chemical principle/revi/ited_Chemical Equilibrium

I. The Thermodynamic Equilibrium Constant

Adon A. GordusThe University of Michigan, Ann Arbor, Ml 48109

Muriel Boyd BishopClemson University

Clemson, SC 29631

In this series of eight articles on chemical equilibrium weexamine how various assumptions, sometimes unstated, re-sult in the simplified equilibrium equations found in mostintroductory texts. In some cases we begin with rigorousthermodynamic expressions and show the conditions underwhich the approximations are valid. In other cases we intro-duce new methods of evaluating the approximations. Thisseries of articles consists of the following:

I. The Thermodynamic Equilibrium ConstantII. Deriving an Exact Equilibrium Equation

III. A Few Math TricksIV. Solutions of Weak Acids or BasesV. Seeing an Endpoint in Acid-Base Titrations

VI. Buffer SolutionsVII. pH Approximations in Acid-Base Titrations

VIII. Precipitates

In this article we consider the general nature of the equi-librium constant: K, Le Chatelier’s principle,1 and the effectof temperature on K.

The Equilibrium ConstantFor the specific case of a chemical reaction consisting of

ideal gases the thermodynamic equations for the Gibbs freeenergy result in a simple expression for chemical equilibrium(the equilibrium constant). The equation is:

AG° = —RT In K (1)

where AG° is the standard-state Gibbs free energy changefor the ideal gas reaction at the chosen absolute tempera-ture, T, with R the gas constant and K the thermodynamicequilibrium constant, for which values are listed in hand-books and texts.

The thermodynamic equilibrium constant for an ideal gasreaction involves terms that are ratios of pressures: P/P°where P° is the standard-state pressure, defined2 as P°= 1.00atm. For instance, K for the idea-gas reaction: 2N02(g) -*

N204(g) is:

pk,oA-°° atmK  -—-- (2)

(PNo2/1.00 atm)2

Because the equilibrium constant is dimensionless, pres-sures2 must be given in the same units as the standard state,which is usually chosen as atmospheres.

If the reaction involves real (i.e., nonideal) gases or, forexample, ionic species, the equation relating AG° to mea-sured equilibrium properties (pressures of gases or concen-

trations of ions) becomes very complicated (if it can even besolved) and the simple form of eq 1 does not apply.

For example, if the gases are only slightly nonideal so thatthey obey the equation: P(V — nb) = nRT, where b is thevolume per mole of gas molecules, then the Gibbs free energyexpression for the 2N02 —  N204 equilibrium becomes:

AG° = —RT In K + 26no„(I.OO atm -

PNC,2)

+ ^Njo/^NjO.,— 1-00 atm) (3)

where K is the same as in eq 2 if the standard state is definedas 1.00 atm. If the gases obey the van der Waals equation, thefinal AG° equation is even messier. And, for substances ormixtures that deviate greatly from nonideality (compared togases, ionic solutions are extremely nonideal), the equilibri-um expression for AG° is even more of an algebraic mess.

Therefore, to maintain the simple form of eq 1, a new

term, activity, assigned the symbol a, is defined,3 *and thethermodynamic equilibrium constant is written in terms ofactivities. For example, the equilibrium constant for thereaction: AgCl(s) -* Ag+ + Cl- would be written as K =

OAg+acr/oAgCi-The next step is to specify how to calculate activities. We

consider four types of chemical species: gases, nonionic sol-utes, solids (e.g., precipitates), and ionic species.

1 There Is some Justification In avoiding discussion of the vaguelyworded Le Chatelier principle, as has been discussed by variousauthors (see for example: Kemp, H. R. J. Chem. Educ. 1987, 64, 482-484, and references therein), and instead use a thermodynamic ap-proach. We have done both because most texts still refer to the LeChatelier principle.

2 The standard state In SI pressure units of Pascal Is usually definedas 1.00 bar = 105 Pa and the pressures are also given In bar. Thedifference Is minor because 1 atm = 1.01325 bar.

3 As Is pointed out In some physical chemistry texts, the Importantstep in the derivation of eq 1 Involves evaluating the Integral of V dPfor each species. Because V = nRT/P for an Ideal gas, the derivativebecomes nRTd In P. To maintain the simple form of eq 1 for nonidealsubstances there i9 defined. VtiP= nRTd In a to parallel the nRTd In Pexpression for an Ideal gas. In Integrating this equation, an activityratio, a/a°, Is obtained, where the activity in the denominator, a°, Iscalled the standard-state activity and Is defined as = 1.00. This Issimilar to the Ideal-gas pressure ratio: PIF6 = P11.00 atm. For Ions, ela° = [X]f«/1.00 where fx Is the activity coefficient for Ion X and a° is a

hypothetically Ideal 1.00 M solution of the Ion. The number 1.00 isalways chosen because it, In effect, drops out of the equations.

138 Journal of Chemical Education

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GasesMost gases around room temperature and at pressures up

to a few atmospheres deviate only slightly from the ideal gaslaw so that their pressure (in atm) can be used in K.

Nonionic SolutesMost nonionic solutes such as undissociated weak acids or

bases, water-soluble organic compounds such as sugar, ethylalcohol, etc,, behave nearly ideally, often at concentrationsup to 0.1 or even 1.0 M, so that their molar concentration canbe used directly in K.

Pure Solids or LiquidsPure solids (precipitates, for instance) or pure liquids are

a special category for which it is generally possible to calcu-late the activity. If the volume per mole, V, of the pure solidor liquid is assumed to be independent of pressure, then:

V(P - 1.00 atm) = R T In (a/<2°) (4)

where a° is the standard-state activity of the pure solid or

pure liquid and is defined as = 1.00; the standard-statepressure on the solid or liquid is defined as 1.00 atm.

The pressure independence of volume is usually a verygood approximation. At room temperature and 100 atm thevolume of a typical solid or liquid differs by less than 1%from the volume at one atm; water, for instance, is about as

incompressible as is steel. As an example, consider the case

of AgCl(s). Its density is 5.56 g/mL, and its molecular weightis 143.3; therefore V = 25.8 mL/mol. Shown in the table arethe values of the activity of AgCl(s) for various pressurescalculated using eq 4.

Activity of Solid AgCI at 298 K and Various Pressures

Pressure, atm. 0.000 1.000 2.000 5.000 10.000 50.000Activity, a 0.999 1.000 1.001 1.004 1.010 1.053

As seen in the table, the calculated activity of AgCl(s),which would be typical of other pure solids or pure liquids,begins to differ markedly from 1.00 only for very high pres-sures. As a result, the activity of a pure solid or liquid, forpressures up to a few atmospheres, can be approximated as

equal to 1.00 in the K expressions. For this reason, the termscorresponding to solids are omitted from solubility productequations. They are present but equal to 1.00. However, forthe K equation to be valid, there must be some pure solidpresent. (Of course, appreciable impurities in the solid willaffect the activity of the solid.)

Ionic SpeciesIonic solutions show the greatest deviations from ideality,

even at concentrations as low as 0.001 M. For any ion, X, theactivity is defined as a\ = [X]/x, where the concentration isgiven by molarity and f\is called an activity coefficient;4 it isdimensionless. Thus, for H+ ions, an+ = [H+]/h+. [Note thatthe precise definition of pH = —log oh+.] As a result, theactivity terms in the thermodynamic equilibrium constantcan be separated into a quotient, Kc, based on the concentra-tion terms and another quotient, Kf, based on activity coeffi-cients so that:

K = KCK, (5)

If a solution is assumed to be ideal, then all/terms as wellas Kf equal 1.00 and K = Kc, which is implicitly assumed inmost introductory texts. There are various theoretical ex-

pressions for / values of ions and these include the extendedDebye-Huckel equation and the Davies equation.5 In gener-al, the greater the charge on an ion or the higher the ionicconcentration of the solution, the greater the degree of noni-deality of the ion.

For instance, consider a solution that is both 0.015 MNa>HP04 and 0.010 M NagPC^. The principal equilibriumcan be described in terms of the third step in the dissociationH;P04 (HPC>42_ —*• H+ + P043-) for which the thermody-namic K value is:

K3 = 4.80 X 10“13 =

 HPOf

[H+][PQ4~] Wpoj-[Hpon

xfHpo-

= K,KV

(6)

For this solution, the extended Debye-Hiickel equationresults in /h+ = 0,824, /po4s- = 0.0911, and /hpo42~ = 0.345, sothat Kzf = 0.218 and K3c = K3/K3f = 4.80 X 10-13/0.218 =

2.20 X 10~12. Although there is some degree of cancellation ofthe / terms in Kf, in general, Kf ^ 1.00 for most ionic solu-tions so that precise equilibrium calculations must take intoaccount nonideal ionic effects. Certain equations, however,do not include activity coefficients no matter how nonidealsolutions may be. Mass balance and charge balance expres-sions are two examples. The reason is that activity is definedto maintain the simplicity of the ideal-gas thermodynamicequilibrium-constant equation and, therefore, activity coef-ficients occur only in equilibrium constant expressions.

Almost all general chemistry texts consider solutions to beideal and the concentrations calculated from equilibriumexpressions can often be in error by as much as 10-100% ormore. In subsequent articles we will usually assume thatsolutions are ideal, although we will occasionally show hownonideality can be included in equilibrium calculations.

Le Chatelier's PrincipleThe principle of LeChatelier1 states that when a chemical

system at equilibrium is subjected to a stress, the reactionwill reequilibrate in the direction that will relieve that stress.The principle is most easily understood by examining thereaction quotient, Q, for a chemical reaction. The quotienthas the same form as the equilibrium constant with concen-trations or pressures of reactants (terms in the denominator)and products (terms in the numerator) each raised to a

power based on the stoichiometry of the balanced equation,as they are in K. However, Q is not necessarily numericallyequal to K since Q represents the actual, not necessarilyequilibrium, concentrations and pressures.6 ***If Q = K, thenthe reaction is at equilibrium. If Q < K, reequilibration mustoccur in the direction that will cause Q to become equal to K.This will happen when the numerator (product terms) in-creases numerically while the denominator (reactant terms)decreases; this corresponds to a reequilibration wherebysome of the reactants are converted to products. Conversely,if Q > K, the numerator (product terms) must decrease whilethe denominator (reactant terms) must increase; this corre-

sponds to a reequilibration whereby some of the productsare converted to reactants. The only terms that are uneffect-ed, if present in Q and K, are activities for solids or pureliquids because their activities equal 1.00 as long as there issome solid or pure liquid present.

4 Unfortunately, there is no way experimentally to measure single fvalues for ions. The reason is that you cannot create a solution havingonly a single ionic species. All Ionic solutions are electrically neutraland have both positive and negative ions. Thus, any measure of thedegree of nonideality of an ionic solution will give information only onthe (weighted) average activity coefficients of the various ions in thesolution. These activity coefficients are often listed for use withconcentrations in molality; they can be converted to / values for usewith molarities.

5 These equations are usually described in undergraduate physicalchemistry texts.

6 For example, K = 7.01 at 25 °C for the 2N02(g) -*  N204(g)equilibrium given by eq 2. If a mixture is prepared so that initially Pn2o4= 0.200 atm and PN02

= 0.300 atm, then Q =Pn2o4/(Pno3)2

= 0.200/(0.300)2 = 2.22.

Volume 68 Number 2 February 1991 139

For example, in the ionization reaction of acetic acid: HAc—* H+ + Ac-, the addition of some NaAc to an equilibriummixture results in excess [Ac-], so that temporarily Q > Kand some of the products are converted to the reactant. Theaddition of NaOH, which will react with H- and thus de-crease its concentration, will result temporarily in Q < K so

that, to reestablish equilibrium, some of the reactant mustdissociate and produce products.

The exception to this general rule exists for the case inwhich there could be precipitation. If Q is less than thesolubility product, Ksp, and if there is no solid present, thenreactant cannot be converted into products because there isno reactant available. However, if Q is greater than Asp thenproducts can be converted into reactant and some solidprecipitate will form.

Volume ChangesA “stress” can also be applied by changing the volume of

the reaction mixture, and the effect can also be determinedby comparing the values of Q and K. The net effect inreestablishing equilibrium is always in the direction towardthe side having the larger number of free particles when thevolume is increased or, conversely, toward the side havingthe smaller number of free particles when the volume isdecreased. For instance, if an equilibrium solution of HAc isdiluted with water (i.e., the volume increased), the reactionHAc —H+ + Ac- will proceed in the forward directionbecause there are two particles as products, H+ and Ac-, butonly one as reactant, HAc. We can show this by defining theoriginal equilibrium concentrations as [H+]e, [Ac ]e, and[HAc]e. When the solution is diluted, the “temporary con-centrations” are x[H+](., x[Ac-]e, and x[HAc]e, where x is anumber less than 1.00. Thus, Q = (x[H+]e)(x[Ac-]e)/x[HAc]e= xK. Because x is less than 1.00, Q < K and the direction ofthe reequilibration is from reactants to products. (If thenumber of free particles is the same on both sides of theequation, the x terms cancel, Q = K, and there is no need toreestablish equilibrium because of a change in volume.) Thiseffect of a change in volume also applies to reactions involv-ing gases. An increase in volume of the reaction vessel resultsin the reequilibration proceeding in the direction toward theside having the larger number of gaseous molecules, and viceversa.

This direction of shift in the acetic acid equilibrium mayappear to contradict what would be suggested intuitively.After all, dilution of an acid with water decreases the acidity,

7 There are possible exceptions to this general rule for the effect ofheat, but there are no common examples. The exceptions wouldoccur in cases where AhP is a very, very small value, so that atemperature increase would result in only a very small change in Kbutwould be more than compensated for by (1) for ionic reactions, thedilution effect of the volume expansion of the solution or (2) for gas-phase reactions, the pressure increases that result from the tempera-ture increase.

8 An "apparent” inconsistency in this rule would appear to exist forthe case of the addition of a substance such as solid NaOH to water inwhich heat is generated when the NaOH dissolves. This suggests thatheat is a product and that the amount of NaOH that dissolves shoulddecrease as the temperature is raised. As everyone knows, thesolubility of most substances, including NaOH, increases with tem-perature. The inconsistency in the reasoning is related to the fact thattwo separate processes are being muddled in the "solubility" descrip-tion given here. What one needs to do is to examine the processwhereby solid NaOH is in contact with a saturated solution of NaOHthat is in equilibrium with Na+ and OH-. This reaction is endothermic,and an increase in temperature does result in an increase in solubility.The solubility of NaOH is discussed by Bodner, G. M. J. Chem. Educ.1980, 57, 117-119, whereas the general concept of the effect oftemperature on solubility is discussed by Brice, L. K. J. Chem. Educ.1983, 60, 387-389.

[H+], of the solution and eventually, with sufficient dilution,the solution simply takes on the properties of water and thepH approaches 7.0. This is true. What the Le Chatelierprinciple says is that it is the total amounts of H+ and Ac- insolution that increase while the total amount of HAc de-creases. The concentrations are another matter. When HAcis diluted, all three concentrations decrease, but after re-

equilibration the HAc concentration has decreased morethan the H+ and Ac- concentrations. In other words, there isa greater degree of ionization of HAc. For example, 0.100 MHAc is about 1.3% ionized, whereas 1.00 X 10-4 M HAc isabout 34.4% ionized.

Temperature ChangesAnother type of stress that can be applied to a chemical

reaction is a change in temperature. The general rule thatcan be followed in evaluating the Le Chatelier stress is tothink of heat as a reactant (if the reaction is endothermic) or

a product (if the reaction is exothermic). For instance, if thetemperature of an endothermic reaction initially at equilib-rium is increased, then some of the reactants will be convert-ed to products. This phenomenon can be justified by exam-

ining the thermodynamic equation that relates the equilibri-um constant to the absolute temperature, T, and the heat ofreaction, AH°. The unintegrated form is: d(ln K)/dT = AH0/RT2 or, by approximating derivatives as difference terms,A(ln K)t AT « AH°!R T-, If the reaction is endothermic, AH0is positive, and the derivative is also positive; therefore, anincrease in T (i.e., AT is positive) results in A(ln K) — a

positive value, which is the same as an increase in K, requir-ing a decrease in reactants and an increase in products. If thereaction is exothermic, AH0 is negative, and the reverseholds.78

Temperature Dependence of KAny reaction that involves the absorption or release of

heat will have an equilibrium constant that changes in valueas the temperature is changed. If the (standard-state) heatof reaction, AH°, is assumed constant in the temperaturerange of interest, then the integrated form of the expressionrelating log K to AH0 and the absolute temperature T is:

log K =AH'1

2.303RT+ C (7)

Consider the case of the ionization of water, which is anendothermic reaction: heat + HaO  * H+ + OH-. Using thehandbook value of AH0 = 56.08 kJ/mol for the temperaturerange 288-313 K (15-40 °C) and the value of K„ = 1.00 X10-14 at 298 K, you can solve for the constant C in eq 7. Theresulting equation is:

log Kw = 4.172 (8)

Because the ionic strength of pure water is very low, thesolution can be considered ideal and, because [H+] = [OH-],we have pH = pOH = (l/2)pXB. = —(l/2)log Xw so that theanalogous expression for the temperature dependence of thepH of pure water (in the range 288-313 K) is:

pH = ~ + 2.086 (9)

Because the water ionization reaction is endothermic, anincrease in temperature causes the reaction to proceed to theright, as predicted by Le Chatelier’s principle, and results inhigher H+ and OH- concentrations. The pH of a neutralsolution, therefore, decreases as the temperature is in-creased. For instance, at 37 °C = 310 K (body temperature),Kv, = 2.4 X 10-14 according to eq 8, and the pH of pure water,according to eq 9, is 6.81.

140 Journal of Chemical Education