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In the Classroom JChemEd.chem.wisc.edu Vol. 76 No. 10 October 1999 Journal of Chemical Education 1391 Why Equilibrium? Understanding the Role of Entropy of Mixing Mary Jane Shultz Department of Chemistry, Pearson Laboratory, Tufts University, Medford, MA 02155; [email protected] Equilibrium is a major topic in both introductory and physical chemistry curricula. This paper grew out of a problem in conveying a fundamental understanding of this important concept to students. Typical introductory courses begin with a discussion of reactions that go to completion, for example, Na with water or displacement of one metal by another. These reactions are complete because one metal (or hydrogen) is more active than another. Then students are introduced to equilibrium: reactions that produce a mixture of reactants and products. Forces that lead to reactions going to completion are put aside and a new quantity, the equilibrium constant, is introduced. Students become expert at utilizing the equi- librium constant via numerous equilibrium calculations, including acid–base, solubility, complex ion formation, and buffer problems. However, the factors that give rise to equi- librium are often not addressed. Physical chemistry courses focus on the connection between equilibrium and thermodynamic quantities, particularly the Gibbs free energy. Equilibrium is recognized as a minimum in the Gibbs free energy vs reaction progress curve for reactions at constant temperature and pressure (1–3). However, the origin of that minimum remains largely unaddressed, although the recent edition of the text by Atkins recognizes the importance of mixing (2). The enthalpic and entropic contributions (the source of the minimum) are not separated; thus an opportunity is missed to reinforce the concept of entropy—specifically entropy of mixing—and its importance. We have found the following example, the dimerization of NO 2 , to be very helpful in making these abstract contributions more concrete. De- pending on local interest, other reactions can be substituted. The two essential characteristics that enable explicit calculation of the Gibbs free energy minimum as well as equilibrium concentrations are that the reaction takes place in the gas phase and that G θ for the reaction is relatively small. It helps simplify the calculation if the number of reactants and prod- ucts is also small. The dimerization of NO 2 fulfills all these requirements and has the added benefit of being familiar to many students either from laboratory exercises (4, 5 ) or lecture demonstrations. A moderate G θ is important because it results in an equilibrium constant on the order of unity and an easily iden- tifiable minimum in the plot of G versus reaction progress. Thermodynamic data for dimerization of NO 2 are given in Table 1. Notice that the reaction’s G θ , { 4.73 kJ, 1 is moderate. Both reactants and products are gases, and to a good approxi- mation, they may be treated as ideal gases as long as the total pressure is on the order of an atmosphere or less. In this case, all the components of G can be calculated explicitly and the origin of the minimum is readily identified. Imagine the reaction as follows. The initial stage consists of two moles of NO 2 at one atmosphere pressure in the left half of a cylinder divided by a transforming partition. The transforming partition merely changes 2NO 2 into N 2 O 4 with no other change. The completion stage has one mole of N 2 O 4 in the right half of the cylinder at one atmosphere pressure. Any intermediate stage has 2 – 2x moles of NO 2 on the left and x moles of N 2 O 4 on the right, both at one atmosphere. As shown below, the equilibrium balance of NO 2 and N 2 O 4 occurs when x = 0.82. Equilibrium actually occurs when the partition is removed and the gases mix. This is illustrated in Figure 1, which also lists results discussed below. Figure 1. Schematic of system discussed in text. 5 2 t a a t a D c i m a n y d o m r e h T . 1 e l b a T 8C e l u c e l o M G f θ / l o m J k {1 H f θ / l o m J k {1 S θ J / K {1 l o m {1 O N 2 ) g ( 1 3 . 1 5 8 1 . 3 3 6 0 . 0 4 2 N 2 O 4 ) g ( 9 8 . 7 9 6 1 . 9 9 2 . 4 0 3 NOTE: Data from ref 6.

Why Equilibrium? Understanding Entropy of Mixing

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Page 1: Why Equilibrium? Understanding Entropy of Mixing

In the Classroom

JChemEd.chem.wisc.edu • Vol. 76 No. 10 October 1999 • Journal of Chemical Education 1391

Why Equilibrium? Understanding the Role of Entropyof Mixing

Mary Jane ShultzDepartment of Chemistry, Pearson Laboratory, Tufts University, Medford, MA 02155; [email protected]

Equilibrium is a major topic in both introductory andphysical chemistry curricula. This paper grew out of a problemin conveying a fundamental understanding of this importantconcept to students. Typical introductory courses begin witha discussion of reactions that go to completion, for example,Na with water or displacement of one metal by another. Thesereactions are complete because one metal (or hydrogen) ismore active than another. Then students are introduced toequilibrium: reactions that produce a mixture of reactants andproducts. Forces that lead to reactions going to completionare put aside and a new quantity, the equilibrium constant,is introduced. Students become expert at utilizing the equi-librium constant via numerous equilibrium calculations,including acid–base, solubility, complex ion formation, andbuffer problems. However, the factors that give rise to equi-librium are often not addressed.

Physical chemistry courses focus on the connection betweenequilibrium and thermodynamic quantities, particularly theGibbs free energy. Equilibrium is recognized as a minimum inthe Gibbs free energy vs reaction progress curve for reactions atconstant temperature and pressure (1–3). However, the originof that minimum remains largely unaddressed, although therecent edition of the text by Atkins recognizes the importanceof mixing (2). The enthalpic and entropic contributions (thesource of the minimum) are not separated; thus an opportunityis missed to reinforce the concept of entropy—specificallyentropy of mixing—and its importance. We have found thefollowing example, the dimerization of NO2, to be very helpfulin making these abstract contributions more concrete. De-pending on local interest, other reactions can be substituted.The two essential characteristics that enable explicit calculationof the Gibbs free energy minimum as well as equilibriumconcentrations are that the reaction takes place in the gasphase and that ∆G θ for the reaction is relatively small. It helpssimplify the calculation if the number of reactants and prod-

ucts is also small. The dimerization of NO2 fulfills all theserequirements and has the added benefit of being familiar tomany students either from laboratory exercises (4, 5) or lecturedemonstrations.

A moderate ∆G θ is important because it results in anequilibrium constant on the order of unity and an easily iden-tifiable minimum in the plot of ∆G versus reaction progress.Thermodynamic data for dimerization of NO2 are given inTable 1. Notice that the reaction’s ∆G θ, {4.73 kJ,1 is moderate.Both reactants and products are gases, and to a good approxi-mation, they may be treated as ideal gases as long as the totalpressure is on the order of an atmosphere or less. In this case,all the components of ∆G can be calculated explicitly and theorigin of the minimum is readily identified.

Imagine the reaction as follows. The initial stage consistsof two moles of NO2 at one atmosphere pressure in the lefthalf of a cylinder divided by a transforming partition. Thetransforming partition merely changes 2NO2 into N2O4 withno other change. The completion stage has one mole of N2O4in the right half of the cylinder at one atmosphere pressure.Any intermediate stage has 2 – 2x moles of NO2 on the leftand x moles of N2O4 on the right, both at one atmosphere.As shown below, the equilibrium balance of NO2 and N2O4occurs when x = 0.82. Equilibrium actually occurs when thepartition is removed and the gases mix. This is illustrated inFigure 1, which also lists results discussed below.

Figure 1. Schematic of system discussed in text.

52taataDcimanydomrehT.1elbaT 88888C

eluceloM ∆Gfθ/ lomJk {1 ∆Hf

θ/ lomJk {1 S θ J/ K {1 lom {1

ON 2 )g( 13.15 81.33 60.042N2O4 )g( 98.79 61.9 92.403NOTE: Data from ref 6.

Page 2: Why Equilibrium? Understanding Entropy of Mixing

In the Classroom

1392 Journal of Chemical Education • Vol. 76 No. 10 October 1999 • JChemEd.chem.wisc.edu

Since the molecules of an ideal gas do not interact,∆Hmixing = 0 and the nonmixing enthalpy change is a linearfunction of reaction progress. Specifically, letting x be the re-action progress (i.e., the number of moles of N2O4), x = 1corresponds to complete conversion of two moles of NO2 toone mole of N2O4. Then enthalpy as a function of reactionprogress is

∆Hinter(x) = [2 × 33.18(1 – x) + 9.16x]kJ (1)

∆Hinter(x) is plotted in Figure 2. This ∆Hinter is ∆Hf of thecomponents of the mixture, not ∆H for conversion of somefraction of reactants to products: ∆Hreaction = ∆H(1) – ∆H(0).Notice that ∆Hinter is a linear function of reaction progress.

The entropic contribution to reaction Gibbs free energyis separated into two parts: that due to the difference in theentropy of the reactants and the products, and that due tomixing. Like enthalpy, the nonmixing contribution to theentropy is a linear function of reaction progress.

S nθonmixing(x) = [2 × 240.06(1 – x) + 304.29x] J K{1 (2)

Since ∆G = ∆H – T∆S, entropy as a function of reactionprogress is multiplied by minus T, {T S n

θonmixing as a function

of reaction progress is plotted in Figure 3.The sum of two linear functions is also linear. Thus the

nonmixing portion of ∆G is a linear function of reactionprogress, as shown in Figure 4, and without mixing, reactantssimply slide down the ∆G slope to products and there is noequilibrium.

This picture changes when mixing is added. The enthalpychange contains no contribution from mixing because reac-tants and products are ideal gases, for which ∆Hmixing = 0.However, the entropy of mixing is nonzero. Entropy of mixingis a classic quantity calculated in most physical chemistry textsand is

∆Smixing = {nR(XA ln XA + XB ln XB) (3)

where n is the total number of moles of gas and XA and XBare the mole fractions of gases A and B. To apply eq 3 to theproblem at hand requires expressing the number of moles ofNO2 and N2O4 in terms of reaction progress

n = nNO2 + nN2O4 (4)

nN2O4 = x; nNO2

= 2 – 2x; n = 2 – x (5)

The entropy of mixing versus reaction progress is

∆Smixing(x) = 2 – x R

2 – 2x2 – x

ln2 – 2x2 – x

+ x2 – x

ln x2 – x

(6)

This function times {T is plotted in Figure 5. Two observa-tions are in order. First, {T∆Smixing is not very large. Second,the position of the minimum in {T∆Smixing depends on thenumber of moles of reactants versus products—that is, thereaction stoichiometry. In this example, the number of molesof reactants is greater than the number of moles of products;hence the minimum is on the product side (>50% completion).

Addition of {T∆Smixing to the nonmixing ∆G results in aminimum in the Gibbs free energy as a function of reactionprogress, as shown in Figure 6. With this plot, students easilysee the minimum in the Gibbs free energy. The minimumoccurs at about 0.8 mol N2O4 produced. This is a veryconcrete illustration of the statement that chemical reactions

0.00 0.50 1.00

Reaction Progress / mol N2O4

–2

0

–4

–T x

∆S m

ixin

g /

kJ

Figure 5. Mixing contribution to reaction entropy × ({T ) (298.15 K)as a function of reaction progress.

0.00 0.50 1.00-85

-80

-75

Reaction Progress / mol N2O4

∆Gno

nmix /

kJ

Figure 4. Nonmixing ∆G as a function of reaction progress.

0.00 0.50 1.00

Reaction Progress / mol N2O4

∆Hin

ter /

kJ

0

35

70

Figure 2. Enthalpy as a function of reaction progress for dimeriza-tion of NO2.

0.00 0.50 1.00-160

-140

-120

-100

-80

Reaction Progress / mol N2O4

–T x

(Sno

nmix) /

kJ

Figure 3. Nonmixing contribution to reaction entropy × ({T )(298.15 K) as a function of reaction progress.

Page 3: Why Equilibrium? Understanding Entropy of Mixing

In the Classroom

JChemEd.chem.wisc.edu • Vol. 76 No. 10 October 1999 • Journal of Chemical Education 1393

roll to a minimum in the chemical potential just as a ball comesto rest at a minimum potential energy.

This visual conclusion about equilibrium can be confirmedby calculating the actual equilibrium concentrations. From

∆G θ = {RT ln Keq (7)

the equilibrium constant is 6.74. Note that Keq is KP, sincethe ∆G θ are for gases. Solving for the number of moles atequilibrium, nNO2 = 0.36 and nN2O4 = 0.82 at equilibrium,just as determined from the graph. NOTE: For the mathemati-cally inclined, the equilibrium amounts can also be calcu-lated explicitly by differentiating ∆G with respect to x andsetting the result equal to zero (equilibrium condition):

0 = d∆Gdx

= ∆Hθ– T ∆S

θ+ RT ln KP = ∆G

θ+ RT ln KP (8)

Doing this explicit calculation with an actual reactionmakes the concept of equilibrium more concrete. Furthermore,

0.00 0.50 1.00-85

-80

-75

Reaction Progress / mol N2O4

∆G /

kJ Minimum

Figure 6. Reaction Gibbs free energy vs reaction progress for dimer-ization of NO2.

the concept of mixing entropy and its role in equilibriumbecomes more tangible. To cement these concepts, studentsare asked to produce graphs such as those in Figures 2–6 atother temperatures. The enthalpy plot is unchanged, thenonmixing entropy slope increases with temperature, and thedepth of the minimum in the mixing entropy plot increases.The net result is that the minimum slides toward the reactants,consistent with the exothermicity of this reaction.

We have found that this calculation for a real systemhelps students understand entropy of mixing and the originof equilibrium. These concepts are further reinforced if studentsboth determine the temperature dependence of the equilib-rium constant in the laboratory and graphically evaluate theminimum in a plot of the Gibbs free energy as a function ofreaction progress at several temperatures.

Note

1. From ∆G θ = ∆H θ – T∆S θ, ∆Gθ = 4.78 kJ/mol. The differencebetween this calculated ∆Gθ and the reported value reflects uncertaintyin the experimental measurements. See ref 6.

Literature Cited

1. Alberty, R. A.; Silbey, R. J.; Physical Chemistry, 2nd ed.; Wiley:New York, 1997.

2. Atkins, P. Physical Chemistry, 5th ed.; Freeman: New York, 1994.3. McQuarrie, D.; Simon, J. Physical Chemistry: A Molecular Ap-

proach; University Science Books: Sausalito, CA, 1997.4. Shooter, D. J. Chem. Educ. 1993, 70, A133–A140.5. Hennis, A. D.; Highberger, C. S.; Schreiner, S. J. Chem. Educ.

1997, 74, 1340–1342.6. Wagman, D.; Evans, W. H.; Parker, V. B.; Schumm, R. H.; Halow,

I.; Bailey, S. M.; Churney, K. L.; Nuttall, R. L. The NBS Tables ofChemical Thermodynamic Properties: Selected Values for Inorganicand C1 and C2 Organic Substances in SI Units; American Chemi-cal Society and the American Institute of Physics: Washington,DC, 1982; Vol. 11.