On the Reunification of Chemical and Biochemical Thermodynamics

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On the Reunification of Chemical and Biochemical Thermodynamics: A Simple Example for Classroom Use Lionel M. Raff* Department of Chemistry, Oklahoma State University, Stillwater, Oklahoma 74078, United States

William R. Cannon* Downloaded via UNIV OF LOUISIANA AT LAFAYETTE on January 9, 2019 at 08:16:44 (UTC). See https://pubs.acs.org/sharingguidelines for options on how to legitimately share published articles.

Pacific Northwest National Laboratory, Richland, Washington 99352, United States

ABSTRACT: For 26 years, it has been assumed by some that the thermodynamics of open-system biochemical reactions must be executed by performing Legendre transformations on the terms involving the species whose concentrations are being held fixed. In contrast, standard nontransformed thermodynamics applies to chemical processes. However, it has recently been shown that such biochemical reactions may be accurately examined using either method. The papers that report this finding use the hydrolysis of ATP at fixed pH and pMg as an example. This biochemical process comprises 14 equilibrium reactions involving 17 chemical species. Consequently, the chemical and mathematical complexity is so high that the underlying principles leading to the equivalence of the two methods tend to become lost. Furthermore, the details of such an example are too complex for classroom presentation. This paper makes these principles abundantly clear by the thermodynamic examination of the simple case of a unimolecular isomerization conducted under both open and closed conditions. For the open system, the analysis is conducted using both Legendre-transformed and nontransformed methods. The results are shown to be identical provided that the chemical potentials of the terms on which the transform is performed are held constant. More importantly, the analysis makes the underlying reasons for the equivalence of the two methods very clear and shows when they will not be equivalent. The model is ideally suited for classroom presentation because of its chemical and mathematical simplicity. KEYWORDS: Upper-Division Undergraduate, Graduate Education/Research, Physical Chemistry, Biochemistry, Misconceptions/Discrepant Events, Biophysical Chemistry, Equilibrium, Thermodynamics



INTRODUCTION

transformed entropy S′.” In his 1992 papers, Alberty contended that the introduction of Legendre transforms of the Gibbs free energy was necessary to have a criterion to establish the spontaneity of biochemical reactions and to determine their equilibrium properties. It was further stated that the use of G was adequate for chemical reactions but inadequate for complex systems such as biochemical reactions with fixed pH. This same assertion appears repeatedly in the literature associated with the thermodynamics of biochemical reactions.

For the past 26 years, the thermodynamics applicable to biochemical reactions under conditions of fixed pH, pMg, etc. has been considered to be separate and different than the thermodynamics of chemical reactions. When the thermodynamics of such biochemical reactions is discussed, it is generally asserted that Legendre-transformed thermodynamic potentials must be employed. The genesis of such assertions dates from 1992, when Alberty1−3 stated, “When pH and pMg are specified, a whole new set of transformed thermodynamics come into play. These properties are different from the usual Gibbs energy G, enthalpy H, entropy S, and they are referred to as the transformed Gibbs energy G′, transformed enthalpy H′, and © XXXX American Chemical Society and Division of Chemical Education, Inc.

Received: September 29, 2018 Revised: December 16, 2018

A

DOI: 10.1021/acs.jchemed.8b00795 J. Chem. Educ. XXXX, XXX, XXX−XXX

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In 1992, in an article published in Biochemistry, Alberty and Goldberg4 stated, “The criterion for chemical equilibrium at specified temperature, pressure, pH, concentration of magnesium, and ionic strength is the transformed Gibbs energy.” An article published in Biophysical Journal in 1993 claimed that three levels of thermodynamic calculations exist,5 which the abstract of the paper summarized as follows: Level 1 is the usual chemical calculation with species at specified temperature and pressure using standard Gibbs energies of formation of species or equilibrium constants K. Level 2 utilizes reactants such as ATP (a sum of species) at specified T, P, pH, and pMg with standard transformed Gibbs energies of formation of reactants or apparent equilibrium constants K′. Calculations at this level can also be made on the enzymatic mechanism for a biochemical reaction. Level 3 utilizes reactants at specified T, P, pH, and pMg, but the equilibrium concentrations of certain reactants are also specified. The fundamental equation of thermodynamics is derived here for Level 3. Equilibrium calculations at this level use standard transformed Gibbs energies of formation of reactants at specified concentrations of certain reactants or apparent equilibrium constants K″. Papers, review articles, and books have continued to assert the necessity of employing Legendre-transformed thermodynamic potentials to treat biochemical processes. The abstract of a 2002 paper6 stated, “Legendre transforms are needed to introduce intensive variables into the description of thermodynamic systems. But when experimental thermodynamic data involve an inconveniently large number of intensive variables, inverse Legendre transforms are needed.” Alberty has written a book entitled Thermodynamics of Biochemical Reactions on the subject of using Legendre-transformed thermodynamic potentials to treat biochemical reactions.7 Courses in biochemical thermodynamics now stress the importance of using Legendre-transformed chemical potentials to treat equilibrium in biochemical reactions. The following is from a course handout citing a statement by Gennis:8 Biochemical reactions are virtually always performed under conditions of constant pH, both in the cell and in vitro. Hence, even if protons are consumed or released by the biochemical reaction itself, the bulk concentration of free protons does not change because of the presence of the buffer.... Because of this, it is necessary to modify the way we express the equilibrium constant and the Gibbs free energy of reaction in reactions in which protons are involved, which is most biochemical reactions. This is handled by using a transformed Gibbs free energy, in place of the Gibbs free energy used traditionally by chemists. An indication of the number of various types of publications that stress the necessity of using Legendre-transformed thermodynamic potentials when treating biochemical reactions can be obtained by performing a Google search on the topic “Transformed Biochemical Thermodynamics”. One such search produced 17 pages of references. Databases reporting Legendre-transformed standard thermodynamic quantities have now been developed. For example, Alberty has utilized the transformed Gibbs free energies to examine the thermodynamic properties of enzyme-catalyzed reactions involving guanine, xanthine, and their nucleosides and nucleotides.9 This study added standard transformed Gibbs free energies for 10 reactants to the BasicBiochemData3 database. Another 15 reactants for enzyme-catalyzed reactions

involving cytosine, uracil, thymine, and their nucleosides and nucleotides were reported in a separate study.10 In 2011, six authors once again advanced the concept that the execution of biochemical thermodynamics requires the use of a Legendre transform and a new thermodynamic potential, the transformed Gibbs free energy, by the submission of recommendations to IUPAC related to the thermodynamic treatment of biochemical reactions.11 The following year, the introduction of an article in Nucleic Acids Research was entitled “Essentials of Biochemical Thermodynamics”. This introduction stated that when equilibrium in biochemical reactions is discussed, the transformed Gibbs energy rather than the change in the usual Gibbs free energy is the important quantity.12 The IUBMB−IUPAC Joint Commission on Biochemical Nomenclature (JCBN) formalized Alberty’s view in their 1994 recommendations. These recommendations were revised and updated in 2011.11 The recommendations stated in part that “Chemical equations are written in terms of specific ionic and elemental species and balance elements and charge whereas biochemical reactions are written in terms of reactants that often consist of species in equilibrium with each other and do not balance elements that are assumed fixed, such as hydrogen at constant pH.” The IUBMB−IUPAC recommendations went on to state that textbooks and research papers often represent biochemical reactions such as the hydrolysis of ATP to ADP by ATP + H 2O V ADP + Pi + H+

but this process is a hybrid of a chemical equation and a biochemical equation and does not have an equilibrium constant. The JCBN recommended that this reaction be written in the form ATP + H 2O V ADP + Pi

where ATP, ADP, and Pi represent equilibrium mixtures of pseudoisomers at the specified pH and pMg. For example, ATP refers to an equilibrium mixture of ATP4−, HATP3−, H2ATP2−, MgATP2−, MgHATP −, and Mg2ATP. The JCBN indicated that the equilibrium constant for this biochemical reaction had to be obtained using the Legendre-transformed Gibbs free energy: ΔG′ = ΔG′° + RT ln Q ′

(1)

with Q′ =

[ADP][Pi ] [ATP]

(2) 11

The 2011 updated recommendations continued to maintain that transformed thermodynamic potentials are needed to treat the thermodynamics of biochemical reactions in which pH and pMg are held constant. As a consequence of the plethora of papers and books and the IUPAC recommendations, two categories of thermodynamics based on different concepts and different formalisms have been established: (1) chemical thermodynamics, which employs conventional thermodynamic potentials to treat chemical reactions, and (2) biochemical thermodynamics, which employs Legendre-transformed thermodynamic potentials to deal with biochemical reactions. However, it has recently been demonstrated13,14 that balancing the biochemical reactions for the hydrolysis of ATP at constant T, P, pH, and pMg allows the biochemical reactions to be treated using standard chemical thermodyB

DOI: 10.1021/acs.jchemed.8b00795 J. Chem. Educ. XXXX, XXX, XXX−XXX

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namics. This result demonstrated that while transformed thermodynamic potentials can be employed, it is not necessary to do so. In fact, the transformed potential changes ΔrG′°, ΔrH′°, and ΔrS′° can be obtained directly from the nontransformed potential changes ΔrG°, ΔrH°, and ΔrS°. The computational requirements of this balanced biochemical equation (BBE) approach are very similar to those required when Legendre-transformed potentials are employed. This development allows the two worlds of chemical and biochemical thermodynamics, which are often treated separately, to be reunified within the same thermodynamic framework. At present, IUPAC has appointed an international committee to bring recommendations for the reunification of chemical and biochemical thermodynamics.15 After months of preparation, the committee met in Florence, Italy, in May 2018. At present, a technical report is being prepared for submission to IUPAC. Although the Legendre transform method and the BBE method have been fully described by Alberty and others1−12 and Iotti and co-workers,13,14 respectively, these papers are not simple to read and understand. The hydrolysis of ATP involves 14 equilibrium reactions among 17 different chemical species. Consequently, the resulting chemical and mathematical complexity tend to obscure the fundamental principles that serve to reunite the two thermodynamic treatments. Furthermore, the presentation of such complex examples in an undergraduate or graduate course is not feasible. What is required for both understanding and a pedagogically clear presentation is a very simple example comprising both a chemical reaction and a corresponding biochemical reaction, each of which is treated using both methods. Comparison of the results will then provide a clear picture of when and, if so, why the two methods are fully equivalent as well as providing a model that is suitable for classroom exposition. The present paper provides such an example.

Let us consider a reaction in which gaseous compound C(g) reacts to yield gaseous compound D(g), that is, νCC → νDD, where the νi are the stoichiometric coefficients. We omit the (g) designation for notational convenience. With no ionic species present, the ionic strength I will be constant at the value of zero throughout the reaction. If the reaction occurs in a fixed-volume container with partial pressures of C and D both well below 1 bar, the gases will be nearly ideal, the fugacity coefficients will approach a constant value of unity, and the chemical potentials will depend only upon the partial pressures of C and D. The use of partial pressures in the chemical potentials instead of total pressure incorporates the contribution of mixing the two ideal gases. We also assume the reactions to be sufficiently slow that T, P, and the thermodynamic potentials may be defined. Table 1 provides a list of symbols and their definitions used the analysis. Where our notation differs from IUPAC, both symbols are given in the table. Table 1. List of Symbols and Definitions Symbol

IUPACa

Pi P0i T V R α A S μi ξ ni(ξ) νi μ°i Δrμ Δrμ° ξeq Kp



A SIMPLE MODEL FOR CHEMICAL AND BIOCHEMICAL REACTIONS Obviously, biochemical reactions are generally much more complex than chemical reactions. However, while this feature serves to make the thermodynamic treatment of the biochemical system more difficult, it does not imply a fundamental difference in the nature of the process. Biochemical reactions at constant pH and pMg differ from chemical reactions primarily because the fixed pH, fixed pMg biochemical reaction is an open system involving mass transfer whereas traditional chemical reactions are almost always closed systems. Both H+ and Mg2+ are reactants or products in the biochemical system. Consequently, if the system is closed, the pH and pMg cannot remain fixed as the reaction proceeds. This can only occur if these ions are either added to or removed from an external source during the reaction process. Therefore, the system is open in a thermodynamic sense. The objective here is to introduce a simple model reaction that can be conducted under either closed or open conditions. The system will be examined thermodynamically under each condition using both transformed and nontransformed methods. The results will show when the two methods are expected to yield identical results as well as the underlying reason for this equivalence. The results will also demonstrate when the two methods will not yield the same results and the reason why this is the case.



3 (yw) G A′ G′ Kp′ γi,k

Δr G ΔrG°

Definition Partial pressure of component i (bar) Initial partial pressure of component i (bar) Temperature (K) Volume (L) Ideal gas constant (L bar mol−1 K−1 or J mol−1 K−1) α = RT/V Nontransformed Helmholtz free energy (J) Entropy (J/K) Chemical potential of component i (J mol−1) Extent of reaction coordinate (mol) Chemical amount of component i as a function of ξ (mol) Dimensionless stoichiometric coefficient of species i Standard chemical potential of component i (J mol−1) Δrμ = ∑pνpμp − ∑rνrμr (p ≡ products, r ≡ reactants) Δrμ° = Σpνpμp° − Σrνrμr° (p ≡ products, r ≡ reactants) Value of ξ at the equilibrium position Nontransformed equilibrium constant in terms of partial pressures Standard pressure of 1 bar Legendre transform operator on conjugate variables y and w Nontransformed Gibbs free energy (J) Legendre-transformed Helmholtz free energy (J) Legendre-transformed Gibbs free energy (J) Transformed equilibrium constant in terms of partial pressures Number of atoms of type k in pseudoisomer i

a

Where the symbol employed in this article is the same as the IUPAC symbol, this column is blank. Where the symbols differ, this column provides the equivalent IUPAC symbol.

For the initial state of our system, we assume a container with a fixed volume of 100 L in a constant-temperature bath at a fixed temperature of 298.15 K. The initial chemical amounts of C and D are nC(0) = 1 mol and nD(0) = 0. The standard chemical potentials of C and D are assumed to be μC° = 4000 J/mol and μD° = 2000 J/mol. Therefore, the initial partial pressure of C is PC0 = C

nC(0)RT = αnC(0) = 0.2479 bar V

(3)

DOI: 10.1021/acs.jchemed.8b00795 J. Chem. Educ. XXXX, XXX, XXX−XXX

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where the constant α has the value of 0.2479 bar/mol. To obtain an open system in which the concentration of C inside the reactant chamber is held constant at its initial value nC(0)/V, we provide an external source of additional C molecules at 298.15 K and pressure P0C. As C reacts to produce D, additional C molecules are introduced reversibly into the system through a membrane permeable only to C at a rate such that the partial pressure of C, PC, is maintained at its initial value PC0 throughout the reaction. Additional C(g) is continuously introduced into the external source to maintain the pressure constant at P0C. That is, we have the situation depicted below:

or

ij ∂A yz jj zz = vDμD − vCμC j ∂ξ z k {T , V

(8b)

− vCRT ln(PC(ξ)/P°)] dξ

(9a)

Division by dξ followed by combination of the two log terms gives É ÄÅ ÅÅ PD(ξ)vD ÑÑÑ jij ∂A zyz ÑÑ Å Å ° ° = (vDμD − vCμC ) + RT lnÅÅ jj zz Ñ ÅÅÇ PC(ξ)vC ÑÑÑÖ k ∂ξ {T , V ÅÄÅ ÑÉ Å P (ξ)vD ÑÑ = Δr μ° + RT lnÅÅÅÅ D vC ÑÑÑÑ ÅÅÇ PC(ξ) ÑÑÖ (9b) where Δrμ° is the standard reaction free energy. The quantity on the left-hand side of 9b is the negative of the chemical force16 that drives the reaction. If −(∂A/∂ξ)T,V is positive, the chemical force will drive the reaction spontaneously toward D. If it is negative, the spontaneous direction will be toward C. It should be noted that for a reaction from state x to state y to be spontaneous, the chemical force must be positive at all points along the path connecting x to y. If this is true, then ΔA for the process will always be negative. That is,

(4)

(5)

∫x

(6a)

Likewise, we also have for reactant C, dnC(ξ) = −vC dξ

(8a)

= [(vDμD° − vCμC°) + vDRT ln(PD(ξ)/P°)

We define the extent of reaction ξ in terms of nD and the unsigned stoichiometric coefficient νD as dnD(ξ) = vD dξ

dA = (vDμD − vCμC ) dξ

− [μC° + RT ln(PC(ξ)/P°)]vC dξ

At constant temperature and pressure, we have dT = 0 and dV = 0, so that

or

(7b)

dA = [μD° + RT ln(PD(ξ)/P°)]vD dξ

CLOSED SYSTEM ANALYSIS: NONTRANSFORMED METHOD Since we are conducting the reaction under conditions of constant temperature and volume, the relevant thermodynamic potential is the Helmholtz free energy A. Along any reversible path, hypothetical or real, the differential change in A is given by

dnD = vD dξ

nD(ξ) = nD(0) + vDξ

The quantity on the right-hand side of 8b is the difference between the chemical potentials of product D and reactant C. As such, the best notation for this difference is Δrμ. IUPAC recommends ΔrG for this quantity. This point has been discussed in detail in previous publications.16,17 For an ideal gas, each chemical potential is given by μi = μ°i + RT ln(Pi(ξ)/P°), where Pi(ξ) is the partial pressure of gas i in bars, and P° is the standard pressure of 1 bar. The use of partial pressures in this expression rather the total pressure incorporates the contribution of mixing to the chemical potential.16 Substitution of this expression into eq 8a yields



dA = μC dnC + μD dnD

(7a)

where nC(0) and nD(0) are the constants of integration, i.e., the numbers of C and D molecules present at the start of the reaction, respectively. Substitution of eqs 6a and 6b into eq 5 gives

where the semipermeable membrane prevents the flow of D out of the reaction chamber. For each of these systems, we wish to obtain the variation of the relevant thermodynamic potential as a function of the extent of reaction ξ, the equilibrium value of the extent of reaction ξeq, the equilibrium constant for the reaction, and the range of ξ values over which the reaction will be spontaneous in the direction of D. For the open system, we also wish to determine how much additional C must be added to maintain a constant partial pressure of C in the reaction vessel. The analysis for both systems will be conducted using both transformed thermodynamics and nontransformed chemical thermodynamics. A previous understanding of thermodynamics at the level of undergraduate physical chemistry is assumed.

dA = −S dT − P dV + μC dnC + μD dnD

nC(ξ) = nC(0) − vCξ

y

jij ∂A zyz dξ = ΔA < 0 jj zz k ∂ξ {T , V

(10)

is a necessary condition for a transformation from state x to state y to be spontaneous. However, it is not a sufficient condition.16 The pressures appearing in eq 9b are given by the ideal gas equation of state:

(6b)

Consequently, the numbers of C and D molecules present as functions of ξ are D

DOI: 10.1021/acs.jchemed.8b00795 J. Chem. Educ. XXXX, XXX, XXX−XXX

Journal of Chemical Education PC(ξ) =

nC(ξ)RT = αnC(ξ) V

PD(ξ) =

nD(ξ)RT = αnD(ξ) V

Article

Reference to the results in Figure 1 shows that over the range 0 ≤ ξ < 0.6914 mol, the reaction is spontaneous in the direction of D. For ξ > 0.6914 mol, the reaction is spontaneous in the direction of C. If we are concerned solely with the equilibrium state and not with spontaneity and the variation of A with ξ, then we do not need Figure 1. Equations 9 and 12−14 show that at equilibrium, we have

(11)

Equation 9b can be further simplified if we take the simple case for which νC = νD = 1, which makes the reaction an isomerization. With nC(0) = 1 mol and nD(0) = 0 mol, eqs 7a and 7b give nC(ξ) = 1 mol − ξ and nD(ξ) = ξ. (For convenience in what follows, the expression “1 mol − ξ” will be written as “1 − ξ”.) Substitution of these results and eqs 11 into eq 9b gives ij ∂A yz i ξ yz jj zz zz = Δr μ° + RT lnjjjj j ∂ξ z z k {T , V k1 − ξ {

i ξ yz i Δ μ° y jij PD zyz zz = expjjj− r zzz = 2.241. jj zz = jjjj z j RT z j PC z k { k {eq k 1 − ξ {eq

(15)

Equation 15 is easily solved for ξeq. The result is ξeq = 0.6914 mol, as expected. It is a simple matter to obtain the total Helmholtz free energy of the system, A(ξ), from the relationships A = G − PV and G = nCμC + nDμD. Using these equations along with the expressions for the chemical potentials and chemical amounts of C and D developed previously, we obtain

for 0 ≤ ξ ≤ 1 (12)

Figure 1 shows a plot of (∂A/∂ξ)T,V versus ξ for Δrμ° = −2000 J/mol. When the chemical force goes to zero, we are at

A(ξ) = {μC° + RT ln[α(1 − ξ)/P°]}(1 − ξ) + [μD° + RT ln(αξ /P°)]ξ − ntotalRT

(16)

since PV = ntotalRT (in the present case, ntotal = 1 mol for all values of ξ). Algebraic rearrangement permits this expression to be written in the form A(ξ) = Δr μ°ξ + RT ln[α(1 − ξ)/P°] + ξRT ln[ξ /(1 − ξ)] + μC° − ntotalRT

(17)

Figure 2 shows a plot of A(ξ) versus ξ. Equilibrium occurs at the minimum value of A(ξ). As expected, this minimum occurs

Figure 1. Variation of the negative of the chemical force, (∂A/∂ξ)T,V, with the extent of reaction, ξ, for the closed unimolecular isomerization reaction C(g) → D(g) under the experimental conditions described in the text.

equilibrium. Therefore, the condition for equilibrium for reactions conducted at constant temperature and volume is jij ∂A zyz =0 jj zz k ∂ξ {T , V

(13) 18

It should be noted that it has been shown that at the equilibrium point the ratio of the reaction probability in the forward direction to that in the reverse direction is unity, but neither reaction probability is zero. This result shows that equilibrium is a dynamic state in which the reactions in the forward and reverse directions are proceeding at the same rate. As can be seen from Figure 1, the system attains equilibrium when ξ = ξeq = 0.6914 mol. This corresponds to an equilibrium pressure of D of (PD)eq = αξeq = 0.1714 bar. At equilibrium, the partial pressure of C is (PC)eq = α(1 − ξeq) = 0.07648 bar. The corresponding equilibrium constant Kp is ÅÄÅ ÑÉ i Δ μ° y Å (P /P°) ÑÑÑ 0.1714 ÑÑ = expjjj− r zzz = K p = ÅÅÅÅ D = 2.241 j RT z Ñ ÅÅÇ (PC/P°) ÑÑÖ 0.07648 k { eq

Figure 2. Variation of the Helmholtz free energy A with the extent of reaction ξ for the closed unimolecular isomerization reaction C(g) → D(g) under the experimental conditions described in the text.

at ξeq = 0.6914 mol. It is important to note that since the finite transition from ξ = 0 to ξ = 0.6914 mol is known to be spontaneous, we expect and observe ΔA for this transition to be negative. However, the transition from ξ = 0 to ξ = 0.9 mol is not spontaneous. The reaction will proceed spontaneously from ξ = 0 to ξ = 0.6914 mol but go no further unless work is done on the system. It should be noted that ΔA for the nonspontaneous transition from ξ = 0 to ξ = 0.9 mol is still negative. Consequently, ΔA < 0 is a necessary condition for spontaneity, but it is not a sufficient condition.16,17

(14) E

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OPEN SYSTEM ANALYSIS: LEGENDRE TRANSFORMATION METHOD Since some readers may not be familiar with Legendre transformations, it will be useful to describe such transformations in simple terms. A Legendre transformation is a simple mathematical method for switching the independent variables in a differential expression. No new science is introduced by the execution of a Legendre transformation. One simply adds or subtracts a differential identity to or from the differential expression to execute the transformation. The decision to add or subtract is made such that the desired variable switch is achieved. Consider the differential d z = v dx + w d y

throughout the reaction. At low pressures, a constant concentration means a constant partial pressure of C and, therefore, a constant chemical potential. Such constraints are often present in biochemical reactions where pH, pMg, etc. are fixed, but they are rarely present in chemical reactions. The second difference is the result of the open nature of the system. Whenever the surroundings participate in a process, we are faced with two calculations. We must consider not only the differential change in the relevant thermodynamic potential for the system but also the same quantity for the surroundings. The most common example of this in introductory thermodynamics occurs whenever the entropy change is being employed to predict spontaneity and equilibrium. In this case, the important quantity is dStotal = dSsys + dSsur, where dSsys and dSsur are the differential changes in the entropy for the system and the surroundings, respectively. For an open chemical reaction occurring under conditions of constant temperature and volume, such as the one under consideration here, we must compute the sum of dAsys and dAsur. From eq 5 at constant T and V, the differential change in the free energy of the system is

(18)

where z, v, and w are functions of the independent variables x and y. Suppose it is useful to express z in terms of the independent variables x and w instead of x and y. This can be achieved simply by subtracting the identity d(wy) = w dy + y dw

(19)

from both sides of eq 18 to obtain dz − d(wy) = d(z − wy) = v dx − y dw

dA sys = μC dnC,sys + μD dnD (20)

where we have added the subscript “sys” to emphasize that this is the result for the system. There is no need to use the subscript “sys” with dnD since compound D is present only in the system. For the surroundings, we have

This procedure is termed a Legendre transformation. The result is a differential expression in which the independent variables are now x and w. The function appearing on the lefthand side of eq 20, z − wy, is called the Legendre transform of z. It may be denoted by any symbol desired, such as z′. Let us define an operator 3 (yw) such that when 3 (yw) operates on a differential of the form given in eq 18, it produces the transformation described by eqs 18−20. Thus, we would have 3 (yw) dz = dz′ = v dx − y dw. Legendre transforms are frequently employed in thermodynamics to switch the independent variables to ones that may be more convenient for treating a particular system. For example, consider the differential change in the internal energy, U, along a reversible path for a system in which there are no variations in the chemical composition: dU = T dS − P dV. This is a convenient expression to employ for processes occurring as a function of volume, but it is not so convenient for processes occurring as a function of pressure. Therefore, we can execute a Legendre transformation to switch the variables P and V: 3(PV ) dU = d(U + PV ) = T dS + V dP

dA sur = μC,sur dnC,sur = μC dnC,sur

(23b)

Since our model assumes that the transfer of C from the surroundings to the system is reversible, we must have μC,sur = μC,sys = μC. Therefore, the total differential change in the Helmholtz free energy is dA total = dA sys + dA sur = μC dnC,sys + μD dnD + μC dnC,sur

(23c)

If pressure were being held constant instead of volume, the sum of dGsys and dGsur would be required. The starting point for a Legendre transform treatment of this open system begins with eq 23c, which provides the total differential change in the Helmholtz free energy. When eq 23c is used, there is no easy way to address the constraint that the chemical potential of C must remain constant since the equation contains no terms for the differential change in the chemical potential of C, dμC. To produce such a term, we need to execute a Legendre transformation of dAtotal using the conjugate variables μC and nC,sys:

(21)

The Legendre transform of U, U + PV, is generally denoted by the symbol H and called the enthalpy. Other well-known Legendre transforms include the following: 3(TS) dH = dG = − S dT + V dP

3(μC nC,sys) dA total = d(A total − μC nC,sys) = dA′total

3(TS) dU = dA = − S dT − P dV 3(PV ) dA = dG = − S dT + V dP

(23a)

= μC dnC,sys + μD dnD + μC dnC,sur

(22)

Obviously, the use of a Legendre transformation does not produce any new thermodynamic information. At most it yields something that may be more computationally or pedagogically convenient. The open system C(g) → D(g) differs from the closed system we have just examined in two important ways. First, we now have an additional constraint. In addition to the constraints of constant temperature and volume, we now require that the concentration of compound C be constant

− (nC,sys dμC +μC dnC,sys) = μD dnD + μC dnC,sur − nC,sys dμC

(24)

where A′total = Atotal − μCnC,sys is the Legendre transform of Atotal. Equation 24 can be simplified further by considering the stoichiometry, which requires that dnC,sur = − dnD F

(25) DOI: 10.1021/acs.jchemed.8b00795 J. Chem. Educ. XXXX, XXX, XXX−XXX

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ij ∂A′total yz jj z = 1457.52 J/mol + RT ln(αξ /P°) j ∂ξ zz k {T , V

since every molecule of C that enters system from the surroundings to keep nC,sys constant reacts to form one molecule of D. Therefore, the total differential change in A′ that determines spontaneity and equilibrium is dA′total = (μD − μC ) dnD − nC,sys dμC

(32)

and Kp′ = (PD/P°)eq = exp(−1457.52/RT) = 0.5555. Figure 3 shows a plot of (∂A′total/∂ξ)T,V versus ξ. Whenever (∂A′total/∂ξ)T,V is negative, the chemical force drives the

(26)

When the chemical potential of C is held constant, eq 26 becomes dA′total = (μD − μC ) dnD

Article

(27)

Insertion of the expressions for the chemical potential of D and the fixed chemical potential of C into eq 27 yields dA′total = [μD° + RT ln(PD(ξ)/P°) − μC° − RT ln(PC0/P°)] dnD

(28)

Combining the log terms and using Δrμ° = μ°D − μ°C produces ÄÅ ÉÑ l o o ÅÅÅ PD(ξ)/P° ÑÑÑ| o ÑÑo dA′total = m Δr μ° + RT lnÅÅÅ 0 } dnD o Ñ o o ÅÅÅ PC/P° ÑÑÑo (29) Ç Ö~ n If we again define the extent of reaction variable ξ using eqs 6a and 7b, eq 28 shows that (∂A′total/∂ξ)T,V is given by ij ∂A′total yz jj z = Δr μ° − RT ln(PC0/P°) + RT ln(αξ /P°) j ∂ξ zz k {T , V

Figure 3. Variation of the negative of the chemical force, (∂A′total/ ∂ξ)T,V, with the extent of reaction, ξ, for the open unimolecular isomerization reaction C(g) → D(g) under the experimental conditions described in the text.

(30)

with α defined by eq 3. The negative of (∂A′total/∂ξ)T,V in eq 30 is the chemical force. It should be noted that the first and second terms on the right-hand side of eq 30 are both constants. In transformed thermodynamic space, these constant terms are combined to form the new reference free energy change, Δrμ′°. In IUPAC notation, this quantity is ΔrG′°. Using eq 14 to write Δrμ° in terms of Kp, we obtain

reaction spontaneously in the direction of D. When this partial derivative is zero, we are at equilibrium. This condition occurs when ξ = ξeq = 2.2407 mol. When ξ > 2.2407 mol, the chemical force is negative. That is, the force drives the reaction in the direction of C, which would require that C be continuously transferred from the reaction chamber to its external source to maintain a constant concentration and partial pressure of C. Once we have the value of ξeq, the simplicity of this model problem makes it trivial to determine the amount of additional C that must have been added to the system from the external source to attain equilibrium. At equilibrium, 2.2407 mol of D is present with the original 1 mol of C. Therefore, an additional 2.2407 mol of C was inserted from the external source. If we are only concerned with the point of equilibrium and not with the detailed variations of (∂A′total/∂ξ)T,V and A′total with ξ, the calculation becomes easier. At equilibrium, the chemical force is zero. If we set the left-hand side of eq 32 to zero, we may easily solve for the equilibrium partial pressure of D and the corresponding value of ξ. The results are (PD)eq = 0.5555 bar and ξeq = (PD)eq/α = 2.2407 mol. The total transformed Helmholtz free energy for the open reaction can be obtained using eq 27 and A′total(ξ) = G′total(ξ) − PV = [G′(ξ)]sys + [G(ξ)]sur − PV = nDμD − nDμC − PV. This gives

ÄÅ ÉÑ | l o o o o o o ÅÅÅÅ (PD/P°) ÑÑÑÑ 0 ° = −RT m ln + ln( P / P ) } Å Ñ C o ÅÅ (P /P°) ÑÑ o o o o ÅÇ C o ÑÖeq n ~ ÄÅ ÉÑ ÅÅ (P /P°) ÑÑ = − RT lnÅÅÅÅ D 0 ÑÑÑÑ ÅÅÅ (PC/PC) ÑÑÑ Ç Öeq

Δr μ′° = − RT ln K p − RT ln(PC0/P°)

= − RT ln K p′

(31)

By carrying out the Legendre transform and subsequently setting nC = nC(0), we have changed the reference state for C from the state at the standard pressure P° to the state at pressure P0C. Since PC is being held constant at the value P0C, we have (PC/P0C)eq = 1. Hence, the denominator of Kp′ becomes unity, and Kp′ = (PD/P°)eq. This is always a feature of transformed equilibrium constants. The chemical species with constant chemical potentials vanish from the expression for the equilibrium constant as well as from the corresponding expression for the reaction quotient, Q. However, it should be noted that the standard chemical potentials for these species still appear in the expression for Δrμ°, as can be seen in eqs 28 and 29. The open system has P0C fixed at 0.2479 bar, T = 298.15 K, and Δrμ° = −2000 J/mol. Insertion of these values into eq 30 gives Δrμ′° = 1457.52 J/mol,

A′total (ξ) = ξ[μD° + RT ln(αξ /P°)] − ξ[μC° + RT ln(PC0/P°)] − (1 + ξ)RT = ξ[Δr μ° + RT ln(ξ /1 mol)] − (1 + ξ)RT (33)

since nD(ξ) = ξ, PD(ξ) = αξ, P0C = αnC(0), and ntotal = nC(0) + nD(ξ) = 1 mol + ξ. Figure 4 shows a plot of the transformed Helmholtz free energy versus ξ. Equilibrium occurs when A′(ξ) attains its minimum value. This takes place when ξ = G

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given by eq 3 as PC(ξ) = P0C = αnC(0), while the analogous equation PD(ξ) = αξ gives the partial pressure of D. Because nC(0) = 1 mol, direct substitution of these results into eq 36 and rearrangement yields the negative of the chemical force, (∂Atotal/∂ξ)T,V, as ij ∂A total yz jj z = (μD° − μC°) + RT ln(PD(ξ)/P°) j ∂ξ zz k {T , V − RT ln(PC0/P°) ÄÅ É ÅÅ P (ξ) ÑÑÑ D Å Å = Δr μ° + RT lnÅÅ 0 ÑÑÑÑ ÅÅÅ PC ÑÑÑ Ç Ö ° = Δr μ + RT ln(ξ /1 mol).

Figure 4. Variation of the total transformed Helmholtz free energy A′total with the extent of reaction ξ for the open unimolecular isomerization reaction C(g) → D(g) under the experimental conditions described in the text.

(37)

This result is easily seen to be identical to the expression for (∂A′total/∂ξ)T,V in eq 30 obtained using Legendre-transformed thermodynamics since α/PC° in eq 30 is equal to 1 mol−1. Thus, Figure 3 provides the negative of the chemical force whether we use transformed or nontransformed thermodynamics. This is always the case for a reaction in which one or more chemical species have fixed chemical potentials. For processes at constant T and V, we always have dA = dA′. For processes at constant T and P, we always have dG = dG′. However, it is very important to note that while we always

2.2407 mol, which is in precise agreement with the result obtained using the chemical force.



OPEN SYSTEM ANALYSIS: NONTRANSFORMED METHOD The analysis presented in the previous section demonstrates how the thermodynamics of an open system with some chemical species held at constant concentration may be accurately treated using a Legendre transformation. However, it should not be presumed that these methods must be employed. We can obtain identical results without the use of a Legendre transformation provided that we are careful to insert the constraint of a constant concentration and partial pressure for C and accurately account for the contribution of the surroundings to the Helmholtz free energy. At constant temperature and volume, the starting point is eq 23a, which we repeat here for convenience: dA sys = μC dnC,sys + μD dnD (34)

brium constant for the transformed analysis is not the same as that for the nontransformed calculation, as shown by eqs 31 and 32. This is not surprising, since eqs 30 and 31 show that Δrμ° ≠ Δrμ′°. From eq 32, we have Kp′ = (PD/P°)eq = exp(−Δrμ′° /RT) = 0.5555, whereas at equilibrium eq 37 yields

The contribution of the surroundings to the total Helmholtz free energy is the same whether we execute a Legendre transformation on Asys or not. Therefore, at a constant chemical potential for C, this contribution is still described by eqs 23b and 25: dAsur = −μC dnD. Addition of this contribution to eq 34 gives the sum of the differential changes in the system and surroundings, which equals the total differential change in the Helmholtz free energy:

In spite of the fact that these equilibrium constants are different, the predicted partial pressures and concentrations of C and D at equilibrium are exactly the same. For the nontransformed calculation, we have (PD)eq = P0C Kp = [α(1 mol)](2.2407) = 0.5555 bar and (nD)eq = (PD)eq/α = 2.2407 mol, which are in agreement with the transformed results given by eq 32. Consequently, the amount of C that must be added to the system from an external source is also the same for both calculations (2.2407 mol).

have

∂A total ∂ξ

=

T ,V

(

∂A′total ∂ξ

)

for such systems, the equili-

T ,V

ji P /P° zyz K p = jjj D z = exp( −Δr μ° /RT ) = 2.2407. j PC/P° zz k {eq

dA sys + dA sur = μC dnC,sys + μD dnD − μC dnD = dA total

Moreover, although we have

(35)

The constraint that the concentration of C in the system remain constant requires that we have dnC,sys = 0. The chemical potentials in eq 35 are still given by the usual ideal gas expressions μi = μ°i + RT ln(Pi(ξ)/P°). Substituting for the chemical potentials in eq 35 gives

( ) ∂A total ∂ξ

=

T ,V

(

∂A′total ∂ξ

(38)

)

, we

T ,V

do not have A(ξ)total = A′(ξ)total. As before, we have Atotal = Gsys + Gsur − PV = nC,sysμC + nDμD − nDμC − PV. Writing the righthand side of this equation in terms of ξ, α, the standard chemical potentials, and T, we obtain A(ξ)total = (1 mol)[μC° + RT ln(PC0/P°)]

dA total = dA sys + dA sur

+ ξ[μD° + RT ln(αξ /P°)]

= [μD° + RT ln(PD(ξ)/P°) − μC° − RT ln(PC(ξ)/P°)] dnD

( )

− ξ[μC° + RT ln(PC0/P°)] − (1 + ξ)RT .

(36)

(39)

Equation 6a still defines the extent of reaction variable, so dnD = dξ and nD = ξ. The (constant) partial pressure of C is still

Direct comparison of eq 39 with eq 33 shows that H

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Journal of Chemical Education ÄÅ É ÅÅ i α yzÑÑÑÑ zzÑÑ A(ξ)total = A′(ξ)total + ÅÅÅ4000 J + (1 mol)RT lnjjj ÅÅÇ k 1 mol {ÑÑÖ = A′(ξ)total + 542.47 J (40)

Article

Equation 40 provides the underlying reason why transformed and nontransformed thermodynamics, when applied to open chemical systems with one or more compounds having fixed chemical potentials, always predict the same results for any observed chemical property. In effect, all that the Legendre transformation does in such a system is alter the reference point for A or G by a constant amount. Consequently, this change has no effect on the slope of the plot of A or G versus ξ that provides the chemical force, the criterion for spontaneity, and the equilibrium states. Furthermore, if we compute the change in A or A′ for any finite change of state, we will always obtain ΔA = ΔA′. The same is true for G and G′ for systems at constant temperature and pressure.

Figure 5. Comparison of the chemical force and equilibrium point predicted by the transformed Helmholtz free energy with the results for the nontransformed potential for the closed C → D system described in the text.



ji μ ° + RT zyz P° ξeq = expjjj− D = 0.6623 bar z j RT zz{ α k

CLOSED SYSTEM ANALYSIS: LEGENDRE TRANSFORMATION METHOD It should not be presumed that a Legendre transformation on a term or terms involving nonfixed species in a chemical reaction will yield the correct results for spontaneity and equilibrium. This can be shown unequivocally by performing such a transformation on the (nC, μC) pair in the closed C → D reaction. If this is done, we obtain dA′ = μD dnD − nC dμC

with

A′ = A − nCμC

which is incorrect. Figure 5 shows a comparison of with

(42)

for the closed system. As can be seen, the

( ) ∂A ∂ξ

T ,V



( ) ∂A′ ∂ξ

is that

T ,V

the quantity we are subtracting from A to obtain the transform A′ in eq 42 is nCμC, which is not a constant but depends upon ξ. We can of course add d(nCμC)/dξ to the right-hand side of eq 44 to obtain the correct result. However, that operation just reverses the Legendre transformation that we executed at step one in the analysis. The essential point is that we cannot investigate equilibrium and spontaneity using a Legendre transform of the relevant thermodynamic potential if the transformation involves a quantity that depends upon the extent of reaction. The same thing occurs when we obtain G by transforming A via 3 (PV) dA = dG, where the transform of A is A + PV = G. If we are interested in investigating equilibrium for a gas-phase reaction under ideal conditions that occurs in a fixed-volume reaction vessel at constant temperature, the relevant thermodynamic potential is A. Can we accurately execute the investigation using the transform of A, the Gibbs free energy G? The answer is equivocal: maybe we can, and maybe we cannot. If the reaction stoichiometry has the change in the total number of moles of gas in going from reactants to products, Δν, equal to zero, then PV is a constant independent of the extent of reaction, and we may use either A or its transform G. However, if Δν ≠ 0, as it is for the reaction 2C(g) → D(g), then PV is not constant but depends upon ξ, and we cannot use the transform to examine spontaneity and equilibrium. For solution reactions, the nearly zero compressibility of the liquid phase and the nearly constant barometric pressure make PV effectively a constant, which means that we may use either

(43)

The differential in the second term on the right-hand side of eq RT 43 is easily shown to be − 1 − ξ dξ since μC° is constant at a

fixed temperature. Therefore, eq 43 yields

(44)

This is clearly not the same as the result shown in eq 12 obtained using the nontransformed method: ij ∂A yz jj zz = Δr μ° + RT ln[ξ /(1 − ξ)] j ∂ξ z k {T , V

T ,V

T ,V

The reason why we now have

dA′ = [μD° + RT ln(αξ /P°)] dξ − (1 − ξ)

ij ∂A′ yz jj z = μD° + RT ln(αξ /P°) + RT j ∂ξ zz k {T , V

∂A ∂ξ

chemical force predicted by the transformed potential is correct only at the crossing point of the two curves. The direction for spontaneous reaction is incorrectly predicted over the range 0.6623 mol ≤ ξ ≤ 0.6914 mol.

From a comparison of eq 42 with eq 5, it is easy to see that in order for the transformed free energy and the untransformed free energy to give equivalent results, one would have to have μC dnC = −nC dμC, or μC dnC + nC dμC = d(nCμC) = 0. However, if this were the case, then the Legendre transform would have no effect at all! This analysis also shows that the requirement that we have dA′ = dA, namely, d(nCμC) = 0, requires that the nCμC product be a constant. To demonstrate, in the closed system we have no need to consider the effect of an external source for compound C. Therefore, (dA′)sur = 0 and eqs 6 are still valid, so eq 42 becomes

d{μC° + RT ln[α(1 − ξ)/P°]}

( ) ∂A′ ∂ξ

( )

(45)

If we erroneously assume that the chemical force is given by the negative of the expression in eq 44 and attempt to find the equilibrium point by setting its value to zero, the result is I

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A or its Legendre transform G to investigate spontaneity and equilibrium in such systems.

the system react to form the pseudoisomers. It should be noted that eq 47 has exactly the same form as eq 25 if that equation is integrated from the starting point to equilibrium since μC is a constant. Therefore, Alberty’s total transformed Gibbs Free energy that includes the contribution of the surroundings is



OVERVIEW OF EQUILIBRIUM CALCULATIONS ON THE HYDROLYSIS OF ATP Alberty1−3 and Iotti et al.14 previously reported thermodynamic calculations on the equilibrium state for ATP hydrolysis using Legendre-transformed and nontransformed methods, respectively. The calculations employed equilibrium constants obtained from standard chemical potentials corrected for constant ionic strength using the extended Debye−Huckel equations. Chemical forces along the 14 reactions involved in the process were not addressed. That is, the calculations did not compute the values of (∂G/∂ξi)T,p or (∂G′/∂ξi)T,p (1 ≤ i ≤ 14) as functions of the ξi or determine the most probable thermodynamic pathway leading from the initial state to the global equilibrium state.17 The emphasis was on the determination of the equilibrium concentrations for a given set of initial conditions at fixed pH and pMg and from those concentrations the associated changes in G, H, and S and the transformed potentials for the initial state going to the equilibrium state. These studies have been fully described in refs 1−3 and 14. Here we will simply note the similarities between the results for this very complex system and those obtained for the simple model system treated here. With both pH and pMg held constant in the ATP hydrolysis at constant T and P, four constraints are present. Consequently, after first omitting the terms in dT and dP, Alberty1−3 proceeded by executing two Legendre transformations, 3 (nHμH) and 3 (nMgμMg), to obtain the transformed Gibbs free energy for the system: G′sys = Gsys − (nHμH + nMg μMg )sys

G′total = Gsys − (nHμH + nMg μMg )sys −

i

∑ [γi ,HniμH + γi ,MgniμMg ] i

(48)

Alberty labeled the left-hand side of 48 as G′ rather than G′total. That is, he did not state that the Legendre transform of Gsys comprises only the first three terms on the right-hand side of the equation while the remaining terms incorporate the contribution from the surroundings. This procedure probably played a role in leading some scientists to conclude that there exist two different types of thermodynamics. Iotti et al.14 proceeded by employing cumulative formation constants to obtain binding polynomials that permitted them to compute the stoichiometric coefficients for all of the processes. This determination permitted them to introduce the constraints of constant pH and pMg by simply balancing all of the chemical equations for mass and charge. In the process, the amounts of H+ and Mg2+ transferred to the system as well as their contribution to the Gibbs free energy were automatically determined along with the equilibrium constants and all of the equilibrium concentrations. Comparison of the results obtained by the two methods shows the following: ΔGtotal = ΔG′total and all of the equilibrium concentrations are the same in spite of the fact that K ≠ K′, ΔrG° ≠ ΔrG′°, and Gtotal ≠ G′total. As noted previously, none of the reported studies1−3,14 determined the thermodynamic most probable pathways and associated chemical forces, −(∂G/∂ξi)T,p or −(∂G′/∂ξi)T,p (1 ≤ i ≤ 14) as functions of the ξi. However, if such studies were performed using the methods described in ref 17 or otherwise, suitably modified to include the contribution from the surroundings, we can be certain that the transformed and nontransformed results would be identical. This is simply a reflection of the fact that the terms subtracted from Gsys to obtain G′sys in eq 46 are all constants when the pH and pMg are fixed. Consequently, we will always have dGsys = dG′sys. The results for the model system have exactly the same characteristics when G and G′ are replaced by A and A′, respectively. .

(46)

It should be noted that G′sys has exactly the same form as eq 24 except for the fact that the relevant thermodynamic potential at constant T and V is A whereas at constant T and P, as in the ATP hydrolysis, G is the relevant potential. To obtain the contribution from the surroundings to G′ at the equilibrium point, Alberty had to determine the amount of H+ and Mg2+ that must be transferred from the surroundings to the system at the equilibrium point. The amounts transferred for an arbitrary set of ξi values were not required. The transferred H+ and Mg2+ ions form the 12 pseudoisomers present in the ATP hydrolysis. These are HATP3−, H2ATP2−, MgATP 2− , MgHATP − , Mg 2 ATP, HADP 2− , H 2 ADP − , MgADP −, MgHADP, HPO 4 2− , H 2 PO 4 − , and MgHPO 4. Consequently, the transferred chemical amounts are given by the summation over the 12 pseudoisomers of the product γi,Hni for H+ and γi,Mgni for Mg2+, where γi,H and γi,Mg are the numbers of H+ and Mg2+ ions, respectively, present in pseudoisomer i (either 0, 1, or 2) and ni is the chemical amount of that pseudoisomer at equilibrium. Since the chemical potentials for H+ and Mg2+ are constant and the transfer from the surroundings is assumed to be reversible, the contribution of the surroundings to the total Gibbs free energy is G′sur = −

∑ [γi ,HniμH + γi ,MgniμMg ]



CONCLUSIONS The fascination afforded by the study of thermodynamics lies, in large measure, in the fact that most classroom and research problems can be addressed successfully by several methods that often appear to be very different but, perhaps surprisingly to some, yield exactly the same correct answers. Such is the case for open-system reactions with one or more chemical potentials held constant. They can be successfully investigated using Legendre transform methods or standard nontransformed thermodynamics with essentially the same degree of computational effort. Therefore, biochemical and chemical thermodynamics have been reunified. This situation was perhaps most elegantly stated by Gertrude Stein (1874−1946) and William Shakespeare (1564−1616) who respectively wrote, Civilization began with a rose. A rose is a rose is a rose is a rose. ...a rose by any other name would smell as sweet.

(47)

where the summation runs over all 12 pseudoisomers and μH and μMg are the constant chemical potentials of H+ and Mg2+, respectively. The minus sign in 47 is a result of the fact that all of the H+ and Mg2+ ions transferred from the surroundings to J

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Biophysical Chemistry Division. https://iupac.org/projects/projectdetails/?project_nr=2017-021-2-100. (16) Raff, L. M. Spontaneity and Equilibrium: Why “ΔG < 0 Denotes a Spontaneous Process” and “ΔG = 0 Means the System is at Equilibrium” Are Incorrect. J. Chem. Educ. 2014, 91 (3), 386−395. (17) Raff, L. M. Spontaneity and Equilibrium II: Multireaction Systems. J. Chem. Educ. 2014, 91 (6), 839−847. (18) Silverberg, L. J.; Raff, L. M. Are the Concepts of Dynamic Equilibrium and the Thermodynamic Criteria for Spontaneity, Nonspontaneity, and Equilibrium Compatible? J. Chem. Educ. 2015, 92 (4), 655−65.

So it is with thermodynamics: thermodynamics is thermodynamics is thermodynamics, which by any other name, transformed or not, smells equally sweet.



AUTHOR INFORMATION

Corresponding Authors

*E-mail: lionel.raff@okstate.edu. *E-mail: William.cannon@pnnl.gov. ORCID

Lionel M. Raff: 0000-0001-5622-8862 Notes

The authors declare no competing financial interest.



ACKNOWLEDGMENTS W.C. was supported by the National Institute of Biomedical Imaging and Bioengineering through Award U01EB022546 and by the U.S. Department of Energy, Office of Biological and Environmental Research, through Projects 69513 and 72228. PNNL is operated by Battelle for the U.S. Department of Energy under Contract DE-AC06-76RLO.



REFERENCES

(1) Alberty, R. A. Equilibrium calculations on systems of biochemical reactions at specified pH and pMg. Biophys. Chem. 1992, 42, 117−131. (2) Alberty, R. A. Calculation of transformed thermodynamic properties of biochemical reactants at specified pH and pMg. Biophys. Chem. 1992, 43, 239−254. (3) Alberty, R. A. The fundamental equation of thermodynamics for biochemical reaction systems. Pure Appl. Chem. 1993, 65 (5), 883− 888. (4) Alberty, R. A.; Goldberg, R. N. Standard thermodynamic formation properties for the adenosine 5′-triphosphate series. Biochemistry 1992, 31, 10610−10615. (5) Alberty, R. A. Levels of thermodynamic treatment of biochemical reaction systems. Biophys. J. 1993, 65 (3), 1243−1254. (6) Alberty, R. A. Inverse Legendre Transform in Biochemical Thermodynamics: Illustrated with the Last Five Reactions of Glycolysis. J. Phys. Chem. B 2002, 106 (25), 6594−6599. (7) Alberty, R. A. Thermodynamics of Biochemical Reactions; John Wiley & Sons: Hoboken, NJ, 2003. (8) Gunner, M. http://www.sci.ccny.cuny.edu/~gunner/Pages-422/ PDF/unt/Gennis-3-thermo-bio-rxn.pdf. (9) Alberty, R. A. Thermodynamic properties of enzyme-catalyzed reactions involving guanine, xanthine, and their nucleosides and nucleotides. Biophys. Chem. 2006, 121 (3), 157−162. (10) Alberty, R. A. Thermodynamic properties of enzyme-catalyzed reactions involving cytosine, uracil, thymine, and their nucleosides and nucleotides. Biophys. Chem. 2007, 127 (1−2), 91−96. (11) Alberty, R. A.; Cornish-Bowden, A.; Goldberg, R. N.; Hammes, G. G.; Tipton, K.; Westerhoff, H. V. Recommendations for Terminology and Databases for Biochemical Thermodynamics. Biophys. Chem. 2011, 155, 89−103. (12) Flamholz, A.; Noor, E.; Bar-Even, A.; Milo, R. eQuilibrator the biochemical thermodynamics calculator. Nucleic Acids Res. 2012, 40 (D1), D770−D775. (13) Sabatini, A.; Vacca, A.; Iotti, S. Balanced Biochemical Reactions: A New Approach To Unify Chemical and Biochemical Thermodynamics. PLoS One 2012, 7, e29529. (14) Iotti, S.; Raff, L.; Sabatini, A. Chemical and biochemical thermodynamics: Is it time for a reunification? Biophys. Chem. 2017, 221, 49−57. (15) Project Details: Chemical and Biochemical Thermodynamics Reunification. IUPAC Project No. 2017-021-2-100, Physical and K

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