Helmholtz Free Energy Minimization for a Chemical Reaction

Transcription

1 Helmholtz Free Energy Minimization for a Chemical Reaction JOSÉ C. ÍÑIGUEZ Retired Instructor of Science and Mathematics 17 1st Street, Douglas, AZ 85607, USA UNPUBLISHED PAPER: Please direct your comments to: iniguez.jose@gmail.com Abstract The equilibrium condition for an isothermal and isometric chemical reaction is located via the minimization of its Helmholtz free energy written in terms of the reaction s degree of advancement. Quantitative confirmation for the different conversions arising in the constant T-V and constant T-P versions of this reaction is obtained from Le Chatelier s Principle. Additional comments on the spontaneity of chemical reactions, the chemical potential, and the connection of thermodynamic work with the Helmholtz and Gibbs free energy changes are also provided. Keywords: chemical education, thermodynamics, chemical equilibrium, ideal gases, Helmholtz free energy, Le Chatelier s Principle, spontaneity, chemical potentials, chemical work, chemical force. 1 Introduction The most general criteria for spontaneity and equilibrium offered by thermodynamics are in terms of entropy (S. In the words of Lewis and Randall [p. 138] any actual, or irreversible, process is characterized by an increase in the total entropy of all the systems concerned On the other hand, a state of equilibrium is one in which the total entropy remains constant Applied to situations dealing with the system of interest evolving either at constant temperature and volume, or at constant temperature and pressure, the entropy-centered criteria find respective re-expression in terms of the Helmholtz (A, and Gibbs (G free energies. In the words of Denbigh The criterion of equilibrium of a system enclosed in a rigid container and held at constant temperature is that A has reached its minimum value [Denbigh, p.67]; The criterion of equilibrium of a system which is held at constant temperature and pressure is that G has reached its minimum value [Denbigh, p. 69]. The shift from entropy to the free energies finds explanation in the following three reasons: 1 that these alternative criteria make exclusive reference to properties of the system of interest; that in the situations described they can be physically interpreted in terms of work, and 3 that the restrictions associated to either A or G are common in the laboratory and industry. Given in reference to the thermodynamic variable known as degree of advancement or reaction coordinate ξ (mol, the said criteria find the following mathematical expressions in which the inequality refers to spontaneity and the equality to equilibrium. ds / d 0 (1 ( A /, 0 ( T V ( / T, P 0 G (3

2 The use of the Gibbs free energy as a spontaneity and equilibrium criterion for chemical reactions (as well as other thermodynamic processes has, over the years, been profusely illustrated in the chemical thermodynamics literature [Bevan Ott & Boerio-Goates, Chapter 9, 000], [Klotz and Rosenberg, Chapter 8, 1986], [Castellan, Capítulo 11, 1974], [Moore, Chapter 9, 197], [Pimentel and Spratley, Chapter 5, 1970], [Hill, Chapter 10, 1966], [Wasser, Chapter 6, 1966]. Similar discussions using the Helmholtz free energy are, however, apparently absent from every textbook on the matter. Even if the omission might be explained on reason of the sole temperature dependence of the thermodynamic equilibrium constant, it is by no means justifiable as it sidelines an outstanding opportunity for students of thermodynamics to get a better understanding of the free energies, of their mutual relationship, and perhaps most important, an opportunity to verify that the magnitude of the thermodynamic equilibrium constant for an isothermal chemical reaction is independent of any other restriction imposed to its path, as well as to understand the reason why these differently constrained yet isothermal paths, characterized by the same equilibrium constant, may bring different equilibrium concentrations for the species involved. It is with these considerations as base that we present here a thermodynamic argument centered on the Helmholtz free energy to identify the equilibrium condition of a gas-phase chemical reaction taking place at constant temperature and volume. For comparison purposes the equilibrium conversion under constant temperature and pressure conditions is also calculated. Quantitative confirmation for the different equilibrium conversions arising in one case and the other is provided via Le Chatelier s Principle. The reaction system Let us consider the stoichiometric, ideal gas (g, isothermal (constant temperature, isometric (constant volume, and irreversible (entropy producing chemical reaction written below bb(g, T, V yy(g, T, V (4 Chemical reaction (4 takes place inside a rigid vessel of volume V immersed in a heat bath of temperature T. Initially the vessel contains b moles of species B. Here T stands for the thermodynamic temperature measured in Kelvin (K. The general equations obtained via reaction (4 will be eventually adapted to the dissociation reaction of molecular chlorine Cl (g = Cl (g at 000 K. The evolution of reaction (4 will be followed with the previously introduced degree of advancement ξ (mol, also known as the extent of reaction, defined for the case at hand, via the following expression [Moore, pp ] Δn B/b Δn Y/y (5 In the previous expression ΔnB and ΔnY stand for the change in number of moles of species B and Y at any given moment in the reaction, and b and y for the respective stoichiometric coefficients of these species. Equation (5 in combination with the facts that b and zero are the respective initial number of moles of B and Y leads to the following equations for the respective numbers of moles of these species, nb, ny, as well as for the total number of moles, ntot, at any given ξ: nb b(1, ny y, ntot b(1 y b n (6

3 Here Δn = yb. The previous equations allows us to realize that while the initial condition of the reaction, given by nb = b and ny = 0, corresponds with ξ = 0, complete conversion, represented by nb = 0 and ny = y, corresponds, on its part, with ξ = 1. For any other condition of the reaction mixture we will have that 0 ξ 1. The vessel and the reaction mixture by it contained constitute the system of interest. The heat bath constitutes, on its part, the surroundings. System and surroundings combined define the universe of the process. Note that no mechanical reservoir is included in the universe. The reasons for this can be traced to the following two facts: 1 that no work different from PV-work can be exchanged by any system evolving in an irreversible fashion, and that no PV-work can be exchanged by a constant volume system such as the one here being considered. Combined these two constraints hinder our system of any exchange of energy in the form of work with its surroundings. 3 Modeling the reaction The thermodynamic argument leading to the construction of the functional relation between the Helmholtz free energy (A for reaction (4 and its degree of advancement (ξ will make reference to the situation depicted in Figure 1. bmol P 1bar T V (i b(1 mol P 1bar T, V B I II III y mol P 1bar T, V Y b( 1 mol y mol T P B PY 1 V V ( j (k B V bar Y b( 1 mol y mol T P P B P Y V V B V Y ( f Figure 1. The irreversible, constant temperature and constant volume evolution of chemical reaction (4 from its initial condition ( i, characterized by ξ = 0, to condition ( f, characterized by any other of its allowed ξ, is shown here taking place, for illustrative and calculation purposes, through the concatenation of processes I, II, and III. Process I corresponds to the ξ-fraction of the standard reaction; process II to a mixing process, and process III to an isothermal, volumechanging process taking the volume of the mixture back to its initial value V. The value of the Helmholtz free energy for the final mixture is thus determined by the combination of the Helmholtz free energy associated to the initial condition represented by b moles of B at unit pressure and temperature T, plus the Helmholtz free energy changes of processes I, II, and III. In (k and (f the pressures and volumes are respectively given in terms of partial pressures and partial volumes. In it state (i represent the initial condition of the reaction system and state (f the condition at any given moment along its evolving path. The problem to be solved with the aid of this reaction scheme is that of obtaining an expression for the Helmholtz

4 energy change of the transition (i to-(f. In order to do this is that processes I, II and III have been introduced. Their combined Helmholtz energy changes will provide that of the said transition. It needs to be pointed out that the reaction scheme detailed in Figure 1 is only an analytical tool devised for calculation purposes and in no way implies that the chemical reaction actually takes place in the way indicated. The validity of the scheme rests on the path-independence of state functions. As long as the initial and final conditions are the same, any path connecting them will bring the same value for the change of the state function being considered. The particularity of this scheme is that the conversion reached in the first step through the standard reaction is kept constant along steps II and III by the presence of a perfect anticatalyst [Kestin, p. 7]; in other words, these steps take place at constant composition. As will be seen below, the particular choice of steps represented in Figure 1 leads to easily and quantifiable changes in A. 4 The minimization of the Helmholtz free energy The step designated as I in Figure 1 corresponds to the ξ fraction of the standard reaction taking place at temperature T. The standard qualifier reflects the fact that starting with pure reactants we get pure products, all at the standard pressure of P = 1 bar at temperature T [Denbigh, p. 148]. It is in reference to these standard hypothetical reactions that the pertinent experimental thermo-chemistry data is actually reported. With it as base the changes experienced by the relevant thermodynamic properties at other conditions can be calculated. As such it is an essential element in the thermodynamic analysis of chemical reactions. In the case at hand, apart from allowing us to link the reaction scheme to the experimental information, it actually sets the stage for the definition of the sequence of processes II and III required to arrive at condition (f. For a given stoichiometric reaction at a given temperature a standard change in the Helmholtz free energy is a constant obtained as the difference of the combined and properly stoichiometric-coefficient weighted values of the standard Helmholtz free energies of formation of products minus the corresponding combination of the reactants. The fact that the standard Helmholtz free energies of formation are solely dependent on the temperature means that for any multiple n of the given reaction the said subtraction will differ from the previous one only in the value of the stoichiometric coefficients, whose magnitudes will now be n times the original ones. Under these conditions the standard Helmholtz free energy change will simply be n times the one previously obtained. The same argument can be applied to the standard changes in internal energy E, enthalpy H, Gibbs free energy G, and entropy S. If the standard Helmholtz free energy change for reaction (4 at temperature T is designated as ΔA o, then the change in A associated to the ξ fraction of this reaction will be the following A A (7 I The fact that it is the standard Gibbs energy change the one commonly quoted in the chemical literature provides us with a simpler path to the ΔA o of a given ideal-gas chemical reaction, this being the substitution in A G RTn of the standard Gibbs energy change at the appropriate temperature. Let us note that Step I is the only one in which there is an interaction between the reaction system and the heat bath. Serves to recall here that by definition

5 A ( E TS. In this expression E S stands for the entropy change of Step I, with, the standard internal energy change for step I, representing, additionally, the heat exchanged by the reaction system with the heat bath. The fact that heat given off by the system ( is heat taken in by the heat reservoir ( or vice-versa, allows us E E / T E to express the entropy change of the heat bath as. If we let represent the total entropy change for step I and as such given by the summation of the entropy changes of the system and the heat bath then S ( E / T, or equivalently, T S I A. In this last expression we see the total entropy change for Step I being re- expressed in terms of its standard energy change. The fact that subsumed in A we find the previously noted change experienced by the heat reservoir is the reason why it wasn t depicted in Figure 1. The mixing step designated as II in Figure 1 is performed by removing the wall separating volumes V B and V Y as represented in condition (j. The final condition of this process, represented as (k in Figure 1, is constituted by a mixture of the indicated amounts of B and Y confined to a volume equal to the combined volumes previously occupied by B and Y at the end of process I. The already noted fact that the Helmholtz free energy is defined as A E TS (8 allows us to write the following expression for the Helmholtz free energy change of step II A E TS (9 II mix The sub-index used in E and S emphasize the fact that the step in question is a mixing process. In order to evaluate this expression we will start by bringing here the fact that no change in internal energy accompanies the formation of ideal mixtures, in other words, that ΔEmix = 0 [Denbigh, p. 117]. On reason of this, equation (9 reduces to S I A II T S mix mix S I (10 The fact that in a non-reactive mixture of ideal gases remember our anticatalyst- every gas is a vacuum to every other gas [J. W. Gibbs, p. 157] will allow us to understand that the mixing process under consideration reduces to the simultaneous, intrinsically isothermal, free-expansions of ideal gases B and Y from their respective initial volumes V B and V Y to a final, common volume V B + V Y. The isothermal nature of ideal gases free-expansions come as a result of the facts that such processes proceed without the performance of work ( W 0 nor the exchange of heat ( Q 0. Under these conditions the first law reduces to E 0. The fact that the internal energy of ideal gases is a sole function of the temperature allows us to infer the constancy of the latter from the constancy of the former. Being this so we can make use of the well known result for the entropy change of an ideal gas free-expansion [Klotz and Rosenberg, p. 136] S nr ln ( V final /( Vinitial (11 to write ΔSmix as the summation of the entropy changes of the two free-expansions constituting process II, as follows

6 S mix R{b(1 ln [( VB VY / VB ] y ln [( VB VY / VY ]} (1 Application now of the ideal gas law to volumes V B, V Y, and V B + V Y allows us to form the following quotients VB /( VB VY nb / ntot xb, VY /( VB VY ny / ntot xy (13 In the previous equation xb and xy represent, respectively, the mole fractions of B and Y in the ideal mixture (k in Figure 1. Furthermore, the fact that in this mixture the total pressure is 1 bar allows us to realize that these mole fractions are numerically identical to the respective partial pressures of species B and Y in the mixture, i.e. The combination of equations (1, (13 and (14 leads to xb ( PB /1, xy ( PY /1 (14 S mix R[ b (1 ln (1/ P B y ln (1/ P Y ] (15 Finally, the substitution of equation (15 in (10 produces the following result for ΔAII: A II RT[b (1 ln ( P B /1 y ln ( P Y /1] (16 Process III is, on the other hand, responsible for bringing, in an isothermal fashion, the volume of mixture (k back to a volume V identical to that characterizing the initial condition of the chemical reaction. Let us in this regard note that depending on the relative magnitudes of the stoichiometric coefficients this step can be an expansion if y b, or a compression if the opposite is true. If this last situation was to be the case, our reaction scheme -which as previously noted is nothing more than a calculation devicedemands of this non-spontaneous process to take place by itself, just as represented in Figure 1, and in doing so to make a negative contribution to the total entropy change of the (i-to-(f transition. The presence of this non-spontaneous step in the reaction scheme doesn t makes the said transition nor the chemical reaction which it represents nonspontaneous, as spontaneity, or the lack of it, is determined by the combination of the entropy changes -or equivalently, of the Helmholtz free energy changes- of all the steps there involved. As long as the combined entropy changes of steps I, II, and III turns out to be positive, or the corresponding Helmholtz free energy change negative, the transition - and the reaction itself- will be spontaneous. This issue is further discussed in Section 5. Just as before done for process II, the Helmholtz free energy change for step III will be written as follows (17 AIII EIII T SIII The fact that the internal energy of ideal gases is a sole function of temperature [Denbigh, p. 113], allows us to conclude that for process III the following holds Under this condition equation (17 reduces to E III 0 (18 (19 AIII T S III The fact that process III reduces in the end to the isothermal changes of partial pressures or partial volumes- of B and Y from the values operating at (k to those they acquire

7 once at (f, means that the calculation of the entropy change associated to this step calls for the use of equation (11 expressed in terms of the respective quotients of partial volumes of B:, and Y:, or, equivalently, given the inverse relation between volume and pressure in ideal-gas isothermal processes, in terms of the (Pinitial/(Pfinal quotients of partial pressures of B: and Y: through an equation similar to equation (15, as follows V B / VB V Y / VY P B / P B P Y / P Y S III R[b (1 ln ( PB / PB y ln ( PY / PY ] (0 The expression of the entropy change of process III in terms of the separate contributions of B and Y demands a non-reactive mixture. This condition is here satisfied, as was done in Step II, by the anticatalyst. Substitution of equation (0 in (19 produces the following expression for the Helmholtz free energy change associated to process III A III RT [ b (1 ln ( PB / PB y ln ( PY / PY ] (1 According to the reaction scheme of figure (1 the Helmholtz free energy (A for the reaction mixture at (f will be given by A A A A A ( i I In the previous equation Ai is a constant representing the Helmholtz free energy of b moles of B at unit pressure and temperature T. The substitution of (7, (16 and (1 in equation ( leads to equation (3 below, and in doing so takes us a step closer to our intended goal of getting an expression for the Helmholtz free energy of the reaction mixture in terms of the degree of advancement ξ II o A Ai A RT [b (1 ln ( P B /1 y ln ( PY /1 ] (3 In order to get the expression sought we need to express the partial pressures of B and Y in equation (3 in terms of ξ. Let us start then by writing the following expressions for PB and PY III P B b (1 RT / V (4 P Y yrt / V (5 A quick inspection of the initial condition of the process, as depicted in figure 1, permits recognition of the fact that This result allows writing equations (4 and (5 as follows brt / V 1 (6 P B 1 (7 P Y y / b (8 Substitution of equations (7 and (8 in (3 leads us to desired functional relation between the Helmholtz energy function ( A of chemical reaction (4 and its reaction coordinate ξ, as follows o A Ai A RT [ b (1 ln (1 y ln (y / b ] (9

8 Defining ΔA as the difference AAi, lead us to the following alternative presentation of equation (9 o A A RT [b (1 ln (1 y ln (y / b] (30 The procedure leading to the identification of the equilibrium condition for reaction (4 starts by taking the first derivative of A, as given by equation (9, with respect to the reaction coordinate ξ. When this is done the following expression is produced o ( A / V, A RT (y b RT [ yln (y / b bln (1 ] (31 T Substitution of difference (yb by Δn, followed by the re-expression of the logarithmic arguments in terms of equations (7 and (8, allows us to write equation (31 as follows y y o (y / b o PY ( A / V, T A RT n RT [ ln ] A RTn RT[ ln ] (3 b b (1 PB If in accord with the equilibrium criterion given in ( we put the previous equation equal to zero, we will be able to identify the extremum of the A-vs-ξ graph as that point in which the following is true A o RT n RT ln K (33 In the previous expression, K, the thermodynamic equilibrium constant, replaces the particular value adopted at the extremum by the arguments of the logarithmic terms of equation (3, i.e. [y( eq T, V / b] y b K ( PY / PB b [1 ( ] eq T, V The (ξeqt,v term appearing in equation (34 represents the abscissa of the extremum. Recognizing now that the equation connecting the standard Gibbs and Helmholtz free energy changes can be written [Denbigh, p. 63] as o o G A (PV and that this equation, in the case of isothermal ideal gas reactions, becomes y eq (34 o o G A R T n (35 allows us, previous substitution of (35 in (33, to write the following expression for the thermodynamic equilibrium constant of reaction (4 o K exp ( G / RT (36 It should be noted that the procedure of imposing to equation (3 the equilibrium condition given in equation ( does not materialize in the identification of the abscissa (ξeqt,v of the extremum of the graph of A-vs.-ξ, but in an expression, which combining the messages of equation (34 and (36, takes the form K f [(, ] exp( G / RT (37 eq T V

9 from whose solution (ξeqt,v is to be identified. Thus obtained, this value allows, via its substitution in equation (9, for the identification of the ordinate A[(ξeqT,V] of the extremum, situation that permits writing the coordinates of this point as ((ξeqt,v, A[(ξeqT,V]. Further substitution of this equilibrium abscissa in equation (31 will confirm, using the prime notation for the first derivative, that A [(ξeqt,v] = 0. We can go back now to Figure 1 and take notice of the fact that the temperature and total pressure conditions of the reactive mixture at the end of process II correspond with those expected along a constant T-P evolution of reaction (4. In other words, only steps one and two are required to describe the isothermal and isobaric evolution of this reaction. This consideration allows us to write the following equation for the Gibbs energy of such a reaction o G G G G G G H T S (38 i I II i ( mix mix In equation (38 Gi stands for the Gibbs energy of the initial condition, and ξδg o for the Gibbs energy change associated to process I. Now, in parallel with our previous discussion of the constant T-V path, the Gibbs energy of mixing (ΔGII associated to process II has been written, according to the defining equation for G, namely G = H TS, in terms of the enthalpy and entropy of mixing The already mentioned fact that no thermal effect accompanies the formation of an ideal mixture takes here the following form ΔHmix = 0 [Denbigh, p. 117]. Furthermore, the fact made evident by figure 1 that the initial and final conditions of process II are the same for both, the constant T-V and the constant T-P paths, leads to the conclusion that the entropy change given by equation (15 holds for both of them, and if so that ΔAII = ΔGII = TΔSmix. Substitution of ΔHmix = 0, as well as of equation (15 in (38 produces o G Gi G RT [b (1 ln ( PB /1 y ln ( PY /1] (39 The fact that a unit total pressure holds for the constant T-P path makes the partial pressures of the reactive species numerically equal to their respective mole fractions, i.e. P B b(1 /(b n (40 P Y y /(b n (41 In the previous equation b + ξδn represents, as previously noted, the total number of moles at ξ, with Δn representing the stoichiometric difference in number of moles between products and reactants, Δn = y b. Based on this consideration, equation (39 can be written as follows o G Gi G RT {b (1 ln [b(1 /(b n] y ln [y /(b n]} (4 In terms of ΔG, defined as G Gi, equation (4 finds the following re-expression o G G RT {b (1 ln [b(1 /(b n] y ln [y /(b n]} (43 In order to identify in the graph G-vs.-ξ the extremum corresponding to the equilibrium condition for this constant T-P evolution of reaction (4, we will proceed to take the first derivative of equation (4 with respect to ξ. Performance of this operation leads to

10 y b(1 ( G / T, P G RT{yln ( bln [ ]} (44 b n b n Substitution of the arguments of the logarithmic terms by P B and P Y, as allowed by equations (40 and (41, permits the following re-expression of equation (44 [y /(b n] ( PY ( G / T, P G RT ln G RT ln [ ] (45 b b [b(1 /(b n] ( P If in accord with the equilibrium criterion given in equation (3 we put the previous equation equal to zero, we will be able to identify the extremum of the G-vs.-ξ graph as that point in which the following is true y o K exp ( G / RT (46 In the previous expression K, the thermodynamic equilibrium constant, replaces the particular value adopted at the extremum by the arguments of the logarithmic terms of equation (45, i.e. {[y( eq T, P ]/[b ( eq T, P n] y b K [( PY /( PB ] b {b[1 ( ]/[b ( n]} eq T, P eq T, P The facts that a common temperature applies to both of the evolving paths above considered, and that ΔG o, as previously noted, is a sole function of T, allows us to conclude, when comparing equations (36 and (46, that the equilibrium condition, as measured by the thermodynamic equilibrium constant K, is one and the same for the constant T-V and constant T-P paths considered. As long as the temperature is the same, the same thermodynamic equilibrium constant will apply for a given chemical reaction with independence of any other constraint imposed to its path. Note that the constancy of K does not necessarily imply the identity of the individual amounts of the participating species present at equilibrium in one case and the other, or equivalently, the identity of the respective equilibrium reaction extents. To illustrate this point let us start recognizing, as previously done in regard to equation (36, that imposition of the equilibrium condition given by equation (3 on equation (45 materializes in a function of the form y eq B y (47 K F[(, ] exp( G / RT (48 eq T P whose solution will produce the abscissa (ξeqt,p of extremum of the G-vs.-ξ graph, a value different than (ξeqt,v on reason of the different nature of the functions f [(ξeqt,v] and F[(ξeqT,P]; difference made evident by the right hand side terms of equations (34 and (47. Let us note here that if in reaction (4 we were to impose the condition b = y, or equivalently Δn = 0, then (ξeqt,p = (ξeqt,v, as the reader can easily prove by substituting the noted conditions in equation (47 and comparing the result with equation (34. In this case, only one equilibrium reaction extent will arise, this on reason of the graphs A-vs.-ξ and G-vs.-ξ coalescing into one. This situation can be understood by recalling that at constant temperature ΔG = ΔA + RTΔn. If Δn = 0, we will have ΔG = ΔA and such a chemical reaction will be simultaneously isobaric and isometric.

11 Substitution of (ξeqt,p in equation (4 leads, in turn, to the ordinate G[(ξeqT,P] of the extremum whose coordinates can now be written as ((ξeqt,p, G[(ξeqT,P]. Further substitution of this equilibrium abscissa in equation (44 will confirm, using the prime notation for the first derivative, that G [(ξeqt,p] = 0. Taking the second derivatives of equations (31 and (44 with respect to ξ, we get y(1 b A/ T V RT [ ] (49 (1 (, by G / T P RT [ ] (50 (1 (b n (, Substitution in the previous equations of the corresponding equilibrium degrees of advancements ξeq > 0, alongside b > 0, and y > 0 leads using the double prime notation for the second derivative- to A [(ξeqt,v] > 0, and G [(ξeqt,p] > 0, respectively. These results, combined with those previously obtained in the form A [(ξeqt,v] = 0 and G [(ξeqt,p] = 0, identify the extrema of the corresponding A-vs.-ξ and G-vs.-ξ graphs, as minimums [Larson, p. 188]. 5 A numerical example A classroom presentation along the lines above described wouldn t be complete without a numerical example in which the constant T-V and constant T-P approaches to equilibrium are contrasted. In order to carry on such a comparison we will be using equations (30 and (43 to follow the evolution at 000 K of the following reaction along the said two paths, starting from a common initial condition represented by one mole of gaseous molecular chlorine at the standard pressure of 1.00 bar Cl (g Cl (g (51 The determination of the equilibrium condition for the constant temperature and pressure evolution of this reaction was worked out by Bevan Ott and Boerio-Goates [p. 467]. For a value of ΔG o 000 = 1016 J, they proceeded to the evaluation of K via equation (46, obtaining K = After this, they expressed K as the ratio of the partial pressures of Cl(g and Cl(g, each raised to its stoichiometric coefficient. The quadratic equation resulting was solved making use of the fact that for the unit total pressure being considered, the following relation applies: P P. The equilibrium pressures obtained this way were: Cl 1 Cl P Cl bar and P Cl bar. The use of the value given by these authors for the standard Gibbs energy change at 000 K, combined with R = JK -1 mol -1, b = 1, y =, and Δn = 1, allows us to re-express equation (43 as follows G {(1 ln [(1 / (1 ] ln [ / (1 ]} (5 The standard Helmholtz energy change for this reaction can now be obtained by substituting in equation (35 the given values for the standard Gibbs energy change, R, T, and Δn, and solving for ΔA o. The result obtained is ΔA o = 6466 J. This value leads to the following re-expression of equation (30

12 A [(1 ln (1 ln ] (53 Table 1 was constructed by performing the indicated operations once each of the values of ξ there shown were substituted in in equations (5 and (53. The evaluation of ΔA and ΔG in these equations at ξ = 0 and ξ = 1 require the use of L Hôpital s rule during the limit-taking procedures [Larson, pp ]. ξ A G Table 1. Values in Joules of ΔA and ΔG for reaction (51 calculated via equations (5 and (53. An approximate graph of equations (5 and (53 is shown in figure. As indicated there, the Gibbs energy change reaches its minimum value of J at a degree of advancement of (ξeqt,p = The equilibrium partial pressures, evaluated through equations (40 and (41, with P = 1.00 bar, amount to P Cl bar and P Cl bar. These values being identical to those of Boerio-Goates previously quoted. 0,000 A A, J G, J 0 G ( 0.306, ( 0.346, ,000 0 Figure Approximate representation of the numerical data given in Table 1. The graph depicts the changes in the Gibbs and Helmholtz energies accompanying the evolution of reaction (51. For curves A and G the minimums correspond to points (0.306, , and (0.346, , and the values at ξ = 1 to J, and 1016 J, respectively The Helmholtz energy change, on the other hand, reaches a minimum value of J at a degree of advancement equal to (ξeqt,v = The equilibrium partial pressures, evaluated through equations (7 and (8, amount to PCl = 0.694, and PCl = The 1

13 corresponding equilibrium constant amounts to K = [(0.61 /0.694] = The agreement between this value and the previously quoted of is also reasonable. The equilibrium total pressure for the constant pressure case, as should be obvious, equals 1.00 bar. The equilibrium total pressure for the constant volume case, calculated by adding the equilibrium partial pressures, amounts to bar. Division of the partial pressures by the corresponding total pressure leads, respectively, to the following equilibrium mole fractions: (XCl = 0.486, XCl = 0.514eq,T,P; (XCl = 0.531, XCl = 0.469eq,T,V. 6 Le Chatelier s Principle The usual rationale, based on a qualitative application of Le Chatelier s Principle, can be offered to explain the difference in mole fractions existing between the constant pressure, and the constant volume equilibrium conditions. The isothermal pressure increase from 1.00 to bar necessary to produce the transit from the former to the latter produces a shift in the relative amounts of the species involved towards the smaller volume side, which in the case at hand is that of the molecular chlorine. This shift is quantified by the quoted mole fraction increase of this species from to The ξ-based approach here introduced allows, however, the opportunity to go beyond the not always precise qualitative approach, into the realm of quantitative applications of the thermodynamic equations embodying this principle, a rarely discussed alternative for the evaluation of the effects brought about in the equilibrium condition when a stress is imposed on it. The equations for the change in equilibrium reaction extent produced by isothermal changes in pressure [Moore, p. 9] and volume [Olivera-Fuentes, 007] required to carry on a numerical analysis of our results via Le Chatelier s Principle, written in that order, are the following eq P T ( V / T, P / ( G / T, P ( / (54 eq V T ( P / T, V / ( A/ T, V ( / (55 The denominators for these two expressions are respectively given by equations (49 and (50. In what follows equation (54 will be used to quantify the equilibrium displacement produced by the isothermal compression from 1 to bar required to transit from the equilibrium condition at constant T-P, to that at constant T-V. As should be obvious this compression proceeds without the presence of the anticatalyst. For reaction (51 the total number of moles at any given moment is (1 + ξ. Therefore, under constant temperature and pressure conditions, the volume at any degree of advancement can be written as V = (1 + ξrt/p. Consequently ( V / T, P RT / P (56 Substitution of b = 1, y =, and Δn = 1 in equation (50 produces Substitution of equations (56 and (57 in (54 gives us ( G / RT / (1 (1 (57 d ( eq / P T (1 (1 / P (58 dp

14 The sole dependence of ξ on P evident in equation (58 justifies the use of the total differentials there shown. Separation of variables, followed by integration by partial fractions, gives us eq 1 / 1 P (59 The constant of integration α can be evaluated by substituting in equation (59 the pair of values corresponding to the constant pressure equilibrium condition, (ξeqt,p = and P = 1. This leads to a value of α = 7.353, which in turn permits writing equation (59 as follows eq 1/ P (60 Introduction in equation (60 of for the pressure corresponding to the equilibrium condition at constant volume and temperature produces the following value for the corresponding equilibrium extent of reaction (, (61 eq T V The agreement between the previous value and the one previously obtained of through the minimum of A in figure, is quite reasonable The equilibrium displacement produced by the transition above referred can also be calculated in terms of the volume change, via equation (55. Under constant temperature and volume conditions the total pressure of reaction (51 at any given moment along its course can be written as P = (1 + ξrt/v. From this equation we get Substitution now of b = 1 and y = in equation (49 leads to (, V ( P / T, V RT / V (6 A/ T RT ( / (1 (63 Further substitution of these two results in equation (55 leads to ( / V T d / dv [ (1 ]/[ V ( ] (64 Integration by partial fractions leads, in turn, to the following expression: /(1 V (65 The integration constant can be evaluated by the appropriate substitution in equation (65 of (ξeqt,v = 0.306, and V = L, this last value being the volume in liters of the constant T-V equilibrium mixture. In doing this we get β = Introduction of this value in equation (65 produces /( V (66 Further substitution in this equation of V = 3.81 L for the volume of the constant T-P equilibrium reaction mixture, a value of (ξeqt,p = is obtained. This value should be compared that previously obtained of magnitude (ξeqt,p = Again, the agreement is quite reasonable.

15 7 A note on the spontaneity of chemical reactions For any two points (A1, ξ1 and (A, ξ, ξ1 ξ located on the interval 0 ξ of the A-vs.-ξ graph of Figure, it is true that (A A1/(ξ ξ1 = ΔA/Δξ 0. The previous expression, the finite-differences equivalent of the spontaneity criterion given in equation (, makes the transition from the former point to the latter, a spontaneous one. In geometrical terms this criterion identifies the spontaneity region as that in which the said graph takes tangents with negative slopes. At precisely (ξeqt,v = the tangent is horizontal, its slope is zero, and also in accord with equation (, the point marks the equilibrium condition for the constant T-V evolution of reaction (51. If the two points above referred are selected from the interval ξ 1, we will get (A A1/(ξ ξ1 = ΔA/Δξ > 0, a result indicating the non-spontaneous nature of the transit from the former point to the latter. In this region the corresponding slope of the tangents taken by the said graph are positive. It is true that Figure depicts the graph of A vs, with A A( A i. The fact that represents the standard molar Helmholtz free energy of formation of at 000 K and that this magnitude along the Gibbs free energy of formation- for any element in its standard state such as - at any temperature is zero, makes. Cl ( g A A A i Cl ( g Looking now at the derivative of equation (, written below as equation (68, we can understand that the sign as well as magnitude that the quantity ( (67 A / T, V acquires at a given ξ depends on the sign as well as magnitude of the respective contributions made by the three processes I, II, and III- involved in the occurrence of reaction (51. ( A / T, V ( AI / T, V ( AII / T, V ( A III / T, V (68 Accordingly, the intervals of spontaneity and non spontaneity of this reaction will be those in which the combined values of the said contributions turn out to be, respectively, negative or positive in sign. In order to apply equation (68 to reaction (51, we will start by identifying the contributing terms in it appearing. From equation (7 we have that Making use of the previously quoted value for the first derivative we can write A ( AI / T, V A (69 A as well as of the prime notation for 1 I ( AI / T, V 6466 J mol (70 From equation (16, with P B = (1 ξ/(1 + ξ and P Y = ξ/(1 + ξ, we get AII ( AII / T, V RT ln [4 /(1 ] (71 Finally, the expression coming from equation (1 takes the following form AIII ( AIII / T, V RT [1 ln (1 ] (7

16 Table was constructed by substituting the indicated values of ξ in equations (71 and (7. In the table A A / A A A. ( T, V I II III x10 4 AI x10 4 AII x10 4 AIII A x Table. The spontaneity of reaction (51 in the interval 0 ξ arise on reason of the combined contributions of spontaneous processes I and II being larger than that of nonspontaneous process III. At ξ = these three contributions, combining to a value of zero, define the equilibrium state for this reaction. The essential features of Table have been graphically represented in Figure 3. Figure 3. Here A 0 and A > 0 define, respectively, planes of spontaneity and nonspontaneity. Processes I and III are, respectively, spontaneous and non-spontaneous for all ξ. Process II, the mixing process, transits from spontaneous to non-spontaneous at ξ = The combined effects of these three processes define the regions of spontaneity, non-spontaneity, as well as the point of equilibrium for reaction (51 as indicated by curve A AI AII AIII. Inspection of Table makes evident that while A I contributes to the buildup of spontaneity along the complete interval of values of ξ, A III undermines it in the same interval. The contribution of A II changes from negative (spontaneous to positive (nonspontaneous at ξ = 0.447, point at which it becomes equal to zero. This value

17 corresponds to the minimum of ΔAII or the maximum of ΔSmix, as given by equations (16 and (15 respectively and was obtained by setting equation (71 equal to zero and solving for ξ. As the numbers in Table indicate the spontaneous effect of the combined A I A II A III magnitudes of and surpass the opposite effect of in the interval 0 ξ 0.306, defining this way the spontaneity interval for reaction (51. At ξ = the three contributions combine to zero defining this way the point of equilibrium of this reaction. For ξ 1, the situation is different. Here the now non-spontaneous effect of, A III A II in combination with, surpasses the opposite effect of, making reaction (51 non-spontaneous in this interval. The balance-tipping role played by the contribution coming from the entropy of mixing should be noted. The combined effects of processes I, II and III are represented in Figure 3 by curve A A A A I II A I 8 A Comment on Chemical Potentials The discussion in the undergraduate classroom of the varied guises available to the chemical potential -evident in the expression below [Denbigh, 1968, p. 80]- usually reduces to the presentation of a thermodynamic argument, not much different from that provided by Denbigh [p. 80], directed to justify the identities there shown, but rarely to exemplify them via concrete situations. The discussion here offered is centered on the identity of the A and G expressions of the chemical potential as they apply to reaction (51, and provides the thermodynamics instructor with some material that might be useful in this direction. A G U H i n i n T V n i n T P n i n,, S V n i j,, j,, j S, P, n j (73 The previous expression makes reference to a homogenous phase of variable composition constituted by moles of substance 1, moles of substance, etc., precisely like that constituting the reaction mixture of ideal-gas reaction (51. Even if other definitions are possible, for example the one originally advanced by Gibbs [Denbigh, p. 78], for the purposes at hand, the chemical potential of substance i, will be taken as the quantifier of the amount by which the capacity of the phase for doing work (other than work of expansion is increased per mole of i added, for an infinitesimal addition at constant temperature and pressure. [Denbigh, p. 79]. The role of this magnitude in chemical thermodynamics has been explained as follows The chemical potential has an important function analogous to temperature and pressure. A temperature difference determines the tendency of heat to pass from one body to another and a pressure difference determines the tendency toward bodily movement (likewise a difference of chemical potential may be regarded as the cause of a chemical reaction or of the tendency of a substance to diffuse from one phase to another. The chemical potential is thus a kind of chemical pressure and is an intensive property of a system, like the temperature and pressure themselves. [Denbigh p. 78] n 1 i n With these antecedents at hand let us express the functional dependence existing between the Helmholtz free energy of reaction (51 with the temperature, volume, and number of moles of molecular and atomic chlorine, as follows III

18 A A( T, V, n Cl, ncl (74 The previous expression, through the rules of partial differentiation, leads to the total differential of A A A A A da dt dv dncl dncl T V, n V i T, n n i Cl n T, V, n Cl Cl (75 T, V, n Cl The usual caveats apply in regard to the sub-indexes of the partial derivatives shown above. When equation (6 is applied to reaction (51 we get, via differentiation: dn Cl 1 d, dn Cl d (76 The proper combination of equations (75 and (76 leads us to A A A A da dt dv d 1 d T, V T, n Cl n (77 Cl V n i n i T, V, n Cl T, V, n cl Division of equation (77 by followed by imposing on it conditions of constant temperature and volume ( dt 0, dv 0 produces d A A T, V n Cl T, V, n Cl A 1 n Cl T, V, n Cl A similar argument as that taking us from equations (74 to equation (78 conducted in terms of the Gibbs free energy leads to G G T, P n Cl T, P, n Cl G 1 n Cl T, P, n Cl Let us now turn our attention to equations (3 and (45. When these equations, in the form they adopt when applied to reaction (51 are combined with the facts that G A RTn, and that G G f, Cl 1G f, Cl, with G f, i representing the standard molar Gibbs energy of formation of species i, we get, after some algebra, the following expressions: (78 (79 A T, V ( G f, Cl RT ln PCl 1( G f, Cl RT ln PCl (80 G ( G f, Cl T, P RT ln PCl 1( G f, Cl RT ln PCl A comparison of equation (80 with (78, and (81 with (79 allows us to make, with the aid of equation (73, the following identifications: (81

19 ( Cl T, V ( A / ncl T, V, G ncl f, Cl RT ln P Cl (8 ( Cl T, P ( G / ncl T, P, G f, Cl RT ln P ncl Cl (83 ( Cl T, V ( A / ncl T, V, G f, Cl RT ln P ncl Cl (84 ( Cl T, P ( G / ncl T, P, G f, Cl RT ln P ncl Cl (85 Recognition of the facts that: 1 the standard Gibbs free energy of formation of is zero at any temperature; that reaction (51 is precisely the formation reaction for atomic chlorine, in which case the standard Gibbs energy change of 1016 J to it associated corresponds to twice the molar standard Gibbs free energy of formation of, i.e. G f, Cl 5081 J mol ; and 3 that for reaction (51 us write the previous equations in the following form Cl(g 1 1 RT 1668 J mol Cl ( g, allows 1 (, ( A / n,, ln P J mol (86 Cl T V Cl T V ncl Cl 1 (, ( G / n,, ln P J mol (87 Cl T P Cl T P ncl Cl 1 ( Cl T, V ( A / ncl T, V, 1668 ln PCl J mol (88 ncl 1 ( Cl T, P ( G / ncl T, P, 1668 ln PCl J mol (89 ncl A comparison between equations (86 and (87 allows us to realize that Cl T, P ( and Cl T, V ( show the same functional dependence with respect to the partial pressure. If so, identical values of these magnitudes ( ( will be produced when Cl T, V Cl T, P quantified for a given value of the partial pressure of chlorine. The previous result tells us that at ( 000 K, p Cl bar the chemical potential of atomic chlorine when part of the reaction mixture evolving at constant temperature and volume is numerically identical to its chemical potential when part of the reaction mixture evolving at constant temperature and pressure. This fact makes the path discriminating-label superfluous. Instead of ( Cl T, V and ( Cl T, P, the chemical potential of Cl (g as part of reaction (51 should be simply denoted as Cl, as its magnitude, in whatever path we might be considering, will be given by Cl ln pcl. This is the meaning of ( A / n ( G / n. Similar considerations apply for Cl. Cl Cl T, V, ncl Cl T, P, ncl In order to get a geometrical interpretation of the previous considerations is that we have included Table 3. In it we find the partial pressures of Cl ( g and Cl(g along the course of the constant T,V, and constant T, P, paths of reaction (51 and the corresponding chemical potentials. The partial pressures were evaluated with equations (7 and (8, and (40 and (41; the corresponding chemical potentials with equations (86-(89. p Cl

20 - P Cl ( P Cl Cl T, V ( Cl T, P - P Cl ( P Cl - ( Cl T, V Cl T, P Table 3. Values for the partial pressures of and at different points along the constant T,V and constant T,P paths of reaction (51, and the corresponding chemical potential. Cl ( g Cl(g Figure 4. The single graph produced by the sets of points, P ] and Cl T, P Cl [( Cl T, V Cl [(, P ] evince the identity given in equation (90. The values for ( Cl T, V and ( Cl T, P of Table 3 were graphed against the corresponding partial pressures. As expected, all the points fall on a single curve. Had Cl been path dependent we would have gotten two curves, one for ( Cl T, V and another for ( Cl T, P.

21 That this is not so confirms the identity previously noted, same that can be expressed as follows (90 Cl ( A / ncl T, V, ( G / n ncl Cl T, P, ln p ncl Cl A similar procedure performed with the data of Table 3 for ( Cl T, V, ( Cl T, P, and the corresponding partial pressures, produces Figure 5. As before noted in regard to the chemical potential of, here we see the data for and coalescing Cl(g ( Cl T, V ( Cl T, P into a single graph, evidence of the fact that the same value for these two magnitudes correspond to a given value of. Let us then reiterate the fact that any position negating the identity between and or equivalently, between ( A / n Cl T, V, ncl and p Cl ( Cl T, V ( G / n Cl T, P, ncl ( Cl T, P implies the existence of two different graphs for ( Cl T, V and ( Cl T, P, a possibility negated by Figure 5 below. These considerations can be summarized via the following equation (91 Cl ( A / ncl T, V, ( G / ncl T, P, 1668 ln p ncl ncl Cl Let me conclude this argument by saying that the development of an argument connecting the A and G expressions of the chemical potential with those given in terms of the internal energy U and the enthalpy H, constitute part of this author s current preoccupations. Figure 5. The single graph produced by the sets of points [( Cl T, V, PCl ] and Cl T, P Cl [(, P ] evince the identity given in equation (91 Given all the information contained by Table 3 it seems only natural to use it to exemplify the spontaneity and equilibrium conditions in terms chemical potentials.