendix C),

(1− x)−1 = 1 + x + x2 + x3 + . . . , 0 < |x | < 1

I Thus the partition function becomes,

qv =e−

βhν2

1− e−βhν



Harmonic oscillatorPart 4: as it is done in the 2nd ed. of the book

I We ignore the zero-point vibrational energy,

εn ≈ nhν

I The partition function becomes

qv ≈∞∑

n=0

e−βnhν =(

1 + e−βhν + e−2βhν + . . .)

=1

1− e−βhν

which is the expression used in most text books.



Rigid rotor model

I The rigid rotor is used to describe rotation of molecules.I The solution to the Schrodinger equation:

I ψl,m (θ, φ) depends on two quantum numbers, l and mI l = 0, 1, 2, . . ., and m = −l ,−l + 1, . . . , 0, . . . , l − 1, lI The energy is

εl =l (l + 1) h2

8πImwhere Im is the moment of inertia.

Im =N∑i

miR2i

I The energy does not depend on the quantum number, m, i.e. the energy isdegenerate with the degeneracy factor,

g(l) = 2l + 1



Rigid rotor modelPart 2

I The rotational partition function for a single particle, qr ,

qr =∞∑l=0

(2l + 1) e−βεl

I For large temperatures, T θr = εlkB

, qr is approximated with an integralas

qr =Tσθr

=8π2ImkBT

σh2

where σ is a symmetry factor that accounts for the number of equivalentorientations.

σ = 1 heteronuclear diatomic molecules (e.g. HF, CO)σ = 2 homonuclear diatomic molecule (e.g. N2), but also water

(H2O)σ = 12 methane (CH4)



Electronic partition functions

I In general,qe = g0 + g1e−β∆ε1 + g2e−β∆ε2 + . . .

where gi is a degeneracy factor and ∆εi is an electronic excitation energy.I Normally, electronic excitations only by interaction with light (no

temperature equilibration) since ∆ε1kB≈ 10000− 100000 K .

I In most cases, g0 = 1I Thus, normally

qe = 1



The molecular partition function

I The translational, vibrational, rotational, and electronic energies are to agood approximation additive.

I Thus the molecular partition function is given as

qm = qtqv qr qe

I Note thatln qm = ln qt + ln qv + ln qr + ln qe

I Also note that it is only qt that depends on the volume, V , whereas qt , qvand qr depends on the temperature, T .

I Also note that the dependence on the number of particles, N, enters inthe total partition function, Q =

qNm

N! , i.e. the partition function for a singlemolecule, qm, is independent of N.



Ideal gas properties from quantum mechanicsHelmholtz free energy and pressure

I Ideal gas (noninteracting particles): Q = qN

N! for indistinguishableparticles.

I Helmholtz free energy, F ,

F = −kBT ln Q = −kBT lnqN

N!= −NkBT ln q + kBT ln N!

I It is noted that the molecular partition function can be written as q = q0V ,(and thus ln q = ln q0 + ln V ) since only the translational part depends onthe volume.

I Pressure, p,

p = −(∂F∂V

)T ,N

= NkBT∂ ln V∂V

=NkBT

V



Ideal gas properties from quantum mechanicsInternal energy

I The internal energy, U, is given as (Eq. 10.34)

U = NkBT 2 ∂ ln q∂T

I For translational motion, qt = c0T32 ,

U =NkBT 2

q∂q∂T

=NkBT 2

c0T 32

32

c0T12 =

32

NkBT

I Similarly, for rotational motion, qr = c1T32 ,

U =32

NkBT

I Each translational and rotational degree of freedom gives 12 NkBT .

I For weak vibrational modes (q 1), gives a contribution kBT for eachmode.



Ideal gas properties from quantum mechanicsEntropy

I Absolute entropies (in contrast to entropy differences) gives afundamental validation of the indistinguishability of gas molecules.

I For a monoatomic gas, only the translational motion contributes. UsingStirling’s approximation,

S = kB lnqN

N!+

UT≈ NkB ln q − kB (N ln N − N) +

32

NkB

= NkB

(ln q − ln N +

52

)= NkB ln

qe52

N= NkB ln

((2πmkBT

h2

) 32 e

52

NV

)which is the Sackur-Tetrode equation.



Ideal gas properties from quantum mechanicsChemical potential

I The chemical potential will be the focus in Ch. 13-16.I First rewrite the Helmholtz free energy using Stirling’s approximation,

F = −kBT lnqN

N!≈ −NkBT (ln q − ln N − 1)

I Derive the chemical potential, µ,

µ =

(∂F∂N

)T ,V

= −kBT (ln q − ln N − 1 + 1) = −kBT lnqN

I We will look at the pressure dependence of µ. Only the translational partof the partition function depends on the volume, so again q = q0V . Usethe ideal gas law, pV = NkBT ,

qN

=q0VN

=q0kBT

p



Ideal gas properties from quantum mechanicsChemical potential, part 2

I q0kBT has the unit of pressure and is denoted point,

point = q0kBT = kBT

(2πmkBT

h2

) 32

qr qv qe

and represents an internal property of the molecule.I Alternatively,

µ = µo + kBT ln p = kBT lnp

point

where point and µo are termed standard state pressure and chemical

potential, respectively.I Note that we take the logarithm of a dimensionless number in this

equation!I The equations for the chemical potential of an ideal gas will be used

repeatedly in the forthcoming chapters.



Summary

I The molecular partition function was derived in terms of a translational,rotational, vibrational and electronic contribution.

I The quantum mechanical model systems (particle-in-a-box, rigid rotorand harmonic oscillator) were employed.

I The ideal gas was studied in detail. In particular an expression for thechemical potential was derived.



Exam June 2011 - Exercise 2

Molecular partition function

a) Give the partition function for an ideal gas with N indistinguishablemolecules. Regard three types of systems: argon atoms, water moleculesand butane (C4H10) molecules. Discuss the relative importance of the variouscontributions to the molecular partition function by comparing the three typesof systems. The electronic ground state is not degenerate for any of of thesystems. Which approximations do we do when we regard a gas as beingideal?

b) If we would like to calculate the pressure, p, from the partition functiondiscussed in (a), which information is required about the molecules and aboutthe system, respectively, to do the actual calculation?



13. Chemical equilibria

Per-Olof Astrand


[email protected]

February 17, 2015


Chemical equilibria

This chapter:I The goal of this chapter is to relate chemical equilibria to atomic structure

through the molecular partition function.I Pressure and temperature dependenceI Perturbation of equilibrium (Le Chatelier’s principle)I The equilibrium between two states A and B,

A K−→ B

I The equilibrium constant, K is the ratio of the numbers (or concentrations) ofparticles in each of the two states at equilibrium.

I Two-state equilibria include: chemical isomerization, folding of biopolymers,binding of ligands to surfaces or macromolecules, condensation of vapour orfreezing of a liquid



Condition for chemical equilibriumI The equilibrium between two states A and B,

A K−→ B

I The direction of the arrow is important. We discuss in terms of initial andfinal state. The equilibrium constant is defined as

K =NB

NA

I Also written in terms of the mol fraction, x , and probability, p,

K =xB

xA=

pB

pAwhere xA = pA =

NA

NA + NB

I The Gibbs free energy is the preferred extremum principle,

dG = −SdT + Vdp + µAdNA + µBdNB

I At constant pressure (dp = 0) and temperature (dT = 0),

dG = µAdNA + µBdNB = 0



The condition for chemical equilibriumPart 2

I The total number of particles, N, is preserved,

NA + NB = N ⇒ dNA + dNB = 0

I Condition for equilibrium rewritten as,

(µA − µB) dNA = 0 ⇒ µA = µB

I So, the chemical potential, µ, has the same role for chemical equilibria asthe temperature, T , for thermal equilibria, i.e. it is equalized.

I Our goal is to relate the chemical potential to the partition function.



Partition functions for chemical equilibria

I To simplify the notation later, denote the partition function q′,

q′ =t∑

j=0

e−βεj = e−βε0 + e−βε1 + . . .+ e−βεt

I Redefine the partition function as,

q = eβε0q′ = 1 + e−β(ε1−ε0) + e−β(ε2−ε0) + . . .+ e−β(εt−ε0)

i.e. for q the ground state energy has been shifted from ε0 to 0.I The chemical potential in terms of the partition function, q′ was given in

Eq. 11.47,

µA = −kBT lnq′ANA

Note that Eq. 11.47 was derived for the canonical ensemble, but theresult for ∂G

∂N will be the same.



Partition functions for chemical equilibriaPart 2

I Thus setting µA and µB equal,

µA = −kBT lnq′ANA

= −kBT lnq′BNB

= µB

gives the equilibrium constant, K as

K =NB

NA=

q′Bq′A

=qB

qAe−β(εB

0−εA0 )



More complex equilibriaI General reaction,

aA + bB K−→ cC

where a, b, and c are the stoichometrics of molecules A, B, C.(E.g. 2H2 + O2 → 2H2O).

I At constant p and T , the condition for equilibrium is,

dG = µAdNA + µBdNB + µCdNC = 0

I Stoichiometric constraints (mass conservation) are phrased in terms of areaction coordinate, ξ,

dNC = cdξ; dNB = −bdξ; dNa = −adξ

I The equilibrium condition is rewritten as,

(cµC − aµA − bµB) dξ = 0



More complex equilibriaPart 2

I Expressing the chemical potential µ in terms of the partition function, q′,

c(−kBT ln

q′CNC

)= a

(−kBT ln

q′ANA

)+ b

(−kBT ln

q′BNB

)which leads to, (

q′CNC

)c

=

(q′ANA

)a( q′BNB

)b

I Define the equilibrium constant, K , as

K =Nc

C

NaANb

B=

(q′C)c(

q′A)a (q′B)b =

qcC

qaAqb

Be−β(cεC

0−aεA0−bεB

0 )

I Define the dissociation energy, D = −ε0, so that

−∆D = ∆ε0 = cεC0 − aεA

0 − bεB0

and

K =qc

C

qaAqb

Beβ∆D



Pressure-based equilibrium constantsI For ideal gases, it is more convenient to express the equilibrium constant

in terms of the pressure.I Employing the ideal gas law,

K =Nc

C

NaANb

B=

(pCVkBT

)c

(pAVkBT

)a (pBVkBT

)b =qc

C

qaAqb

Beβ∆D

I Multiply with(

VkBT

)a+b−c, to get a new equilibrium constant, Kp,

Kp =pc

C

paApb

B= (kBT )c−a−b qc

0C

qa0Aqb

0Beβ∆D

whereq0 =

qV

=qt

Vqr qv qe



Le Chatelier’s principle

I Describes the response to a perturbation, i.e. a fluctuation of species Bby dNB.

I Return to the two-state equilibrium,

A K−→ B

I The resulting change in the free energy, dG,

dG = (µB − µA) dξ

I dG ≤ 0 implies that the system is driven back to equilibrium.I One of the two terms, (µB − µA) or dξ, has to be negative.I If a system is perturbed, it is driven back towards equilibrium.



Temperature dependence of equilibrium constantsvan’t Hoff equation

I Measure the equilibrium constant at different temperatures, K (T ).

I Example: two state equilibrium, A K−→ BI At constant T and p, the condition for equilibrium is µA = µB.I Thus for an ideal gas and recalling Eq. 11.50,

µoA + kBT ln pA = µo

B + kBT ln pB

which gives

ln Kp = lnpB

pA=−(µo

B − µoA

)kBT

= −∆µo

kBTI Adopt partial molar properties, Eq. 9.32, (µj = hj − Tsj ),

∆µo = ∆ho − T ∆so



Temperature dependence of equilibrium constantsvan’t Hoff equation, part 2

I The temperature dependence of Kp,

∂ ln Kp

∂T= − ∂

∂T

(∆µo

kBT

)

= − ∂

∂T

(∆ho − T ∆so

kBT

)I Approximating ∆ho and ∆so as

independent of temperature,

∂ ln Kp

∂T=

∆ho

kBT 2 ⇒∂ ln Kp

∂ 1T

= −∆ho

kB

which is the van’t Hoff equation. See figurefor the dissociation of water.



Temperature dependence of equilibrium constantsGibbs-Helmholtz equation

I Generalize the van’t Hoff equation to any dependence G(T ). RearrangeG = H − TS as,

H = G + TS = G − T(∂G∂T

)p

I Use the following mathematical rearrangement,(∂ G

T∂T

)p

=1T

(∂G∂T

)p− G

T 2 = − 1T 2

(G − T

(∂G∂T

)p

)

I This leads to Gibbs-Helmholtz equation,(∂ G

T∂T

)p

= −H (T )

T 2

i.e. the temperature dependence of a Gibbs free energy can be obtainedby measuring the enthalpy as a function of the temperature.



Temperature dependence of equilibrium constantsGibbs-Helmholtz equation, part 2

I Similarly, for Helmholtz free energy,(∂ F

T∂T

)V

= −U (T )

T 2



Pressure dependence of the equilibrium constant

I Follow the procedure for the temperature dependence,

∂ ln K (p)

∂p=

∂

∂p

[−µo

B − µoA

kBT

]= − 1

kBT∂∆µo

∂p

I Use either Gibbs-Duhem equation, or, as here, a Maxwell relation,(∂µ

∂p

)T ,N

=

(∂V∂N

)T ,P

= v

where v is the partial molar volume.I Thus,

∂ ln K (p)

∂p= − 1

kBT∂(µo

B − µoA

)∂p

= −vB − vA

kBT= − ∆v

kBT



Pressure dependence of the equilibrium constantExample 13.6. Two state equilibrium

I Example: The partitioning of an anestheticdrug, halothane, between water (state A)and lipid bilayer membrane (state B), seethe figure.

I Using the values of (p, ln K ) of (0,7.84) and(280,7.6),

∆v = vB − vA = −RTln K2 − ln K1

p2 − p1

= 21 cm3 mol−1

I Since vB − vA > 0, increasing the pressureshifts the equilibrium from B towards A.



Summary

I Equilibrium constants from atomic structure through the partition function.I The temperature and pressure dependence of equilibrium constants.I In Chapter 19, we will extend this to chemical kinetics, i.e. rates of

reactions.



Exam June 2012 - Exercise 1Solved a-c in chapter 10

Boltzmann distribution lawa) How is an ensemble defined in statistical thermodynamics? Discuss brieflythe differences between the microcanonical, canonical andisobaric-isothermal ensembles. Which are the variables, fundamental functionand extremum principle for each ensemble, respectively? Which ensemblesare preferred experimentally (motivate the answer)?

b) For which ensemble is the Boltzmann distribution,

pj =gje−Ej/kBT

Q(1)

derived? What is gj (we have also used the notation Wj ) in eq. (1)? How is Qdefined? What does the magnitude of Q tell us?

c) Assume that we have three molecules, N = 3, and also assume that onlythe two lowest molecular states, ε0 = ε and ε1 = 2ε, can be occupied. What isthe probability for that the total energy, E , is 5ε? The temperature, T = 300 Kand ε = 10 kJ/mol.Per-Olof Astrand Statistical Thermodynamics in Chemistry and Biology February 17, 2015 18 / 19


Exam June 2012 - Exercise 1Part 2 - to be solved now

Boltzmann distribution law

d) The equilibrium constant, K ,

K =x1

x2

is investigated. Show that it can be written both in terms of the energydifference, ∆E = E1 − E2, as

K = Ae−∆E/kBT

and in terms of the Helmholtz free energy difference, ∆F = F1 − F2, as

K = e−∆F/kBT

Use eq. (1) as a starting point. Define A in terms of properties used in eq. (1).



14. Equilibria between liquids, solids, and gases

Per-Olof Astrand


[email protected]

February 17, 2015


Phase equilibria are described by the chemicalpotential

This chapter:I A lattice model of pure solids and liquids is explored.I Vapor pressures over liquidsI Processes of boiling and sublimationI Surface tension from the equilibrium between bulk (interior) and the

surface.I Foundation for treating mixing and solvation (Chapters 15 and 16)



Why do liquids boil?

I Vapor is treated as an ideal gas.I Balance between two driving forces.

I Attractive intermolecular interactions hold the molecules together in theliquid phase.

I The particles gain translational entropy when they escape to the vaporphase.

I The free energy is minimized,

∆G = ∆H − T ∆S or ∆F = ∆U − T ∆S

I The molecules prefer the liquid phase at low temperatures, and the vapor athigh temperatures.



Why do liquids boil?Part 2

I Exactly the same procedure as for chemical equilibria.I Nc particles in the condensed phase, and Nv particles in the vapor phase.I Regard p, T and N as constant,

dG = µv dNv + µcdNc ; dNv + dNc = dN = 0

I At equilibrium,

dG = (µv − µc) dNv = 0 ⇒ µv = µc

I The vapor is regarded as an ideal gas, Eq. 11.50,

µv = kBT lnp

point



Why do liquids boil?Part 3: solids and liquids

I Regard a liquid or a solid as a lattice ofsingle-type particles, see figure.

I Gases are around a factor 1000 less densethan a liquid; a liquid is around 10% lessdense than a solid.

I Here we neglect the distinction betweenliquids and solids.

I Several approximations: liquids do not havelong-range order (periodicity), the numberof nearest neighbours fluctuate in a liquid,molecules diffuse in a liquid.

I Use a lattice model for the condensedphase.



Lattice model of a condensed phase

I For the vapor, G(T ,p,N), whereas for the condensed phase we useF (T ,V ,N). For the condensed phase, we can ignore the distinction to agood approximation. When the pressure is kept constant, the fluctuationsin the volume are small, and vice versa.

I Only include short-range (nearest-neighbour) interactions. Two particlesof type A have the bond energy, wAA < 0, which is attractive.

I Regard the interaction energy as independent of the temperature. Verygood approximation for covalent bonds, (c.f. we only regarded the groundstate in the calculation of the electronic contribution to the partitionfunction). Thus, good model for solids.

I The multiplicity, W = 1, since if a pair of particles swap position, wecannot distinguish the new arrangement from the old one(indistinguishable particles). Thus, S = kB ln W = 0.

I Contributions from vibrations (qv ) and rotations (qr ) in the liquid areignored.



Lattice model of a condensed phasePart 2

I Each particle on a lattice has z nearest neighbours, denoted thecoordination number.

I Each bond has an energy of wAA. Since a bond is shared between twoparticles, assign wAA

2 to each particle.

I The total internal energy, U, is thus

U =NzwAA

2

I The free energy (since S = 0),

F = U − TS = U =NzwAA

2

and the chemical potential, µc ,

µc =

(∂F∂N

)T ,V

=zwAA

2



Vapor pressureI At equilibrium, µc = µv ,

kBT lnp

point

=zwAA

2⇒ p = po

intezwAA2kB T

where p is denoted the vapor pressure.

I Since (from the ideal gas law),

p =

(NV

)kBT

the vapor pressure may be regarded as ameasure of the density of the gas phase.

I If the bonds become stronger (note,wAA < 0), the vapor pressure decreases(see figure).

I Also, if the temperature is increased, thevapor pressure increases.



Cavities in liquids and solidsI The energy to remove a particle (upper figure),

∆Uremove = −zwAA

which is a positive energy.I Note that half of ∆Uremove is assigned to to the

particle leaving and half to the particles aroundthe cavity.

I Another process of interest is the removal of aparticle and the subsequent closure of thecavity (both figures),

∆Uremove+close = U(N − 1)− U(N) = −zwAA

2

I This will be useful to model solvation (Ch. 16),

Uclose =zwAA

2; Uopen = −zwAA

2



p(T ) for phase equlibria

I Upper figure shows a phase diagram. Eachline (phase boundary) shows a point (p,T )where two phases are equally stable (are inequilibrium).

I solid-gas (sublimation), solid-liquid (meltingor freezing), liquid-gas (boiling orcondensation)

I The lower figure shows the liquid-gasphase boundary for water.

I We will derive the slope of p(T ).



p(T ) for phase equlibriaClapeyron equation

I Two points, (T1,p1) and (T2,p2), atequilibrium,

µL (T1,p1) = µG (T1,p1)

µL (T2,p2) = µG (T2,p2)

I Regard the chemical potential at point 2 asa perturbation of point 1,

µL (T2,p2) = µL (T1,p1) + dµL (T ,p)

µG (T2,p2) = µG (T1,p1) + dµG (T ,p)

I This leads to

dµG (T ,p) = dµL (T ,p)



p(T ) for phase equlibriaClapeyron equation, part 2

I We regard µ as a function of (T ,p),

dµ (T ,p) =

(∂µ

∂T

)p,N

dT +

(∂µ

∂p

)T ,N

dp

I Use Maxwell relations (Table 9.1),(∂µ

∂T

)p,N

= −(∂S∂N

)T ,p

= −s ;

(∂µ

∂p

)T ,N

=

(∂V∂N

)T ,p

= v

where s is the partial molar entropy and v is the partial molar volume.I This leads to

dµG = −sGdT + vGdp = dµL = −sLdT + vLdp




I Rearrange as,dpdT

=sG − sL

vG − vL=

∆s∆v

I At equilibrium: ∆µ = ∆h − T ∆s = 0,

dpdT

=∆h

T ∆v

which is the Clapeyron equation.




I Since vG vL,

∆v = vG − vL ≈ vG =RTp

which leads to Clausius-Clapeyronequation,

d ln pdT

=∆hRT 2

I Alternative form by integration,∫ p2

p1

d ln p =

∫ T2

T1

∆hRT 2 dT

lnp2

p1= −∆h

R

(1T2− 1

T1

)I See figure for vapor pressure of benzene.



Refrigerators and heat pumps

I A refrigerator or a heat pump absorbenergy from a cold place and releases it ina warmer place.

I Absorbs energy by boiling (breaking bonds)and releases energy by condensation.

I A fluid can be boiled a low temperature andrecondensed at high temperature bycontroling the pressure (compressing andexpansion)

I It is repeated as a thermodynamic cycle.



Equilibrium between molecules at the surface and inthe bulkSurface tension

I Surface - boundary between a condensedphase and a gas (or vapor)

I Interface - boundary between any twophases.

I Surface tension, γ,

dF = · · ·+ γdA

I Consider the lattice model in the figure: nmolecules on the surface and N − nmolecules in the bulk.



Equilibrium between molecules at the surface and inthe bulkSurface tension, part 2

I Surface atoms have z − 1 nearest neighbours,

U =zwAA

2(N − n) +

(z − 1) wAA

2n =

wAA

2(Nz − n)

I The multiplicity, W = 1, thus the entropy S = 0 and F = U.I The surface tension, γ,

γ =

(∂F∂A

)T ,V ,N

=

(∂F∂n

)T ,V ,N

dndA

=

(∂U∂n

)T ,V ,N

dndA

I The total area, A = na, where a is the area per particle,

dndA

=1a

;∂U∂n

=−wAA

2



Equilibrium between molecules at the surface and inthe bulkSurface tension, part 3

I The surface tension, γ, for this lattice model becomes,

γ =−wAA

2a

I γ is a positive quantity (wAA < 0).I γ is the free energy required to move a particle from the bulk to the

surface increasing the area with dA.I Strong intermolecular forces gives high surface tensions, and the

minimum possible surface (e.g. spherical drops).



Summary

I Lattice model for vaporization.I For the question: why we have a vapour phase? (or why we have two

coexisting phases?), we have to wait until chapter 25 on Phasetransitions.

I Clausius-Clapeyron equation describes how phase equilibria depends onp and T .

I Refrigerator and heat pumps as examples.I Lattice model for surface tension.



Exam Dec 2008 - Exercise 1 a-b

Vapour pressure

a) The vapour pressure for a pure solvent may be written as

p = pointe

βzwAA2

Explain the various variables and constants in the equation. Which are thetwo fundamental approximations in this model for the vapour pressure, andwhat are the main limitations in each of the approximations?b) What happens with the vapor pressure at high and low temperatures, andat strong and weak interactions between the liquid particles, respectively?Explain the results.



15. Solutions and mixtures

Per-Olof Astrand


[email protected]

March 10, 2015


Lattice model for mixtures and solutions

This chapter:I A lattice model for mixturesI Ideal and regular solutionsI The mean-field approximationI Basis for later chaptersI Foundation for treating polymers, colloids and biomolecules

Per-Olof Astrand Statistical Thermodynamics in Chemistry and Biology March 10, 2015 2 / 17


Entropy of mixing

I Lattice model with NA and NB molecules ofspecie A and B, respectively.

I N = NA + NB fill the lattice completely, seefigure.

I A and B have the same size (see Ch. 32 fordifferent sizes, Flory-Huggins).

I The multiplicity for the mixture, WAB, isgiven as,

WAB =N!

NA!NB!

I What is the multplicity of the pure systemsA and B?

∆Smix = SAB−(SA+SB) = kB ln WAB−(kB ln WA + kB ln WB) = kB ln WAB

since WA = WB = 1.



Entropy of mixingPart 2

I The entropy becomes (using Stirling’s formula)

∆Smix = kB ln WAB = kB (N ln N − NA ln NA − NB ln NB)

= kB (NA ln N + NB ln N − NA ln NA − NB ln NB)

I Use mole fractions, xA = NAN , xB = NB

N , xA + xB = 1.

∆Smix = −kB (NA ln xA + NB ln xB)

I Furthermore, use x = xA and xB = (1 − x),

∆Smix

NkB= −x ln x − (1 − x) ln (1 − x)

I Note that x is fix in this model (noreactions), so it does not change to reachthe maximum in the figure.



Ideal solutions

I For an ideal solution, the interactions between the particles are ignored(analogy to ideal gas),

∆Fmix = −T ∆Smix

I This model (ideal solutions) will not be used. Only included for theanalogy to ideal gases.



Energy of mixingI Same lattice model as in the previous

chapter (only nearest-neighbourinteractions),

U = mAAwAA + mBBwBB + mABwAB

where mXY is the number of bonds and wXYis the interaction energy between species Xand Y .

I Express m in terms of NX . Each lattice sitehas z sides,

zNA = 2mAA + mAB ; zNB = 2mBB + mAB

I Thus

mAA =zNA − mAB

2; mBB =

zNB − mAB

2



Energy of mixingPart 2

I The energy expression becomes,

U =zNA − mAB

2wAA +

zNB − mAB

2wBB + mABwAB

=zwAA

2NA +

zwBB

2NB +

(wAB − wAA + wBB

2

)mAB

which needs to be simplified by a suitable approximation.



The mean-field (Bragg-Williams) approximation

I There are many different possibilites of mAB that fulfills the macroscopicconstraints (F (T ,V ,N)), which all should be included in a sophisticatedmodel.

I In the mean-field approximation, the particles are mixed as randomly andas uniformly as possible.

I What is the probability that a B occupies a neighbouring site to an A?I In the Bragg-Williams approximation, the B particles are distributed

randomly.I The probability, pB , that any site is occupied by B is,

pB ≈ NB

N= xB = 1 − x

I In reality, however, pB , for a single site depends on the interaction energies,wXY as well as which type of particle that occupies the neighbouring site.



The mean-field approximationPart 2

I Since there are z neighbours to a particular A molecule, it has theaverage number of AB contacts,

zpB =zNB

N

I Since there is a total of NA A molecules,

mAB =zNANB

N= zNx (1 − x)

I The internal energy becomes,

U =zwAA

2NA +

zwBB

2NB + z

(wAB − wAA + wBB

2

)NANB

N

=zwAA

2NA +

zwBB

2NB + kBTχAB

NANB

Nwhere χAB is termed the exchange parameter.



The mean-field approximationPart 3

I The exchange parameter, χAB, is given as

χAB =z

kBT

(wAB − wAA + wBB

2

)I The Bragg-Williams model fail when there are big differences in the

interaction energies, w , but it serves as a good first approximation.I Note the unfortunate dependence of χAB on kBT .



Free energy of mixingI Since F = U − TS,

F (NA,NB)

kBT= NA ln

NA

N+ NB ln

NB

N+

zwAA

2kBTNA +

zwBB

2kBTNB + χAB

NANB

N

I We are normally interested in the free energy difference, ∆Fmix betweenthe mixed and the initial pure states,

∆Fmix = F (NA,NB) − F (NA,0) − F (0,NB)

I The free energies of the pure states only consist of an internal energyterm,

F (NA,0) =zwAANA

2; F (0,NB) =

zwBBNB

2I So the final result for ∆Fmix becomes,

∆Fmix

NkBT= x ln x + (1 − x) ln (1 − x) + χABx (1 − x)

which is termed the regular solution model.



The chemical potential

I The chemical potential for molecule A becomes

µA

kBT=

1kBT

(∂F∂NA

)T ,V ,NB

= ln xA +zwAA

2kBT+ χAB (1 − xA)2

I In thermodynamics, the chemical potential is often expressed as

µA = µoA + kBT ln γx

where γ (here!) is called the activity coefficient.I Note the inconsistency in the dependency on x .



Interfacial tension

I Regard the boundary between twocondensed phases, the interface.

I The interfacial tension, γAB is the cost infree energy to increase the interfacial area.

I So if γAB is large the interfacial area will besmall.

I Let us extend the lattice model for surfacetension,

U = (NA − n)zwAA

2+ n

(z − 1) wAA

2

+ (NB − n)zwBB

2+ n

(z − 1) wBB

2+ nwAB



Interfacial tensionPart 2

I The entropy is 0 (i.e. no mixing) in this lattice model, leading to

γAB =

(∂F∂A

)NA,NB ,T

=

(∂U∂A

)=

(∂U∂n

)dndA

I Calculating the derivatives (A = na),(∂U∂n

)= wAB − wAA + wBB

2;

dndA

=1a

the interfacial tension becomes,

γAB =1a

(wAB − wAA + wBB

2

)=

kBTza

χAB

I Depending on the sign of χAB, γAB can have any sign. However, there is acompetition between mixing and creating an interface, and mixing willoccur if ∆Fmix < 0.

I Is the interfacial tension, γAB, temperature-dependent? No



What have we left out?I We should have evaluated the multiplicity W for each mAB (number of AB

interactions), and evaluated the partition function,

Q =∑mAB

W (NA,NB,mAB) e−βE(NA,NB ,mAB)

where W is here also interpreted as a degeneracy factorI Secondly, we have ignored the contributions from the molecular partition

functions, qA and qB,

F (NA,NB)

kBT= NA ln

NA

N+ NB ln

NB

N+

zwAA

2kBTNA +

zwBB

2kBTNB + χAB

NANB

N−NA ln qA − NB ln qB

where qX includes molecular vibrations, rotations, etc.I However, regarding the free energy of the pure states,

F (NA,0)

kBT=

zwAANA

2kBT− NA ln qA ;

F (0,NB)

kBT=

zwBBNB

2kBT− NB ln qB

the molecular partition functions, qX , do not contribute to the mixing freeenergy, ∆Fmix, if qX do not depend on the interactions with theneighbours.



Summary

I Lattice model for mixing and solutionsI Introduced the mean-field (Bragg-Williams) approximation.I Calculated the free energy of mixing, ∆Fmix.I Discussed interfacial tension.



Exam Aug. 2012 - Exercise 2 a-b

Mixture and surface tension

a) Assume that we have a cluster (small drop) with e.g. methanol andchloroform. It is found that methanol has a higher concentration on thesurface, whereas chloroform has a higher concentration in the centre of thecluster. Explain (not derive) this behaviour in terms of a lattice model (thecurvature of the drop may be ignored) and a regular solution. What is aregular solution? Which are the driving forces? What would happen if allintermolecular interactions would be the same?

b) Use the same example as in a). Would methanol or chloroform have thehighest surface tension (motivate the answer)? Explain what the surfacetension is in terms of a lattice model.



16. Solvation and the transfer of molecules between phases

Per-Olof Astrand


[email protected]

March 11, 2015


The chemical potential describes the tendency ofmolecules to exchange and partition

This chapter:I Some examples of colligative properties.

I vapor pressure depression by solutesI freezing temperature depression (e.g. salt melts ice)I boiling temperature elevation (e.g. salt condenses boiling water)I osmotic pressure, partition coefficients, dimerization in a solvent

I The high-concentration component is called the solvent; thelow-concentration component is called the solute.

I Two competing driving forces:I Molecules tend to move from regions with high concentration to low

concentration to increase the entropy.I Also, molecules tend to move to regions where they have high chemical

affinity.



SolvationSolvation is the transfer of molecules between vapor and liquid phases

I Model system:I Liquid mixture of two components, A and B.I Suppose B is volatile, but A is not.I Constant T , p, NA, NB ,

NgasB + N liquid

B = NB

I Two typical examples:I A gas, e.g. carbondioxide (B) dissolved in water (A).I A salt, e.g. sodium chloride (A) dissolved in water (B).

I At constant p and T , the condition for equilibrium is

µgasB = µliquid

B



Lattice model for solvation

I Ideal gas (Eq. 11.50),µgas

B = kBT lnpB

poB,int

and for the mixture (Eq. 15.17),

µliquidB = kBT ln xB +

zwBB

2+ kBTχAB (1− xB)2

I At equilibrium,

pB

poB,int

= xB exp(χAB (1− xB)2 +

zwBB

2kBT

)



Lattice model for solvationPart 2

I Rewrite as

pB = poBxBeχAB(1−xB)

2

where p0B is the vapor pressure of the pure

substance (Eq. 14.9),

poB = po

B,intezwBB2kB T

I See the figure for graphs with different χAB.



Two limiting cases

I Case 1: Salt in waterI B is the solvent and is volatile.I xB ≈ 1,

pB = poBxB

which is Raoult’s law.I It is interpreted as an entropy effect (ideal solution).

I Case 2: Carbondioxide in waterI B is the solute and is volatile.I xB ≈ 0,

pB = poBeχAB xB

which is Henry’s law.I It is interpreted as an energy effect.I Typical laboratory experiment for measuring χAB experimentally.



Henry’s lawI Henry’s law is in the literature written as

pB = kHxB

where kH is the Henry’s law constant.I Start at Eq. 16.2 (and xB ≈ 0),

pB

poB,int

= xB exp(χAB +

zwBB

2kBT

)= xB exp

(z

kBT

(wAB −

wAA

2

))which gives

kH = poB,int exp

(z

kBT

(wAB −

wAA

2

))= po

B,int exp(

∆hosolution

kBT

)I ∆ho

solution (to be exact ∆U/N) describes the process in the figure below



Solutes raise the boiling temperature of a solventI The boiling point, Tb, is the temperature where the vapour pressure

equals the atmospheric pressure (normally 1 atm). (Not shown in thiscourse).

I First assume that we have a pure liquid (no salt) with a boiling point, Tb0at a pressure, patm,

patm = poB,int exp

(zwBB

2RTb0

)(it does not matter if we use R or kB, but could have been usedconsistently.).

I Secondly, we have a solution with a salt concentration, xA. Using an idealsolution (or Raoult’s law), xB ≈ 1,

patm = poB,intxB exp

(zwBB

2RTb1

)I Putting these two equations equal,

ln xB =zwBB

2R

(1

Tb0− 1

Tb1

)Per-Olof Astrand Statistical Thermodynamics in Chemistry and Biology March 11, 2015 8 / 21


Solutes raise the boiling temperature of a solventPart 2

I First, a Taylor expansion around Tb0,

1Tb0− 1

Tb1≈ 1

Tb0−(

1Tb0

+ (Tb1 − Tb0)−1T 2

b0

)=

∆TT 2

b0

I Secondly (see exercise E16.1),

ln xB = ln(1− xA) ≈ −xA −x2

A2−

x3A3− . . .

I Finally, introducing

∆hovap = ∆ho

gas −∆holiq =

−zwBB

2gives the final result commonly found in text-books. (Why is ∆T > 0?)

∆T =RT 2

b0xA

∆hovap

I Note that the result is for an ideal solution, i.e. it is explained as anentropy effect. The result is generic for any solute; it does not depend onwAA and wAB. Easy to generalize to a more sophisticated model.



Units of concentrationI Sometimes other units are used than molar fractions, xA

xA =nA

nA + nB≈ nA

nB

I The molarity or concentration, cA (in mol solute/liter solvent) is

cA =nA

VI The molality, mA, (in mol solute/kg solvent) is,

xA ≈ mAMB

where MB is the molar mass (g/mol).I If only the linear term is included for the change in boiling point,

∆T = KbmA =RT 2

b MB

∆hovap

mA

where Kb is a constant for boiling that only depends on properties for thepure solvent.



Solutes lower the freezing point of a solvent

I Assuming that the salt is only present in liquid water and not in ice, weget similar expressions,

∆Tf = Tf − T =kBT 2

f∆ho

fus

(xA +

(12− χAB

)x2

A + . . .

)and

Tf − T = Kf mA

where Tf is the freezing point of the pure liquid.I Not derived. We also indicate how non-ideal corrections enter, which

would show up analogously for the boiling point.



Osmotic pressureSemipermeable membranes

I Model system: pure liquid B separated froma mixture of A and B by a semipermeablemembrane that lets B but not A pass freely.

I Molecules B will be drawn from the pureliquid to the mixture to increase the entropy.

I This can be countered byI Increasing the pressure of the mixtureI or accumulate extra volume on the top,

which will give an extra hydrostaticpressure because of its weight.

I This additional pressure is termed theosmotic pressure, π.



Osmotic pressureSemipermeable membranes, part 2

I The condition for equilibrium becomes (atconstant T ),

µpureB (p) = µmixture

B (p + π, xB)

I Use that the free energy is a state function,and divide the process into two steps (seefigure).

I First step p → p + π,

µpureB (p + π) = µpure

B (p) +

∫ p+π

p

∂µB

∂pdp

I Use the Maxwell relation: ∂µB∂p = ∂V

∂NB= vB,

µpureB (p + π) = µpure

B (p) + πvB



Osmotic pressureSemipermeable membranes, part 3

I For the second step (creating the mixture) using an ideal solution,

µmixtureB (p + π, xB) = µpure

B (p + π) + RT ln xB

I For both steps,

µmixtureB (p + π, xB) = µpure

B (p) + πvB + RT ln xB

I Thus at equilibrium (µmixtureB (p + π, xB) = µpure

B (p))

−πvB = RT ln xB

I Again use,

ln xB = ln(1− xA) ≈ −xA −x2

A2−

x3A3− . . .

I Finally, the osmotic pressure is given in terms of the concentration as(using the linear term only),

π =RTvB

xA ≈nARTnBvB

=nART

V= cART

I Again, a text-book result is obtained, that we can extend systematicallywith the lattice model.



Partition coefficientsSolutes can transfer from one solvent to another

I Model system: Two immiscible solvents A and B (e.g. oil and water). Asolute s can transfer between the two solvents.

I The partition coefficient, K BA , is defined as

K BA =

xBs

xAs

where xAs is the molar fraction of s in solvent A.

I Condition for equilibrium,µA

s = µBs

which gives (for µAs

kBT =µB

skBT ) using Eq. 15.17 for µ of a solution,

zwss

2kBT+ ln xA

s + χsA

(1− xA

s

)2=

zwss

2kBT+ ln xB

s + χsB

(1− xB

s

)2

I The final result becomes,

ln K BA = ln

xBs

xAs

= χsA

(1− xA

s

)2− χsB

(1− xB

s

)2≈ χsA − χsB

where we in the last have used that xAs → 0 and xB

s → 0.Per-Olof Astrand Statistical Thermodynamics in Chemistry and Biology March 11, 2015 15 / 21


Model for dimerizationI Dimerization is the equilibrium between a solvated

dimer, AB, and two solvated monomers A and B (seefigure). The equilibrium constant is defined as

Kdimer =xAB

xAxB

I Approximation: All three species are at infinite dilution ina solvent s.

I At constant p and T ,

dG = µAdNA + µBdNB + µABdNAB = 0

I Conservation of A and B leads to

dNAB = −dNA = −dNB

I At equilibrium, dG = 0,

(−µA − µB + µAB) dNAB = 0 ⇒ µAB = µA + µBPer-Olof Astrand Statistical Thermodynamics in Chemistry and Biology March 11, 2015 16 / 21


Model for dimerizationPart 2

I Rewrite the equilibrium condition as(µo

ABkBT

+ ln xAB

)=

(µo

AkBT

+ ln xA

)+

(µo

BkBT

+ ln xB

)so that

ln Kdimer = ln(

xAB

xAxB

)= − 1

kBT(µo

AB − µoA − µo

B) = − 1kBT

∆µo

I Using the notation for µoA (and µo

B) (in line with Eq. 15.17),

µoA

kBT=

zwAA

2kBT+ χsA − ln qA =

zkBT

(wsA −

wss

2

)− ln qA

where the molecular partition functions, qA, has been included andχsA ≈ χsA(1− xA)2




I We also need µoAB, which is obtained from the free

energy of the process in the figure.I Consider a two-site cavity surrounded by a first

shell of 2 (z − 1) solvent molecules.I The energy cost of opening a two-site cavity is

thus −wss (z − 1) (see ch. 14)I Insertion of an A molecule gives (z − 1) wsA;

insertion of B molecule gives (z − 1) wsB; a dimeris formed with the energy wAB,

µoAB

kBT=

z − 1kBT

(wsA + wsB − wss) +wAB

kBT− ln qAB




I The equilibrium constant becomes,

ln Kdimer = −∆µo

kBT=

wsA + wsB − wss − wAB

kBT+ ln

(qAB

qAqB

)

=1z

(χsA + χsB − χAB) + ln(

qAB

qAqB

)I Note that qAB may be very different from qAqB. For example, if A and B

are rare gas atoms, they only include a translational contribution to q.However, qAB includes also vibrational and rotational contributions.

I Dimerization in a solvent can be driven by many different ”effects”. Stronginteractions between A and B or between s and s will favour dimerization.Strong interactions between A and s or between B and s will disfavourdimerization.

I The hydrophobic effect (the dimerization of two unpolar (oil) molecules inwater) as an example. In this case, water likes water, i.e. a large wssfavour dimerization.



Summary

I Many different phenomena have been considered based on the latticemodel:

I SolvationI How solvation can raise the boiling temperature and lower the freezing

temperatureI Osmotic pressure - semipermeable membranesI Partition of solute molecules between two phases - partition coefficientsI Model for dimerization in solution



Exam Dec 2008 - Exercise 1We looked at a) and b) in chapter 14

Vapour pressurea) The vapour pressure for a pure solvent may be written as

p = pointe

βzwAA2

Explain the various variables and constants in the equation. Which are thetwo fundamental approximations in this model for the vapour pressure, andwhat are the main limitations in each of the approximations?

b) What happens with the vapor pressure at high and low temperatures, andat strong and weak interactions between the liquid particles, respectively?Explain the results.

c) What is the difference between an ideal and a regular solution? For aregular solution with components A and B, where only A forms a vapour,discuss the condition for that the vapour pressure, pA, increases or decreasesas compared to an ideal solution?



17. Physical kinetics: diffusion, permeation and flow

Per-Olof Astrand


[email protected]

March 17, 2015


Forces drive molecules to flow

This chapter:I Many systems are out of equilibrium and are governed by transport

equations.I Molecules diffuse, transporting matter and heat.I What are the rates at which molecules flow from one place to another?I What forces drive them?I See Appendix G for repetition of mathematics related to this chapter.



Definition of flux

I The flux J is defined as the amount of materialpassing through a unit area per unit time,

J =cV

A∆t=

cA∆xA∆t

=c∆x∆t

= cv

where c is the concentration and v is the velocity.I Fluxes may arise from applied forces, here we

assume a linear relation between the force, f andthe velocity, v ,

f = ξv

where ξ is the friction coefficient. This leads to

J = cv =cfξ

= Lf

where L is a general proportionality constant.



Fick’s lawI At equilibrium for a single-phase system: the concentration is the same

everywhere.I Systems with concentration gradients are not in equilibrium.I A fundamental empirical law, Fick’s law, relates the flux to the

concentration gradient,

J = −Ddcdx

; ~J = −D~∇c

or in general a force to the flow of particles.I The proportionality constant, D, is termed the diffusion coefficient.I Similarly, Ohm’s law relates forces to electrical current, and Fourier’s law

relates forces to flow of heat,

Jq = −κdTdx

where κ is the thermal conductivity.



Fick’s second law

I The flux into a volume element does notneed to be the same as out of the volumeelement since particles may be depleted oraccumulated in the volume element.

I The increase of the amount of particles canbe written as the difference between thenumber of particles entering and leavingthe volume element,

A∆t (J(x , t)− J(x + ∆x , t))

or as the change in number of particles withtime at the center of the volume element,

A∆x(

c(x +∆x2, t + ∆t)− c(x +

∆x2, t))



Fick’s second lawPart 2

I Taking the limits ∆x → dx and ∆t → dt gives,

∂c∂t

= −∂J∂x

which is a fundamental equation, useful to remember.I Substituting into Fick’s law,

∂c∂t

=∂

∂x

(D∂c∂x

)= D

∂2c∂x2

which is called Fick’s second law or the diffusion equation.I In three dimensions,

∂c∂t

= −~∇ · ~J = D∇2c



Example 17.2: Diffusion through a membrane

I Many applications: cell membranes,polymer membranes in industry, etc., etc.

I The concentration to the left, cl , is higherthan to the right, cr .

I cl and cr are kept constant by addition andremoval, respectively, of the species.

I Note the effect of the partition coefficient,K , in the graph.

I There is a driving force from left to right.I Assume steady-state,

∂c∂t

= 0



Example 17.2: Diffusion through a membranePart 2

I Fick’s second law becomes∂2c∂x2 = 0

I Integration givesc(x) = A1x + A2

where A1 and A2 are constants of integration.I The partition coefficient, K , gives us two concentrations in the membrane,

c(0) = Kcl ; c(h) = Kcr

I For the concentration profile, we get

c(x) =K (cr − cl )

hx + Kcl



Example 17.2: Diffusion through a membranePart 3

I The flux, J, becomes

J = −D∂c∂x

=KDh

(cl − cr ) =KDh

∆c

I The permeability, P, of a membrane is defined as

P =KDh

=J

∆c

and the resistance as 1/P.



Diffusion of particles toward a sphereI Example: micelle or a proteinI Assume steady-state.I Spherical polar coordinates with no angular

dependence (see Appendix G),

∇2c =1r

d2 (rc)

dr2 = 0

I Integration gives,

c(r) = A1 +A2

r

I Two boundary conditions: A1 = c∞, and theabsorbing boundary condition, c(a) = 0,

c(r) = c∞(

1− ar

)Per-Olof Astrand Statistical Thermodynamics in Chemistry and Biology March 17, 2015 10 / 19


Diffusion of particles toward a spherePart 2

I The flux, J, becomes

J(r) = −Ddcdr

=−Dc∞a

r2

I The current, I(a), is the number of collisions per second at r = a (i.e fluxtimes area),

I(a) = J(a)4πa2 = −4πDc∞a



The Smoluchowski equation

I Sometimes particle flow is driven by both a concentration gradient and anapplied force.

I The fluxes are additive,

J = −D∂c∂x

+cfξ

where ξ is a friction coefficient.I Again combining with (see how we obtained Fick’s second law),

∂c∂t

= −∂J∂x

gives the Smoluchowski equation,

∂c∂t

= D∂2c∂x2 −

fξ

∂c∂x



The Einstein-Smoluchowski equationI Start with the flux from the Smoluchowski eq.,

J = −Ddcdx

+cfξ

and assume equilibrium, J = 0, (e.g. as in the figure),

Ddcdx

=cfξ

⇒ Ddcc

=fdxξ

I Assume a reversible work, w = −∫

fdx ,

D lnc(x)

c(0)= −w(x)

ξ⇒ c(x)

c(0)= e−

w(x)ξD

I An equilibrium system must also follow a Boltzmann distribution, whichgives the Einstein-Smoluchowski equation

c(x)

c(0)= e−

w(x)kB T ⇒ D =

kBTξ



Ex. 17.7. Diffusion coupled to a chemical reactionI Assuming a chemical reaction (empirical

relation),dcdt

= −krxc

which works as a sink (in contrast to asource).

I Fick’s second law becomes,

dcdt

= D(∂2c∂x2

)− krxc = 0

where we have assumed steady state,dcdt = 0.

I The general solution,

c(x) = A1e−ax + A2eax

where a =√

krxD



Diffusion coupled to a chemical reaction, part 2

I Boundary condition: c∞ = 0 (everything has disappeared in the reaction)gives A2 = 0,

c(x) = c(0)e−x√

krxD

I The flux at the surface, J(0), may be calculated as,

J(x) = −Ddcdx

= c(0)D

√krx

De−x

√krxD

andJ(0) = c(0)

√Dkrx

I Using the example of a drug tablet, the drug is released faster if D or krxare high.



Onsager reciprocal relations

I If two ”gradients” are applied to a system, e.g. a temperature gradientdrives a heat flow and a voltage difference generates an electricalcurrent, these processes are not independent. In general,

J1 = L11f1 + L12f2 ; J2 = L21f1 + L22f2

I The coupling elements, L12 and L21, tell us that e.g. a voltage differencealso generates a heat flow.

I Onsager showed that L12 = L21, and are called Onsager reciprocalrelations.

I The coupling elements, Lij , are probably small (negligable) in manycases, but when they are substantial they may lead to novel phenomenaand functional materials.

I Many potential applications, e.g. thermoelectricity



Summary

I Introduction to nonequilibrium statistical thermodynamics.I Primarily discussed diffussion (Fick’s laws), but many other transport

processes exist.I Extended to coupled processes through Onsager reciprocal relations.I For the future: TKJ4200 Irreversible thermodynamics (Prof. Signe

Kjelstrup)




Transport processes

a) When studying transport processes, we often use the approximation ofsteady-state. Explain what we mean by a system being in steady-state. Howwould you in a few sentences define what a flux of particles is? What is thedistinction between a system being in steady-state or being at equilibrium?

b) Assume that we have a two-phase system and we add a solute, s, that maypartition between the two phases to reach equilibrium. What is the definitionof the partition coefficent in terms of the molar fractions, x , of the solute in thetwo phases? What is the condition for equilibrium expressed in terms ofchemical potentials? Is the partitioning of the solute between the two phasesan entropy-driven or an energy-driven process (motivate the answer)?




Transport processes - part 2

c) Particles flow from a reservoir to the left with a concentration, c1, through amembrane, and leaves the system to the right with an imposed concentration,c2 (see figure). Assuming steady-state, draw the concentration profile, c(l),where l is the length of the tube (from left to right). Explain each part of thegraph with a few sentences.



19. Chemical kinetics and transition states

Per-Olof Astrand


[email protected]

March 19, 2015


Rates of chemical reactions

This chapter:I Rate constants for reactions are connected to equilibrium constants

(Chapter 13)I Add one additional concept: transition state or activation barrierI Chemical reactions are strongly temperature-dependent



Reaction rates are proportional to concentrationsI A reaction,

Akf

kr

B

where kf and kr are rate coefficients for the forward and reversereactions, respectively.

I The rates are given as

d [A(t)]

dt= −kf [A(t)] + kr [B(t)]

d [B(t)]

dt= kf [A(t)]− kr [B(t)]

where [A(t)] is the concentration (or number of particles) at time t .I Rate constants kf have the unit of inverse time.I Coupled differential equations that can be solved by standard matrix

techniques.



Reaction rates are proportional to concentrationsPart 2

I In many cases, the reverse reaction may be neglected kr kf ,

d [A(t)]

dt= −kf [A(t)]

which has the solution,

[A(t)] = [A(0)]e−kf t

I An exponential decay.I [B] changes as,

[B(t)] = [B(0)] + [A(0)](

1− e−kf t)



At equilibrium

I At equilibrium, the change in concentrations are zero,

d [A(t)]

dt=

d [B(t)]

dt= 0

I It results in the principle of detailed balance,

kf [A]eq = kr [B]eq

which may be rewritten as

K =[B]eq

[A]eq=

kf

kr

where K is the equilibrium constant.



The mass action law – Guldberg-Waage law

I Suppose we have a chemical reaction,

aA + bB + cC → P

I The law of mass action:

d [P]

dt= kf [A]a[B]b[C]c

I The stoichiometric coefficients (the mechanisms of reactions) are thusavailable from kinetics experiments.



Temperature dependence on reaction rates

I Consider the binary reaction,

A + B k−→ P ;d [P]

dt= k [A][B]

I k is assumed to be independent of the concentrations, but may bestrongly dependent on the temperature.

I Based on the van’t Hoff equation (Eq. 13.35),

d ln KdT

=∆ho

kBT 2

Arrhenius suggested for the rate constants,

d ln kf

dT=

Ea

kBT 2 ;d ln kr

dT=

E ′akBT 2

I Ea and E ′a are termed activation energies.



Transition states

I The concept of transition state or activationbarrier is introduced.

I Integrating the equation above gives theArrhenius equation,

kf = Ae−βEA

where A is a constant that can bedetermined by kinetics experiments atdifferent temperatures.

I In experiment, A and EA are determined byplotting ln kf against 1/T (see figure to theright).



Transition state theoryOnly sketched briefly

I In transition state theory, the reaction is divided into two steps,

A + BK‡−−−− (AB)‡

k‡−→ P

I (AB)‡ denotes the complex formed at the transition state.I The first step is the equilibrium between the reactants and the transition

state described by the equilibrium constant K ‡.I The second step, downhill from the transition state, is described by the

rate coefficient, k‡.I The equilibrium constant, K ‡, is given as

K ‡ =[(AB)‡]

[A][B]

I The overall rate, k , may be introduced as

d [P]

dt= k‡[(AB)‡] = k‡K ‡[A][B] = k [A][B]



Transition state theoryPart 2

I Expressing K ‡ in terms of the partition function (eq. 13.17),

K ‡ =q(AB)‡

qAqBe

∆D‡kB T

and factorizing q(AB)‡ = q‡qξ where qξ is the partition function for themotion along the reaction coordinate.

I Skipping the details...

K ‡ = K ‡qξ ≈ K ‡kBThνξ

where νξ is a vibrational frequency along the reaction coordinate. Alsoskipping the details, the rate constant for the second step,

k‡ ≈ νξI The entire rate constant k is given as

k = k‡K ‡ =kBT

hq‡

qAqBe

∆D‡kB T =

kBTh

K ‡

where K ‡ is termed the activation equilibrium constant.Per-Olof Astrand Statistical Thermodynamics in Chemistry and Biology March 19, 2015 10 / 18


Thermodynamics of the activated stateI Relate the activation equilibrium constant, K ‡, to the activation free

energy, ∆G‡,−kBT ln K ‡ = ∆G‡ = ∆H‡ − T ∆S‡

where ∆H‡ is the activation enthalpy and ∆S‡ is the activation entropy.I Using the result for the rate constant, k , from transition state theory,

k =kBT

hK ‡ =

kBTh

e−∆G‡kB T =

kBTh

e−∆H‡kB T e

∆S‡kB

I Comparing to the Arrhenius equation,

kf = Ae−βEA

leads to that the A-term in Arrhenius equation can be identified

A =kBT

he

∆S‡kB



The primary isotope effectI Isotope substitution is useful for determine reaction mechanisms.I For example, at room temperature the C-H bond cleaves 8 times faster

than a C-D bond.I This will be explained by an example. Comparing

CH (CH)‡ → C + H

with the rate constant kH, and

CD (CD)‡ → C + D

with the rate constant kD and D denotes deuterium 2H.I Use previous result for the rate constant, k ,

k =kBT

hq‡

qAqBe

∆D‡kB T



The primary isotope effectPart 2

I The ratio of the reaction rates becomes

kH

kD=

q‡CHqCH

e∆D‡CHkB T

q‡CDqCD

e∆D‡CDkB T

≈ e∆D‡CH−∆D‡CD

kB T

where we now use the notation q‡CH instead of q‡.I The difference in dissociation energies is the difference in ground-state

energies. The electronic states are the same for CH and CD, whereas thevibrational ground-state energy, the zero-point vibrational energy differs.

I At the transition state, the bond breaks (not vibrating), so it is only thezero-point vibrational energy for the reactants that are affected,

∆D‡CH −∆D‡CD = −12

h(νCD − νCH)

where ν is a vibrational frequency.



The primary isotope effectPart 3

I Regard the reduced masses,

µCH =mCmH

mC + mH≈ mH ; µCD =

mCmD

mC + mD≈ mD ≈ 2mH ≈ 2µCH

I The force konstant, ks is the same for CH and CD, giving the frequency,

νCD =1

2π

√ks

µCD≈ 1

2π

√ks

2µCH=νCH√

2; νCD − νCH =

(1√2− 1)νCH

I This leads to the final result

kH

kD= exp

(−hνCH

kBT

(1√2− 1))

I Example 19.3: For νCH = 2900 cm−1 and at T = 300 K, we get kHkD

= 7.68.



Catalysis

I Pauling suggested that catalysts work bystabilization of the transition state.

I A catalyzed reaction (C is not consumed),

A + B + C (ABC)‡ → P + C

I Two reaction rates, kc for the catalyzedreaction and k0 without a catalyst,

kc

k0=

K ‡cK ‡0

=[(ABC)‡]

[(AB)‡][C]= K B

where the equilibrium constant, K B isreferred to as a binding constant.



The Brønsted law. . . of acid (base) catalysis - used in chapter 28

I The stronger the acid, the faster the reaction it catalyzes.

AH Ka−→ H+ + A−

R + H+ ka−→ P + H+

with

Ka =[H+][A−]

[AH]

I Let’s write (alternative ∝ [H+][R]),

d [R]

dt= −ka[AH][R]



The Brønsted lawPart 2

I Brønsted law is an observation,

ln ka = α ln Ka + ca

where ca and α > 0 are constants.I With pKa = − ln Ka, (analogous to pH)

ln ka = −αpKa + ca

I Implies a linear free-energy relationship of the form

Ea = a∆G + b

where Ea is the activation barrier and ∆G = −kBT ln Ka is for the aciddissociation. a and b are constants.

I Motivated in the Evans-Polanyi model (in the book, but not included in thecourse)



Summary

I Introduction to chemical kineticsI Concepts like rate constants and their relation to equilibrium constants.I Temperature-dependence on rate constants introduced through

Arrhenius equations.I Concepts like activation energies and transition states.I Briefly sketched transition state theory.I The primary isotope effect for determining reaction mechanismsI Brief introduction to catalysis and the Brønsted law.



Introduction to electrostatics

Per-Olof Astrand


[email protected]

March 10, 2014



I These notes are an alternative (as compared to the book) introduction toconcepts like electrostatic potential, electric field and dipole moment.



Introduction to electrostaticsPart 2

I The interaction between two charges, qi andqj , are given by an empirical law, Coulomb’slaw, which is the starting point in electrostatics,

V (R) =qiqj

4πε0R

where ε0 is the permittivity of vacuum and R isthe distance between the charges.

I Regard the interaction between a set ofcharges, qi , i = 1, . . . ,N, (e.g. a molecule) anda test charge qt . See figure.

V =N∑i

qiqt

4πε0Rit

where interactions within the molecule hasbeen ignored.

~RO

~ri

qt



Electrostatic potential

I Define the electrostatic potential, ψi , at chargei , as

V =N∑i

qiqt

4πε0Rit=

N∑i

qiψi

so thatψi =

qt

4πε0Rit

I We note that the electrostatic potential isadditive, since we can also write

V = qtψt

since

ψt =N∑i

qi

4πε0Rit

~RO

~ri

qt



Multipole expansion

I The next step is to carry out a multipoleexpansion, i.e. a Taylor expansion of theelectrostatic potential, ψi , around the origin ofthe molecule, ~RO . In one dimension, x , itbecomes,

ψi = ψO + ri,x∂ψO

∂x+ · · ·

where ψO is the electrostatic potentialcalculated at the origin and ∂ψO

∂x is its gradient,also calculated at the origin.

~RO

~ri

qt



Multipole expansionPart 2

I Putting the multipole expansion into the expression for the energy,

V =N∑i

qiψi =N∑i

qiψO + qi ri,x∂ψO

∂x+ . . .

=

(N∑i

qi

)ψO +

(N∑i

qi ri,x

)∂ψO

∂x+ . . .

I Define the molecular charge, qmol, as

qmol =N∑i

qi

I the molecular dipole moment, ~µmol, as

~µmol =N∑i

qi~ri



Multipole expansionPart 3

I the electric field, ~EO calculated at the origin O as

~EO = −~∇ψO ; In one dim.:Ex,O = −∂ψO

∂x

I The end result for the energy becomes,

V = qmolψO − ~µmol · ~EO + · · ·

i.e. the interaction can be obtained from molecular multipole moments(charge, dipole moment, etc.) and the electrostatic potential, electric field,etc. calculated at the origin of the molecule.

I This energy term can be added to the internal energy, U, and can thus beincluded the thermodynamics machinery.



Summary

I An introduction to electrostatics starting at Coulomb’s law.I Define electrostatic potential.I A multipole expansion gives concepts like the electric field and dipole

moment.I Energy expression to used in statistical thermodynamics (to be added to

dU).I Trivial to extend to charge-dipole and dipole-dipole interactions,

quadrupole moments, etc.



20. Coulomb’s law of electrostatic forces

Per-Olof Astrand


[email protected]

March 10, 2014


Interactions between charges

This chapter:I Electrostatic interactions govern many properties in physics, chemistry

and biology.I Salt concentrations regulate many processes in biology: transport

through membranes; nerve systems are regulated by ion fluxI Technical applications in electrochemistry: batteries, corrosion, etc.I Electrostatic interactions are long-range.I Definition of the electric field.

I The interaction between two charges, q1 and q2, are given by anempirical law, Coulomb’s law,

V (r) =q1q2

4πε0r

where ε0 is the permittivity of vacuum.



Ionic interactions are weaker in a medium

I Ionic interactions are weakened in a medium,since the liquid is polarized.

I Basically three effects:I The electrons are polarized (increasing the dipole

moment), described by the electronic polarizability.I The atoms within the molecule are moved further

apart, described by the vibrational polarizability.I The molecular dipole moments are reoriented.

I Described by the dielectric constant, D (alsocalled the relative permittivity εr ),

V (r) =q1q2

4πε0Dr

I Typical values: D ≈ 1 (air); D ≈ 2 (hydrocarbons,proteins); D ≈ 78 (water)



Ionic interactions are strong and long-range

I Macroscopically, charge neutrality is a very strong condition.I r−1 is the most long-range interactions.I Nearest-neighbour interactions in a lattice (Bragg-Williams model) are not

sufficient.I Illustrated by the Bjerrum length, lB, i.e. the distance where the

elecrostatic energy equals the thermal energy RT ,

RT =q1q2

4πε0DlB⇒ lB =

q1q2

4πε0DRT

I Typical values at room temperature: lB = 56 nm (in air); scales with 1D :

lB = 0.7 nm (in water).



Electrostatic forces are pair-wise additive

I The force, ~f is given from the gradient of the energy,

~f = −~∇ q1q2

4πε0Dr=

q1q2~r4πε0Dr3

I The Coulomb’s law is pair-wise additive,

V (r) =12

∑I

∑J 6=I

qIqJ

4πε0DrIJ

so the force on a charge qI is the sum of the force from all other chargesqJ

I The force on a test charge of unity size, qtest = 1, is defined as theelectrostatic field,

~E(~r)

=q~r

4πε0Dr3



Electric fields



Fluxes and electric fieldsGauss’s law

I The electric field flux, Φ, through a surface isdefined as

Φ =

∫surface

D~E · d~s

I Special case of a sphere:I The electric field in sperical polar coordinates (and

no angular dependence),

E(r) =q

4πε0Dr 2

I The area is 4πr 2.I The flux,

Φ = DE(r)

∫surface

ds = DE(r)4πr 2 =qε0

I A simple and useful expression (independent ofthe radius of the sphere).



Generalization of electric field flux

I Skipping the derivation...I For any shape of the surface and for many

charges,

Φ =

∫surface

D~E · d~s =1ε0

n∑i=1

qi

which is called Gauss’s law.I Instead of a set of point charges, we may have a

charge distribution, ρ(r),

Φ =

∫surface

D~E · d~s =1ε0

∫V

ρdV



Summary

I Coulomb’s lawI Electric fieldsI Gauss’s law



21. The electrostatic potential

Per-Olof Astrand


[email protected]

March 10, 2014


Electrostatic potential

This chapter:I Connection between the electrostatic potential and the electric field.I Poisson’s equation



What is the electrostatic potential?I Regard the work, δw , required to move a charge, q, a small distance, d~l ,

in a fixed electrostatic field, ~E ,

δw = −~f · d~l = −q~E · d~l

where the minus sign arises from that the work is carried out against thefield and not by the field.

I To move a charge from point A to B,

wAB = −q

B∫A

~E · d~l

I The difference in electrostatic potentials, ψB and ψA is defined as thework, wAB of moving a unit test charge, qtest, from point A to point B,

ψB − ψA =wAB

qtest= −

B∫A

~E · d~l

I The electrostatic potential multiplied by a charge is an energy.Per-Olof Astrand Statistical Thermodynamics in Chemistry and Biology March 10, 2014 3 / 15


Electric field and electrostatic potentialI To relate the electric field to the electrostatic potential,

∆ψ = ψB − ψA = −B∫

A

~E · d~l = −xB∫

xA

Exdx −yB∫

yA

Ey dy −zB∫

zA

Ezdz

I Now convert from an integral to a differential equation. AssumeA = (x , y , z) and B = (x + ∆x , y , z),

∆ψ = −x+∆x∫x

Exdx = −Ex ∆x

I In the limit ∆ψ∆x →

∂ψ∂x , the electric field is identified as minus the gradient

of the electrostatic potential,

Ex = −∂ψ∂x

;(~E = −~∇ψ

)



Electrostatic potential surfaces

I Point charge (upper figure)I Equipotential surfaces at a constant distances r .I The electric field (vectors in the figure) is

perpendicular to the equipotential surface.I Two positive point charges (lower figure)

I At distances far away, the electrostatic potentialbehaves as if we have a point charge equal to 2q.



Electrostatic interactions are conservative forces

I Electrostatic work is a reversible work andtherefore a path-independent quantity that sumsto zero around a cycle (upper figure).

I The electric field in spherical polar coordinates,

E(r) =q1

4πDε0r2

I The work of moving a charge q2 from point A to C,

w = −q21

4πDε0

rb∫ra

q1

r2 dr =q1q2

4πDε0

(1rb− 1

ra

)

I Any path can be approximated by sequence ofradial and equipotential segments (example of aprotein in the lower figure).



Ex. 20.7: The field from a charged planar surface

I Calculate the field from a charged planar surface(plate of a capacitor, an electrode, or a cellmembrane).

I Assuming a “thin” plane, a field both upwards anddownwards: the total flux: 2DEA.

I The surface charge, σ, gives the total charge, σA.I Gauss’s law gives,

2DEA =Aσε0

⇒ E =σ

2ε0D



Ex. 21.2: The electrostatic potential in a parallel platecapacitor

I The potential difference, ∆ψ, is given by the work of moving a unit testcharge from one side to the other.

I The electric fields E+ and E− are given by the previous example,

E± = ± σ

2ε0D

I The force driving a positive unit charge has twoidentical contributions,

Einside =σ

ε0D

I The potential difference becomes,

∆ψ = −d∫

0

Einsidedx =−σdε0D



The electrostatic potential in a parallel plate capacitorPart 2

I The capacitance, C0 is defined as,

C0 =Aσ|∆ψ|

=Aε0D

d

I What is the electric field outside the plates (answer to Exercise 20.10)?

Eoutside = 0



Dipole moment

I The dipole moment, µ, is defined from oppositecharges, ±q, separated by a distance, l ,

~µ = q~l

I Leave as exercises:I The energy of a dipole moment in an electric field.I The interaction between a point charge and a

dipole moment.



Poisson’s equationI Using Gauss’s theorem (Appendix G, Eq. G.15),∫

surface

~v · d~s =

∫volume

~∇ · ~vdV

I Substituting ~v = D~E , ∫surface

D~E · d~s =

∫volume

D~∇ · ~EdV

relates the flux of the electrostatic field through a closed surface with thedivergence of the field throughout its volume.

I Substituting Gauss’s law (Eq. 20.19),∫surface

D~E · d~s =

∫volume

ρ

ε0dV

gives a the differential form of Gauss’s law,

D~∇ · ~E =ρ

ε0



Poisson’s equationPart 2

I Substituting ~E = −~∇ψ gives

∇2ψ = − ρ

ε0Dwhere ~∇ · ~E = −∇2ψ

which is Poisson’s equation.



Summary

I Discussed the electrostatic potentialI Derived Poisson’s equation



Exam August 2011 - Exercise 2Molecular electrostatics

a) The trans and gauche conformations of dibromoethane are given in thefigure. Dibromoethane has a 3-fold rotation axis around the C-C bond, i.e. arotation of 120 of one of the -CH2Br groups moves the molecule from oneenergy minimum to the next. What is the multiplicity of the trans and gaucheconformation, respectively? Assuming that the energy is the same for allconformation minima, what is the difference in Helmholtz free energy, ∆F ,between the trans and gauche conformation? Which conformation isfavoured? The temperature is constant at 300 K.

Figure: trans-dibromoethane and gauche-dibromoethane, respectively



Exam August 2011 - Exercise 2Molecular electrostatics - part b)

b) Assume that the energy is determined by the atomic charges of the Bratoms, qBr = −0.4 e. The distance between the two Br atoms is 4.6 A for thetrans conformation and 3.7 A for the gauche conformation, respectively. Whatis ∆F including the Coulomb interaction energy between the two Br atoms inthe gas phase (the dielectric constant, D = 1)? How does a solvent change∆F? Is it possible to change the favoured conformation from trans to gauche(or the other way around) by shifting the solvent?

Figure: trans-dibromoethane and gauche-dibromoethane, respectivelyPer-Olof Astrand Statistical Thermodynamics in Chemistry and Biology March 10, 2014 15 / 15


22. Electrochemical equilibria

Per-Olof Astrand


[email protected]

March 26, 2015


Electrochemical equilibria

This chapter:I Introduction to electrochemistryI Combine the laws of electrostatics and (statistical) thermodynamics



Driving forces

I The internal energy, U, is a function of the extensive variables U(S,V ,N),which is now extended with the charges, q of the system: U(S,V ,N,q).

I The fundamental equation for U, augmented with M charges, becomes

dU = TdS − pdV +N∑

j=1

µjdNj +M∑

i=1

ψidqi

where ψi is the electrostatic potential at ion i .I The Gibbs free energy, (G = U + pV − TS), becomes

dG = −SdT + Vdp +N∑

j=1

µjdNj +M∑

i=1

ψidqi



Driving forcesPart 2

I For molecular ions, qi = zieNi , where zi is the ion charge in unit chargese, (electron charge: −e)

dG = −SdT + Vdp +N∑

i=1

(µi + zieψi ) dNi

I The electrochemical potential is defined as

µ′i = µi + zieψi

I At constant p and T , equilibrium occurs when the electrochemicalpotential is equal.

I An example of a general approach to extend thermodynamics with anextra energy term.



The Nernst equationI The electrostatic potential is a function of the

position, ψ(x) or ψ(~r).I For a single ionic species, the condition for

equilibrium becomes,

µ′(x1) = µ′(x2)

I Using µ(x) = µo + kBT ln c(x), (for an idealsolution) gives for the electrochemical potential,

µ′(x) = µo + kBT ln c(x) + zeψ(x)

I At equilibrium,

lnc(x2)

c(x1)=−ze (ψ(x2)− ψ(x1))

kBT

which is the Nernst equation



The Nernst equationPart 2

I Alternatively (see figure when ψ is a linearfunction in x),

c(x2) = c(x1)e−ze(ψ(x2)−ψ(x1))

kB T

I Note that electroneutrality is a very strongmacroscopic condition. Whereas the chemicalpotential µ measures the free energy difference toinsert a particle, the electrochemical potentialmeausures the free energy difference of insertingan electroneutral ion pair.

I A general approach for extendingthermodynamics with an energy that is dependenton the spatial coordinate, x .



Neutral salts

I For a salt, e.g. NaCl, that ionizes,

µNaCl = µ′Na+ + µ′Cl−

I or

µNaCl = µoNa+ + kBT ln cNa+ + eψNa+ + µo

Cl− + kBT ln cCl− − eψCl−

I If ψ is spatially uniform, ψNa+ = ψCl− ,

µNaCl = µoNaCl + 2kBT ln cNaCl

where the standard chemical potnetials have been grouped together, andcNaCl = cNa+ = cCl− .



On the relationship between the chemical potentialand the equilibrium constantA recapitulation of previous material

I For the reaction,aA + bB K−→ cC

we had at equilibrium (Eq. 13.16),

cµC = aµA + bµB

I For a component A, in a liquid (Eq. 15.19),

µA = µoA + kBT ln γAxA

i.e. the chemical potential is divided into a standard (intrinsic) term and aterm dependent on the composition.

I For an ideal solution, the activity γA = 1,

µA = µoA + kBT ln[A]

where we equally well can express µA in terms of the concentration [A] asin terms of the molar fraction, xA.



On the relationship between the chemical potentialand the equilibrium constantPart 2

I For the reaction (at equilibrium),

cµoC − aµo

A − bµoB = −kBT (c ln[C]− b ln[B]− a ln[A]) = −kBT ln

[C]c

[A]a[B]b

I Rewritten as,µo

liquid = −kBT ln KC

or in general (not assuming equilibrium),

µliquid = µoliquid + kBT ln

[C]c

[A]a[B]b= µo

liquid + kBT ln Q

where KC is the equilibrium constant expressed in concentrations and Qis the reaction quotient also expressed in concentrations. At equilibiriumQ = KC .



The Nernst equation for an electrodePhrased slightly different as compared to the book

I Consider a reaction, Mz+ + ze− → MI Condition for equilibrium,

µ′solid = µ′liquid

I For the liquid,

µ′liquid = µoliquid + kBT ln Q + zeψliquid

and for the solid,

µ′solid = µosolid + zeψsolid

I At equilibrium,

µoliquid + kBT ln Q + zeψliquid = µo

solid + zeψsolid



The Nernst equation for an electrodePart 2

I The unmeasurable materials properties are grouped together,

zeψ0 = µosolid + zeψsolid − µo

liquid

where ψ0 is termed the half-cell potential and is a property of theelectrode reaction (and can be found in tables).

I Thus,zeψliquid = zeψ0 − kBT ln Q

I Using the definition for Faraday’s constant, F = NAe and R = NAkB,

ψliquid = ψ0 −RTzF

ln Q

which is more general than eq. (22.18), but in principle the same.



The Nernst equation for an electrochemical cellThe combination of two electrodes (not in the book)

I Consider the cell reaction,

A(s) + Bz+ → Az+ + B(s)

I The potential difference, ∆ψ, between two electrodes, A and B, in a cellbecomes

∆ψ = ψ0,B − ψ0,A −RTzF

ln Q

I The half-cell potentials are measured relative to each other, and thehydrogen electrode is used as the reference ψ0,H2 = 0 V .

I Example: What is the half-cell potential for the following reactionZn2+ + 2e− → Zn(s)?

I Given: The half-cell potential for Cu2+ + 2e− → Cu(s) is 0.34 V.I Given: The potential difference in the cell for Zn(s) + Cu2+ → Cu(s) + Zn2+

is measured to 1.10 V at standard conditions (i.e. the concentrations are1 M).

I The cell reaction is obtained by the given half-cell reaction minus therequested half-cell reaction: 1.10 = 0.34− ψ0,A, which gives a half-cellpotential of −0.76 V.



Summary

I An introduction to electrochemistry: Nernst equationI Example of how to add an extra energy term to thermodynamics.I Introduced how to treat energies/potentials that depend on the spatial

coordinate.I Introduced half-cell potentials and Nernst equation for an electrode.



Task from exam 2008Exercise 3

a) Describe in a few sentences what the chemical potential is. If we have ionsin the solution, what is the electrochemical potential?b) We have a container with SF6 gas with the height, h0 = 50 cm. Include theeffect of gravitation in the Gibbs free energy, and derive an expression for theconcentration dependence on the container height, c(h), at constant pressureand temperature. Is the concentration highest at the top or at the bottom ofthe container? Is the concentration gradient significant for molecularsystems? The temperature is 300 K. (Atomic masses: mS = 32.06 g/mol;mF = 18.99840 g/mol)



24. Intermolecular interactions

Per-Olof Astrand


[email protected]

March 17, 2014


Intermolecular interactions

This chapter:I Brief background of intermolecular interactionsI Foundation for more complex statistical thermodynamical models with

more accurate models for the interaction between particles.



Typical potential surfaces

I Intermolecular interaction energies are 0 at longdistances (definition/convention).

I Molecules normally (but not always) attract eachother at intermediate distances.

I Molecules always repel each other at shortdistances.

I Coulomb’s law behave as r−1 and is the mostlong-range interaction.

I r−6 is a typical distance-dependence of ashort-range interaction.



Example: Argon dimerI An argon atom is spherically symmetric (no dipole

moment) and uncharged.I At intermediate range they attract each other

because of London dispersion forces. Dispersionenergies can be derived from quantummechanics, and arises from fluctuations in thecharge distribution (fluctuation dipole moments).

I At short range, the atoms repel each otherbecause of the Pauli exclusion principle, i.e. twoelectrons cannot be in the samequantum-mechanical state.

I The sum of these two terms are normally termedvan der Waals interactions, and are normallymodelled with a Lennard-Jones potential,

VLJ =∑I,J>I

(aij

r12ij

−cij

r6ij

)

where aij and cij are atom-type parameters.Per-Olof Astrand Statistical Thermodynamics in Chemistry and Biology March 17, 2014 4 / 12


Electrostatic interactionsI For charged and polar molecules, the electrostatic

interactions dominate. Coulomb’s law,

VCoul =∑I,J>I

qiqj

4πε0rij

I Molecules are characteriszed by the leadingnon-zero electric moment: monopole (charge),dipole moment, quadrupole moment, octupolemoment, etc.

I The electrostatics is often represented by atomiccharges that gives the correct molecular electricmoments.

I Electrostatic interactions can be either attractiveor repulsive (for example depending on theorientation of the dipole moment), which oftendetermines if the overall interaction energy isattractive or repulsive.



Electrostatic interactionsPart 2

I For dipole-dipole interactions to be attractive in the condensed phase, itoften leads to an ordering of the system (which decreases the entropy).

I Distance-dependence of electrostatic interactions:I charge-charge: r−1

I charge-dipole: r−2

I charge-quadrupole and dipole-dipole: r−3

I charge-octupole and dipole-quadrupole: r−4

I etc.I Still more long-range than dispersion interactions (r−6).



Polarization energiesNot good in the book. Polarization and dispersion are confused with each other

I We make a distinction between permanent dipole moments (include inthe electrostatic term) and induced dipole moments (included in thepolarization energy).

I An induced dipole moment arises from that the electrons are polarized(shifted away from the nuclei) in an electric field.

I The induced dipole moment, ~µind, is given as

~µind = α~E

where α is a polarizability and ~E is the electric field.I The electric field arises from external sources, the molecular electric

moments of the other molecules, and from the induced dipole momnts ofthe other molecules (leads to a set of coupled equations)

I Electronic polarization gives an extra energy contribution that is alwaysattractive (the molecules polarize in such a way that the total energy islowered).



Hydrogen bonds

I Hydrogen bonds are crucial in many systems (aqueous solutions,proteins, etc.) but are difficult to model accurately.

I Example: N-H· · ·O=CI The interaction is strongly orientation-dependenent and increases the

“structure” of the system.I To a good degree, the hydrogen bond is described by electrostatics and

may be modelled by local dipole moments.



Force fields - empirical energy functions

I In addition to van der Waals, electrostatic and polarization contribution, aforce field contains terms to model intramolecular energies:

I Bond stretches (by a harmonic term),

kb

2(b − b0)

2

I Bond angles (also by a harmonic term)

kθ

2(θ − θ0)

2

I Dihedral angles (by a periodic term)

kφ (1 + cos (nφ− δ))

where kb, kθ and kφ are force-field parameters.



The van der Waals gas model

I The van der Waals gas model is an extension of the ideal gas law thatincludes some interatomic interactions.

I The van der Waals equation of state,

p =NkBT

V − Nb− aN2

V 2 =ρRT

1 − bρ− aρ2

where a and b are parameters.I The b parameter gives a “volume” for each particle.I The a parameter includes a long-range attractive interaction.I Not derived here.



Summary

I Introduction to intermolecular forcesI Basically four contributions:

I Dispersion and repulsion (grouped as van der Waals interactions)I Electrostatics (most important for charged or polar systems)I Polarization energies

I Introduction to equation of states with the van der Waals gas model as anexample



Task from exam Aug. 2012Exercise 3

a) In our models, we normally use an interaction parameter wXY to describedintermolecular interactions. For more sophisticated models of intermolecularinteractions, the interaction energy is often partitioned into four terms:electrostatic energy, polarization energy, repulsion energy, and dispersionenergy. Discuss the physical origin of each energy term, and discuss therelative distance-dependence of each term (e.g. which terms is mostimportant at long distances between the molecules?).

b) Compare liquid argon and liquid water. Which energy terms in a) would bemost important for respective liquid at long distances? Discuss the relativeimportance of each energy term for the two cases.



25. Phase transitions

Per-Olof Astrand


[email protected]

April 8, 2015


Phase transitionsThis chapter:

I Introduction to phase transitionsI Two states can be stable at the same timeI As usual, the free energy is minimizedI The free energy curve may have several minima

Per-Olof Astrand Statistical Thermodynamics in Chemistry and Biology April 8, 2015 2 / 12


Temperature dependence

I Oil and water form two phases in equilibrium.I A phase diagram is shown in the figure.I The temperature dependence is demonstrated.



Experiment to create phase diagram

I Six bottles with different oil fraction, x , a temperature, T0.



Experiment to create phase diagramPart 2

I Exp. 1. Small fraction of oil. Everything dissolved in water.I Exp. 2. Happens to be exactly where a second phase appears. It marks

the phase boundary. The fraction of oil in the water-rich phase is x ′.I Exp. 3. Two phase solution. The fraction of oil in the water-rich phase is

x ′. The fraction of oil in the oil-rich phase is x ′′.I Exp. 4. Two phase solution. The fraction of oil in the water-rich phase is

x ′. The fraction of oil in the oil-rich phase is x ′′. The amount of thevarious phases have shifted.

I Exp. 5. Happens to be exactly where the second phase vanishes. Itmarks the phase boundary. The fraction of oil in the oil-rich phase is x ′′.

I Exp. 6. Large fraction of oil. All water dissolved in oil.



Experiment to create phase diagramPart 3

I The phase boundary is also called the solubility curve or coexistencecurve.

I In the coexistence region we have two phases in equilibrium.I We have a critical temperature, TC . Above TC the two components are

miscible at all proportions.



Lattice model for phase separation

I The free energy of mixing for a lattice model (Eq. 15.16),

∆Fmix

N= kBTx ln x + kBT (1− x) ln (1− x) + kBTχABx (1− x)



Lattice model for phase separationPart 2

I Note that the energy term is not temperature dependent (in themean-field model) since

χAB =z

kBT

(wAB −

wAA + wBB

2

)I The way to think about it: (left figure, previous slide)

12

F (x0 + ∆x) +12

F (x0 −∆x) > F (x0)

i.e. two phases with compositions x0 + ∆x and x0 −∆x have a higherfree energy than one phase with composition x0.

I On the other, for the figure to the right, it is possible to find a compositionof two phases, F (x ′) and F (x ′′), with lower free energy than F (x0).

I Also note that we have a criteria for stability,(∂2F∂x2

)x0

> 0



Predict the compositions

I The compositions x ′ and x ′′ can be found bydrawing the tangent to F (x) at two points.

I Why? Regard the free energy

F = N (µAxA + µBxB) = NxB (µB − µA) + NµA

I We also have∂F∂xB

= N (µB − µA)

I At a two-phase equilibrium, the chemical potential of each componentmust be the same in both phases,

µ′A = µ′′

A , µ′B = µ′′

B ⇒ µ′B − µ′

A = µ′′B − µ′′

A

I This leads to(

∂F∂xB

)x′

=(

∂F∂xB

)x′′

, which is close to, but not necessarilythe same as, the local minimum points.



The lever rule: the amount of the two stable phasesI Take Exp. 3 as an exampleI We would like to get the fraction, f ,

f =number of molecules in A-rich phase

total number of molecules in both phasesI We also have

x ′ =number of B molecules in A-rich phasenumber of molecules in A-rich phase

I sofx ′ =

number of B molecules in A-rich phasetotal number of molecules in both phases

I Using the same procedure,

(1− f )x ′′ =number of B molecules in B-rich phase

total number of molecules in both phasesI Combining the last two expressions,

fx ′ + (1− f )x ′′ = x0 =number of B molecules in both phases

total number of molecules in both phases



The lever rule: the amount of stable phasePart 2

I This leads to the lever rule

f =x0 − x ′′

x ′ − x ′′

To summarize: we know how to obtainI the compositions of each phase, x ′ and x ′′

I the amount of each phase



Summary

I Brief introduction to phase transitions and phase diagramsI Predict compositions of each phaseI Lever rule: the amount of each phase

Not included ... but perhaps you will hear about it in the futureI Spinodal curvesI More about critical points and critical temperatures



27. Adsorption, Binding and Catalysis

Per-Olof Astrand


[email protected]

April 14, 2015


Adsorption, binding and catalysis

This chapter:I Molecules adsorb to surfacesI Important in heterogeneous catalysis, filtration, corrosion,

chromatography, etc.I Molecules that bind to surfaces are called adsorbates or ligands.I The Langmuir model is a simple model:

I adsorbates on macroscopic solid surfaces.I pH titrationI kinetics of enzyme reactionsI transport of particles through biological membranes



Langmuir modelI Regard a system consisting of a gas (ideal gas)

and a surface (described by a lattice model),where the gas molecules may bind to the surface.

I The Langmuir model describes the balancebetween the energetic tendency of the moleculesto stick to the surface and the entropic tendencyof the particles to gain translational freedom byescaping from the surface.

I The condition for equilibrium,

µbound = µgas

I The density, θ, of the adsorbate on the surface is,

θ =NA

where N is the number of adsorbates and A is thenumber of surface sites.



Langmuir modelPart 2

I The multiplicity becomes (as usual),

W =A!

N! (A− N)!

I The entropy becomes (using Stirling’s approximation),

SkB

= ln W = −N lnNA− (A− N) ln

A− NA

⇒ SAkB

= −θ ln θ − (1− θ) ln (1− θ)

I Appeared before:I Eq. 6.10., in the derivation of the ideal gas law from a lattice modelI Eq. 15.3., for an ideal mixture

I Note that the entropy term does not depend on the arrangement of thebinding sites, i.e. as well as it could be a surface it could also be thebinding sites in a protein.




I To get an expression for the free energy, F , the internal energy, U, is

U = Nw

where w < 0 is the particle-surface interaction energy.I In addition, the molecular partition function of the surface molecules,

qbound, are different from the gas phase molecules (fewer rotationaldegrees of freedom, shifted vibrational frequencies). More detaileddiscussion in the end of Ch. 15.

I The free energy, F , of the adsorbed gas becomes,

FAkBT

=U − TSAkBT

= θ ln θ + (1− θ) ln (1− θ) +

(w

kBT

)θ − θ ln qbound

and the chemical potential for the surface molecules becomes,

µbound

kBT=

(∂ F

kBT

∂N

)A,T

=

(∂ F

AkBT

∂θ

)A,T

= lnθ

1− θ+

wkBT

− ln qbound




I For the gas phase (vapor pressure for an ideal gas),

µgas = kBT lnp

point

where point is the standard state pressure (material property).

I Recall that (see Eqs. 11.48 and 11.49)

point = q0kBT and qgas = q0V

I Using the condition for equilibrium (µbound = µgas),

p =q0kBTqbound

θ

1− θe

wkB T

I (Note that q′gas has been used in the book to denote q0 as introduced inch. 11. q′ was used for a different purpose in ch. 13)




I The binding constant, K , is defined as (see Eq. 13.10)

K =qbound

q0kBTe

−wkB T

to get the Langmuir adsorption equation,

θ =Kp

1 + Kp

I For small p, analogy to Henry’s law for the vapor pressure, θ ≈ Kp.I For experimental observations, rearrange to a linear relationship

1θ

=1 + Kp

Kp⇒ p

θ=

1K

+ p

and find 1/K as the intercept when plotting p/θ against p.



Langmuir model for binding in solution

I Regard the binding of a ligand to a particle (e.g. protein, polymer, micelle,etc.) in solution,

X + PK−−−− PX

I The binding or equilibrium constant is given as,

K =[PX ]

[P][X ]

I The fraction of filled binding sites, θ,

θ =[PX ]

[P] + [PX ]=

K [P][X ]

[P] + K [P][X ]=

K [X ]

1 + K [X ]



The Michaelis-Menten modelI Regard the reaction,

E + SK−−−− ES k2−→ E + P

where E is an enzyme, S is a substrate, and P is the product.I The equilibrium constant, K ,

K =[ES]

[E ]x

where x = [S].I The velocity, v , of the reaction (where k2 is a rate constant),

v =d [P]

dt= k2[ES] = k2K [E ]x

I The total enzyme concentration, ET , is

ET = [E ] + [ES] = [E ] (1 + Kx)

I which gives the velocity per enzyme molecule as,

vET

=k2K [E ]x

[E ] (1 + Kx)=

k2Kx1 + Kx



The Michaelis-Menten modelPart 2

I The maximum velocity, vmax = k2ET , appears when all enzyme sites arefilled with a substrate. So the Michaelis-Menten equation becomes,

vvmax

=Kx

1 + Kx

I which is often written in terms of a dissociation constant or Michaelisconstant, Km = 1

K , asv

vmax=

xKm + x

I To measure Km and vmax experimentally, it is rephrased as,

1v

=Km

vmax

1x

+1

vmax

where 1v is plotted against 1

x .



Sabatier’s principleCatalysts should bind neither too tightly nor too weakly

I For heterogeneous catalysis (reaction catalyzedby a surface), we use the Michaelis-Mentenmodel.

A + SK−−−− AS k2−→ B + S

where S indicates a surface.I Regard the reaction rate, r , (see Eq. 19.2)

r =d [B]

dt= k2[AS] = k2θAs

where As is the number of surface sites.I Use the Langmuir model for the binding to the

surface,

θ =Kp

1 + Kp

where p is the pressure of A.



Sabatier’s principlePart 2

I This leads tor

As=

k2Kp1 + Kp

I The rate constant, k2, is given as (see Eq. 19.11),

k2 = c1e−EakB T

and according to the Brønsted relationship (Eq. 19.40),

Ea = a∆G2 + b

I From the figure,

∆G2 = ∆G −∆G1 ⇒ Ea = −a∆G1 + a∆G + b︸︷︷︸constant



Sabatier’s principlePart 3

I We get for the rate constant

k2 = c2ea∆G1kB T

I Using the relationship between equilibriumconstant and free energies, K = e

−∆GkB T ,

k2 = c2K−a

which gives the final relation (see three topfigures)

rAs

=c2K 1−ap1 + Kp

I Termed volcano curves (see bottom figure).



Summary

I Discussed two models with a common feature,I Langmuir model for surface adsorptionI Michaelis-Menten model for enzyme kinetics

I The common feature is saturation, i.e. a model of the form

f (x) =Kx

1 + Kx

i.e. where we havef (x) ≈ Kx ; Kx 1

andf (x) ≈ 1 ; Kx 1

I Sabatier’s principle (in heterogeneous catalysis)



Exam Aug. 2011: Exercise 33a and 3ba) The Langmuir adsorption equation may for a gas be written as

θ =Kp

1 + Kp

and for a solution asθ =

Kx1 + Kx

What does the different variables denote? What is the range of values for θ?What happens at high and low K , respectively? In heterogeneous catalysis, itis desired to have a high or a low θ?

b) For the Langmuir adsorption equation for a gas, K is given by

K =qbound

q′gaskBTe

−wkB T

Also here explain what the different variables and constants denote. Whathappens if w has a positive or negative number, respectively?



Exam Aug. 2011: Exercise 33c

c) In the derivation of the Langmuir adsorption equation, we have included theintramolecular terms in the free energy expression. How does that appear inthe equations above? In most chapters in the book, the intramolecular termsto the free energy expression has, however, been ignored. What is themotivation for that? Why is it important for interactions with surfaces?



32. Polymer solutions

Per-Olof Astrand


[email protected]

April 16, 2015


Polymer solutions

This chapter:I A brief introduction to polymer solutionsI The Flory-Huggins modelI Polymers may be linear or branchedI Polymer solutions are described by distributions of

chain lengths and conformations



ConformationsI Rotations around the C − C single bonds for ethane and butane are

shown in the figureI For ethane (left), we have three identical minimaI For butane (right), we have one global minimum (trans conformation) at

180 , and two local minima (gauche+ and gauche− conformations) at60 and 300.

I N rotatable C − C bonds give 3N conformations. Will entropy beimportant?



Polymer solutionsI Polymer molecules are typically much larger than solvent molecules

(perhaps 10000 monomers).I For a polymer, the mole fraction is very different from volume fraction.I Figure to the left: very different from an ideal solution when mole fraction

is consideredI Figure to the right: if we consider volume fraction, the polymer behaves

more like an ideal solutionI We cannot treat a polymer on an equal footing as a solvent molecule.

Instead, each solvent-sized segment is treated similarly as a solventmolecule



The Flory-Huggins model

I Lattice of M sites. Each site has z neighbours.I ns solvent moleculesI np polymer molecules, each with N chain

segments, termed beads.I The volume fractions become

φs =ns

M; φp =

Nnp

M



The Flory-Huggins modelThe entropy of mixing

I Counting the number of arrangements of a polymer molecule by thegrowth method.

I The second bead can be placed in z positions relative to the first bead.I The third bead can be placed in z − 1 positions relative to the second

bead (since one position is taken by the first bead), etc.I The total number of conformations for polymer 1 becomes,

v1 = Mz(z − 1)N−2 ≈ M(z − 1)N−1

I Incomplete, since it does not include excluded volume effects



The Flory-Huggins modelThe entropy of mixing, part 2

I To account for the excluded volume effect, Flory suggested anapproximation: the excluded volume for each bead is proportional to theamount of space already filled, as if the space had been filled randomly.The total number of conformations for polymer 1 becomes in the Florymodel

v1 = M(

zM − 1

M

)((z − 1)

M − 2M

). . .

((z − 1)

M − N + 1M

)

≈(

z − 1M

)N−1 M!

(M − N)!




I Now we will put all the np polymer chains on the lattice.I We start by putting the first bead for all np chains on the lattice,

vfirst = M (M − 1) (M − 2) . . . (M − np + 1) =M!

(M − np)!

I Secondly, we count the number of arrangements of the np (N − 1)remaining beads of the np polymers.

I M − np sites are available for the first subsequent bead; M − Nnp + 1sites are available for the last bead.

I Following the results for v1 on the previous slide,

vsubsequent =

(z − 1

M

)np(N−1)(M − np)!

(M − Nnp)!




I The total number of arrangements (multiplicity), W (np,ns), is

W (np,ns) =vfirstvsubsequent

np!=

(z − 1

M

)np(N−1) M!

(M − Nnp)!np!

=

(z − 1

M

)np(N−1) M!

ns!np!

I The extra factor np! in the denominator accounts for the indistinguishibilityof the polymers.

I Likewise, (M − Nnp)! (or ns!) accounts for the indistinguishibility of thesolvent molecules.




I The mixing entropy, ∆Smix, is the difference of theentropy for the mixture and for the puresubstances,

∆Smix = kB lnW (np,ns)

W (0,ns) W (np,0)

I W (0,ns) = 1 for the pure solventI W (np,0) is obtained from W (np,ns) by setting

M = Nnp,

W (np,0) =

(z − 1Nnp

)np(N−1)(Nnp)!

np!




I Calculating,

W (np,ns)

W (0,ns) W (np,0)=

(M!

np!ns!

)(z − 1

M

)np(N−1) ( np!

(Nnp)!

)(Nnp

z − 1

)np(N−1)

=M!

(Nnp)!ns!

(Nnp

M

)np(N−1)

Stirling’s approx.≈ MM

(Nnp)Nnp nnss

(Nnp

M

)npN (M

Nnp

)np

=

(Mns

)ns ( MNnp

)np

I The mixing entropy becomes,

∆Smix

kB= −ns ln

ns

M− np ln

Nnp

M⇒ ∆Smix

MkB= −φs lnφs −

φp

Nlnφp

I If N = 1, the Flory-Huggins model is reduced to the mixing entropy of alattice model for a solution.



The Flory-Huggins modelThe energy of mixing

I Follow the approach for a solutions: number of contacts (mss, msp, andmpp) and pair interaction energies (wss, wsp and wpp). The total internalenergy, U, becomes

U = msswss + mppwpp + mspwsp

I Conservation of the number of particles gives,

zns = 2mss + msp

andzNnp = 2mpp + msp

where (z − 2) Nnp would have been a better approximation for zNnp. Thereason is that the polymer has internal bonds between the beads.



The Flory-Huggins modelThe energy of mixing, part 2

I Use the Bragg-Williams approximation,

msp ≈znsNnp

M

so thatU

kBT=

zwss

2kBTns +

zwpp

2kBTNnp + χsp

nsnpNM

where as usual

χsp =z

kBT

(wsp −

wss + wpp

2

)I Note that the book is not consistent here: we have obtained U for the

solution, not ∆Umix.



The Flory-Huggins modelThe free energy

I The free energy, F = U − TS, becomes,

Fmix

kBT= ns lnφs + np lnφp +

zwss

2kBTns +

zwpp

2kBTNnp + χsp

nsnpNM

I Generalize the Flory-Huggins theory to the mixture of two polymers Aand B,

Fmix

kBT= nA lnφA + nB lnφB +

zwAA

2kBTNAnA +

zwBB

2kBTNBnB + χAB

nAnBNANB

M

I Not consistent here: we actually look at U − T ∆Smix, i.e. a mixture of twoterms.



The Flory-Huggins modelThe chemical potential

I The chemical potential, µB, (either polymer or small molecule) becomes

µB

kBT=

(∂

∂nB

Fmix

kBT

)nA,T

= lnφB + 1− nBNB

M− nANB

M+ χABNB (1− φB)2 +

zwBB

2kBTNB

[Note in the differentiation that M = nANA + nBNB and φB = nBNB/M]I Simplify to

µB

kBT= lnφB + (1− φB)

(1− NB

NA

)+ χABNB (1− φB)2 +

zwBB

2kBTNB

where we have used

nANB

M=

nANA

MNB

NA= φA

NB

NA= (1− φB)

NB

NA



The Flory-Huggins modelThe vapour pressure

I If A is the polymer and B is the small molecule, the vapour pressure of Bis determined from setting the chemical potential of B equal in the vapourphase and in the polymer solution

pB

poB,int

= φB exp[

(1− φB)

(1− NB

NA

)+ NBχAB (1− φB)2 +

zwBB

2kBTNB

]where the first term in the exponential accounts for nonideality arisingfrom the difference in size between the particles



Summary

I Brief introduction to statistical mechanics of polymer solutions.I Introduction to the elements of Flory-Huggins theory, another example of

the “standard model” in our course.I The success of the model for polymers (in contrast to for example

proteins) depends on that we in many cases do not have strongintermolecular and inter-bead interactions (e.g. hydrogen bonding)



Exam June 2013: Exercise 4

What is the most important difference between a lattice model for a regularsolution of molecules (with about equal size) and a lattice model for a polymersolution? Why do we use volume fractions instead of mole fractions forpolymer solutions? Let us next regard a solution between a polymer A and amolecule B, where the volume fraction of the molecule B is 99%. How will thevapour pressure of the molecules B be affected by i) the length of the polymerA, and ii) the interaction energies wAA, wBB and wAB, respectively? Explainwhy and discuss the relative importance of the contributions at the givenvolume fraction. Finally, what would you regard as an ideal polymer solution incomparison to a regular polymer solution?


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Statistical Thermodynamics in Chemistry and Biology - …Statistical Thermodynamics in Chemistry and Biology Introduction Per-Olof Astrand˚ D3-119 Realfagsbygget, Department of Chemistry,