chemical energetics_physical chemistry

Chemistry 180-213B

Introductory PhysicalChemistry

David Ronis

© 2002

McGill University

All truths are easy to understand once they are discovered; the point is to discoverthem.

Philosophy is written in this grand book-I mean the universe-which stands contin-ually open to our gaze, but it cannot be understood unless one first learns to com-prehend the language and interpret the characters in which it is written.It is writ-ten in the language of mathematics, and its characters are triangles, circles, andother geometrical figures, without which it is humanly impossible to understand asingle word of it.

Opere Il SaggiatoreGalileo Galilei (1564 - 1642)

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The Founders of Thermodynamics

Joule

Clausius

Maxwell

Carnot

Kelvin

Gibbs

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Table of Contents

1. General Information. . . . . . . . . . . . . . . . . . . . . . 52. Kubo’sDivertissements. . . . . . . . . . . . . . . . . . . . . 8

2.1.Divertissement1: Founders of the first law of thermodynamics . . . . . . . . 82.2.Divertissement2: Why do we hav ewinter heating? . . . . . . . . . . . 82.3.Divertissement3: NicolasLeonard Sadi Carnot. . . . . . . . . . . . . 102.4.Divertissement4: AbsoluteTemperature . . . . . . . . . . . . . . . 102.5.Divertissement8: Onthe names of thermodynamic functions. . . . . . . . 113. Probability and Statistics . . . . . . . . . . . . . . . . . . . . 134. Maxwell-Boltzmann Distribution . . . . . . . . . . . . . . . . . . 165. Collision Rates, Mean Free Path, and Diffusion . . . . . . . . . . . . . 226. Collision Theory of the Rate Constant. . . . . . . . . . . . . . . . 277. Notes on the First Law of Thermodynamics . . . . . . . . . . . . . . . 297.1. ZerothLaw of Thermodynamics . . . . . . . . . . . . . . . . . 297.2. SomeDefinitions . . . . . . . . . . . . . . . . . . . . . . 297.3. Euler’s Theorem, Partial Molar Quantities,. . . . . . . . . . . . . . 317.4. Work and Heat in Thermodynamics. . . . . . . . . . . . . . . . 327.5. Energy in Classical Mechanics. . . . . . . . . . . . . . . . . . 357.6. TheFirst Law of Thermodynamics: . . . . . . . . . . . . . . . . 368. Thermochemistry. . . . . . . . . . . . . . . . . . . . . . . 388.1. Enthalpy Calculations: Chemical Reactions and Hess’ Law . . . . . . . . . 388.2. Measuring∆H0

f . . . . . . . . . . . . . . . . . . . . . . . 398.3. Reactions at Different Temperatures: Kirchoff’s Law . . . . . . . . . . . 408.4. Bond Energies . . . . . . . . . . . . . . . . . . . . . . . 418.5. Some Manipulations Involving Thermodynamic Functions. . . . . . . . . 438.5.1. The relationship betweenCP andCV . . . . . . . . . . . . . . . 438.5.2. The Joule-Thompson Experiment. . . . . . . . . . . . . . . . . 449. The Second Law of Thermodynamics . . . . . . . . . . . . . . . . 4610. Carnot Engines, Efficiency, and The Second Law . . . . . . . . . . . . 4911. Efficiency Of A Real Carnot Engine. . . . . . . . . . . . . . . . . 5412. The Clausius Inequality and the Mathematical Statement of the Second Law . . . . 5713. Systems With Variable Composition: The Chemical Potential. . . . . . . . 5914. Exact Differentials and Maxwell Relations. . . . . . . . . . . . . . 6115. Thermodynamic Stability: Free Energy and Chemical Equilibrium. . . . . . . 6715.4.1. Free Energy and Entropy of Mixing . . . . . . . . . . . . . . . 7515.4.2. Determination of Free Energies of Formation . . . . . . . . . . . . 7615.4.3. Determination of the Extent of a Reaction. . . . . . . . . . . . . 7615.5. Coupled Reactions . . . . . . . . . . . . . . . . . . . . . 7715.6. Temperature Dependence ofK p . . . . . . . . . . . . . . . . . 7716. Eigenvalues and Thermodynamic Stability . . . . . . . . . . . . . . 7917. Entropy of Mixing: A Statistical View . . . . . . . . . . . . . . . . 8118. Phase Equilibrium . . . . . . . . . . . . . . . . . . . . . . 8419. Minimum and Maximum Boiling Azeotropes: The Gibbs-Konovalov Theorem . . . 8820. Ideal Eutectic Phase Diagrams. . . . . . . . . . . . . . . . . . 91

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21. Capillary Rise and Depression. . . . . . . . . . . . . . . . . . 9322. Problem Sets. . . . . . . . . . . . . . . . . . . . . . . . 9522.1. Problem Set 1. . . . . . . . . . . . . . . . . . . . . . . 9522.2. Problem Set 2. . . . . . . . . . . . . . . . . . . . . . . 9722.3. Problem Set 3. . . . . . . . . . . . . . . . . . . . . . . 9922.4. Problem Set 4. . . . . . . . . . . . . . . . . . . . . . . 10022.5. Problem Set 5. . . . . . . . . . . . . . . . . . . . . . . 10122.6. Problem Set 6. . . . . . . . . . . . . . . . . . . . . . . 10222.7. Problem Set 7. . . . . . . . . . . . . . . . . . . . . . . 10322.8. Problem Set 8. . . . . . . . . . . . . . . . . . . . . . . 10422.9. Problem Set 9. . . . . . . . . . . . . . . . . . . . . . . 10522.10. Problem Set 10. . . . . . . . . . . . . . . . . . . . . . 10623. Past Exams . . . . . . . . . . . . . . . . . . . . . . . . 10723.1. Quiz I: 1999 . . . . . . . . . . . . . . . . . . . . . . . 10723.2. Quiz I: 2000 . . . . . . . . . . . . . . . . . . . . . . . 10923.3. Quiz I: 2001 . . . . . . . . . . . . . . . . . . . . . . . 11223.4. Quiz II: 1999 . . . . . . . . . . . . . . . . . . . . . . . 11423.5. Quiz II: 2000 . . . . . . . . . . . . . . . . . . . . . . . 11723.6. Quiz II: 2001 . . . . . . . . . . . . . . . . . . . . . . . 12023.7. Final Exam:1999 . . . . . . . . . . . . . . . . . . . . . . 12223.8. Final Exam:2000 . . . . . . . . . . . . . . . . . . . . . . 12523.9. Final Exam:2001 . . . . . . . . . . . . . . . . . . . . . . 128

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CHEMISTR Y 213B: Intr oductory Physical Chemistry I.

1. GeneralInformationLectures: MWF 9:30 - 10:30

Otto Maass 10Course Web Site: http://ronispc.chem.mcgill.ca/ronis/chem213/Notes.html

(Username: chem213; Password: Gibbs)

Professor: David RonisOffice: Otto Maass Room 426E-mail: r [email protected]

Text

G. W. Castellan,Physical Chemistry.

Supplementary Texts

1. P. A. Rock,Chemical Thermodynamics.2. GordonM. Barrow, Physical Chemistry.3. R.Kubo,Thermodynamics(Physics orientation, advanced)

Grades

There will be approximately one problem set per week, two quizes and a final exam.The quizes will be given at 18:00 on:

Wednesday, February 6, 2002andWednesday, March 13, 2002(in Otto Maass 112)

Completion of the homework is mandatory. Most of the problems will not be graded, althoughone or two problems may be chosen at random and graded.Solutions to the problem sets will beposted on the course web page.

You are strongly encouraged to do the homework. Theproblems will cover manydetails not done in class and will prepare you for the exams. The exams will involve extensiveproblem-solving and, in part, will contain problems from the homework! The typical studentwill not do well in (or even pass) this course unless the problems are done. The course gradingscheme is:

Winter Term 2001-2002

General Information -6- ChemistryCHEM 213W

Grade Distribution

Problems 10%

1st Quiz 25%

2nd Quiz 25%

Final 40%



CHEMISTR Y 213B: TENTATIVE OUTLINE

Date Topic Text Chapter

Lecture 1. Introduction: Thermodynamics,an overview 2Lecture 2. Empirical properties of gases 2Lecture 3. Empirical properties of liquids and solids 5

Lecture 4. Molecular basis: Kinetic theory of gases 4Lecture 5. Collision rates, effusion, and diffusion 30Lecture 6. Viscosity 30

Lecture 7. Temperature: thezeroth law of thermodynamics 6Lecture 8. Mechanics, Work, and Heat 7Lecture 9. Reversible and irreversible changes 7

Lecture 10. The First Law of Thermodynamics: Energy 7Lecture 11. Enthalpy, Hess’s Law 7Lecture 12. Heat Capacities, Kirchoff’s Law 7

Lecture 13. Estimating Enthalpy Changes: BondEnthalpies 7

Wednesday, February 6, 2002: First Quiz

Lecture 14. The Carnot Engine/Refrigerator 8Lecture 15. The Second Law of Thermodynamics: Entropy 8

Lecture 16. Entropy Calculations 8Lecture 17. The Third Law of Thermodynamics: AbsoluteEntropies 9Lecture 18. Conditions for Stable Equilibrium: Free Energies 10

Lecture 19. Equilibrium Conditions (continued) 10Lecture 20. Maxwell Relations and applications 9.4Lecture 21. Chemical equilibrium 11

February 25 - March 1: STUDY BREAK

Lecture 22. Chemical equilibrium calculations 11Lecture 23. Heterogeneous Equilibrium: The phase rule 12Lecture 24. 12Phase diagrams of Simple systems: Clapeyron and Clau-

sius-Clapeyron equations

Lecture 25. Ideal Solutions 13Lecture 26. Colligative properties 13

Wednesday, March 13, 2002: Second Quiz

Lecture 27. Colligative properties (continued) 13

Lecture 28. Solutions of volatile components 14Lecture 29. Lever-principle and fractional distillation 14Lecture 30. Henry’s Law and solubility. 14

Lecture 31. Solubility calculations 14Lecture 32. Electrolyte solutions 17Lecture 33. Electrochemistry: Electrochemicalcells 17

Lecture 34. Redox reactions and half-cell potentials 17Lecture 35. Nernst equation and applications1 17Lecture 36. Summary


2. Kubo’s Divertissements

From Thermodynamics(North Holland, 1976)

Ryogo Kubo

University of Tokyo

2.1. Divertissement1: Founders of the first law of thermodynamics

If a tomb of the Unknown Scientists had been built in the 1850’s, the most appropri-ate inscription would have been "In memory of the grief and sacrifice of those who fought torealize a perpetuum mobile". But the law of conservation of energy, or the first law of thermody-namics, is associated primarily with three great names, Mayer, Helmholtz and Joule.

Julius Robert Mayer (1814-1878) was really a genius who was born in this worldonly with the errand to make this great declaration. Hermann Ludwig Ferdinand von Helmholtz(1821-1894) gav ethis law the name "Erhaltung der Kraft" or "the conservation of energy". LikeMayer, he started his career as a medical doctor but lived a glorious life as the greatest physiolo-gist and physicist of the day. James Prescott Joule (1818-1889) worked over forty years to estab-lish the experimental verification of the equivalence of work and heat.

Among the three, Mayer was the first who arrived at this law and the last whosework was recognized. His life was most dramatic. A lightening stroke of genius overtook him, aGerman doctor of the age of twenty six, one day on the sea near Java when he noticed thatvenous blood of a patient under surgical operation appeared an unusually fresh red. He consid-ered that this might be connected with Lavoisier’s theory of oxidation in animals, which processbecomes slower in tropical zones because the rate of heat loss by animals will be slower there. Agreat generalization of this observation lead him to the idea of the equivalence of heat andmechanical work. For three years after his voyage, while he was working as a medical doctor athome, he devoted himself to complete the first work on the conservation of energy "Bemerkun-gen uber die Krafte der unbelebten Natur" which was sent to the Poggendorf Annalen and wasnever published by it. In 1842 Liebig published this paper in his journal (Annalen der Chemieund Pharmacie) but it was ignored for many years.

Mayer wrote four papers before 1851. During these years of unusual activity hecared for nothing other than his theory. In 1852 he became mentally deranged and was hospital-ized. He recovered after two years but never returned to science.

2.2. Divertissement2: Why do we hav ewinter heating?

The layman will answer: "To make the room warmer." The student of thermody-namics will perhaps so express it: "To import the lacking (inner, thermal) energy." If so, then thelayman’s answer is right, the scientist’s is wrong.

We suppose, to correspond to the actual state of affairs, that the pressure of the air inthe room always equals that of the external air. In the usual notation, the (inner, thermal) energyis, per unit mass,*

u = cvT.

Divertissement2 -9- ChemistryCHEM 213W

(An additive constant may be neglected.) Thenthe energy content is, per unit of volume,

u = cvρT,

or, taking into account the equation of state, we have

P

ρ= RT,

we have

u = cvP/R.

For air at atmospheric pressure,

u = 0. 0604cal/cm3.

The energy content of the room is thus independent of the temperature, solely determined by thestate of the barometer. The whole of the energy imported by the heating escapes through thepores of the walls of the room to the outside air.

I fetch a bottle of claret from the cold cellar and put it to be tempered in the warmroom. It becomes warmer, but the increased energy content is not borrowed from the air of theroom but is brought in from outside. Then why do we hav eheating? For the same reason thatlife on the earth needs the radiation of the sun. But this does not exist on the incident energy, forthe latter apart from a negligible amount is re-radiated, just as a man, in spite of continualabsorption of nourishment, maintains a constant body-weight. Our conditions of existencerequire a determinate degree of temperature, and for the maintenance of this there is needed notaddition of energy but addition of entropy.

As a student, I read with advantage a small book by F. Wald entitled "The Mistressof the World and her Shadow". These meant energy and entropy. In the course of advancingknowledge the two seem to me to have exchanged places. In the huge manufactory of naturalprocesses, the principle of entropy occupies the position of manager, for it dictates the mannerand method of the whole business, whilst the principle of energy merely does the bookkeeping,balancing credits and debits.

R. EMDEN

Kempterstrasse 5,Zurich.

The above is a note published in Nature141 (l938) 908. A. Sommerfeld found it sointeresting that he cited it in his bookThermodynamic und Statistik(Vorlesungen uber theoretis-che Physik, Bd. 5, Dietrich’sche Verlag, Wiesbaden; English translation by F. Kestin, Academic

* The author has assumed that the specific heat of the gas is independent of temperature;a reasonable approximation for the oxygen and nitrogen around room temperature.



Press Tic., New York, 1956). R. Emden is known by his work in astrophysics and meteorology asrepresented by an article in der Enzyklopadie der mathematischen WissenschafteThermody-namik der Himmelskorper(Teubuer, Leipzig-Berlin, 1926).

2.3. Divertissement3: NicolasLeonard Sadi Carnot

In the first half of the last century, the steam engine, completed by introduction ofthe condenser (the low-temperature heat reservoir), due to James Watt (1765) had come to pro-duce more and more revolutionary effects on developments in industry and transportation. Manyeminent physicists like Laplace and Poisson set about to study the Motive Power of Fire. SadiCarnot (1796-1832) was a son of Lazare Carnot, Organizer of Victory in the French Revolution,and was born and died in Paris. He probably learned the caloric theory of heat, in which heat wasassumed to be a substance capable either of flowing from body to body (heat conduction) or ofmaking chemical compound with atoms (latent heat). He wrote a short but very important book,Reflexions sur la puissance motrice du feu et sur les machines propres a developper cette puis-sance(Paris, 1824), which was reprinted by his brother (1878) together with some of Carnot’sposthumous manuscripts.

Carnot directed his attention to the point that, in the heat engine, work was done notat the expense of heat but in connection with the transfer of heat from a hot body to a cold body,and thus heat could not be used without a cold body, in analogy of water falling from a highreservoir to a low reservoir. In his book he assumed the law of conversation of heat, namely thatthe quantity of heat was a state function, although he later abandoned this law and arrived at thelaw of equivalence of heat and work: he actually proposed many methods to estimate themechanical equivalent of heat. He introduced what came to be known as Carnot’s cycle, andestablished Carnot’s principle.

Carnot’s book had been overlooked until B. P. E. Clapeyron (1834) gav eCarnot’stheory an analytical and graphical expression by making use of the indicator diagram devised byWatt. The law of conservation of heat assumed by Carnot was corrected by R. Clausius (1850),based on the work of J. R. von Mayer (1841) and J. P. Joule (1843-49), into the form that notonly a change in the distribution of heat but also a consumption of heat proportional to the workdone is necessary to do work, and vice versa. Clausius named this modification the First Law ofThermodynamics. H. L. F. van Helmholtz (1847) and Clausius generalized this law to the princi-ple of the conservation of energy. W. Thomson (Lord Kelvin), who introduced Kelvin’s scale oftemperature (1848) based on Carnot’s work, also recognized the law of equivalence of heat andwork. The Second Law of Thermodynamics was formulated by Thomson (1851) and Clausius(1867).

A sketch of the history of early thermodynamics is given by E. Mendoza, PhysicsToday 14 (1961) No. 2, p. 32. See also E. Mach:Principien der Warmelehre(vierte Aufl. 1923,Verlag von Johann Ambrosius Barth, Leipzig).

2.4. Divertissement4: AbsoluteTemperature

The absolute temperature scale means that temperature scale which is determined bya thermodynamic method so that it does not depend on the choice of thermometric substance, the



zero of the scale being defined as the lowest temperature which is possible thermodynamically.Absolute temperature, which is now used in thermal physics, was introduced by Lord Kelvin(William Thomson) in 1848 and is also called the Kelvin temperature.

For the complete definition of the scale, we have two choices; one is to use two fixedpoints above zero and assign their temperature difference and the other is to use one fixed pointand assign its numerical value. Until recently the calibration of the Kelvin temperature scale wasperformed using two fixed points: the ice pointT0

oK and the boiling pointT0 + 100oK of purewater under 1 standard atm (= 101325 Pa). We can measureT0 by a gas thermometric method.At low pressures, the equation of state of a real gas can be written in the form

pV = α +κ p.

We measure the values of pV, α andκ at the above two fixed points.Considering thatα is equalto nRT, we hav e

T0 =100α 0

α 100 − α 0

If we put T0 = 0, we get the thermodynamic Celsius temperature scale.Hence,−T0oC means

absolute zero as measured by this scale.

The precise gas thermometric investigations of the Frenchman P. Chap- puis from1887 to 1917 gav ethe value ofT0 between 273.048 and 273.123. Inspired by this work, morethan one hundred determinations ofT0 were performed until 1942. Among them, the results ofW. Heuse and J. Otto of Germany, W. H. Keesom et al. of the Netherlands, J. A. Beattie et al.ofthe U.S.A. and M. Kinoshita and J. Oishi of Japan are noted for their high precision. Their valuesare found to lie between 273.149 and 273.174.

Considering these results and the fact that the triple point of pure water is very nearto 0.0100oC, the 10th General Conference on Weights and Measures in 1954 decided to use thetriple point of the water as the fixed point and to assign the value 273.16 as its temperature. Italso redefined the thermodynamic Celsius temperature toC as t = T-273.15, where T is the valueof the absolute temperature determined from the above decision. The zero of the new thermody-namic Celsius temperature differs by about 0.0001o from the ice point.

For ordinary purposes, the differences in these new and old scales are quite negligi-ble. However, for cases where a precision of 1O−4 degree in the absolute value is required, wemust take the differences into consideration.

2.5. Divertissement8: On the names of thermodynamic functions

The word "energy εν ε ργ εια " can be seen in the works of Aristotle but "internalenergy" is due to W. Thomson (1852) and R. J. E. Clausius (1876). The portion "en" meansinhalt=capacity and "orgy", like the unit "erg", derives from ε ργ oν =work. "Entropy" is also at-tributed to Clausius (1865) who took it from {εν τ ρε πειν =verwandeln and meansverwand-lungsinhalt=change quantity. "Enthalpy" was introduced by H. Kamerlingh Onnes (1909) from{ εν θα λ π ειν =sich erwarmen which meanswarmeinhalt. J.W. Gibbs called it the heat function



(for constant pressure). "Free energy" is due to H. van Helmholtz (1882), and means that part ofthe internal energy that can be converted into work, as seen in the equation dF=d’A for anisothermal quasi-static process. It was customary to call the remaining part, TS, of the internalenergy, U = F + TS, thegebundene energie (bound energy), but this is not so common now. TheGibbs free energy (for constant pressure) was introduced by Gibbs, but German scientists used tocall it die freie enthalpie. Thus the thermodynamic functions often have different names in Ger-man and in English.

Further, on the equation of state: Kamerlingh Onnes gav ethe names,thermische zus-tandsgleichungto p = p(T,V) and the namekalorische zustandsgleichung to E = E(S,V). M.Planck (1908) called the latterkanonische zustandsgleichung.


Probability -13- Chemistry CHEM 213W

3. Probability and Statistics

As we have stressed in class, when dealing with a macroscopic sample of a materialit is practically impossible to measure or calculate all the microscopic properties of the∼ 1023

atoms or molecules in the system.Fortunately, many kinds of phenomena do not depend on theprecise behavior of any single particle, and only average properties of the constituent particlesare important. In order to quantify these averages, we must consider some simple ideas in proba-bility and statistics.

We all encounter probabilistic concepts in daily life. Results of opinion polls, lifeexpectancy tables, and grade distributions are but a few examples. Shown below, are two gradedistributions for two classes taking an hourly exam in some course.

How would you use this information?

Perhaps the simplest quantity to compute is the average grade; i.e.,

AVERAGE GRADE=100

i=0Σ N(i)

Ntotali ,

whereN(i) is the number of students with grade i andNtotal is the total number of students ineach class. Notice that even though the two distributions are very different, they hav ethe sameroughly the same average grade.

How much information is contained in the average grade and how relevant is theav erage to any individual in the class? In the hypothetical distribution the average tells the wholestory. The distribution is extremely narrow, and thus practically everyone got the average grade.The same is not true in the real distribution; there is a significant spread in the scores.Thisspread can be quantified by defining the probability, P(i), of finding any student with grade i.Forthis example,

P(i) ≡N(i)

Ntotal



which is simply the faction of students with grade i. Notice that probabilities are "normalized" inthe sense that

iΣ P(i) = 1.

This is just a complicated way of stating the fact that every student taking the exam receivessome grade.

A measure of the width of the distribution can be obtained by computing the stan-dard deviation. If we denote the average grade by <i>, then the standard deviation,σ , is definedas:

σ 2 ≡iΣ P(i) [ i − < i > ]2

(Note,σ is the Greek letter sigma).

When we consider atomic and molecular systems the situation becomes somewhatmore complicated, although the basic ideas are the same as those introduced in the grade exam-ple discussed above. You have already used probability when you learned about atomic andmolecular orbitals. In the kinetic theory of gasses, a different sort of question is being asked;namely, how do the molecules move?

To be specific, suppose we want the distribution of velocities in the x direction for asample of gas containing 1023 molecules. Even if we ignore the experimental impossibility ofmeasuring the velocities of all of the molecules, what would we do with this huge amount ofinformation? Itwould be useful to make the kind of histogram used in discussing grade distribu-tions; however there is an additional complication.For the exam discussed above, no fractionalgrades were assigned, and thus there is a natural bin width of 1 grade point.On the other hand,nature does not assign discrete values to the x components of the molecular velocities. Hence,ifwe were to make our bins too small in constructing our histogram, most of them would containonly 0 or 1 particles, even for a sample containing 1023 particles.

In order to get around this difficulty, we must include some information about thesize of the "bin" in our introduction of probability. This is done by definingprobability density,F(vx):

F(vx) ∆vx ≡ The probability that a molecule hasvelocity betweenvx andvx + ∆vx.

Notice that the bin size is explicitly contained in our definition.Also note that in general proba-bility densities have units (in this example the units are inverse velocity).

Once we know the probability density, averages can be computed just as in the gradeexample considered above. For example,

< vnx > =

vx

Σ vnx F(vx) ∆vx. (1)

Throughout this part of the course, we will denote average quantities by surrounding them with



"< >". What should this average be for n=0 or 1? What is the physical significance of the aver-age for n=2?

Of course, typical samples of the gas contain a large number of particles, and thus,the bins can be taken to be very small.What does the sum in eq. (1) become in this limit?Weknow that this is an integral, and we therefore rewrite eq. (1) as

< vnx > =

∞

−∞∫ vn

x F(vx) dvx.

Finally, one more aspect of probability must be considered. In the molecular veloc-ity example, we discussed only the x component.However, velocity is a vector and there arealso the y and z components.How would we describe the probability that any molecule is travel-ing in some specific direction?Clearly, for molecules in a gas, the probability that the x compo-nent of the velocity lies in some interval should be independent of what the other two compo-nents are doing.For such situations, the probability that a molecule has velocity with compo-nents in the intervalsvx to vx + ∆vx, vy to vy + ∆vy, and vz to vz + ∆vz is

F(vx, vy, vz)∆vx∆vy∆vz = F(vx)F(vy)F(vz)∆vx∆vy∆vz.

If you are having trouble believing this, consider the probability that three coins come up headsor tails in any set of three of tosses. This is a model the velocity probability distribution in a uni-verse were all the magnitudes of the velocity components in the three directions are equal.

Further aspects of probability densities in gas kinetics will be discussed in class, butthis material should get you started.


Boltzmann Distribution -16- Chemistry CHEM 213W

4. TheMaxwell-Boltzmann Distribution

The molecular description of the bulk properties of a gas depends upon our knowingthe mathematical form of the velocity distribution; That is, the probability,F(vx, vy, vz)∆vx∆vy∆vz, of finding a molecule with velocity components in the rangevx tovx + ∆vx , vy to vy + ∆vy, and vz to vz + ∆vz (see the last handout). This can be found by makingtwo very simple assumptions:

1. Thethree components of the velocity are independent of each other. This impliesthat the velocity probability density can be written as:

F(vx, vy, vx) = f (vx) f (vy) f (vz) (1)

2. All directions are equivalent (space is isotropic). This implies that the probability offinding a molecule moving with a certain velocity cannot depend on the direction; itis equally probable to find a molecule with any speed v moving in any direction.Mathematically, this implies that the probability density can only depend on themagnitude of the velocity, or the molecular speed. Hence,

F(vx, vy, vx) = F([v2x + v2

y + v2z]

1/2). (2)

By comparing Eqs. (1) and (2), we have

f (vx) f (vy) f (vz) = F([v2x + v2

y + v2z]

1/2). (3)

Very few functions can satisfy Eq. (3); in fact, the only one is:

f (vi ) = A e−bv2i , i = x, y, z (4)

where A and b are, as yet, arbitrary constants. Verify that f given by Eq. (4) satisfies Eq.(3). Toshow that Eq. (4) is the only possible function, requires some mathematics which might bebeyond the level of Chemistry 213; however, the proof is contained in the appendix for those ofyou who have this extra background.

How do we determine the values of the constants A and b?The function f, is a prob-ability density and every molecule must have some velocity. From the previous section, weknow that this means

1 =∞

vx=−∞Σ f (vx)∆vx →

∞

−∞∫ f (vx)dvx, (5)

where the integral is obtained when we make ∆vx very small. By going to a table of integrals wefind that

∞

−∞∫ A e−bv2

dv = A

πb

1/2

, (6)



which when used in Eq. (5) gives

A =

b

π

1/2

. (7)

The parameter b is found as follows: We will compute the pressure that a dilute gasexerts on the walls of its container and then compare the result with experiment (i.e., the idealgas equation of state).What is the pressure and how do we get it from the velocity distribution?The pressure is the force exerted by the gas per unit area.Moreover, the force is the rate ofchange of momentum per unit time due to collisions with the wall.

Imagine the molecules which comprise our gas as billiard balls and assume that thewalls are perfectly smooth. What happens when a molecule collides with the wall? Actuallyrel-atively little; the normal component of the velocity changes sign (see Fig. 1).

|v dt|x

��

��

Figure 1

x

If the wall is taken to be the y-z plane, the momentum change is

∆p = − 2mvx, (8)

where m is the mass of the molecule.A typical velocity is 105 cm/sec; what is a typical momen-tum change for argon? How many molecules with velocity vx will collide with the wall per unitarea in time∆t? From Fig. 1, it should be clear that any molecule within a distancevx∆t will hitthe wall. Thenumber per unit area is therefore:

n(vx)vx∆t, (9)

where n(v) is the number of molecules per unit volume withx component of velocity in therangevx to vx + ∆vx. This is related to the molecular velocity distribution by

n(vx) = n0 f (vx)∆vx, (10)

wheren0 is the number of molecules per unit volume. Bymultiplying Eqs. (8) and (9), dividingby ∆t and by using Eq. (10), we arrive at the pressure exerted by those molecules with x velocitycomponent in the rangevx to vx + ∆vx:

2 m n0 v2x f (vx)∆vx. (11)



All that remains is to include the contributions from all velocities corresponding to moleculesmoving towards the wall. Thepressure, P, thus equals:

P =∞

vx≥0Σ 2 m n0 vx2 f (vx)dvx =

∞

0∫ 2 m n0 v2

x f (vx)dvx

=∞

0∫ 2 m n0 v2

x

b

π

1/2

e−bv2x dvx, (12)

where the last equality comes from using the explicit form of the probability density, [see Eqs.(4) and (7)]. The value of the integral is:

∞

0∫ v2

xe−bv2x dvx =

1

4b

πb

1/2

.

If we use this result in Eq. (12), we find that

P =m n0

2b. (13)

Next we write the ideal gas equation of state in terms of the number density:

P = n0 kB T,

wherekB is Boltzmann’s constant:

kB ≡R

6. 0225× 1023= 1. 38× 10−16 erg/deg.

By comparing this with Eq. (13) we see that

b =m

2kBT.

The velocity probability density for the x component can now be written as

f (vx) =

m

2πkBT

1/2

e−

mv2x

2kBT (14)

or for the full velocity as

F(vx, vy, vx) =

m

2πkBT

3/2

e−

E

kBT , (15)



whereE = m(v2x + v2

y + v2z)/2 is the energy of the molecule. This is referred to as the Maxwell-

Boltzmann distribution function and is illustrated in the following figure:

Notice that increasing the temperature makes it more probable to find moleculeswith higher velocities and that the probability of observing any giv en velocity will decrease asthe energy associated with that velocity increases. This last comment is not limited to our simplebilliard ball model of the molecules.

For many applications, the Maxwell-Boltzmann velocity distribution gives too muchdetail. Inparticular, remember that it is a probability density for the vector velocity. Suppose weare interested in some property which depends only on the speed of the molecules and not theirdirection. Whatis the probability density which describes the distribution of molecular speeds?

The speed distribution function,F(c) dc, is the probability of observing a moleculewith a speed in the interval c to c + dc irrespective of i ts direction. It can be obtained from Eq.(15) as follows: Eq. (15) describes the distribution of velocities. Ifwe are not interested in thedirection of the velocity, all we need do is to average it out; i.e.,

F(c)dc =c≤|→v|≤c+dc

Σ F(vx, vy, vz)∆vx∆vy∆vz = F(vx, vy, vz) |v|=c c≤|→v|≤c+dcΣ ∆vx∆vy∆vz. (16)

The sums in these last equations are over all velocities such that the speed is betweenc andc + dc. The second equality comes from noting that F only depends on the magnitude of thevelocity (see Eq.(15)). Whatis this last sum?∆vx∆vy∆vz is a volume element in a coordinatesystem whose axes are the components of velocity. The sum represents the volume between twoconcentric spheres of radiusc and c+dc, respectively. Thus

c≤|→v|≤c+dcΣ ∆vx∆vy∆vz =

4π3

(c + dc)3 − c3

≈ 4πc2dc, (17)

where the last expression was obtained by expanding the products and dropping nonlinear termsin dc (remember that we will consider infinitesimally small dc).If we use Eq. (17) in (16), we



find that

F(c) = 4πc2F(vx, vy, vz) |v|=c= 4πc2

m

2πkBT

3/2

e−

mc2

2kBT . (18)

This is the speed distribution and is shown below. Notice that it vanishes when c=0; why?

4.1. Appendix: Proof of Eq. (4)

You will not be responsible for this proof. Differentiate both sides of Eq. (3) withrespect tovx and divide byvx. You get:

df (vx)

dvxf (vy) f (vz)

vx=

dF([v2x + v2

y + v2z]

1/2)

dvx

[v2x + v2

y + v2z]1/2

.

By repeating this forvy, equating the results and carrying out a little algebra, we find that:

df (vx)

dvy

vx f (vx)=

df (vy)

dvy

vy f (vy)

The right hand side of the equation depends only onvy and the left hand side depends only onvx. Since these two components of velocity can be chosen independently, the only way that thislast equation can be satisfied is if both sides are equal to the same constant. Hence,

df (vx)

dvx= −

1

2bvx f (vx).

This is a first order differential equation. It has a general solution which will depend on a single



multiplicative constant. Thissolution is given by Eq. (4).


Collision Rates and Diffusion -22- Chemistry CHEM 213W

5. CollisionRates, Mean Free Path, and Diffusion

In the previous section, we found the parameter b by computing the average forceexerted on the walls of the container. Suppose, instead, that the rate of collisions (i.e., the num-ber of collisions per unit area per unit time) was desired.This is important for a number of prac-tical considerations; e.g., if a chemical reaction takes place every time a molecule hits the sur-face, then the rate of the reaction will just be the collision rate.

We obtain the collision rate by repeating the analysis which determined the force onthe wall in the previous section. Thus the number of molecules with velocity vx which collide intime interval∆t is

n(vx)vx∆t, (1)

where we are using the same notation as in the preceding section. The total number of collisionsthus becomes:

Zwall∆t ≡0

−∞∫ n(vx)|vx|∆tdvx,

=0

−∞∫ dvx n0

m

2πkBT

1/2

e−mv2x/(2kBT)|vx|∆t

=

kBT

2πm

1/2

n0∆t (2)

Hence the wall collision rate,Zwall , is giv en by

Zwall =

kBT

2πm

1/2

n0 (3)

Aside from the previously mentioned example of chemical reaction on a surface, another applica-tion of this expression is ineffusionthough a pinhole. If there is a hole of areaA in the surface,the rate that molecules escape the system is justZwall A. Notice that heavier molecules or iso-topes will escape more slowly (with a 1/√ massdependence); hence, effusion through a pinhole isa simple way in which to separate different mass species.



RA + RB

|v| t∆

Fig. 1

Next consider the number of collisions which a molecule of type A makes withthose of type B in a gas. We will model the two molecules as hard spheres of radiiRA and RB,respectively. Moreover, in order to simplify the calculation, we will assume that the B moleculesare stationary (this is a good approximation ifmB >> mA). An A molecule moving with velocity→v for time ∆t will collide with any B molecule in a cylinder of radiusRA + RB (cf. Fig. 1) andlength |→v|∆t. Hence, volume swept out will just be

|v|∆tπ(RA + RB)2, (4)

and the actual number of collisions willnB × volume swept out The average A-B collision rate, isobtained by multiplying by the probability density for A molecules with velocity →v and averag-ing. Moreover, since only the speed of the molecule is important, we may use the molecularspeed distribution function obtained in the preceding section to average over the speed of A.Thus, we get

ZA∆t ≡∞

0∫ 4πc2

mA

2πkBT

3/2

e−mac2/(2kBT)nBc∆tπ(RA + RB)2dc (5)

which, when the integral is performed, gives:

ZA = π(RA + RB)2cAnB, (6)

wherecA is the average molecular speed of A; i.e.,

cA ≡ 8kBT

πmA

1/2

. (7)



This is the number of collisions one A suffers in time∆t. The number of collisionswith B’s that all the A molecules in a unit volume suffer per unit time,ZA,B, is

ZA,B = π(RA + RB)2cAnAnB (8)

As was mentioned at the outset, our expression is correct if the B’s are not moving.It turns out that the correction for the motion of B is rather simple (and involves going to what iscalled the center-of-mass frame, which is not); specifically, we replace themA in the definition ofthe mean speed byµ A,B, thereduced massfor the A,B pair. The reduced mass is defined as:

µ A,B ≡mAmB

mA + mB. (9)

With this correction, our expression becomes

ZA,B = π(RA + RB)2cA,BnAnB, (10)

where

cA,B ≡

8kBT

π µA,B

1/2

(11)

is the mean speed of A relative to B. A special case of this expression it the rate of collision of Amolecules with themselves. Inthis case, Eq. (10) becomes

ZA,A =1

2π σ2

AcA21/2n2A, (12)

whereσ A is the molecular diameter of A and where we have divided by 2 in order to not counteach A-A collision twice.

Next, we will consider how far a molecule can move before it suffers a collision.For simplicity, we will only consider a one component gas. Accordingto Eq. (6), the mean timebetween collisions is

τ collision ≈1

21/2π σ2AcAnA

.

Hence, the typical distance traveled by a molecule between collisions,λ is approximatelyτ collisioncA or

λ =1

21/2π σ2AnA

.

This distance is called the mean free path. Note that it only depends on the density of the gasand the size of the molecules.



We can use these results to obtain an approximate expression for the way in whichconcentration differences equilibrate in a dilute gas. Consider a gas containing two kinds of mol-ecules in which there is a concentration gradient; i.e., the density of molecules per unit volume,ni (z) depends on position. One way in which the concentration becomes uniform is via diffu-sion.

To quantify diffusion, we want to find the net number of molecules of a given typethat cross a plane in the gas per unit area per unit time; this is known as the diffusion flux.To seehow the diffusion flux depends on molecular parameters, consider the following figure.

z − λ

z + λ

z

Fig. 2

We want the net flux through the plane at z. From our preceding discussion of mean free paths,clearly any molecule that starts roughly from a mean free path above or below z, moving towardsz, will not suffer any collisions and will cross. The number that cross the planes atz ± λ per unitarea, per unit time is the same as the wall collision rates we calculated above, that is

kBT

2πm

1/2

n(z ± λ ) =1

4cn(z ± λ ),

where we have rewritten Eq. (3) in terms of the average molecular speed, Eq. (7).Since all themolecules that wont collide and will thus cross the plane at z, the net flux (upward) is just

J =1

4cn(z − λ ) −

1

4cn(z + λ )

= −1

4c[n(z + λ ) − n(z − λ )].

Since, for most experiments the density barely changes on over a distance comparable to themean free path, we use the Taylor expansion to write



n(z ± λ ) ≈ n(z) +dn(z)

dzλ +

1

2

d2n(z)

dz2λ 2 + ...,

which gives Fick’s Law of diffusion,

J = −Ddn(z)

dz,

where

D ≡1

2λ c

is known as the diffusion constant and has units oflength2/time. (Actually the factor of 1/2 inour expression forD is not quite right, but the other factors are).

Next consider the total number of molecules inside of some region bounded byplanes atz andz + L. The lengthL is small compared to the scales that characterize the concen-tration nonuniformity, but large compared with the mean free path. Clearly the only way for thetotal number of molecules in our region, n(z, t)L, to change is by fluxes at the surfaces atz andz + L. Thus,

∂n(z, t)L

∂t= −J(z + L, t) + J(z, t) ≈ −L

∂J(z, t)

∂z,

where we have again used a Taylor expansion, now for the flux itself. Finally, by using our resultfor the diffusion flux and canceling the factors of L, we see that

∂n(z, t)

∂t= D

∂2n(z, t)

∂z2,

which is a kinetic equation for the relaxation of the concentration and is known as the diffusionequation. Although value of the diffusion constant is quite different, the diffusion equation actu-ally is more general than our derivation might suggest and holds in liquids and solids as well.Finally, note that we have only considered systems where the concentration is only nonuniformin one spatial direction. Should this not be the case, then some simple modifications of ourexpressions must be introduced, but these will not be considered here.

There are standard ways in which to solve diffusion equations. However, there is onegeneral feature of systems relaxing diffusively. Imagine introducing a small droplet of an impu-rity into a system.Clearly the droplet will spread in time, and you might naively think that theav erage droplet size would grow like ct. Of course, this ignores collisions and is incorrect.From the diffusion equation dimensional analysis correctly suggests that

1

t≈

D

R2

or R ≈ √ Dt, which is much slower than the linear behavior that would arise in the absence of col-lisions. Thisis what is observed.



6. CollisionTheory of the Rate Constant

The simplest collisional model for the rate constants in bimolecular elementary reac-tions, e.g.,

A + B → Products,

is to assume that reaction occurs every time an A and B molecule collide. Thus the ratebecomes:

RATE= ZA,B (1)

whereZA,B is the number of AB collisions occurring per unit time, per unit volume of the sys-tem. Thekinetic theory of gases gives

ZA,B = π(RA + RB)2cABnAnB, (2)

whereRA andRB are the van der Waals radii of A and B, respectively, nA andnB are their num-ber densities, andcAB is the mean relative speed of A with respect to B; i.e.,

cAB ≡ 8kBT

π µAB

1/2

. (3)

In Eq. (3), kB is Boltzmann’s constant andµ AB ≡ mAmB/(mA + mB) is the AB reduced mass(mA,B denotes the mass of A or B).

Equation (2) is obtained by considering how many B molecules will lie in the cylin-der swept out by an A molecule in time dt. The radius of the cylinder is the maximum distanceand A and B can be in order to collide, i.e.,RA + RB. Thus the volume of the cylinder for mole-cules moving with relative speedv is

V(v) ≡ π(RA + RB)2vdt. (4)

The number of B molecules in the cylinder with relative speedv to v + dv is V(v)nB f (v)dv,where

f (v) = 4πv2

µ AB

2πkBT

3/2

e−

µ ABv2

2kBT (5)

is the Maxwell-Boltzmann speed distribution function governing the relative motion of B withrespect to A.Finally, our expression for the rate is obtained by adding (integrating) the contribu-tions for the different relative velocities, multiplying by the number of A molecules per unit vol-ume and dividing out the factor of dt; i.e,

RATE= nAnBπ(RA + RB)2∞

0∫ dv f(v)v, (6)


Collision Rate Constants -28- ChemistryCHEM 213W

which gives the expression presented above.

As was mentioned in class, Eqs. (1) - (3) in general do not correctly describe thetemperature dependence of rate constants since the activation energy does not appear. In order tocorrect for this, we will modify our initial hypothesis; specifically, now assume that only thosemolecules whose relative speed is greater thanvmin can react. In this case, Eq. (6) becomes:

RATE= nAnBπ(RA + RB)2∞

vmin

∫ dv f(v)v, (7)

which upon evaluating the integral gives

RATE= π(RA + RB)2cABe−EA/kBT 1 +

EA

kBTnAnB, (8)

whereEA ≡ µ ABv2min/2 is the activation energy. This expression has the correct Arrhenius form

except for the extra nonexponential factors containing temperature.However, compared with theexponential factor, these usually do not change very rapidly in temperature and can approxi-mately be treated as constant. Thus we have derived an approximate expression for the bimolec-ular rate constant which has at least some of the qualitative features of those found in gas reac-tions.


Notes on the First Law -29- ChemistryCHEM 213W

7. Noteson the First Law of Thermodynamics

7.1. Zeroth Law of Thermodynamics

If two bodies at equilibrium are brought intothermal(i.e., no mechani-cal, electrical, magnetic, gravitational, etc., work is performed) andnothing happens, then they are at the same temperature.

Given this, we can always measure the temperature of any system by bringing it intothermal contact with some standard thermometer. As we shall see, a very convenient choice isthe ideal-gas thermometer. Here a manometer is used to measure the pressure of a fixed amountof gas in a fixed volume and the relation

V =NRT

P

is used to calculate the temperature. Needless to say, other temperature standards can (and are)used.

7.2. SomeDefinitions

Intensive Doesn’t depend on the size of the system; e.g., P, T, partial molar quan-tities.

Extensive The opposite of intensive; e.g., mass, volume, energy (but not energyper unit volume or mass), heat capacities (but not specific heats).

System Thepart of the universe under investigation. Systemscan be:

a) Isolated:no interaction of any kind with surroundings.Note that real systems cannot be truly isolated, but can beapproximately so on the time-scale of relevance.

b) Closed:energy can be exchanged with surroundings, butmatter cannot.

c) Open:matter and energy can be exchanged.

Surroundings Thepart of the universe not under investigation.

Boundary Whatdivides the system from the surroundings (and controls whetherthe system is open, closed, or isolated).

State Asystems state is specified when all measurable properties have defi-nite valuesto the accuracy of the experiment.

State Variables Aset of measurable quantities, which when known, completely specifythe state of the system. In classical or quantum mechanics there are onthe order of 1023 state variables; however, in thermodynamics, experi-ence tells us that themacroscopicstate of the system is specified after asmall set of measurements are made (e.g., T, P, V, x1, ..., xr ).



Process Somethingwhereby the state of a system is changed.A process has twoparts:

a) Initial and final states (i.e., where the system starts andends).

and a

b) Path. Thepath describes how the change was effected. Inorder to specify the path, intermediate changes in the sys-tem, surroundings and boundary must be specified. This isclearly something which we would like to avoid in manycases.

Reversible A process is said to be reversible if it is possible to returnboth the sys-tem and the surroundings to their original state. If not, it is irreversible(even if the system can be returned to the original state).

State Function A property of the system which only depends on the current state of thesystem. Hence,changes in state functions do not depend on the pathtaken. Statefunctions play a key role in thermodynamics and allowmacroscopic properties of matter to be studied in a rigorous, systematicmanner. Examples of state functions are: energy, entropy (to be intro-duced later), P, V, T, etc. A one-component ideal gas has a pressure,P(T,N,V), given by PV=NRT no matter what--how the T, V, or Nattained their current values is irrelevant.



7.3. Euler’s Theorem, Partial Molar Quantities, and the Gibbs-Duhem Relations

Next consider any extensive quantity in a mixture containingr components; i.e.,A(T, P, N1, ..., Nr ). Realexamples could be the energy, volume, mass, heat capacity, etc.. Con-sider the small change inA associated with changes in its arguments, namely

dA = ∂A

∂T P,N1,...,Nr

dT + ∂A

∂P T,N1,...,Nr

dP

+

∂A

∂N1

T,P,N2,...,Nr

dN1 + ...+

∂A

∂Nr

T,P,N1,...,Nr−1

dNr , (1)

Now, by assumption,A is extensive; hence,

A(T, P, λ N1, ..., λ Nr ) = λ A(T, P, N1, ..., Nr ). (2)

If we differentiate both sides of this equation with respect toλ and evaluate the answer atλ = 1 itfollows that

A(T, P, N1, ..., Nr ) =

∂A

∂N1

T,P,N2,...,Nr

N1 + ...+

∂A

∂Nr

T,P,N1,...,Nr−1

Nr (3)

≡r

i=1Σ Ai Ni , (4)

where

Ai ≡

∂A

∂Ni

T,P,N j≠i

(5)

is called apartial molar quantity. Note that the partial molar quantities are intensive. In obtain-ing Eq. (3) you may use Eq.(1) for dT = dP = 0 and d(λ Ni ) = Ni dλ for i = 1, ..., r ). Also notethat Eq. (3) is a special case of Euler’s theorem for homogeneous functions in calculus.

Equations (3) or (4) allow us to explicitly express the nontrivial features of an exten-sive quantity in terms of intensive ones, thereby reducing the number of dependencies we mustworry about. It also turns out that the partial molar quantities (or more specifically changes inthem) are not all independent.To see this, we calculatedA from Eq. (4):

dA =r

i=1Σ Ai dNi + Ni dAi , (6)

where we have used the calculus resultd(xy) = xdy+ ydx. Of course,dA could have been com-puted from Eq. (1); i.e.,



dA = ∂A

∂T P,N1,...,Nr

dT + ∂A

∂P T,N1,...,Nr

dP +r

i=1Σ Ai dNi , (7)

where we have rewritten the derivatives with respect to the numbers of moles in terms of the par-tial molar quantities, cf. Eq. (5). By equating the right hand sides of Eqs. (6) and (7) it followsthat

∂A

∂T P,N1,...,Nr

dT + ∂A

∂P T,N1,...,Nr

dP −r

i=1Σ Ni dAi = 0 (8)

and hence, the changes in the partial molar quantities and other derivatives are not all indepen-dent. Equation(8) is known as a Gibbs-Duhem relation and can be used to relate seemingly dis-parate thermodynamic derivatives.

As an exercise, what are the partial molar volumes for an ideal gas mixture obeyingDalton’s law of partial pressures? Do they obey the Gibbs-Duhem relation?

7.4. Work and Heat in Thermodynamics

Tw o central concepts in thermodynamics are work and heat.You probably haveseen examples of the former in your freshman physics class, and the latter is something youexperience in daily life. In thermodynamics, both have very precise definitions.

Work: anything which crosses the boundary of the system and is completely convertibleinto the lifting of a weight in the surroundings.

Note that work only appears at the boundary of a system, during a change of state, isextensive, and is manifested by an effect in the surroundings. From mechanics, we know that

dW = Fdx = mgdx,

where dW is the incremental work done by the system, F is the force, and dx is the distance tra-versed. Thesecond equality is for moving a mass m a distance dx in a gravitational field (g is thegravitational acceleration constant). Consider the following apparatus:



dx

M (mass)

A (area)

The inside of the piston is filled with some gas at pressure Pand is maintained at constant temperature T. Instead ofcharacterizing the work done in terms of the mass, it ismore convenient to introduce the pressure exerted on thetop of the piston (i.e., force per unit area, A)

Pop ≡F

A=

mg

A.

Note thatPop need not equal P.

Thus

d− W = PopAdx = PopdV

where dV is the incremental change in the volume of the system. The work involved in thesesorts of processes is known as pressure-volume work.

Note that the work done isnot a state function--it depends on the pressure exertedon the piston (the path) and is not simply a function of the state of the gas in the piston.To stressthis fact, the notationd− will be used for infinitesimal changes in quantities which depend on thepath.

For an process whereby the gas is expanded against some pressure,dV > 0, andhence,d− W > 0. Conversely, in a compression process,d− W < 0, i.e., negative work is done bythe system. The surroundings do positive work on the system.

This diagram shows a process for the isothermalexpansion of a gas (in this case an ideal gas). Thesolidcurve giv es the pressure of the gas (i.e., its equation ofstate). Thedotted curve giv es the opposing pressure actu-ally used in the expansion. Notethat the latter lies com-pletely below the former. If at any point this were not thecase, then the expansion would not proceed spontaneously(i.e., the opposing pressure would be too large and the gaswould contract).

The total work done by the system is just the areaunder the dotted curve:

W =V final

Vinitial

∫ Pop(V)dV ≤V final

Vinitial

∫ P(V)dV ≡ Wrev.

If the gas inside the piston is ideal, then the ideal-gas equation of state is valid and



Wrev =V final

Vinitial

∫NRT

VdV = NRT ln(V final /Vinitial ).

Note that the maximum work you can get out of a spontaneous expansion isobtained when the opposing pressure is infinitesimally less than the pressure being exerted by thegas in the piston. In this case,W = WR. Unfortunately, the rate of such an expansion would bezero, as would be the power delivered by the system. On the other hand, it is easy to show thatthe path given by Pop = P is the only reversible one for the isothermal expansion of an ideal gas.

Consider the following apparatus (from L. K. Nash,Elements of Chemical Thermo-dynamics, Addison-Wesley, 1970):

The spring is assumed to obey Hooke’s law (i.e, the force is proportional to the elongation). Aseries of experiments are performed whereby weights are moved to pan from platforms at variousheights in the surroundings.In doing so, the system (the spring and pan) move from state I to II.How much work is performed in each of the cases (a)−(c) (assume that there is a total 1cm elon-gation of the spring and ignore the mass of the pan and spring)?

In order to reverse the process (i.e., the expansion of the spring) the weights aremoved back to the adjacent platforms.However, it is easy to see that while the spring will befully compressed at the end of the experiment, the surroundings will not be restored to their ini-tial state; specifically, in the best case, the topmost weight will be transferred to the lowest plat-form, no matter how many platforms are used. Clearly the biggest change in the surroundingswill happen in case (a) and the smallest in (c).Moreover, the smaller the individual weights weuse, the more reversible the process becomes (i.e., the change to the surroundings decreases).The process is reversible in the limit where the applied force (the weight) is only infinitesimallymore than the force exerted by the spring.

This argument can easily be extended to our discussion of pressure-volume work orto other kinds of work. Hence,the maximum work in a P-V expansion is obtained for areversible path.



Another key quantity in thermodynamics is heat.

Heat: Anything which flows across the boundary of a system by virtue of a temperaturedifference between the system and the surroundings.

Heat is quantified by measuring the temperature rise (or fall) in a standard material(e.g., a calorie corresponds to the amount of heat required to raise the temperature of 1 g of water1 oC). Like work, heat appears only at the boundary of a system, during a change of state, isextensive, and is manifested by an effect in the surroundings. It is also not a state function sinceit depends on the nature of the thermal contact allowed at the boundary of the system.

In a classic set of experiments in the 19’th century, J. P. Joule showed that the samechanges in the state of a system could be achieved by either adding work or heat to the system.One of his devices is shown below.

As the weight falls, the paddles turn andheat up the liquid in the container by friction(viscous heating). The same temperaturerise can be achieved by directly heating thecontainer using a known amount of heat.

The amounts of heat and work weredefinite and Joule concluded that work andheat were simply two different ways inwhich energy could be added to a system.Specifically, Joule showed that

1 calorie = 4. 184kg m2/sec2.

We are now ready to state the first law of thermodynamics. Beforedoing so, it isillustrative to consider energy in classical mechanics.

7.5. Energy in Classical Mechanics

You probably have heard the statement that "energy is conserved." Whatdoes thismean exactly? Considera system comprised of N point particles of massm, at positions→r 1, ..., →r N , and moving with velocities →v1, ...→vN . The system is not subjected to any externalforces. Inaddition, assume that Newton’s laws of motion are valid and that the particles interactvia pairwise additive forces which are derivable from a potential; i.e., the force particle i exertson j,

→Fi , j is given by

→Fi , j ≡ −

∂ui , j

∂→r i.

Consider the energy function, E, defined as



E ≡N

i=1Σ m

2→v2

i +1

2i≠ j

N

i=1Σ

N

j=1Σui , j .

How does E change as the particles move around under the action of Newton’s Laws?

dE

dt=

N

i=1Σ m→vi ⋅

→Fi −

1

2i≠ j

N

i=1Σ

N

j=1Σ

→Fi , j ⋅ (→vi − →v j ),

where→Fi ≡

j≠iΣ

→Fi , j is the total force acting on the i’th particle.Using this definition of

→Fi in our

expression for the rate of change of E gives:

dE

dt=

1

2i≠ j

N

i=1Σ

N

j=1Σ

→Fi , j ⋅ (→vi + →v j )

=1

2i≠ j

N

i=1Σ

N

j=1Σ(

→Fi , j +

→F j ,i ) ⋅ →vi

where the dummy summation indices, i and j, were exchanged for the terms in→v j in obtainingthe last equality. Newton’s third law states that

→Fi , j = −

→F j ,i ; i.e., the force i exerts on j is equal

and opposite to that j exerts on i. Using this in our last expression immediately shows that

dE

dt= 0.

In other words, the energy of our classical system of particles doesn’t change in time or is con-served. Thesame is true under the laws of quantum mechanics. What happens if, in addition tothe forces acting between the particles, the particles are subjected to an external forces?

The first law of thermodynamics also has something to say about changes in energy,although not in precisely the same way as in classical or quantum mechanics

7.6. TheFirst Law of Thermodynamics:

In any cyclic process (i.e., one where the system returns to its initialstate) the net heat absorbed by the system is equal to the work pr o-duced by the system.

Suppose this were not the case. Then you could presumably find a process whichproduced more work than it absorbed heat. This extra work could be used to run a generator,which in turn could be used to produce more heat, which could run more of process, producingev en more excess work, and so on.The energy crisis, electric bills, etc. would be things of thepast. Unfortunately, no such device has ever been built and the first law still stands.



In mathematical terms, the first law implies that there is a state function, called theinternal energy of the system, defined up to an arbitrary additive constant through its differential

dE ≡ d− Q − d− W,

whered− W is the work done by the system (the negative of the work done on the system).For afinite change of state, the change in the internal energy,∆E, is giv en by

∆E =final state

initial state∫ d− Q − d− W.

The first law states that

∫odE = ∫od− Q − ∫od− W = 0.

The first law implies that the energy change computed along different paths mustgive the same answer. If not, two such paths could be used to to build the energy-creating devicediscussed above (i.e., by reversing one of the paths).


Thermochemistry -38- Chemistry CHEM 213W

8. THERMOCHEMISTR Y

8.1. EnthalpyCalculations: Chemical Reactions and Hess’ Law

The enthalpy change for a process,∆H , is equal to the heat absorbed by the systemif that process is done under constant pressure conditions (and assuming that only P-V work ispossible). Sincethe enthalpy of a system, H = E + PV, is a state function, we can systematizeenthalpy calculations by considering a path whereby the compounds first turn into their con-stituent elements in their standard states (by convention at 25oC and 1 atm pressure) and thenrecombine to form the products. The enthalpy change in the latter step is just the enthalpy of for-mation of the products and the former is the enthalpy of destruction (i.e., the negative of theenthalpy of formation) of the reactants. Hence,

∆H = Σ ∆H0

f (products) − ∆H0f (reactants)

. (1)

Since we are interested in calculating a difference, the absolute enthalpy of the ele-ments in their standard states is unimportant [it cancels out of Eq. (1)], and we adopt the conven-tion that the enthalpy of formation of an element in its standard state is zero.

Consider the following example (reduction of iron oxide):

Fe2O3(s) + 3H2(g)25 oC, 1 atm

→ 2Fe(s) + 3H2O(l ).

A table of thermochemical data gives:

Enthalpies of Formation at 1 atm and 25 C

Compound ∆H0f (kJ/mole)

Fe2O3(s) -824.2H2(g) 0.0Fe(s) 0.0H2O(l) -285.830

By using these in Eq. (1), we find that

∆H = [3(−285. 830)− (−824. 2)]kJ/mole

= −33. 29kJ/mole.

Note that the calculated enthalpy change depends on how the reaction was written.For example,if we wrote



1

2Fe2O3(s) +

3

2H2(g)

25 oC, 1 atm→ Fe(s) +

3

2H2O(l ),

then∆H = −16. 65kJ/mole.

8.2. Measuring∆H0f

There are a number of ways in which to measure the enthalpy of formation of acompound; here are two. Themost obvious is to simply carry out the formation reaction fromthe constituent elements in their standard states in a constant pressure calorimeter (recall that∆H = Qp).

For example, consider the combustion of graphite to form carbon dioxide

C(graphite) + O2(g)25 oC, 1 atm

→ CO2(g).

The heat released in this reaction is−∆H0f (CO2), since the standard enthalpy of formation of the

reactants is zero.

For this method to work, two conditions must be met:1) thereaction goes to com-pletion and 2) only one product is formed. Thus, the reaction

C(graphite) + 2H2(g)25 oC, 1 atm

→ CH4(g)

is not suitable for this method since it doesn’t readily go to completion and we get a complicatedmixture of hydrocarbons.

In order to get around this, note that it is often possible to burn something to com-pletion (and measure∆Hcombustion, the heat released). Thus consider

CH4(g) + 2O225 oC, 1 atm

→ CO2(g) + 2H2O(l ).

Equation (1) gives

∆Hcombustion= ∆H0f (CO2(g)) + 2∆H0

f (H2O(l )) − ∆H0f (CH4(g)).

The standard enthalpies of formation of carbon dioxide and water can be measured using the firstmethod; hence, once we measure the heat of combustion, the only unknown is the standardenthalpy of formation of methane (CH4) and a little algebra gives:

∆H0f (CH4(g)) = ∆H0

f (CO2(g)) + 2∆H0f (H2O(l )) − ∆Hcombustion.

= [−398. 51+ 2(−285. 83)− (−890. 36)]kJ/mole

= −74. 81kJ/mole.



In general, in order to measure the enthalpy of formation, all you need to to is findany reaction where all but one of the standard enthalpies of formation are known and where thereaction goes to completion. These sorts of manipulations are valid because the enthalpy is astate function, and are referred to as Hess’s law. Also note that the same arguments could bemade for the energy changes (under constant volume conditions).

8.3. Reactionsat Different Temperatures: Kir choff ’s Law

What happens if the temperature at which you perform the reaction (either at con-stant P or V) is different than that of your table of enthalpies of formation. Since the enthalpy isa state function, an alternate path can be found whereby the enthalpy change, calculated usingthe temperature of your table, can be used. Consider the constant pressure case depicted below

Reactants Products

Reactants Products

I II

(T1

(T )0

)

The enthalpy change for the reaction atT1 is equal to the enthalpy change atT0 plus the enthalpychange for paths 1 and 2.However, on 1 or 2, only the constant pressure heating or cooling ofthe reactants or products is performed (i.e., no chemical reaction takes place). Since the constantpressure heat capacity,CP was defined as

CP ≡ ∂H

∂T P,N

,

the incremental heat absorbed by the system on 1 or 2 isCPdT. Integrating gives:

∆H1 =T0

T1

∫ CP(reactants)dT = −T1

T0

∫ CP(reactants)dT

and

∆H2 =T1

T0

∫ CP(products)dT.

Adding the contributions together gives



∆H(T1) = ∆H(T0) +T1

T0

∫ CP(products) − CP(reactants)

.

This is known as Kirchoff’s law. What changes must be made for the energy?

Consider our example of the reduction ofFe2O3. What is the enthalpy change at358K? We will assume that the heat capacities are constant over the temperature range 298 - 358K.

Constant Pressure Heat Capacities at 1 atm and 25 C

Compound CP (J/mole/K)

Fe2O3(s) 103.8H2(g) 28.8Fe(s) 25.1H2O(l) 75.3

Note that elements in their standard states donot have zero heat capacities. Using the data in thetable, and the result of our earlier calculation gives

∆Hrxn(358K ) = −33. 29+[2(25. 1)+ 3(75. 3)− 103. 8− 3(28. 8)][358− 298]

1000

= −28. 1kJ/mole.

Note that no change in phase occurred when we cooled (heated) the reactants (prod-ucts). Whatchanges would have to be made if the reaction was carried out at 400 K?

8.4. BondEnergies

A useful, albeit very approximate, way to calculate enthalpy changes in chemicalreactions is through the concept of bond energies. Considerthe following reaction

CH4 → CH3 + H ;

i.e., one C-H bond is broken. Experimentally, ∆H for this reaction is 102 kcal/mole.Similarly,∆H = 96 kcal/mole for

C2H6 → C2H5 + H .

A survey of such reactions will show that the heat required to break a single C-H bond is in therange 96-102 kcal/mole.We can thus assign 98 kcal/mole as an average bond energy for the C-Hbond. Similartends are observed in the bond strength of other types of bonds, and the results aresummarized in the following table



Av erage Bond Energies

Energy Energy(kcal/mole) (kcal/mole)

Bond Bond

H-H 103 C-H 98C-C 80 N-H 92C=C 145 O-H 109

C ≡ C 198 Cl-H 102N-N 37 Br-H 87

N ≡ N 225 I-H 71O-O 34 C-Cl 78O=O 117 C-N 66Cl-Cl 57 C ≡ N 210Br-Br 45 C-O 79

I-I 35 C=O 173

How can this be used? Consider the hydrogenation of ethlyene:

H2C = CH2 + H − H → H3C − CH3.

At the molecular level, we break one H-H and one C=C bond, and form one C-C and two C-Hbonds. Theenergy change is just the net energy left in the molecule in such a process.From thetable, the bond breaking steps take 145+103=248 kcal/mole. The bond formation will give off80+2(98)=276 kcal/mole.Hence the net energy change in the system is 248-276 = -28kcal/mole. To get the enthalpy change, note that at constant pressure,

∆H = ∆E + P∆V.

For this reaction, all reactants and products are gases. If we assume that the gases are ideal, wecan compute∆V; i.e.,

∆V =RT

P∆N.

Here∆N = −1 and thus

∆H = −28− 1. 9872× 10−3kcal/mole /K × 298K

= −28. 6kcal/mole.

(Note that here the difference between∆H and∆E is relatively small). The correct answer is-32.7 kcal/mole. Thus while, the bond energy method is not exact it gives a reasonable estimate.The reason for the discrepancy is the assumption that the bond energy doesn’t depend on whatother bonds are present in the molecule--in general this is not true.



8.5. SomeManipulations Involving Thermodynamic Functions

8.5.1. Therelationship betweenCP and CV

We know that

CV = ∂E

∂T N,V

and that

CP = ∂H

∂T N,P

.

How these two quantities are related is a good exercise in manipulating thermodynamic func-tions. SinceH=E+PV,

CP = ∂E

∂T N,P

+ P∂V

∂T N,P

,

where the last derivative should be recognized asVα , whereα is the thermal expansion coeffi-cient,

α ≡1

V∂V

∂T N,P

.

If we view the energy as a function of N,V,T,

∂E

∂T N,P

= ∂E

∂T N,V

+ ∂E

∂V N,T

∂V

∂T N,P

= CV + α V∂E

∂V N,T

.

Hence,

CP − CV =P +

∂E

∂V N,T

α V

=VTα 2

β,

where



β ≡ −1

V∂V

∂P T,N

is the isothermal compressibility, and where the last equality will be proven later. The compres-sitility must be positive (i.e., things get smaller when squeezed) and this implies thatCP ≥ CV .For an ideal gas, Joule showed that the internal energy per mole did not depend on the volume.In this case,

CP − CV = α PV = R.

8.5.2. TheJoule-Thompson Experiment

Consider the following adiabatic (i.e, Q=0) process, whereby a gas is squeezedthrough a porous, rigid plug

P1 P2

T1 T 2 211 2 P P V V

Porous Plug

Fig. 1. The Joule-Thompson Experiment

Since, by assumption, Q=0,

∆E = E(P2,V2) − E(P1,V1) = −W = P1V1 − P2V2.

By rearranging this expression we show that

E(P1,V1) + P1V1 = E(P2,V2) + P2V2;

i.e., the enthalpy, H is constant in the Joule-Thompson expansion.

In practice, large temperature changes can be obtained in this type of expansion(which can be used in designing a refrigerator).The key parameter is the so-called Joule-Thompson coefficient

µ JT ≡ ∂T

∂P H ,N

. (2)

In order to expressµ JT in terms of more readily measurable quantities, note that



∂y

∂x f

= −

∂ f

∂x y

∂ f

∂y x

,

which is sometimes known as the "cyclic rule" or "implicit function differentiation." Itis provedby noting that

df = ∂ f

∂x y

dx + ∂ f

∂y x

dy,

setting df=0, and by solving for the ratio dy/dx.

By using the cyclic rule on Eq. (2), we find that

µ JT = −

∂H

∂P T,N

∂H

∂T P,N

= −1

CP

∂H

∂P N,T

= −V

CP

1 − β

P +

∂E

∂V N,T

= −V

CP

1 −

βα V

(CP − CV)

= −V

Cp[1 − α T],

where the second to last equality follows when the definition of the enthalpy in terms of theenergy is used and the manipulations used in calculatingCP − CV are repeated.β is the isother-mal compressibility. Note that the Joule-Thompson coefficient vanishes for an ideal-gas.



9. TheSecond Law of Thermodynamics

The second law of thermodynamics gives information concerning the direction ofspontaneous change. If the second law says that a certain process is impossible, you will not beable to get the process to go.On the other hand, note that if the second law says that a process ispossible, you still have to worry about kinetics--you have to find a way in which to carry out theprocess in a reasonable amount of time.

There are a number of equivalent physical statements of the second law of thermo-dynamics. Accordingto Kubo (Thermodynamics, North Holland Publishing Co., 1976) they are:

1. Clausius principle: A process which involves no change other than the transfer ofheat from a hotter body to a cooler body is irreversible; or, it is impossible for heatto transfer spontaneously from a colder to a hotter body without causing otherchanges.

2. Thompson’s (or Kelvin’s) principle: A process in which work is transformed intoheat without any other changes is irreversible; or, it is impossible to convert all theheat taken from a body of uniform temperature into work without causing otherchanges.

3. Impossibility of perpetual motion of the second kind: (due to Max Planck) It isimpossible to devise an engine operating in acyclewhich does work by taking heatfrom a single heat reservoir without producing any other change.

4. Caratheodory’s principle : For a given thermal equilibrium state of a thermallyuniform system, there exists another state which is arbitrarily close to it, but whichcan never be reached from it by an adiabatic change.

Any of these physical statements can be used to prove the others, and to finallyprove the mathematical statement of the second law of thermodynamics:

5. Thereexists a state function, the entropy (denoted by the letter S) which for anyspontaneous process satisfies the Clausius inequality:

∆S ≥ ∫d/Q

T

where the equality holds when the process is reversible.

Before showing how 1.-4. imply 5., let’s first consider how the different physicalstatements imply one another. For example, how does one show that Clausius’ principle impliesThompson’s? Supposeit didn’t; i.e., Clausius’s principle is correct, but Thompson’s is not. Thismeans that you can build an engine which produces work, and which is connected to a singleheat reservoir. If so, consider the following device:



Q

Q’L

TH

TL

W

=QH L

QE=W

+W

RE

where, E is the Thompson violator, and R is a Carnot refrigerator.

If we adjust the sizes of E and R such that all the work is used to run the Carnotrefrigerator, and view the combined E-R apparatus as the system, we have succeeded in creatinga device, which spontaneously pumps heat from cold to hot without any work input from the sur-roundings. Thisviolates Clausius’ principle and thus we have proved Thompson’s principle bycontradiction.

Similarly, we can use Thompson’s principle to prove Clausius’. Again, the proof isby contradiction. If Clausius’s principle is untrue, then you can find a device which sponta-neously (i.e., without any work input) transfers heat from a colder body to a hotter one.Considerthe following apparatus:

T

T

ClausiusViolator

+ QQ

Q1

Q1

1 2

W= Q2

C

H

C

where C is a Carnot engine and where the sizes of the Carnot engine and our Clausius violatorare adjusted such that the heats transferred are as indicated.



What is the net result after one cycle? Work has been produced in the surroundings,but there is no net change in the heat content in the cooler reservoir. Hence, it is as if the systemwere operating in contact with a single reservoir and producing work in the surroundings, in con-tradiction to Thompson’s principle.

These kind of arguments can be used to prove the equivalence of the other physicalformulations of the second law.


Second Law of Thermodynamics -49- Chemistry CHEM 213W

10. Carnot Engines, Efficiency, and The Second Law

Fig. 1. The Carnot Cycle

The Carnot engine is a useful construction for relating the mathematical (∆S ≥ 0 fora spontaneous process in an isolated system) and the physical statements of the Second Law ofThermodynamics (heat spontaneously flows from hot to cold). This handout goes through theanalysis of the amounts of work and heat produced in the isothermal and adiabatic parts of theCarnot cycle for an ideal gas.

10.1. Energy in an Ideal Gas: Joule’s Experiment

In his study of the thermal properties of gases, Joule considerer the isothermalexpansion of dilute gases using the apparatus depicted below:

��

��

P=0

Thermometer

P=0/ ��

��

Fig. 2. Joule’s Experiment

The bulb on the right was evacuated initially and heat was allowed to exchange with the heatbath, whose temperature was measured.For sufficiently dilute gases, Joule found that the tem-perature of the bath didn’t change. Thishas some important consequences.



For this process,Pop = 0, and hence,W = 0. Moreover, since the temperature of thebath remained constant, no heat was absorbed by the system, and thus∆E = 0. Whatdoes thissay about the functional form of the energy of an ideal gas?

We know that we can write the energy of a one-component system as

E(N,V,T) = NE(T, N/V),

where density was chosen instead of pressure as an independent variable (as can always be doneif the equation of state is known). In Joule’s experiment T and N were held fixed but V, andhence N/V, changed. Nonetheless,E didn’t change. Therefore, for an ideal gas,

E(N,V,T) = NE(T);

i.e., the energy per mole of an ideal gas depends only on the temperature of the gas. Similarly,the heat capacity will only be a function of temperature (as it turns out, the heat capacity of anideal gas is usually only weakly dependent on temperature).

10.2. Reversible, Adiabatic Expansion or Compression of an Ideal Gas

When an ideal gas is reversibly expanded (compressed) adiabatically, its temperaturefalls (rises). In order to relate the temperature and volume changes, we note that the energy of aideal gas depends only on the temperature; hence,

dE = CV dT = d− Q − d− W = d− Q − NRTdV

V, (1)

whereCV is the heat capacity at constant volume and where the last equality is obtained by usingthe ideal gas equation of state.For an adiabatic process,d− Q = 0. After a little algebra, Eq. (1)can be rearranged to give:

CV

NR

dT

T= −

dV

V.

If we assume thatCV is independent of temperature (this a good approximation for gases of sim-ple molecules such as Ar, CO2 etc.), this equality can be summed (or integrated) over the entireadiabatic expansion; that is,

CV

NR

T final

Tinitial

∫dT

T=

CV

NRln

T final

Tinitial

= −V final

Vinitial

∫dV

V= − ln

V final

Vinitial

, (2)

where the integrals have been evaluated and "ln" is the natural logarithm function.Finally, bothsides of Eq. (2) are exponentiated, and we find that

Vinitial

V final

=

T final

Tinitial

CV /NR

. (3)



This shows how volume and temperature changes are related along an adiabatic path. (No viola-tion of Charles’ law is implied; why?) Notice,if Vinitial < V final , then the gas is cooler after theexpansion. Thisis to be expected since the expansion removes energy from the system, energywhich is not replaced by the addition of heat from a heat reservoir.

Finally, for an adiabatic change in an ideal gas,

∆E = −W = +∆Umechanical=T final

Tinitial

∫ CV(T)dT ≈ CV∆T

whether or not the path is reversible!

10.3. Reversible, Isothermal Expansion or Compression of an Ideal Gas

Since the energy of an ideal gas depends only on the temperature T, it remains con-stant during any isothermal process (i.e., dE = 0). From the first law, this implies that

d− Q = dW = P dV = NRTdV

V. (4)

If the dQ’s giv en by Eq. (4) are summed over the entire expansion (i.e., between the initial vol-ume,Vinitial and the final volume,V final) we find that

W = Qisothermal = NRT

V final

Vinitial

∫dV

V= NRT ln

V final

Vinitial

, (5)

10.4. Entropy Changes in the Carnot Cycle

Next we use Eq. (5) for the isothermal portions of the Carnot cycle (see Fig. 1, part 1and 3); it is easy to show that

QH

TH+

QC

TC= NR ln

V ′2 V ′

1

V1V2

. (6)

However, if we use Eq. (3) on part 2 of the cycle, it follows that

V ′2

V2=

TC

TH

CV /NR

.

Similarly, for the adiabatic compression,

V ′1

V1=

TH

TC

CV /NR

.



Thus, if we use these two expressions in Eq. (6) the argument of the logarithm becomes equal to1, and

QH

TH+

QC

TC= 0. (7)

This is an explicit demonstration of the Second Law of Thermodynamics (i.e., that the entropy isa state function).

Note that the efficiency, (denoted by the Greek letter eta,η ), of the Carnot cycle (i.e.,how much work is produced per unit heat absorbed) is easily obtained using Eq. (7):

η ≡W

QH=

QH + QC

QH= 1 −

TC

TH,

where the second equality follows form the first law and where the last equality follows from Eq.(7). As is shown on the following pages of the handout, this efficiency formula must hold nomatter what the working fluid in the Carnot engine, and hence, Eq. (7) must hold for materialsother than ideal gases.

10.5. Entropy Changes in Arbitrary Cycles

What if we have a process which occurs in a cycle other than the Carnot cycle, e.g.,the cycle depicted in Fig. 3.

Figure 2

Fig. 3. An arbitrary cycle.

If entropy is a state function,

cycle∫ dS= 0

no matter what the nature of the cycle. In order to see that this is true, break up the cycle into



sub-cycles, each of which is a Carnot cycle, as shown in Fig. 3. If we apply Eq. (7) to eachpiece, and add the results, we get zero for the sum.

In summary, we hav eshown that the entropy

S ≡ ∫dQreversible

T

is a state function. Moreover, sinced− Q ≤ dQreversible,

dS≥d− Q

T;

the equality holds only for reversible processes.


Efficiency of Carnot Engines -54- ChemistryCHEM 213W

11. Why the Efficiency of a Carnot Engine is Independent of the Kind of Working Material

Here are two proofs that the efficiency of a rev ersible Carnot engine depends only onthe temperatures of the heat baths.

11.1. MethodI using Thompson’s Principle

W

Q

HQ

HQ’

Q’L

L

TH

TL

R EW’

where

W′ = the work produced by the engine,Q′

H = the heat absorbed by the engine from the hot reservoir,Q′

L = the heat given off by the engine to the cooler reservoir,W = the work used to run the Carnot refrigerator,QH = the heat given off by the refrigerator to the hot reservoir,QL = the heat absorbed by the refrigerator from the cooler reservoir.

The sizes of the engine and refrigerator are adjusted such that no net heat is takenfrom the cool reservoir in one cycle (i.e.,QL = QL ′). Accordingto Thompson’s principle, no netpositive work can be realized in the surroundings from any device which takes heat from a singleheat source. Thus,

W′ − W ≤ 0. (1)

However, from the First Law, the net work produced must equal the net heat absorbed by the sys-tem; i.e.,

W′ − W = QH ′ − QH . (2)

If we denote the "engine efficiencies" of the refrigerator and engine asη andη ′, respectively, Eq.(2) can be rewritten as:



1 − η ′

= [1 − η ]

QH

QH ′(3)

From Eqs. (1) and (2) it follows that

QH

QH ′≥ 1;

which when used in Eq. (3) gives:

η ≥ η ′. (4)

Note that this is true even if one or both of the engines is not rev ersible. If both engines arereversible, then the roles of engine and refrigerator can be interchanged and we conclude that

η ′ ≥ η .

In light of Eq. (4) this is possible only ifη = η ′. Note that for this case, the net work is zero!.Finally, if the engine is irreversible, an upper bound to its efficiency can be obtained by using areversible refrigerator. In this case, Eq. (4) implies that

η IR ≤ η REV = 1 −TL

TH.

11.2. MethodII using Clausius’ Principle

Q

HQ

HQ’

Q’L

L

TH

TL

WE R

where

W = the work produced by the engineQH = the heat absorbed by the engine from the hot heat bath



QL = the heat released by the engine to the cool heat bathQ′H = the heat released by the refrigerator to the hot heat bathQ′L = the heat absorbed by the refrigerator from the cool heat bath

We will assume that the Carnot engine is more efficient than the Carnot refrigeratorand that the size of the engine is adjusted so that the work output of the former equals thatneeded to run the latter; i.e.,

η E =W

QH> η R =

W

Q′H. (1)

By assumption therefore,

QH < Q′H . (2)

Since the engine and the refrigerator run in a cycle, the first law tells us that (∆E = 0):

W = QH − QL = Q′H − Q′L , (3)

which when combined with (2) shows that

QL − QL ′ = QH − Q′H < 0. (4)

What does this mean? The quantityQH − Q′H is the net heat takenOUT of the hot-ter heat source.However, cf. Eq. (4), we have just shown that it is negative; i.e., a net amount ofheat (>0) has been transferredINTO the hotter heat source from the cooler one. No net work hasbeen done by the surroundings (i.e., by us), and thus our initial assumption violates Clausius’statement of the second law of thermodynamics. Hence,we must conclude thatη E ≤ η R. Thisreverses the inequality in Eq. (4); i.e., now the net heat takenOUT of the hotter body and goingINTO the colder one is positive, and does not violate Clausius’ statement of the second law.

If both the engine and refrigerator are reversible, then they can both be run inreverse; hence the engine now acts as a refrigerator and the refrigerator as an engine.All thesigns on work and heat flip and we conclude thatη R ≤ η E (where the R and E refer to the origi-nal devices). Theonly way out of this contradiction is for the efficiencies of all reversible Carnotcycles to be equal no matter what the nature of the material in the engine or refrigerator.


The Clausius Inequality -57- ChemistryCHEM 213W

12. TheClausius Inequality and the Mathematical Statement of the Second Law

Here is a summary of the proof of the Clausius inequality given in class.

To

−dQS

S

C

} dW

dW

Total dW

System, T

Carnot Engine

where C is a reversible Carnot engine which takes heat−d− Qs from a part of the system at tem-perature T, produces work, d− Wc, in the surroundings, and gives the remaining heat to a reservoirat temperatureT0. While this happens, the system absorbs heatd− Qs and produces workd− Ws.

From our definition of efficiency and the fact that the efficiencies of all reversibleCarnot engine are the same (cf. previous handout), we have:

d− Wc = −η cd− Qs = To

T− 1

d− Qs. (1)

From Thompson’s principle and Eq. (1), it follows that

WTotal = ∫o d− Ws + d− Wc = ∫o d− Ws + To

T− 1

d− Qs ≤ 0. (2)

This inequality must hold if the process is to proceed as written.Now we integrate Eq. (2)around one cycle of the system (reversible or not) and use the fact that the system’s energy isconserved; i.e.,

∫od− Ws = ∫od− Qs.

After a little algebra, this gives:

∫od− Qs

T≤ 0, (3)

where the constant, positive multiplicative factor, T0, has been dropped. This is the Clausiusinequality.


The Clausius Inequality -58- ChemistryCHEM 213W

Equation (3) holds forany spontaneous process which can occur in the system;although, all irreversible processes will require a net work input in order to run in the configura-tion depicted above. The inequality in Thompson’s principle, cf. the last handout, becomes anequality only for reversible processes (no matter what the path) and thus Eq. (3) becomes:

0 = ∫od− Qrev

T≡ ∫odS (4)

where the entropy is defined along any rev ersible path (Eq. (4) is a proof that it is a state func-tion) through

dS≡d− Qrev

T(5)

Proof that dS≥d− Q

T

Consider some process whereby the system changes from state A to B along path I;it may be irreversible. Afterthe process is finished, the system is restored to its initial state (A)along a reversible path II. If we apply the Clausius inequality to this cycle, we have

B

A,pathI∫

d− QI

T+

A

B,pathII∫

d− Qrev

T≤ 0. (1)

Since path II is reversible, the process can be carried out in reverse (This is not necessarily truefor path I) and thus:

A

B,pathII∫

d− Qrev

T= −

B

A,pathII∫

d− Qrev

T= −∆SA→B (2)

By using Eq. (2) in Eq. (1) we conclude that

∆SA→B ≥B

A∫

d− QI

T,

or for infinitesimal changes

dS≥d− Q

T.

Note that the equality holds only for reversible processes.


Chemical Potential -59- ChemistryCHEM 213W

13. SystemsWith Variable Composition: The Chemical Potential

Up to now, we hav enot seriously considered the consequences of changing the com-position of a thermodynamic system.In practice, this can happen in two ways: 1) by externallyadding or removing material or 2) by changing the composition through chemical reaction.Nonetheless, as far as state functions are concerned, the same results must be considered.

There is an energy change associated with changing the composition of the system.For example, you may add compounds with different types of bonds and this will change theenergy available to carry out other processes (e.g., via combustion).

In order to account for the energy change associated with adding material to the sys-tem we introduce a new kind of work done by the system:

d− Wi ≡ −µ i ,opdNi , (1)

wheredNi is the change in the amount of the i’th component (i.e., mass, number of moles, etc.)and whereµ i ,op is called the opposing chemical potential (and is analogous to the opposing pres-sure). For reversible processes, the opposing chemical potential equals the chemical potential,µ i , of component i in the system.In general, the chemical potential for component i in a systemis intensive and is a function of T,P, composition and phase.Moreover, like the equation of state,it must be measured or calculated from a microscopic theory, and we will consider some specificexamples later.

The energy change of the system can now be written as:

dE = d− Q − PopdV +iΣ µ i ,opdNi (2)

or for reversible paths as

dE = TdS− PdV +iΣ µ i dNi . (3)

Henceforth, we will restrict ourselves to processes where the addition of matter is reversible andthus drop the subscript "op" onµ.

Equation (2) has an interesting consequence. If we view the energy of a system as afunction of S, V, and theNi (all of which are extensive), then Euler’s theorem implies that

E(S,V, Ni ) = ∂E

∂S V,Ni

S+ ∂E

∂V S,Ni

V +iΣ

∂E

∂Ni

S,V,N j≠i

Ni .

The partial derivatives immediately follow from Eq. (4), and we find that

E = TS− PV +iΣ µ i Ni . (4)

We also obtain the Gibbs-Duhem equation


Chemical Potential -60- ChemistryCHEM 213W

0 = SdT− VdP+iΣ Ni dµ i . (5)

By using Eq. (4) andH ≡ E + PV, it follows that

H = TS+iΣ µ i Ni . (6)

As we will now see, the chemical potentials play a key role in any quantitative anal-ysis of chemical equilibria.


Maxwell Relations -61- ChemistryCHEM 213W

14. ExactDifferentials and Maxwell Relations

14.1. ATheorem From Calculus

Consider the differential form:

df ≡ M(x, y)dx + N(x, y)dy. (1)

If can we define asingle-valued, differentiable function f(x,y) which satisfies Eq. (1), thenM(x, y)dx + N(x, y)dy is said to be an exact differential. Ofcourse, we can always define f(x,y)by integrating the right hand side of Eq. (1) along some path; however, we require that the func-tion be single-valued (i.e., that it be a state function), hence different paths must give the sameanswer.

THEOREM:

If M and N have continuous first partial derivatives at all points of some open rectan-gle, the differential form, (1), is exact at each point of the rectangle if and only if the condition

∂M

∂y x

= ∂N

∂x y

(2)

is satisfied throughout the rectangle. When this holds, the function f(x,y) is given by the lineintegral

f (x, y) =C∫ M(s, t)ds+ N(s, t)dt (3)

along the path from (a,b) to (x,y) shown in Fig. 1.

s

t

(x,y)

(a,b)C

C’

Figure 1

Proof:



The necessity of the condition is shown by noting that if f is exact, then

∂ f

∂x y

= M(x, y)

and

∂ f

∂y x

= N(x, y).

However, if the second partial derivatives of a function are continuous then the order of differen-tiation is immaterial and

∂2 f (x, y)

∂x∂y=

∂2 f (x, y)

∂y∂x,

which when expressed in terms of derivatives of M and N gives Eq. (2).

To show that the condition is sufficient is slightly more complicated. First, considera point infinitesimally close to (a,b); i.e., (a+dx,b+dy). In this case, we can make linear approxi-mations for the behaviors of M and N and the function defined on C,fC becomes

fC =dx

0∫ M(a + s, b)ds+

dy

0∫ N(a + dx, b + t)dt

≈dx

0∫

M(a, b) +

∂M

∂x x=a;y=b

sds+

dy

0∫

N(a, b) +

∂N

∂x x=a;y=b

dx + ∂N

∂y x=a;y=b

tdt

= M(a, b)dx + ∂M

∂x x=a;y=b

dx2

2+

N(a, b) +

∂N

∂x x=a;y=b

dxdy +

∂N

∂y x=a;y=b

dy2

2(4)

which is valid up to terms of third order in dx and/or dy. Next we repeat the preceding argumenton the path C’ shown in Fig. 1. This gives

fC′ ≈ N(a, b)dy + ∂N

∂y x=a;y=b

dy2

2+

M(a, b) +

∂M

∂y x=a;y=b

dydx +

∂M

∂x x=a;y=b

dx2

2. (5)

If f is single-valued,fC = fC′ . By equating the right hand sides of Eqs. (4) and (5) we see that

fC − fC′ ≈ dxdy

∂N

∂x−

∂M

∂y

(6)

which vanishes if Eq. (2) holds.Incidentally, note that the left hand side of Eq. (6) is just a line



integral around a closed path and the quantity on the right hand side is an approximate surfaceintegral.

Of course, Eq. (6) is valid only for (x,y) infinitesimally close to (a,b).For arbitrarypaths, and (x,y) we break up the interior of the path into small rectangles as shown in Fig. 2.

Figure 2

If the rectangles are small enough, then Eq. (6) can be used on each one and the results addedtogether. This leads to cancellations of all the line integrals on the edges of adjacent rectanglesinside the path (because of the directions of the integrations) and finally gives

∫o M(x, y)dx + N(x, y)dy = ∫ ∫ dxdy

∂N

∂x−

∂M

∂y

.

In calculus, this result is called Green’s theorem in the plane.Finally, Since we’ve assumed that(2) holds,

∫o M(x, y)dx + N(x, y)dy = 0

which is just what we need to show that f defined by Eq. (3) is single-valued (i.e., it is a statefunction).

14.2. Applications to Thermodynamics: Maxwell Relations

We hav ebeen able to combine the first and second laws of thermodynamics to write

dE = TdS− PdV.

Since E is a state function, Eq. (2) must hold and thus

∂T

∂V S

= −∂P

∂S V

.



This is called a Maxwell relation and is a powerful tool for relating different quantities thermo-dynamics.

Another Maxwell relation can be obtained from the enthalpy for which

dH = TdS+ VdP. (7)

Hence, Eq. (2) gives

∂T

∂P S

= ∂V

∂S P

Clearly every state function will generate one or more Maxwell relations. The trick is to knowwhich ones to use in any giv en application.

14.3. AComplicated Example

Suppose we want to express the change in the entropy as a function of T and P. Tobegin, note that

dS=

∂S

∂T P,N

dT +

∂S

∂P T,N

dP + SdN,

whereS is the partial molar entropy (remember that entropy is extensive and Euler’s theoremmust hold). The term in dT is easy to reexpress. Whenthe pressure is constant, the change inthe entropy is

dS=d− QP

T=

CP

TdT

and thus

dS=CP

TdT +

∂S

∂P T,N

dP + SdN.

A new entropy derivative still remains to be reexpressed in terms of something moreconventional. To do so we will use a Maxwell relation.We know that

∂2H

∂T∂P=

∂2H

∂P∂T; (8)

however, from Eq. (7)

LHS =∂

∂T

T

∂S

∂P T,N

+ V



=

∂S

∂P T,N

+ T∂2S

∂T∂P+ Vα ,

where recall thatα ≡ V−1(∂V/∂T)P,N . Similarly,

RHS=∂

∂P

T

∂S

∂T N,P

= T∂2S

∂P∂T.

Since the entropy is a state function, its mixed second derivatives must be equal; hence, equatingthe LHS and RHS of Eq. (8) and carrying out some algebra gives:

∂S

∂P T,N

= −∂V

∂T P,N

= −Vα . (9)

Thus, we have succeeded in expressing the change in the entropy in terms of readily measurablequantities; namely

dS=CP

TdT − Vα dP + SdN. (10)

(Actually, there is a much simpler route to Eq. (9) using the Maxwell relation for the Gibb’s Freeenergy, see below). For for finite changes in state, Eq. (10) gives

S(T, P) = S(T0, P0) +(T,P)

(T0,P0)∫

CP

TdT − Vα dP,

where the choice of path is unimportant. Note that no change in phase must occur along thepath. If not, corrections for the enthalpy change associated with the transition (i.e., the heat)must be included.

With Eq. (10), we can finish our discussion of the difference between the heat capac-ities. InSec. 5.1 of the Thermochemistry handout we showed that

CP − CV = α VP +

∂E

∂V N,T

. (11)

But, dE = TdS− PdV; hence,

∂E

∂V N,T

= T

∂S

∂V N,T

− P



= T

∂S

∂P N,T

∂P

∂V N,T

− P,

where the second equality follows from the chain rule. Equation (10) gives the entropy deriva-tive, and the pressure derivative is equal to −V/β , where β is the isothermal compressibility.Thus,

∂E

∂V N,T

=TVα

β− P

and

CP − CV =VTα 2

β.

Note thatCP ≥ CV for all materials! Express the Joule-Thompson coefficient in terms ofα , β ,andCP/V.


Free Energy -67- Chemistry CHEM 213W

15. Thermodynamic Stability: Free Energy and Chemical Equilibrium

All the criteria for thermodynamic stability stem from the Clausius inequality; i.e.,for any possible change in nature,

dS≥d− Q

T. (1)

Conversely, if

dS<d− Q

T(2)

for every allowed change in state, then the system cannot spontaneously leave the current stateNO MATTER WHAT ; hence the system is in what is called stable equilibrium.

The stability criterion becomes particularly simple if the system is adiabaticallyinsulated from the surroundings. In this case, if all allowed variations lead to a decrease inentropy, then nothing will happen. The system will remain where it is.Said another way, theentropy of an adiabatically insulated stable equilibrium system is a maximum.Notice that theterm allowedplays an important role.For example, if the system is in a constant volume con-tainer, changes in state or variations which lead to a change in the volume need not be consideredev en if they lead to an increase in the entropy.

What if the system is not adiabatically insulated from the surroundings?Is there amore convenient test than Eq. (2)? The answer is yes.To see how it comes about, note we canrewrite the criterion for stable equilibrium that using the first law as

d− Q = dE + PopdV − µdN > TdS, (3)

which implies that

dE + PopdV − µdN − TdS> 0 (4)

for all allowed variations if the system is in equilibrium. Equation (4) is the key stability result.As discussed above, if E, V, and N are held fixed d− Q = 0 and the stability condition becomesdS< 0 as before.

What if S,V,N is held constant?From Eq. (4), the system will be stable ifdE > 0;i.e., the energy is a minimum.This has a nice mechanical analogy. Consider a ball rolling on africtionless parabolic surface in a gravitational field. Clearly, if we place the ball at rest at thelowest point then it will stay there forever. This is the point which minimizes the energy.

Of course, it is not always easy to see how to hold the entropy constant in real exper-iments. (Whenis the entropy constant?) Amore common situation is when the temperature ofthe system is held fixed. Whatis the stability criterion?The problem and its solution are similarto those which led to the introduction of the enthalpy. If (N,T,V) are held fixed, Eq. (4) becomes

(dE)N,T,V − T(dS)N,T,V > 0, (5a)



or since T is constant,

d(E − TS)N,T,V > 0. (5b)

Thus, we see that a new state function,A ≡ E − TS, is a minimum for a stable equilibrium where(N,T,V) are not allowed to vary. This new state function, is defined via a Legendre transforma-tion on the energy and is called the Helmholtz free energy.

From the definition of A, for a general change in state (i.e., not necessarily withdT = 0, etc.)

dA = dE − SdT− TdS

= (d− Q − TdS) − d− W − SdT+ µdN. (6)

The Clausius inequality implies that the quantity in the parenthesis is negative (or zero for areversible process) for any spontaneous change in the state of the system.Moreover, if we con-sider systems where T and N are held fixed

dA ≤ −d− W

or

W ≤ −∆A. (7)

This means the−∆A is the maximum work you can get out of a process run under constant T andN conditions (hence the name "free energy"). In addition, since A is a state function, you can getthe bound without knowing anything about the path (or device)--just by knowing the initial andfinal states and how to carry out a calculation similar to those we did in thermochemistry.

Since A is a state function, we can always compute changes along reversible paths.In this case,

dA = −SdT− PdV + µdN. (8)

In addition, we pick up some new Maxwell relations, e.g.,

∂S

∂V T,N

= ∂P

∂T V,N

. (9)

Clearly, there are many different choices of which state variables can be held con-stant. We will only consider two more. Firstsuppose (S,P,N) is held fixed. Thisis analogous towhat we encountered with the enthalpy. In this case, Eq. (4) becomes

d(E + PV)S,P,N = (dH)S,P,N > 0 (10)

for stable equilibrium; i.e., the enthalpy is a minimum.



Finally, and perhaps most importantly, suppose T, P, and N are held fixed. Thisisthe most commonly encountered case. Now Eq. (4) becomes

d(E + PV − TS)T,P,N > 0. (11)

Thus a new state function,G ≡ E + PV − TS= H − TS, is a minimum for a stable equilibriumwith fixed temperature, pressure, and mass. This state function is called the Gibbs free energy.

As was the case with the Helmholtz free energy, ∆G has a direct physical interpreta-tion. Fromits definition, for a constant (T,P,N) processes,

dG = dE − TdS+ PdV (12a)

= (d− Q − TdS) − (d− W − PdV) ≤ −(d− W − PdV), (12b)

where the last inequality follows from the Clausius inequality. For finite changes in state, wefind

W − ∫ PdV ≤ −∆G. (13)

What does this mean? Up to now, we hav emainly considered PV work. Of course, there areother kinds (magnetic, electrical but to name two). Hence,−∆G provides an upper bound to thenon-PV work that can be obtained from a constant T,P,N process. If you are manufacturing elec-tric batteries you probably don’t care about the amount of PV work which is wasted if the batteryexpands or contracts--all you want is the electrical work.

As in the case of the Helmholtz free energy, we can consider arbitrary changes in theGibbs free energy along reversible paths. From its definition

dG = −SdT+ VdP+ µdN. (14)

As before, this gives additional Maxwell relations, for example

∂S

∂P T,N

= −∂V

∂T P,N

= −Vα , (15)

which we obtained in a very complicated way in the last handout.

One final point, note that the partial molar Gibbs free energy is

G =

∂G

∂N T,P

= µ, (16)

where Eq. (14) was used.Hence, in a one component system, the Gibbs free energy per mole isjust the chemical potential. More generally, Gi , the partial molar quantity for the i’th componentis µ i , and hence, from Euler’s theorem



G =iΣ Ni µ i . (17)

As we discussed earlier, in order that Eqs. (17) and (14) be consistent, a Gibbs-Duhem relationmust hold; i.e.,

0 = SdT− VdP+iΣ Ni dµ i , (18)

which shows that the changes in temperature, pressure and chemical potentials are not all inde-pendent.

The various stability results are summarized in the following table.

Criteria for Stable Equilibrium

State StableEquilibriumFunction Criterion

Held Fixed Definition Differential

E,V,N S - dS=dQrev

Tmaximum

S,V,N E - dE = TdS− PdV + µdN minimum

S,P,N H H ≡ E + PV dH = TdS+ VdP+ µdN minimum

T,V,N A A ≡ E − TS dA= −SdT− PdV + µdN minimum

T,P,N G G ≡ H − TS dG= −SdT+ VdP+ µdN minimum

15.1. Examplesof free energy calculations

Free energy calculations are carried out in much the same as enthalpy calculations.There are tables of standard free energies of formation of compounds.Elements in their stan-dard states are assigned zero as their Gibbs free energy of formation.

Consider the following chemical reaction:

H2(g) + Cl2(g) → 2HCl(g).

Will the reaction proceed as written under constant T and P conditions?The free energy changeis simply

Σ∆G0f (products) − ∆G0

f (reactants) (19)

which for the case at hand is just 2∆G0f (HCl, g) or -184.62 kJ/mole (from Barrow). Hence,a

mixture of hydrogen and and chlorine can lower its free energy (by a substantial amount) byreacting to form HCl.



This is an interesting example for another reason; if you mix stoichiometric amountsof H2 andCl2, you will not see any perceptible reaction--the rate of reaction (no matter what thethermodynamics says) is in this case extremely slow. On the other hand, a small amount of lightat the right frequency will catalyze the reaction which then proceeds explosively!

Next consider the reaction between graphite and diamond,

C(graphite, s) → C(diamond, s).

Now ∆G = 2.90 kJ/mole. The reaction does not proceed as written (too bad).What is perhapsmore troubling is that the reverse reaction should proceed spontaneously at STP. (So why inv estin diamonds?)

What happens at other temperatures or pressures.To answer this note that from Eq.(14), for any compound,

∆G f (T, P) = ∆G0f +

(T,P)

(298K, 1 atm)∫ −S(T, P)dT + V(T, P)dP,

where any convenient path can be chosen.

Thus if we raise the pressure,

∆Grxn(T, P) = ∆G0rxn +

(298K ,P)

(298K, 1 atm)∫ ∆VrxndP. (20)

At STP, ∆Vrxn = -1.9 cm3/mole. Hence, increasing the pressure decreases the Gibbs free energychange. Ifwe assume that the molar densities of carbon are roughly independent of pressure, wecan calculate the pressure at which the reaction will proceed as written. In this approximation,

∆Grxn(T, P) ≈ 2. 90− 1. 9× 10−9∆P (kJ/mole)

Hence, the reaction begins to be possible whenP ≈ 1. 530× 109Pa or about 15,000 atm.

15.2. ChemicalEquilibrium

A very important example of thermodynamic equilibrium is that of chemical equilib-rium at constant pressure and temperature. Consider the following general chemical reaction:

2A + B ←→ 3C + D.

The chemical equation imposes a strong constraint on the changes in the numbers of moles ofeach component; for the forward reaction, each time a mole of B reacts, 2 of A are used up and 3of C and one of D are produced. Mathematically,

dNA

−2=

dNB

−1=

dNC

3=

dNA

1≡ dξ ,



where the extent of the reaction is characterized by the quantityξ (the Greek letter, pronouncedkse) called the progress variable. For an arbitrary chemical reaction involving r chemical com-ponents, the last expression generalizes to

dN1

ν 1= ... =

dNr

ν r≡ dξ , (22)

whereν i is the stoichiometric coefficient for the i’th component in the reaction (by convention, itis negative for reactants).

For constant temperature and pressure and total mass (for each element) conditions,the reaction can proceed until the Gibbs free energy is a minimum with respect to allallowedvariations in the state of the system.For fixed total mass, temperature, and pressure, the onlyvariations which can be considered are those which changeξ . That is the reaction moves eitherto the right or left untilG(ξ ) is a minimum.

G can be a minimum with respect to changes in the progress variable only if

∂G

∂ξ T,P,Ntotal

= 0 (23)

and

∂2G

∂ξ 2 T,P,Ntotal

> 0. (24)

By using the differential form for the change in the free energy together with Eq. (22) we findthat

dG = −SdT+ VdP+r

i=1Σ ν i µ i dξ , (25)


r

i=1Σ ν i µ i = 0 (26)

at equilibrium. Since theµ i are the partial molar Gibbs free energies, Eq. (26) is equivalent to∆G = 0. At equilibrium the free energy change in the reaction per mole vanishes. (Indeed,this is the principle we applied in the "reaction" between graphite and diamond).From the defi-nition of the Gibbs free energy (G = H -TS), it follows that

∆S =∆H

T

at equilibrium.

What happens if we change temperature or pressure by a small amount? Which waywill the equilibrium shift?To answer this note the following Maxwell relations:



∂µ i

∂T P,N j

= −

∂S

∂Ni

P,T,N j≠i

= −Si (27a)

and

∂µ i

∂P T,N j

=

∂V

∂Ni

P,T,N j≠i

= Vi (27b)

which follow from the Gibbs free energy. Thus the changes in the chemical potential associatedwith temperature or pressure are related to the partial molar entropies or volumes, respectively.Next consider

d(iΣν i µ i ) =

iΣν i

∂µ i

∂T P,N j

dT +iΣν i

∂µ i

∂P P,N j

dP + ∂(Σν i µ i )

∂ξ T,P,Ntotal

dξ

= −∆SdT + ∆V dP +∂∆G

∂ξdξ , (28)

where, cf. Eqs. (27),∆S ≡ Σν i Si and∆V ≡ Σν iVi are the entropy and volume changes per moleor reaction. Equation (28) shows how the free energy change per mole of reaction changes whenwe change T, P, or ξ .

What happens if we change, T or P in a system where chemical reaction is possible?The progress variable will change until Eq. (26) is again valid. Sinceboth the initial and finalstates satisfy Eq. (26), the change in∆G must vanish. FromEq. (28) this implies that

dξ =∆SdT − ∆V dP

∂∆G

∂ξ T,P,Ntotal

. (29)

Moreover, the denominator of the right hand side of the equation is positive, cf. Eq. (24). Equa-tion (29) can be rewritten by noting [cf. Eq. (26)] that∆S = ∆H /T; i.e.,

dξ =

∆H

TdT − ∆V dP

∂∆G

∂ξ T,P,Ntotal

. (30)

Equation (30) is a mathematical statement ofLeChatellier’ s principle . For reac-tions which lead to an increase in the volume (∆V > 0, increasing (decreasing) the pressure willshift the equilibrium to the reactant (dξ < 0) [product (dξ > 0)] side of the equation.The reverseis true if the volume change is negative. Similarly, increasing the temperature shifts the equilib-rium to the reactant side for reactions which are exothermic (∆H < 0) and to the product side forreactions which are endothermic.



15.3. Chemicalequilibria in dilute gases

For a one-component gas, the pressure dependence of the chemical potential (freeenergy per mole) is easily obtained from Eq. (20); i.e.,

µ = µ0(T) +P

1∫ V(T, P′)dP′.

Whereµ0(T) is the standard Gibbs free energy of formation at one atm and temperature T. If thegas is ideal, then V=RT/P and we find that

µ(T, P) = µ0(T) + RT ln(P), (31)

where P is the pressurein atmospheres.

This result can be generalized to gas mixtures if we recall our discussion of Dalton’slaw of partial pressures. There we considered a gas mixture where one of the components coulddiffuse in and out of the system through a selective, porous film into a container containing apure sample of that component. At equilibrium, the pressure in the pure sample wasPi , the par-tial pressure of the i’th component in the mixture.

If we view the process as the following "chemical reaction"

Component i in mixture←→ Component i in pure sample.

The equilibrium condition becomes:

µ i ,mixture(T, P, x1,..., xr−1) = µ i ,pure(T, Pi ) = µ0

i (T) + RT ln(Pi ). (32)

Hence, the form of the chemical potential in a gas mixture is very similar to that in a pure sam-ple, with the exception that the pressure is not the total pressure of the gas, but is the partial pres-sure of the component in question.

Knowing this, we are ready to discuss chemical equilibria in gases. FromEq. (26),the equilibrium condition becomes:

0 =iΣν i µ

0i (T) + RT ln(Pν 1

1 Pν 22

...Pν rr ) (33a)

or

Pν 11 Pν 2

2...Pν r

r = KP(T) ≡ exp −Σν i µ

0i (T)/RT

= e−∆G/RT, (33b)

whereKP(T) is called the pressure equilibrium constant. Notice that it is only a function of tem-perature, the stoichiometric coefficients, and properties of the pure (i.e., unmixed) gases.



15.4. Examplesof Chemical Equilibrium Calculations

15.4.1. Free Energy and Entropy of Mixing

Perhaps the simplest process were two samples of different pure gases are isother-mally mixed as depicted below

A+B (P,T,2V)B(P,T,V)A(P,T,V) +As you might expect, this process always occurs spontaneously. The total pressure and tempera-ture remain constant during the process. What is the Gibbs free energy change?

From Eq. (32), it follows that the free energy of the final state is

G final = N1µ0

1(T) + RT ln(P1)

+ N2µ0

2(T) + RT ln(P2). (34a)

Similarly, the Gibbs free energy of either of the pure samples is

Gi = Niµ0

i (T) + RT ln(P)

(34b)

and hence, the free energy of mixing per mole of mixture,∆Gmix, is

∆Gmix = RTx1 ln(x1) + RTx2 ln(x2), (35)

where xi is the mole fraction of i and where Dalton’s law of partial pressures,Pi = Pxi , wasused. Itis easy to generalize this result to arbitrary mixtures of ideal gases

∆Gmix = RTiΣ xi ln(xi ). (36)

Since 0 <xi < 1, the free energy change is negative and the mixing occurs spontaneously.

Equation (36) can be used to calculate the entropy and enthalpy of mixing. By usingEq. (14) it follows that

∆Smix = −∂∆Gmixing

∂T P,xi

, (37)


∆Smix = −RiΣ xi ln(xi ) > 0. (38)

Moreover, since



∆Hmix = ∆Gmix + T∆Smix,

it follows that the heat of mixing associated with the mixing of ideal gases is zero.No heat isabsorbed or released for the mixing of ideal gases.The process is driven entirely by entropy.As we shall see later, a similar result holds for the mixing of dilute solutions.Finally, note thatthere is no volume change for mixing ideal gases (see Eq. (14) and take a pressure derivative). Ina binary mixture, what composition has the most negative free energy of mixing?

15.4.2. Determinationof Free Energies of Formation

There are a number of ways in which to measure the standard free energies of for-mation of a compound. Consider the formation of ammonia,

1

2N2(g) +

3

2H2(g) ←→ NH3(g),

at STP. The free energy change for the reaction is∆G0f (NH3, g). If we measure the equilibrium

constant,

PNH3

P1/2N2

P3/2H2

= e−∆G0f (NH3,g)/RT,

then we can easily compute the free energy of formation of ammonia.

15.4.3. Determinationof the Extent of a Reaction

Reconsider the reaction

N2O4(g) ←→ 2NO2(g).

The extent of the reaction is easily measured by measuring the apparent deviation from the idealgas law. As before, letα be the fraction ofN2O4 dissociated. Ifthere wereN0 moles ofN2O4initially, then there are (1− α )N0 and 2α N0 moles of N2O4 and NO2 at equilibrium, respec-tively. The corresponding partial pressures can be computed from Dalton’s law.

When the result is used in the equilibrium constant condition we find that

KP =P2

NO2

PN2O4

=4α 2

1 − αN0RT

V=

4α 2P

1 − α 2,

where P is the total pressure on the system.This can be solved for the fraction dissociated, withthe result that

α =

KP

KP + 4P

1/2

Thus, if we calculate the equilibrium constant from a table of free energies, the degree of dissoci-ation is easily found. Note that the result depends on both T and P.



15.5. CoupledReactions

In some cases, the direct formation of a certain compound by direct reaction is ther-modynamically forbidden. An example is the formation of titanium tetrachloride from commonTiO2 ore; i.e.,

TiO2(s) + 2Cl2 → TiCl4(l ) + O2(g).

It turns out that∆G = +152. 3kJ/mole. Nonetheless, we can make the reaction go by coupling itto one which pulls it along.For example, suppose we use the produced oxygen to burn carbon;i.e.,

C(s, graphite) + O2(g) → CO2(g),

where here∆G = −394. 36kJ/mole. The free energy change for the coupled processes is -394.36+ 152.3 = -242.1 kJ/mole, and thus the coupled reaction can proceed. The burning carbon sup-plies the needed free energy to make the desired reaction work.

15.6. Temperature Dependence ofK p

The equilibrium constant,K p, is only a function of the temperature. From its defini-tion, cf. Eq. (33b),

d ln(KP)

dT= −

d(∆G/RT)

dT

= −1

RT2T

d(∆G)

dT− ∆G

=T∆S+ ∆G

RT2

=∆H

RT2, (39)

where the second to last equality follows when Eqs. (27a) and (33b) are used. Thus,

ln

K p(T2)

K p(T1)

=T2

T1

∫∆H

RT2dT (40)

≈∆H

R

1

T1−

1

T2

. (41)

Equation (41) follows if we assume that∆H is approximately constant. Note that as in our



discussion of LeChatellier’s principle, the equilibrium will shift to the product side when thetemperature is raised if∆H > 0


Thermodynamic Stability -79- ChemistryCHEM 213W

16. Eigenvalues and Thermodynamic Stability

When considering thermodynamic stability for an isolated system, we had to showthat

∆E(2) ≡1

2∂2E

∂S2 V,Ni

∆S2 +

∂2E

∂S∂V Ni

∆S∆V +1

2∂2E

∂V2 S,Ni

∆E2 + ... (1)

was positive, where terms involving changes ofNi have been dropped.Moreover, by using theexpressions for the derivatives of the internal energy, Eq. (1) can be rewritten as:

∆E(2) ≡T

2CV∆S2 +

∂T

∂V S,Ni

∆S∆V −1

2∂P

∂V S,Ni

∆V2 + ..., (2)

or in matrix notation:

∆E(2) =1

2

∆S

∆V

†

⋅

T

CV

∂T

∂V S,Ni

∂T

∂V S,Ni

−∂P

∂V S,Ni

⋅

∆S

∆V

. (3)

The matrix in Eq. (3) is symmetric.From linear algebra, we know that symmetric matrix can bediagonalized; i.e., a basis in the [∆S, ∆V] space can be found where the matrix elements are zeroexcept for the diagonal ones which are equal to the eigenvalues (denoted asλ ±). In this basis,Eq. (3) becomes

∆E(2) =1

2(λ +c2

+ + λ −c2−),

wherec± are the expansion coefficients of [∆S, ∆V] in the special basis.

In order that the right hand side of Eq. (3) is greater than zero for all possible varia-tions of the system, it is necessary that the eigenvalues,λ ±, be positive. The eigenvalues,λ ±, sat-isfy the characteristic equation, i.e.,

0 = λ 2± − λ ±

T

CV−

∂P

∂V S,Ni

−T

CV

∂P

∂V S,Ni

− ∂T

∂V

2

S,Ni

. (4)

This quadratic equation is easily solved. It thus follows that in order that the eigenvalues be pos-itive,

CV > 0, (5)

−∂P

∂V S,Ni

> 0, (6)


Thermodynamic Stability -80- ChemistryCHEM 213W

and

∂T

∂V

2

S,Ni

< −T

CV

∂P

∂V S,Ni

. (7)

The analysis for variations involving the number of moles of a given species follows in exactlythe same manner.


Entropy & Randomness -81- Chemistry CHEM 213W

17. Entropy of Mixing: A Statistical View

As was mentioned in class, it is possible to give a simple quantitative microscopicderivation of the expression for the entropy of mixing for an ideal solution.To begin, considerthe following simple lattice model for a solution:

1 1 1 1 18 8 8

1 12 28 88 8

1 2 358 87 7

71 2 24 4 86

8 113 8 8 8 8

41 12 34 8 8

Fig. 1

The volume occupied by the solution has been divided into M equivalent cells, and each is ran-domly occupied by a single molecule of a given type in the solution.Let Ni be the number ofmolecules of the i’th species (the "solvent" counts as a species).

How many states are available for the solution in this model?Specifically, howmany ways can the molecules occupy the cells? Consider species 1: The first molecule canchoose M cells, the second M-1, etc..Finally, the last species 1 molecule can choose M-N1-1different cells to occupy. Thus the number of ways of assigning the species 1 molecules to thecells is

M(M − 1)...(M − N1 − 1) ≡M !

(M − N1)!, (1)

whereN! = N(N − 1)(N − 2)...1 is called the factorial function.

The other species must still be added to the lattice. Consider species 2.Now thereare onlyM − N1 cells to choose from; by repeating the preceding argument, it is easy to showthat the number of ways of adding species 2 is

(M − N1)!

(M − N1 − N2)!. (2)

The total number of ways of adding both species 1 and species 2 to the lattice is the product ofthe ways of adding each; i.e.,

M !

(M − N1 − N2)!. (3)



Finally, we can repeat the argument until the entire lattice is filled.The total number of ways ofadding the molecules to the solution is

M !. (4)

Does each of these ways correspond to a state of the system? The answer is no, notbecause we have made an error in our calculation, but because we have ignored a basic propertyof nature; namely, the Heisenberg Uncertainty Principle.Equation (4) would be correct if wecould distinguish the different molecules of each species. The uncertainty principle makes thisimpossible, and thus each state of the system cannot depend on which of the equivalent mole-cules are in the specific cells -- it is impossible to tell.

Thus Eq. (4) overcounts the number ofdifferent states available to the system.Byhow much? Again consider species 1. After theN1 cells are chosen one still has the freedom topermute species 1 molecules between the different chosen cells; there areN1! ways of permutingthe species 1 molecules, and these permutations are included in Eq. (2).However, as we hav ejust argued, quantum mechanics makes these permutations irrelevant to the calculation of thenumber of inequivalent states available to the system, and hence, Eq. (2) should be divided by thenumber of ways of rearranging the equivalent molecules on the same set of lattice cells.Byrepeating this argument for all species, it follows that the number of inequivalent states of thesolution are:

M !

N1!N2!.... (5)

The calculation of the entropy of mixing now follows by using the statistical (Boltz-mann) expression for the entropy:

S = kB ln(number of states), (6)

where kB is Boltzmann’s constant (kB = R/NA = 1. 38× 10−23J K−1). By using Eq. (5) in Eq.(6), we obtain

∆Smixing = kB

ln(M !) −

iΣ ln(Ni !)

. (7)

This still doesn’t look like the expression we obtained in class. Note, however, that the factorialswhich appear in Eq.(7) are factorials of huge numbers (≈ 1023). Thereis an accurate approxi-mation for the natural logarithm of a large factorial known as Sterling’s formula; it is:

ln(N!) ≈ N[ln(N) − 1] (8)

(try it for N=50). If Eq. (8) is used in Eq. (7) and we remember thatM =iΣ Ni , a little algebra

shows that



Smixing = − kBiΣ Ni ln

Ni

M. (8)

Since,xi , the mole fraction of species i is

Ni

M,

Eq. (8) is equivalent to the expression we obtained by examining solutions that obey Raoult’sLaw. (Recall that the gas constantR = kBNA, whereNA is Avogadro’s number).


Phase Equilibrium -84- ChemistryCHEM 213W

18. PhaseEquilibrium

An important class of equilibria involves the different equilibria established betweendifferent states of matter, e.g., solids, liquids, and gases. Aswe shall now show, such equilibriaare governed by very simple principles, essentially the same as those we worked out for chemicalequilibria.

1 2N N

Figure 1

Consider a system containing an arbitrary number of non-reacting componentsplaced in a constant pressure and temperature container (the space enclosed by the heavy lines inthe figure). When equilibrium is attained, each part of the system will have a certain number ofmoles in each part.For example, if we divide the system up according to the dashed lines thefigure, there will beN1 moles of one component in the first part,N2 in the second, etc.

Since the entire system is at constantT, P, and number of moles of each component,the Gibbs free energy, G, must be at a minimum at equilibrium. If so, consider what would hap-pen if we changed the number of moles in the first two compartments by lettingN1 → N1 + dNandN2 → N2 − dN. Clearly, we hav enot changed the total number of moles. However,

dG = (µCompartment 1− µCompartment 2)dN (1)

which will be nonzero (and hence G cannot be a minimum) unless

µCompartment 1= µCompartment 2, (2)

or, since the division into compartments was completely arbitrary, unless the chemical potentialof each component is the same in each part of the system.Note, that this is just a special exam-ple of our criterion for chemical equilibrium for the trivial "chemical" reaction:

NCompartment 1→←NCompartment 2. (3)



Since, our preceding discussion only depended on the ability to have matterexchanged between different parts of the system (but not on transfers to and from the surround-ings), it will also hold for systems in which phase separation has occured, for example if therewas a liquid-vapor interface in our system, or if there was a precipitate at the bottom of the con-tainer. Would it hold if there was a semipermiable membrane separating different parts of ourcontainer?

In order to study some of the general consequences of these equilibrium conditions,consider a system containingC components that have phase separated intoP phases. ApplyingEq. (2) to each component we see that

µ(1)1

⋅⋅⋅

µ(1)C

=⋅⋅⋅=

µ(2)1

⋅⋅⋅

µ(2)C

=⋅⋅⋅=

⋅⋅⋅⋅⋅

=⋅⋅⋅=

µ(P)1

⋅⋅⋅

µ(P)C ,

(4)

where µ( j )i is the chemical potential of componenti in phasej . It is easy to see that there are

(P − 1)C conditions. Eachchemical potential depends onT andP (the same throughout the sys-tem, WHY?) and on the composition of the phase in question (e.g., as we saw for dilute gases).Since, we aren’t currently interested in extensive properties, we can characterize the compositionof each phase by the mole fractions,xi , of which there areC − 1 independent ones in each phase,or P(C − 1) in total. Thus, including temperature and pressure, there are a total ofP(C − 1) + 2unknowns to be determined by the (P − 1)C equations given in (4), and in general the system ofequations will be either over- or under-determined. Theextent to which this happens is obtainedby looking at the number of degrees of freedom,f , the difference between the number ofunknowns and equations; i.e.,

f = P(C − 1) + 2 − C(P − 1) = 2 + C − P. (5)

This simple result is known as the Gibbs Phase Rule.

As an example, suppose we have a one component system (C=1). According to Eq.(5), if there is only a single phase,f = 2; i.e., we can pickT andP arbitrarily. If there are twophases in equilibrium, thenf = 1; i.e., we can pick at most eitherT or P, but not both.Finally, fvanishes if there are three phases in equilibrium, which implies that we can’t pick anything at theso-called triple point.For pure water, the triple point occurs at 0.01C and 611 Pa. Finally, notethat while our simple counting arguments rule out over-determined solutions, they don’t alwaysguarantee the existence of solutions, e.g., as is found above the critical temperature in one-com-ponent systems.

We next turn to a more quantitative analysis of the equilibrium between phases.Suppose we have a one component system containing two phases (for concreteness, a liquid andits vapor). Accordingto Eq. (5) we can pick one intensive quantity, e.g., the temperature,T, andthe pressure at coexistence will be determined.How does the coexistence pressure change if wechange the temperature? Since,

µ liquid = µvapor (6)



both before and after the change in temperature,

dµ liquid = dµvapor. (7)

However, since for a one component system,µ = G/N,

dµ i = −Si dT + Vi dP, (8)

whereSi andVi are the molar entropy and volume, respectively in phasei . By substituting Eq.(8) into (7) and rearranging the result, it follows that

dP

dT Coexistence

=S1 − S2

V1 − V2=

∆S

∆V=

∆H

T∆V, (9)

where the last equality follows by noting that∆H = T∆S at equilibrium. Of course holds for anytwo-phase, one-component equilibrium, and is known as the Clapeyron equation.For the liquid-vapor equilibrium example,∆H is just the heat of vaporization. Whatis it for solid-liquid equi-libria?

There are several approximations that follow from Eq. (9). For example, if we canignore the temperature dependences of∆H and∆V, then we can integrate both sides of (9) andobtain

P2 − P1 ≈∆H

∆Vln(T2/T1) ≈

∆H∆T

T1∆V, (10)

where the last approximation follows if ∆T/T1 is small. This is a reasonable approximation forcases where both phases are solids and/or liquids (and hence, their molar volumes don’t changemuch with temperature).

Another approximation is obtained if one of the phases is a dilute gas and the other asolid or liquid, then

∆V = VVapor − V2 ≈ VVapor ≈RT

P, (10)

which allows us to rewrite Eq. (9) as

d ln(P)

dT Coexistence

=∆H

RT2, (11)

which is known as the Clausius-Clapeyron equation. Compare this with the expression weobtained for the temperature dependence of the equilibrium constant,KP. They are the same,perhaps as expected since, we can view the process as a chemical reaction, e.g.,

liquid→←vapor ,

for which the equilibrium would be determined fromP(T) = K p.



Finally, With the further assumption that that∆H is independent of temperature, wecan integrate the Clausius-Clapeyron equation and show that the equilibrium vapor pressure of aliquid or solid obeys (approximately)

P(T) = P(T0) exp

∆H

R

1

T0−

1

T, (12)

again consistent with thechemical equilibriumpoint of view. How would you use this result todetermine the normal boiling point of the material?


Azeotropes -88- Chemistry CHEM 213W

19. Minimum and Maximum Boiling Azeotropes: TheGibbs-Konovalov Theorem

As we discussed in class, a system where the temperature-composition phase dia-gram has a minimum or maximum for some composition is called an azeotrope.Here we willshow that the composition of the vapor and liquid phases of azeotropes must be the same.Thisresult is called the Gibbs-Konovalov theorem, and is an sophisticated exercise in Maxwell rela-tions, equilibrium conditions, and partial molar quantities.

For what follow, we will call the two components 1 and 2. Quantities pertaining tothe two phases will be distinguished by having a prime in the case of the vapor. Finally, the molefraction of 1 in the liquid and vapor phases arex1 andy1, respectively.

The phase rule tells us that a two-phase, binary mixture hasF = 2 + 2 − 2 = 2degrees of freedom. One will be taken to be the pressure, which is held fixed. Theother will besome convenient concentration variable. Ingeneral, the chemical potentials in either phase canbe viewed as functions of T,P, and a mole fraction (e.g.,x1 or y1). Hence,for arbitrary changesof state:

dµ i = ∂µ i

∂T P,x1

dT + ∂µ i

∂P T,x1

dP + ∂µ i

∂x1

P,T

dx1. (1)

The first two partial derivatives may be rewritten by using Maxwell relations obtained from theGibbs free energy (recall,dG = −SdT+ VdP+ Σ µ i dNi ). Thus,

dµ i = −Si dT + Vi dP + ∂µ i

∂x1

P,T

dx1, (2)

where

Si ≡

∂S

∂Ni

T,P,N j≠i

,

and

Vi ≡

∂V

∂Ni

T,P,N j≠i

are the partial molar entropies and volumes, respectively.

Equation (2) must hold for either component in either phase (in the vapor phase, justadd primes to all quantities and changex1 to y1). Moreover, for changes in state along the coex-istence curve,

dµ i = dµ ′i i = 1, 2. (3)

If we use Eq. (2) in (3), two equations relating changes in T, P, x1 and y1 are obtained. In addi-tion, for the case at handdP = 0, and thus, we find that



−(Si − S′i )dT +

∂µ i

∂x1

T,P

dx1 =

∂µ ′i

∂y1

T,P

dy1, i = 1, 2. (4)

Next, we divide through bydx1, multiply the equation for component 1 byx1, the equation forcomponent 2 byx2 and add the results. This gives

−[x1(S1 − S′1) + x2(S2 − S′

2)]

∂T

∂x1

P,coex

+ x1∂µ1

∂x1

T,P

+ x2∂µ2

∂x1

T,P

=

x1

∂µ ′1

∂y1

T,P

+ x2

∂µ ′2

∂y1

T,P

∂y1

∂x1

P,coex

.

(5)

The changes in the chemical potentials at constant T and P are not independent, butare related through the Gibbs-Duhem equations,

x1dµ1 + x2dµ2 = 0 (6a)

and

y1dµ ′1 + y2dµ ′

2 = 0. (6b)

Dividing Eqs. (6a) and (6b) bydx1 anddy2, respectively, and using the results in Eq. (5) showsthat

−[x1(S1 − S′1) + x2(S2 − S′

2)]

∂T

∂x1

P,coex

=

∂µ ′1

∂y1

T,P

∂y1

∂x1

coex

x1 −

x2y1

y2

(7)

=

∂µ ′1

∂y1

T,P

∂y1

∂x1

coex

(x1 − y1)

y2, (8)

where the last equality is obtained by noting that the sum of the mole fractions in either phase isunity.

The left hand side of the equation vanishes at a minimum or maximum of the T-Xphase diagram; hence, one of the factors on the right hand side must be zero.Thermodynamicstability requirements (i.e., G is a minimum) can be used to show that



∂µ1′∂x1

T,P

> 0.

It is also easy to show, using the definitions of the mole fractions and the fact that the total num-ber of moles of each type is constant, that

∂y1

∂x1

coex

≠ 0.

Hence, at an azeotropic point,x1 = y1; i.e., the compositions of the liquid and vapor phases arethe same. This is called the Gibbs-Konovalov theorem. Notethat the derivation did not requireany specific properties of liquids or gases and our result will hold for any two-phase coexistencein binary mixtures.

Finally, note that essentially the same manipulations can be used to show that

∂P

∂x1

T,coex

=

∂µ ′1

∂y1

T,P

∂y1

∂x1

coex

(x1 − y1)

y2[x1(V1 − V ′1) + x2(V2 − V ′

2)].

Hence, the P-X phase diagram will also have a minimum or maximum at the azeotropic compo-sition.


Eutectics -91- Chemistry CHEM 213W

20. IdealEutectic Phase Diagrams

The ideal lead-antimony eutectic phase diagram is shown in the figure, and was cal-culated assuming that the solids are completely immiscible and that the solution was ideal.Theliquid-solid coexistence lines are computed from the equilibrium conditions:

µ(0)i (solid,T) = µ(0)

i (liquid, T) + RT ln(xi ), i = A, B. (1)

or using the fact that∆G(0)fus(i) ≡ µ(0)

i (liquid, T) − µ(0)i (solid,T) = 0 at the normal melting point,

T = T (i)MP, of materiali , we can rewrite Eq. (1) as

ln(xi ) = ∫T

T (i)MP

∆H (i)fus

RT2≈

∆H (i)fus

R

1

T (i)MP

−1

T

, (2)

where the last approximation follows by assuming the the heat of fusion is independent of tem-perature.

The point where the two curves cross is called the Eutectic point. The question iswhat happens below the eutectic temperature?From the point of view of the phase rule, or theequilibrium conditions given in Eqs. (1) or (2), we could still have L+A, L+B, or A+B (i.e., amixture of the solid phases). Indeed, it is the last possibility which happens; WHY?

In order to answer this question, we must go back a few steps and remember thatunder constant T,P,N conditions, the equilibrium state is the one with thelowest Gibbs freeenergy; specifically if there are multiple local minima in G, it is the one that is lowest that shouldbe observed at equilibrium. This is what happens below the eutectic temperature.To see this,consider the total free energy for the system (again for ideal solutions and immiscible solids):

G = n(s)A µ(0)

A (solid,T) + n(s)B µ(0)

B (solid,T)

+ n(l )A [µ(0)

A (liquid,T) + RT ln(xA)] + n(l )B [µ(0)

B (liquid,T) + RT ln(xB)] (3a)


Eutectics -92- Chemistry CHEM 213W

= NAµ(0)A (solid,T) + NBµ(0)

B (solid,T)

+ n(l )A [∆G(0)

fus(A,T) + RT ln(xA)] + n(l )B [∆G(0)

fus(B,T) + RT ln(xB)], (3b)

wheren(s)i , n(l )

i , and Ni are the number of moles ofi in the solid, liquid, and total, respectively.Remembering that the total number of moles of each material remains constant, the equilibriumstates must be local minima ofG, and these are found by taking derivatives of Eq. (3b) withrespect ton(l )

A or n(l )b and setting the result to zero. It is straightforward to show that this results

in Eq. (1).

As was discussed above, when deciding between different possibilities, it is neces-sary to go back to Eq. (3) and find the local minimum with the lowest free energy. For our exam-ple, consider the following two possibilities L+A or A+B. The latter obeys the liquid-solid equi-librium condition for A, cf. Eq. (1) and hence, from Eq. (3b) we see that the states have free ener-gies:

GA+B = NAµ(0)A (solid,T) + NBµ(0)

B (solid,T) (4a)

and

GL+A = NAµ(0)A (solid,T) + NBµ(0)

B (solid,T) + NB[∆G(0)fus(B,T) + RT ln(xB)], (4b)

where we have set n(l )A = n(l )

B = 0 for the A+B state andn(l )B = NB for the L+A state.Thus, L+A

state will be the equilibrium one as long as

∆G(0)fus(B,T)

RT+ ln(xB) < 0 (5)

or, approximately, as long as

ln(xB) <∆H fus

R

1

T (B)MP

−1

T

, (6)

which, cf. Eq. (2), breaks down below the L+B coexistence line. Thus, since all points on theL+A coexistence line are below the L+B coexistence line below the eutectic temperature, therethe L+A state has a higher free energy than A+B state.Clearly, there is nothing special about thenotation for A, and hence, exactly the same conclusion can be drawn for B. Thus we see that thesolid mixture is the minimum free energy solution below the eutectic temperature.Finally, whilewe have used ideal solutions in this example, you can show that the same argument still holds fora nonideal solution* .

* To do this consider how Eqs. (1) and (3b) would have to be modified for the nonidealcase, and what are the signs of the various terms in different parts of the phase diagram.


Capillary Rise -93- ChemistryCHEM 213W

21. Capillary Rise and Depression

Consider a two phase system with a flat interface as shown in Fig. 1.

Phase I

Phase II

z

FIG. 1Since the two phases are in equilibrium in the presence of a gravity field, the pressure varies withheight,z; namely,

p1(z) = p1(0) − ρ1gz (1)

in phase 1, and

p2(z) = p2(0) − ρ2gz (2)

in phase 2, whereρ1(2) is the density in phase 1(2). Of course, at the interface (here assumedflat) the pressures must be equal and

p1(0) = p2(0) ≡ p0. (3)

Now put a capillary tube into the system as shown in Fig. 2.



h

Phase I

Phase II θ

FIG. 2The surface tension,γ , between phase 1 and phase 2 wants the pressures between the two phasesto be unequal at the curved interface (why?). Specifically, for the convex surface shown in Fig.2, we know that

p2(−h) − p1(−h) =2γR

. (4)

The only way in which the pressure difference can be achieved is by moving thecurved interface to a new position in the fluid. (Changing the pressure in the gas or liquid fromthat of the surrounding material at the same height would cause material to flow from the high tolow pressure regions until the pressures were equal at all points at the same height).Thus, byusing our expressions for the pressure dependence in the phases at any giv en heights [Eqs. (1),(2), and (3)] in Eq. (4) we find that

p0 + ρ2gh− (p0 + ρ1gh) =2γR

(5)

or

h =2γ

Rg(ρ2 − ρ1)≈

2γRgρ2

for ρ2 >> ρ1, (6)

where the approximate expression is useful for liquid-vapor interfaces. (Ifthe interface is con-cave then the pressures in Eq. (4) must be switched and the sign ofh changes.)

Finally, if we write R, the radius of curvature of the interface, in terms ofd, theradius of the tube and the contact angleθ , it follows that

h = −2γ cos(θ )

dg(ρ2 − ρ1)

For a 45o contact angle in a capillary tube with a 1mm radius containing water(γ = 72. 75dynes/cm the capillaryrisewill be 1. 05cm



22. Problem Sets

22.1. Problem Set 1

DUE: Wednesday, January 17, 2001

1. a)If

α ≡1

V∂V

∂T P,N

,

show that

α = −1

ρ∂ρ∂T

P,N

,

whereρ is the mass density.

b) More generally, show that

dρρ

= −α dT +κ dP,

where

κ ≡ −1

V∂V

∂P T,N

is the isothermal compressibility. Note: Donot use the ideal gas equation in proving a) and b).

2. At 25C a sealed, rigid container is completely filled with liquid water. If the temperature israised by 10C, what pressure will develop in the container?For water,α = 2. 07× 10−4/K andκ = 4. 50× 10−5/atm.

3. Castellan,problem 3.3 (Read Secs. 3.7 and 3.8 first).

4. Castellan,problem 4.2.


6. A simple model for gas reactions on a surface can be obtained as follows: Assumethat everymolecule that collides with the surface will react with probabilityP(vx) where

P(vx) ≡

0 if |vx| < v0

1 otherwise,

wherevx is the velocity normal to the surface. Use the kinetic theory of gases to derive a molecu-lar expression for the rate of reaction (i.e., the number of reactions per unit time per unit area).


Problem Set 1 -96- ChemistryCHEM 213W

7. Computethe number of collisions an Argon atom has per second at 1 atm pressure and 25C.

Assume that the argon atom has a 3Ao

diameter. What is the mean free path under these condi-tions?



22.2. Problem Set 2

DUE: Friday, January 26, 2001

1. How much work will be produced in the isothermal, reversible expansion fromV1 to V2 of agas with the equation of state:

pV = RT +(bRT− a)

V

2. (Barrow, Problem 5-1) The acceleration due to gravity on the earth’s surface is about 9.8m sec−2.

a) Whatis the force of gravity on a 1-kg mass?

b) How much mechanical energy could be obtained by fully harnessing the downwardmovement of a 1-kg mass through a distance of 58 m (the height of Niagara Falls)?

c) How much thermal energy would be produced if the mass were allowed to fall freelythrough this distance?

d) If the mass consisted of water and all the thermal energy were absorbed by thewater, How much would the temperature of the water rise?

3. a) Consider a process where the heat absorbed by the system per mole is given by:

d− Q ≡ −3a

4T3/2VdT −

RT

V − b+

a

2T1/2V2dV

Evaluate the heat absorbed by the system along the following paths:

i) T1V1 → T2V1 → T2V2

ii) T1V1 → T1V2 → T2V2

b) Show that 1/T is an integrating factor ford− Q (i.e., d− Q/T becomes the differentialof a state function) by evaluating the integral ofd− Q/T along paths i) and ii).

4. An average man (mass = 70 kg, specific heat - same as water) produces about 104 kJ of heateach day through metabolic activity.

i) If he were a closed system, what would his temperature rise be in oneday?

ii) He is, of course, really an open system--losing heat through evaporationof water. How much water must he evaporate per day to maintain hisconstant temperatureof 37oC? You must first calculate∆Hvap at 37o.

iii) The heat of combustion of cane sugar is 3.95 kcal/g.How many gramsof sugar will furnish energy for one day’s metabolism, assuming thetransfer of heat from cane sugar bonds to metabolic heat to be perfectlyefficient?



5. In the thermite reaction:

2Al(s) + Fe203(s) → 2Fe(l ) + Al203(s),

what is the maximum temperature attainable by the products? Assume that the reactants are at 1atm pressure and 25C, and that all heat capacities are constant over the required temperatureranges. You will need to go to tables of thermodynamic constants; a good place to look is theCRC handbook. Is the assumption thatAl203 is solid in writing the reaction reasonable?

6. Considerthe following two isomers ofC3H6:

CH2

CH2

CH2

Cyclopropane

CHH C2 CH3

Propene

a) Calculate∆H for the interconversion of cyclopropane and propene using a table ofstandard heats of formation.

b) Calculate∆H for this reaction using a table of bond energies.

c) Whichanswer is more reliable? Why? Whatare the sources of error?

d) Whichcompound would yield more heat upon complete combustion in oxygen?



22.3. Problem Set 3

DUE: Wednesday, February 13, 2002

This material is based on the First Law.

1. Castellan,Problem 7.28.





6. A chemistry 213 student was overheard arguing with a friend in management about why wehave winter heating. The latter stated that it was to make the air in the room warmer, while thechemistry 213 student claimed that it was to increase the energy content of the air in the room.Which is right? Why?

This next material covers aspects of the second law.






12. Considera system comprised of two 1000g blocks of copper (cp = 0. 1cal/g/oK). If oneblock is at 300K and the other is at 400K, what is the maximum amount of work that can beextracted from the system if no additional heat is allowed to flow into or out of the system.Whatwill the final temperature be after the work is extracted? Describea process whereby you couldextract the maximum work. In working out this problem, ignore any PV work associated withthe expansion of the blocks.



22.4. Problem Set 4

Due: Friday, February 22, 2002





5. Castellan,Problem 10.7-10.9.

6. Castellan,Problem 10.23




22.5. Problem Set 5

Due: Monday, March 4, 2002

1. Castellan,Problem 11.4 (you might also try 11.7, which is similar).









22.6. Problem Set 6

Due: Wednesday, March 21, 2001.









22.7. Problem Set 7

Due: Wednesday, March 28, 2001.









22.8. Problem Set 8

Due: Wednesday, April 4, 2001









22.9. Problem Set 9

Due: Wednesday, April 11, 2000.

1. Completeand balance the following reactions:

ClO−3 + As2S3 → Cl− + H2AsO−

4 + SO=4 (acidic solution)

and

Cu(NH3)++4 + S2O

=4 → SO=

3 + Cu+ NH3 (basic solution).

In each case, identify the oxidized and reduced species and compute the standard state cell poten-tial.

2. Considerthe cell

Pt(s)|H2(g, 1 atm), H+(aq, 1M)||Fe3+(aq), Fe2+(aq)|Pt(s),

given that∆ε o = 0. 771V for

Fe3+ + e− → Fe2+.

a) If the measured potential of the cell is 0.712 V, what is the ratio of concentrations ofFe2+ to Fe3+?

b) Whatis the ratio if the potential of the cell was 0.830V?

c) Plotthe fraction of total iron present asFe3+ over the voltage range 0.65 V to 0.9 V.






22.10. Problem Set 10

Due: Wednesday, April 12, 2000.


2. Completeand balance the following reactions:

ClO−3 + As2S3 → Cl− + H2AsO−

4 + SO=4 (acidic solution)

and

Cu(NH3)++4 + S2O

=4 → SO=

3 + Cu+ NH3 (basic solution).

In each case, identify the oxidized and reduced species and compute the standard state cell poten-tial.

3. Considerthe cell

Pt(s)|H2(g, 1 atm), H+(aq, 1M)||Fe3+(aq), Fe2+(aq)|Pt(s),

given that∆ε o = 0. 771V for

Fe3+ + e− → Fe2+.

a) If the measured potential of the cell is 0.712 V, what is the ratio of concentrations ofFe2+ to Fe3+?

b) Whatis the ratio if the potential of the cell was 0.830V?

c) Plotthe fraction of total iron present asFe3+ over the voltage range 0.65 V to 0.9 V.





Past Exams -107- ChemistryCHEM 213W

23. Past Exams

23.1. Quiz I: 1999

23.1.1. INSTRUCTIONS

1. Nobooks or notes are permitted.

2. Calculatorsare permitted.

3. Answerall questions and show all work clearly.

4. Thereare 4 questions and each is of equal value.

5. Besure to indicate the total number of exam books handed in on book 1.

6. You may need the following data:

Useful Constants

Constant Value

Gas Constant, R 8.31442J K−1 mole−1

Boltzmann’s Constant,kB 1. 381× 10−23J/KFaraday,F 96,484.6 Coul/moleStandard Atmosphere 1. 01325× 105PaAv ogadro’s Number 6.0225× 1023

1 cal 4.184J

7. You may need the following results from calculus:

∫∞0

xe−ax2/2dx =1

a,

B

A∫

dx

x= ln(B/A),

B

A∫

dx

x2=

1

A−

1

B,

and

ln(1 + x) ≈ x for |x| << 1.

8. GoodLuck.

February 3, 1999

Quiz I -108- ChemistryCHEM 213W

(1. 25% From the Homework) A simple model for gas reactions on a surface can be obtained asfollows: Assume that every molecule that collides with the surface will react with probabilityP(vx) where

P(vx) ≡

0 if |vx| < v0

1 otherwise,


(2. 25%) Give expressions for the following (derivations are not required, but be sure to defineyour symbols):

a) Themean speed of a gas molecule.

b) Themean free path.

c) Thenumber of collisions per second a gas molecule makes.

d) Thefirst law of thermodynamics.

e) Theenthalpy.

(3. 25%) Computethe heat of reaction for

C(graphite)+ O2(g)1500K ,1atm→ CO2(g).

Is the reaction exothermic or endothermic?You will need the following data:

Cp

(J K−1 mol−1)Compound

C(graphite) 8.53O2(g) 29.35CO2(g) 37.11

You also need∆Hof (CO2, g) = −393. 51kJ/mol.

(4. 25%) One mole of an ideal gas expands isothermally at temperature T from volumeV1 to vol-umeV2.

a) Calculatethe amount of work done if the expansion is performed against a constantpressureP0.

b) Calculatethe amount of work done if the expansion is done reversibly.

c) Whatis the energy change in a) and b)? (Remember that the gas is ideal).

d) Calculatethe amount of heat absorbed by the system in parts a) and b).

February 3, 2000


23.2. Quiz I: ChemistryCHEM 213W








Useful Constants

Constant Value



1 cal 4.184J


∫∞0

xe−ax2/2dx =1

a,

B

A∫

dx

x= ln(B/A),

B

A∫

dx

x2=

1

A−

1

B,

and

ln(1 + x) ≈ x for |x| << 1.

8. GoodLuck.

February 3, 2000


(1. 20% From the Homework) A simple model for gas reactions on a surface can be obtained asfollows: Assume that every molecule that collides with the surface will react with probabilityP(vx) where

P(vx) ≡

0 if |vx| < v0

1 otherwise,


(2. 20%) Thereaction

HCl(g) + NH3(g) → NH4Cl(s)

is carried out by placing 2 cotton balls, one dipped in concentratedHCl and the other in concen-tratedHN4OH, at opposite ends of a 1m long piece of glass tubing.Given the molecular weightsin the table, and other things being equal, where will theNH4Cl precipitate first appears (as mea-sured from the from theHCl cotton ball)? What assumptions are you making?

Molecular WeightsMolecular Weight

(g/mol)Compound

HCl 36.46NH3 17.03

NH4Cl 53.49Air * 28.85

* This is the average molecular weight in air (approximately 79%N2, 21%O2).

(3. 20%)Derive an expression for the diffusion flux caused by a concentration gradient in adilute gas.

(4. 20%,From the Homework) How much work will be produced in the isothermal, reversibleexpansion fromV1 to V2 of a gas with the equation of state:

pV = RT +(bRT− a)

V

(5. 20%,Castellan, 7.43) At 25C, the following enthalpies of formation are given:

Compound SO2(g) H2O(l )∆Ho

f (kJ/mol) -296.81 -285.83

For the reactions at 25C:

2H2S(g) + Fe(s) → FeS2(s) + 2H2(g), ∆Horxn = −137. 0kJ/mol,

February 3, 2000


and

H2S(g) +3

2O2(g) → H2O(l ) + SO2(g), ∆Ho

rxn = −562. 0kJ/mol.

Calculate the heats of formation ofH2S(g) and FeS2(s).

February 6, 2001


23.3. QuizI: Chemistry CHEM 213W








Useful Constants

Constant Value



1 cal 4.184J


∫∞0

xe−ax2/2dx =1

a,

B

A∫

dx

x= ln(B/A),

B

A∫

dx

x2=

1

A−

1

B,

and

ln(1 + x) ≈ x for |x| << 1.

8. GoodLuck.

February 6, 2001


(1. 20%)Show that

CP − CV =P +

∂E

∂V T,N

∂V

∂T P,N

. (1)

What does the right hand side of Eq. (1) become for an ideal gas?

(2. 20%)

a) Describethe Joule-Thompson expansion; in particular, show what thermodynamicquantity is constant during the expansion.

b) Definethe Joule-Thompson coefficient,µ JT, and show that

µ JT = −1

CP

∂H

∂P T,N

. (2)

(3. 20%)(Castellan, 7.64) Calculate the final temperature of the system if 20g of ice at -5C isadded to 100g of water at 21C in a Dewar flask (an insulated flask); for the transformation

H2O(s) → H2O(l ), ∆Ho = 6009J/mol,

at 0C,CP(H2O, s) = 37. 7J/(K mol), andCP(H2O, l ) = 75. 3J/(K mol) .

(4. 20%)

a) Writedown the expression for the Maxwell-Boltzmann velocity distribution for oneof the components of velocity (e.g.,vx) in a gas at temperatureT comprised of mol-ecules of massm.

b) Useyour result to derive the distribution of molecular speeds.

c) How would your result in part c) change if you lived in Flatland (i.e., in a 2 dimen-sional world)?

d) Derive expressions for the most probable speed in the real world and in Flatland.

(5. 20%)(Castellan, 30.4) Consider a gas at constant temperature. If the pressure is doubled,what effect does this have on:

a) thenumber of collisions per second made by any one molecule;

b) thetotal number of collisions per second occurring in 1m3 of gas;

c) themean free path of a gas molecule;

d) thediffusion constant.

March 16, 1999

Quiz II -114- ChemistryCHEM 213W

23.4. Quiz II: 1999








Useful Constants

Constant Value



1 cal 4.184J


∫∞0

xe−ax2/2dx =1

a,

B

A∫

dx

x= ln(B/A),

B

A∫

dx

x2=

1

A−

1

B,

and

ln(1 + x) ≈ x for |x| << 1.

8. GoodLuck.

March 16, 1999


1. (20%)

a) Usethe second law of thermodynamics to obtain the formula for the maximum effi-ciency of a Carnot engine (not necessarily using an ideal gas).

b) Prove that for any constant temperature process,

W ≤ −∆A.

2. (20%)Fill in the required information in the following table (don’t do this on the exam sheet,copy it to your exam book) giving the state functions etc., that govern equilibrium under thestated conditions.

Constant StateFunction Differential MinimumorConditions Name Change Maximumat Equilibrium

Definition

E,V,NS,V,NT,V,NT,P,N

3. (20%) Show that

a)

dS=CP

TdT − Vα dP (N constant )

b)

CP − CV =VTα 2

κ,

whereα is the thermal expansion coefficient andκ is the compressibility.

4. (20%,Castellan, Problem 11.27) Consider the reaction

FeO(s) + CO(g)→←Fe(s) + CO2(g)

for which we have

T (oC) 600 1000KP 0.900 0.396

a) Calculate∆Ho, ∆Go, and∆So at 600oC.

March 16, 1999


b) Calculatethe mole fraction ofCO2 in the gas phase at 600oC.

5. (20%)

a) Derive the Gibbs phase rule.How does your result change if there areR chemicalreactions that can take place between the components in all phases and that come toequilibrium.

b) Derive the Clapeyron equation (the equation giving the slope of the coexistence lineseparating two phases in equilibrium with each other).

March 9, 2000


23.5. Quiz II: 2000





4. Therearefive (5) questions and each is of equal value.



Useful Constants

Constant Value



1 cal 4.184J


∫∞0

xe−ax2/2dx =1

a,

B

A∫

dx

x= ln(B/A),

B

A∫

dx

x2=

1

A−

1

B,

and

ln(1 + x) ≈ x for |x| << 1.

8. GoodLuck.

March 9, 2000


1. (20%) State and derive the Clausius inequality.

2. (20%)

a) (5%)State the third law of thermodynamics.

b) (5%) Show that the third law of thermodynamics implies that heat capacities mustvanish atT → 0.

c) (10%. Castellan, 9.24) In the limitT → 0 it is known empirically that the value ofthe coefficient of thermal expansion of solids approaches zero.Show that, as a con-sequence, the entropy is independent of pressure at 0K.What does this imply foryour answer in part a)?

3. (20%)

a) Show that

dS=CP

TdT − Vα dP, (for N constant),

whereα is the thermal expansion coefficient. [NOTE: You don’t hav e to use thederivation first presented in class].

b) (Fromthe Homework, Castellan, 10.28). Use the result in a) to show that:

i)

∂S

∂P V,N

=κ CV

Tα,

ii )

∂S

∂V P,N

=CP

TVα,

and

iii ) −1

V∂V

∂P S,N

=κγ

,

whereγ ≡ CP/CV and whereκ is the isothermal compressibility.

4. (20%. Castellan,9.27) In an insulated flask (Dewar flask) 20g of ice at -5C are added to 30gwater at 25C. The heat of fusion,∆HFusion, is 334J/g and heat capacities areCP(H2O, l ) = 4. 18J/(gK) andCP(H2O, s) = 2. 09J/(gK). Pressuredoes not change.

a) Whatis the final state of the system?

b) Calculate∆S and∆H for the process.

March 9, 2000


5. (20%)

a) (5%)Show that

iΣν i µ i = 0

at chemical equilibrium. Under what conditions does this relation apply?

b) (15%. Castellan, 11.15) At 298.15K we have the following data:

∆Gof ∆Ho

f(kJ/mol) (kJ/mol)

Compound

C2H4(g) 68.1 52.3C2H2(g) 209.2 226.7

a) CalculateKP at 298.15K for the reaction:

C2H4(g)→←

C2H2(g) + H2(g)

b) Whatmust the value ofKP be if 25% of theC2H4 is dissociated intoC2H2 andH2 at a total pressure of 1atm?

c) At what temperature willKP have the value determined in b)?

March 12, 2001


23.6. Quiz II: 2001





4. Therearefive (5) questions and each is of equal value.



Useful Constants

Constant ValueGas Constant, R 8.31442J K−1 mol−1

Boltzmann’s Constant,kB 1. 381× 10−23J/KStandard Atmosphere 1. 01325× 105PaAv ogadro’s Number 6.0225× 1023

1 cal 4.184J


∫∞0

xe−ax2/2dx =1

a, ln(1+ x) ≈ x for |x| << 1,

∫B

A

dx

x= ln(B/A), and ∫

B

A

dx

x2=

1

A−

1

B.

8.

Themochemical Data at 298.15K and 1 atm.

∆H0f ∆G0

fkJ/mol kJ/mol

Substance

NH3(g) -45.94 -16.5

9. GoodLuck.

March 12, 2001


1. (20%)

a) Stateand prove the Clapyeron equation governing the slope of a coexistence line inthe P-T phase diagram.

b) Useyour result in a) to derive the Clausius-Clapeyron equation. Clearly state whatapproximations you are making and why they are valid.

c) Useyour results in parts a) and b) to discuss the shape of the P-T phase diagram forwater.

d) Is the following phase diagram ever possible? Ifso, for what kind of system? Whyor why not?

�� α

β

χ

δ

T

P

2. (20%) For fixedN show that :

dS=CV

TdT +

αβ

dV,

whereα ≡1

V∂V

∂T P,N

andβ ≡ −1

V∂V

∂P T,N

3. (20% From the Homework) Consider a system comprised of two 1000g blocks of copper(cp = 0. 4184J/g/oK). If one block is at 200K and the other is at 400K, what is the maximumamount of work that can be extracted from the system if no additional heat is allowed to flow intoor out of the system. What will the final temperature be after the work is extracted? Describeaprocess whereby you could extract the maximum work. In working out this problem, ignore anyPV work associated with the expansion of the blocks.

4. (20%) Prove LeChatellier’s principle.

5. (20%) A representative of Crimson Chemical Corporation (CCC) has just visited you andencouraged you to invest in his company. The reason for his optimism is a new catalyst devel-oped by CCC, which he maintains yields a minimum 80% conversion of hydrogen in the ammo-nia synthesis reaction

N2(g) + 3H2(g) → 2NH3(g)

at 500oK and one atmosphere pressure when the reactants,N2 andH2, are mixed in a 1:3 ratio.By assuming that the gases are ideal, and using the data given in the table on the cover sheet,evaluate the validity of his claim. Be sure to show your work.

March 12, 2001


23.7. Final Exam: 1999








Useful Constants

Constant Value



1 cal 4.184J


∫∞0

xe−ax2/2dx =1

a,

B

A∫

dx

x= ln(B/A),

B

A∫

dx

x2=

1

A−

1

B,

and

ln(1 + x) ≈ x for |x| << 1.

8. GoodLuck.

April 20, 1999

Final Exam -123- ChemistryCHEM 213W

1. (From the Homework. 20%) Consider a system comprised of two 1000g blocks of copper(cp = 0. 1cal/g/oK). If one block is at 300K and the other is at 400K, what is the maximum amount ofwork that can be extracted from the system if no additional heat is allowed to flow into or out of the sys-tem. Whatwill the final temperature be after the work is extracted? Describea process whereby youcould extract the maximum work. In working out this problem, ignore any PV work associated with theexpansion of the blocks.

2. (Fromthe Homework. Castellan-11.820%) At 500K we have the following data

∆Ho500 So

500(kJ/mol) (J/Kmol)

Substance

HI (g) 32.41 221.63H2(g) 5.88 145.64I2(g) 69.75 279.94

One mole ofH2 and one mole ofI2 are placed in a vessel at 500K. At this temperature only gases arepresent and the equilibrium

H2(g) + I2(g)→←2HI (g)

is established.CalculateKP and the mole fraction ofHI present at 500K and 1atm. What would themole fraction ofHI be at 500K and 10atm?

3. (From the Homework. Castellan-14.1. 20%)Benzene and toluene from nearly ideal solutions.At283K, Po

toluene= 30mmHgandPobenzene= 100mmHg.

a) A liquid mixture is composed of 3 mol of toluene and 2 mol of benzene.If the pressureover the liquid is reduced, at what pressure does the first vapor form?

b) Whatis the composition of the first trace of vapor?

c) If the pressure is reduced further, at what pressure does the last trace of liquid dissapear?

d) Whatis the composition of the last trace of liquid?

e) Whatwill the pressure, the composition of the liquid, and the composition of the vapor bewhen 1 mol of the mixture has been vaporized?

4. (20%)

a) Derive an expression for the freezing point depression of an ideal solution.

b) Derive the van’t Hoff equation for the osmotic pressure.

c) TheClausius-Clapeyron equation.

d) TheNernst equation.

In all parts, be sure to clearly state any assumptions or general principles you use.

5. (Castellan-12.8. 20%)Iodine boils at 183.0 C; the vapor pressure of the liquid at 116.5C is 100 Torr.If ∆Ho

fus = 15. 65kJ/mol and the vapor pressure of the solid is 1 Torr at 38.7C, calculate

a) Thetriple point temperature and pressure.

April 20, 1999


b) ∆Hovap and∆So

vap.

c) ∆Gof (I2, g) at 298.15K.

April 28, 2000







4. Thereare five (5) questions and each is of equal value.



Useful Constants

Constant Value



1 cal 4.184J


∫∞0

xe−ax2/2dx =1

a,

B

A∫

dx

x= ln(B/A),

B

A∫

dx

x2=

1

A−

1

B,

ln(1 + x) ≈ x −x2

2+. . . for |x| << 1,

and

ex ≈ 1 + x +x2

2!+. . . for |x| << 1.

8. GoodLuck.

April 28, 2000


1. (20%) Derive or prove the following. Besure to clearly state any approximations or assumptions youmake.

a) TheGibbs phase rule.

b) An expression for the temperature dependence of the cell potential (EMF); i.e., for(∂∆ε /∂T)P,N .

c) Thelever rule.

d)

Cp − CV =TVα 2

β,

where

α ≡1

V∂V

∂T P,N

and β ≡ −1

V∂V

∂P T,N

e) The expression for the boiling point elevation of an ideal solution containing a singlevolatile component.

2. (20%, Castellan, 13.16) The melting point of iodine is 113.6oC and the heat of fusion is 15.64kJ/mol.

a) Whatis the ideal solubility of iodine at 25oC?

b) How many grams of iodine dissolve in 100g of hexane at 25oC? [The molecular weights ofI2 and hexane are 253.81 and 86.18 g/mol, respectively].

3. (20%, Castellan, 14.14) The Henry’s law constant for argon in water is 2.17× 104 at 0 oC and3. 97× 104 at 30oC. Calculate the standard heat of solution of Argon in water. What assumptions areyou making?

4. (20%)

a) Derive the equation(s) governing the coexistence lines in the ideal eutectic phase diagramand describe how to use your result to construct the phase diagram.

b) Sketch the phase diagram, clearly indicating where the eutectic point is.

c) Choosea composition either to the right or left of the eutectic point, and describe what hap-pens as you cool a liquid with that composition.Discuss at least four different temperatures(high, at the coexistence line, intermediate and low) and be sure to indicate (at least approxi-mately) the compositions and amounts of the phases you have present.

d) Briefly repeat your analysis in part c) for a liquid whose mean composition is the eutecticcomposition.

5. (20%,Castellan, 17.9) The standard potentials at 25oC are:

Pd2+(aq) + 2e− → Pd(s) ε o = 0. 83V

and

April 28, 2000


PdCl2−4 (aq) + 2e− → Pd(s) + 4Cl−(aq) ε o = 0. 64V.

a) Calculatethe equilibrium constant for the reaction

Pd2+ + 4Cl−→←PdCl2−

4 .

b) Calculate∆Go for the reaction.

Wednesday, April 18, 2001

Final Exam -128- 180-213B



1. Nobooks or notes are permitted. Calculators are permitted.

2. Answerall questions in the answer book and show all work clearly. The exam hasof two (2) pages including this one and five (5) questions, each is of equal value.



Useful Constants

Constant Value

Gas Constant, R 8.31442J K−1 mol−1

Boltzmann’s Constant,kB 1. 381× 10−23J/KFaraday,F 96,484.6 Coul/molStandard Atmosphere 1. 01325× 105PaAv ogadro’s Number 6.0225× 1023

1 cal 4.184J

5. You may keep the exam.

6. GoodLuck.


∫∞0

xe−ax2/2dx =1

a,

∫B

A

dx

x= ln(B/A), ∫

B

A

dx

x2=

1

A−

1

B,

and

ln(1 + x) ≈ x −x2

2+. . . for |x| << 1,

Wednesday, April 18, 2001

Final Exam -129- 180-213B

1. (20%) Derive or prove the following. Besure to clearly state any approximations or assump-tions you make.

a) TheNernst Equation.

b) Theexpression for the saturation solubility (mole fraction solute) in an ideal solu-tion.

c) Theexpression for the temperature dependence of the equilibrium constantKP(T).

d) Themaximum work you can get out of a constant temperature process.

2. (20%, Castellan, 11.16) At 25oC, for the reaction

Br2(g)→←2Br(g)

we have∆Go = 161. 67kJ/mol and ∆Ho = 192. 81kJ/mol.

a) Computethe mole fraction of bromine atoms present at equilibrium at 25oC andP = 1atm.

b) At what temperature will the system contain 10 mole percent bromine atoms in equi-librium with bromine vapor at a total pressure of 1 atm?

3. (20%, Castellan, 12.20)For the phase transition

Srhombic → Smonoclinic,

the value of∆S is positive. The transition temperature increases with the increase in pressure.Which is denser, the rhombic or monoclinic form? Prove your answer mathematically.

4. (20%, Castellan, 14.2) Two liquids, A and B, form an ideal solution. At the specified tempera-ture, the vapor pressure of pure A is 200 mmHg while that of B is is 75 mmHg.If the vapor overthe mixture consists of 50 mole percent A, what is the mole percent A in the liquid?

5. (20%, Castellan, 17.19) H.S. Harned and W.J. Hamer [J. Amer. Chem. Soc.57, 33 (1935)]present values for the potential of the cell,

Pb(s)|PbSO4(s)|H2SO4(aq, a)|PbSO4(s)|PbO2(s)|Pt(s),

over a wide range of temperature and concentration ofH2SO4. In 1M H2SO4, they found,between 0oC and 60oC,

∆ε = 1. 91737+ 5. 61× 10−5t + 1. 08× 10−6t2,

wheret is the temperature in Celsius.What are∆G, ∆H and∆S for the cell reaction at 0oC and25oC?

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