Uttam,

What standard states are you using, and what are the reaction conditions of interest? Free energy numbers are meaningless without that information.

It is piroxicam (a drug).

Have a look at that thread: https://www.researchgate.net/post/Is_it_possible_to_calculate_the_solubility_of_some_molecule_in_water_by_MD_simulation_If_so_how

I know the value of the free energy of solvation. I want to know whether there is a relation between the free energy of solvation and the solubility so that I can calculate the solubility in mg/ml.

I do not know that there is a direct relationship between free energy of solvation and solubility.

Solvation can be treated in exactly the same way as a chemical reaction.

When you say “Free energy of solvation” you have to also specify the thermodynamic states of the reactants and products. The specification has to be rather detailed. I will assume the temperature is constant and at a specified value. If the starting point is a gas then you have to specify the pressure or something with equivalent information, such as the number density. The same goes for the material in solution. You have to specify the concentration or some other equivalent information.

Anyway, there is a direct and rigorous relationship between the Gibbs free energy of a process and the equilibrium constant. You can find the equation in many places, such as most any book on chemical thermodynamics. Here is one link to the equation.

http://www.chem.purdue.edu/gchelp/howtosolveit/Thermodynamics/K_from_DelG.html

You will see in that equation a quantity delta G with a superscript of zero. That quantity is the Gibbs free energy of the reaction when all the reactants and products are in their "standard states." This means that you will have to learn the meaning of “standard states”, something that can sometimes be surprisingly subtle and tricky. Even chemists, who are the ones who should be experts on this topic, sometimes get confused. You will also have to learn the concept of “activity” as the term is used in chemical thermodynamics.

Anyway, in the case of solubility you are usually talking about a solid in equilibrium with a saturated solution. The activity of the solid has a value of 1. (It is 1 by definition.) The activity of the solute is related to the concentration. If the concentration of the solute is very low and the solute is otherwise pure then the activity can normally be set equal to the concentration. If the concentration of the solute is high then the solution becomes “non-ideal” and one has to deal with some quantities known as activity coefficients.

If you are interested in the solubility of a solid, and the information on free energy that is available to you is for a gas-phase starting point then you will need some additional information, namely the free energy of vaporization.

You can work out the correct expressions by considering thermodynamic cycles for chemical reactions. I won’t try to write them here, but hopefully this is enough to get you started.

Your best source of information to deal with this in more depth would be in textbooks, either chemical thermodynamics (e.g. authored by Klotz or Lewis and Randall, or other texts) or physical chemistry. The chemical thermodynamicists and physical chemists have worked this out in excruciatingly rigorous detail.

COSMO-RS and COSMOtherm provide an excellent theoretical background and practical solution for your problem. You can do without gas phase calculations. In short, unimolecular DFT calculations for the solute and the solvent molecules provide two chemical potentials, namely for the solute in the solvent and for the solute in the solute. The difference of chemical potentials basically gives you the thermodynamic solubility constant of the solute in the solvent. Works for arbitrary solvents, solvent mixtures, and solutes.

For an introductory article, please see

http://onlinelibrary.wiley.com/doi/10.1002/jcc.1168/abstract

and check

http://www.cosmologic.de/index.php

for more information.

A good book to check is

Solvation Thermodynamics by Arieh Ben-Naim

Please check this paper it may help you to answer your question

J. Phys. Chem. B, 2001, 105 (32), pp 7775–7782

DOI: 10.1021/jp0036902

If I understood you right, you calculated the free Gibbs energy of solvation of the pure substance. Hence, it is related to a equilibrium constant for solubility and this again to a equilibrium concentration which can be converted into solubility.

ln K = -Delta_solv G / (RT) = ln c(aq); c(aq) = beta (aq) / M

I put it the other way round. With a rough value of 15-20 mg/L as solubility (or beta (aq)) at 25°C and 1 bar I found on the web, the concentration is around 0,05 mM, leading to a free energy of solvation of around 25 kJ/mol. Does this somehow fit to your value?

Hi Michael, I am getting a very high value of Delta_G ( around 60 kJ/mol) from simulation. I have found in the literature that the material is sparingly soluble in water.

Uttam,

What standard states are you using, and what are the reaction conditions of interest? Free energy numbers are meaningless without that information.

The answers Alan gave to you are right, and as far as I know, the only way you have to calculate the equilibrium constant of your process, i.e. the solubility of your material. The equation reported by Michael:

ln K = -Delta_solv G / (RT)

is only partly correct, as the Delta_solv G^0 has to be used instead of the "simple" Delta G. This tiny but fundamental different may completely change your results.

One thing I don't see in any of the answers so far is mention of the form of solid state. A thermodynamic solubility calculation will require a calculation of the enthalpy of formation of the crystal form you wish to dissolve,as this is one of the reference states. So you will need an appropriate crystal structure. Different crystal structures for a single material (called polymorphs) often have very different solubilties (much to the discomfort of, for instance, the pharmaceutical industry) because of differences in molecular packing. If you have a reference to a determined crystal structure for piroxicam you can access an electronic file of the structure as submitted, from the Cambridge Crystallographic Data Centre for free.

I am not an expert in energy calculations on crystal structures. Maybe others can advise on how best to do that.

John Liebeschuetz is correct. The form of the solid state needs to be specified, and any free energies relevant to that state need to be taken into account.

This is a specific example of the general principle mentioned earlier that the thermodynamics states of all reactants and products need to be carefully and fully specified.

Here is a link that elaborates on the point that John Liebeschuetz made, including some diagrams to illustrate some priniciples.

http://particle.dk/stability-of-crystalline-structure/

Let me elaborate on the role of thermodynamic states. This may take more than one post. During all of this discussion it will be assumed that the processes are occurring at constant temperature and constant total pressure.

Let me start by discussing the concept of “escaping tendency”. This is a general term that refers to the tendency of a component to leave a phase. For example, a volatile component of a phase tends to “escape” from a phase via evaporation. If a compound exists in two different phases, such as two solutions that at different concentrations, the compound in the solution of greater concentration will have a greater tendency to vaporize from solution.

This greater tendency to vaporize is reflected in a higher vapor pressure. It is a general principle that when comparing two phases with different vapor pressures, the phase of greater vapor pressure is the one with greater escaping tendency with regard to the compound of interest.

If two phases with different vapor pressure are placed in proximity with each other, such that the vapors from the two phases can freely intermingle, the phase of higher vapor pressure will lose mass and the phase of lower vapor pressure will gain mass via vaporization of the volatile component, and that mass will be transferred to the phase of lower escaping tendency.

I have illustrated this using transfer of mass via vapor, but the principle is a general principle, regardless of the mechanism of transfer. For example, if there are two solutions that are able to exchange a solute by any mechanism, be it vaporization or some other mechanism, the solution of higher vapor pressure will lose solute molecules, and the solution of lower vapor pressure will gain solute molecules.

This principle is also general in the sense that the phase of high escaping tendency of a particular component will lose mass to a phase of low escaping tendency. Thus, vapor pressure is a valid measure of escaping tendency, but as we shall see later it is not the only measure of escaping tendency, nor is it necessarily the most convenient measure.

One important conceptual feature of the relationship between escaping tendency and vapor pressure is that it is a monotonic relationship, i.e. higher vapor pressure always implies higher escaping tendency and higher escaping tendency always implies higher vapor pressure. However, the relationship between escaping tendency and vapor pressure is not otherwise defined. For example, an escaping tendency scale could be defined as equal to the vapor pressure, or it could be defined as the square root of vapor pressure, or any other monotonic relationship.

However, by convention the term “escaping tendency” is often taken to be synonymous with the term “fugacity”. The root of the word “fugacity” comes from a Latin word that means to “flee”. It has the same root as the word “fugitive”. You can therefore see that the term “fugacity” is conceptually connected to the term “escaping tendency”.

In order to keep the length of these posts at least somewhat manageable I will defer the definition of fugacity to a later post, as well as a discussion of how this relates to the specification of thermodynamic states and free energy calculations.

In an earlier post I introduced the term “fugacity”, but I did not define it, other than to say that it is sometimes used as synonymous with “escaping tendency”.

To define fugacity it is useful to start by discussion the relationship between Gibbs free energy and pressure for an ideal gas. For an ideal gas of a specific material (such as hydrogen or helium) at two different pressures, P1 and P2, the Gibbs free energy for the process of compressing or expanding the gas under isothermal conditions is given by the expression

Delta(G)=RTln(P2/P1) (1)

where Delta(G) is the Gibbs free energy of the process, R is the gas constant, T is the absolute temperature, ln() is the natural logarithm function, P2 is the final pressure, and P1 is the initial pressure. If R is expressed in units of moles then equation (1) gives the Gibbs free energy for the process applied to one mole. If R is expressed on a per-molecule basis then it equals k, Boltzmann’s constant, and in that case equation (1) refers to the Gibbs free energy per single molecule. We will assume R is given in units of moles.

Equation (1) is a general equation in the sense that applies to all ideal gases, as long as the temperature is high enough and the pressures are low enough so that one can ignore the Boson or Fermion character of the gases. In other words, they are not “degenerate” in the statistical mechanical sense.

Real materials at finite pressure are never ideal gases, so equation (1) never applies to any real gas at finite pressure. (This concept that a real gas never obeys equation (1) is important in the definition of fugacity.) However, equation (1) becomes true in the limit of low zero pressure for P2 and P1 because gases become ideal in the limit of zero pressure.

It turns out that it is convenient to define a new function that looks just like equation (1) using different symbols in place of P1 and P2. The new function is the fugacity, f, and the related equation is

Delta(G)=RTln(f2/f1) (2)

And f is defined so that equation (2) is always true for real gases.

This is not quite enough information to fully define the fugacity. One must also specify that f approaches P as P approaches 0. In other words, in the limit of zero pressure the fugacity equals the pressure.

The beauty of equation (2), which is a very simple relationship, is that it always gives the Gibbs free energy of a process in which a material starts out with a fugacity of f1 and ends with a fugacity of f2. It is made to be true by definition.

Let us take a specific example, Suppose one were to compress a sample of carbon dioxide gas (i.e. CO2), starting at pressure P1 and ending at pressure P2. The starting fugacity would be f1, and the ending fugacity would be f2. The Gibbs free energy for the process is given by equation (2). We don’t even have to know what the actual values of f1 and f2 are to know that this is true, nor do we need to know what the values of P1 and P2 are to know that equation (2) is true. It is simply true because we defined fugacity in a way to guarantee that equation (2) is true. At this point I am not saying that we can calculate the actual value of the Gibbs free energy of the process without knowing more information, but I am saying that Gibbs free energy of the process is always related to fugacity via equation (2).

Let me break for now and come back later with more discussion.

I am actually very interested in how Alan is explaining it. Looking forward to further "answers".

The sentence: "It is made to be true by definition." somehow encloses the poetry of thermodynamics, which gives you true answers by definition.

Things become more interesting when one starts to deal with systems of variable composition, such as solutions.

Let us start by considering the transfer of CO2 between two aqueous solutions, the two solutions being at different concentrations. Let us suppose that we have one sample of each solution, and that each of the two solutions is in equilibrium with CO2 vapor. This does not mean that the two solutions are in equilibrium with the same sample of gas. In fact, we need to assume that the two systems are isolated from each other. Solution 1 is in equilibrium with a sample of gas, and solution 2 is in equilibrium with a different sample of gas. For convenience we will assume that both phases are subjected to the same total pressure.

Let us now calculate the Gibbs free energy for the transfer of a mole of gas from solution 1 to solution 2. We will assume that the solutions are large, so that the transfer of a mole of gas does not make a significant change in concentration of either solution.

For conceptual purposes we break this overall process down into several different process, such that when we combine the individual steps it equals the overall process.

The first process is the vaporization of a mole of CO2 from solution 1. Because the solution is in equilibrium with CO2 vapor the Gibbs free energy of the process is zero. (It is always true that the Gibbs free energy of an isothermal process taking place at constant pressure is zero. The function was defined so that this is true.)

The CO2 that is to be transferred is now in the gas phase at pressure P1. The fugacity is f1.

Let us now assume that there is a semipermeable membrane in the gas above Solution 1. The membrane is permeable to CO2 but not to other gases. Let us further assume that there is a piston that is initially in contact with the membrane, and that the piston is raised slowly and synchronously with the transfer of CO2 from solution 1 to the gas phase. The rate of raising the piston is chosen to match the rate at which CO2 leaves solution 1. The net effect is that all of the CO2 ends up on the piston-side of the semi-permeable membrane. This process is performed slowly enough so it is reversible, and the Gibbs free energy of the process is therefore zero.

Next, let us separate the piston chamber from the membrane, and then compress the CO2 under isothermal conditions until the pressure is equal to P2. The fugacity of the CO2 is then at f2. The Gibbs free energy of this process is given by

Delta(G)=RTln(f2/f1) (2)

I have transferred the equation number (2) from an earlier post rather than restarting the numbering scheme.

Next, transfer the CO2 into solution 2 using a process that is analogous to the reverse of the process of transferring CO2 out of solution 1. The Gibbs free energy of that process is zero.

The Gibbs free energy for the overall process is just the sum of the Gibbs free energies for the individual steps, or in other words

Delta(G)=RTln(f2/f1) (3)

This of course is identical to equation (2).

This equation tells us that we can calculate the Gibbs free energy of the transfer of CO2 from one solution to another by using equation (3). The same applies to any other solute, such as naphthalene, or sucrose, or cytochrome C.

Notice that I have not said anything about how to obtain actual values of f1 and f2. This is not always easy. In fact, it is almost never easy to obtain values of fugacities, and for most solutes it is a practical impossibility.

However, this need not deter us. Stay tuned later for a discussion of thermodynamic activity.

Continuing the discussion of fugacity, so far I have mostly framed the issue in terms of the Gibbs free energies of compressing of gases and transfer of solutes between solutions with solutes of different concentrations. However, in framing the discussion in terms of solutions and solutes nowhere did I make use of the fact that the materials are solutions. They can be any sort of phases.

For example, if one of the phases were pure crystalline naphthalene and the other phase were an aqueous solution of naphthalene at some specified concentration exactly the same analysis would apply. The free energy for the transfer of naphthalene from the crystalline state to a solution can be broken down into the same set of steps as discussed above for the transfer of a solute between to solutions. The only difference is that on one side of the process naphthalene vapor is considered to be in equilibrium with solid naphthalene, and the vapor pressure of solid naphthalene is P1 and the fugacity of solid naphthalene is f1. On the other side of the process, naphthalene vapor is in equilibrium with a solution of specified concentration.

The same applies to solids of two different crystalline structures. For example, at room temperature carbon exists in at least two different crystalline structures, graphite and diamond. Graphite is the more stable form at ordinary temperatures and pressures, and diamond is a metastable form, but both forms can exist under ordinary ambient conditions. There is a Gibbs free energy associated with the transfer of carbon from diamond to graphite forms. That free energy is given by equation (3):

Delta(G)=RTln(f2/f1) (3)

where f1 is the fugacity of carbon vapor in equilibrium with diamond and f2 is the fugacity of carbon vapor in equilibrium with graphite.

As a side comment, the vapor pressure of carbon in equilibrium with either of those two phases is so low that one could safely substitute P1 and P2 in place of f1 and f2 in the expression. Also, as a practical matter, one can’t directly measure the vapor pressure of either phase, so one would never apply equation (3) directly to diamond and graphite, but it illustrates the principle, and we will return latter to the question of how to deal with systems for which the vapor pressure is too low to measure.

Now let us bring a few of these ideas together. Let us discuss a concrete example, ranatidine hydrochloride, which is an important drug. For short I will just call ranitidine hydrochloride by the term “ranitidine”, even though it would be more correct to use the full name, “ranitidine hydrochloride”.

Crystalline ranitidine has a certain vapor pressure and a certain fugacity. Similarly, aqueous ranitidine of a particular concentration has a certain vapor pressure and a certain fugacity. Based on the discussion presented above, the Gibbs free energy of transfer of ranitidine from the solid crystal to the solution is given by the following equation, designated as equation (3) in some earlier posts

Delta(G)=RTln(f2/f1) (3)

where f1 is the fugacity of ranitidine in crystalline form, and f2 is the fugacity of ranitidine in the solution of a specified concentration.

The quantity f1, being the fugacity of the pure crystalline solid at a specified temperature, is a constant. The quantity f2, being the fugacity of a solution, is not a constant but depends on the concentration of the solution. This has an extremely important implication that relates to some comments in earlier posts. The value of Delta(G) in equation (3) is not a single unique value, but rather it depends on the f2, the fugacity of the ranitidine solution, which in turn depends on the concentration of the solution. In fact, there is no lower bound to this function. If we pick f2 to be extremely dilute then Delta(G) can be an extremely negative value, and in the limit of zero concentration the value of Delta(G) approaches negative infinity. This is at the heart of earlier comments that one must carefully and fully specify the thermodynamic states of all reactants and products. Otherwise “Gibbs free energy” has no meaning.

An interesting case occurs when f1=f2. In that case the Gibbs free energy is zero. It is a universal truth that equilibrium occurs when Delta(G) is zero. Therefore, equilibrium occurs when the fugacity of the reactant equals the fugacity of the product. In this case it means that the fugacity of the crystalline solid equals the fugacity ranitidine in the solution.

As a side comment, in this case the vapor pressures are so low that one could probably substitute vapor pressures in place of fugacities in expression (3).

In a more general sense, when a component is in equilibrium between two phases the escaping tendency of that component is the same with respect to the two phases. Fugacity just happens to be the most convenient measure of escaping tendency. Vapor pressure is also the same for a component in equilibrium between two phases, though there are a few subtleties about vapor pressure equivalence that I won’t discuss here.

Ranitidine is one of the examples I used in my last post. This is an interesting example because, like carbon, it can exist in more than one crystalline form, i.e. there are polymorphs. Whenever a substance can exists as polymorphs, at any specific temperature and pressure one of the forms is almost always more stable than the other, i.e. the Gibbs free energy of the two forms is different, and there is a thermodynamic driving force to convert the less stable form to the more stable form. This is manifest in a difference in escaping tendency of the two forms. In other words, the vapor of the less stable form is higher than that of the more stable form, and more importantly the fugacity of the less stable form is higher than the fugacity of the more stable form.

One implication of this is that if a crystal of each form were placed in the same container there would be, in principle, a net transfer of matter from the less stable form to the more stable form via net evaporation of matter from the less stable form and net deposition of matter on the more stable form. This process may be fast or slow. Ordinary equilibrium thermodynamics says nothing about the rate of the process, only the overall energetics. In fact, the process could be too slow to observe, in which case we would consider the less stable form to be a metastable form. The less stable polymorph of ranitidine likely falls into this category.

The difference in fugacity has an interesting effect on solubility. Recall that when a solubilization reaction reaches equilibrium the fugacity of the solute in solution equals the fugacity of the undissolved solute. Consider then two cases, one in which a saturated solution is in contact with the more stable crystalline form of ranitidine and one in which a saturated solution is in contact with the less stable crystalline form of ranitidine.

For sake of discussion let us assume that ranitidine is unable crystalize from solution into a specific crystalline form in the absence of a seed crystal of that same crystalline form. This means that the sample discussed above that is saturated by dissolution of the less stable crystalline form of ranitidine does not crystalize out into the more stable form unless a seed crystal of the more stable form is added.

One of the properties of solutions is that the fugacity of a solute increases as the concentration of the solute increases. As noted above, the less stable crystalline form of ranitidine has a higher fugacity than the more stable form, and the saturated solution has the same fugacity as the crystal. The three facts noted earlier in this paragraph imply that the concentration of ranitidine solution in equilibrium with the less stable crystalline form of ranitidine is higher than the concentration of ranitidine in equilibrium with the more stable form. In other words, the two forms have different solubilities. This fact has brought us around to an earlier comment by another person about differing solubilities of polymorphs and puts it in terms of rigorous thermodynamic argument.

A quick note: I am getting an unfavorable vote on each of most of my recent posts in this thread. I hope they are placed by mistake, but if not I at least hope that other people are finding the comments useful. I do have some more information to give and expect to make a few more posts on this topic.

In my post just previous to my last post I discussed differing solubilities for different crystalline structures of the same compound, and I gave a thermodynamic explanation for how this occurs. This amplifies the point made by John Liebeschuetz in this thread that the solubility of a compound depends on which crystalline form is being considered.

This is all to emphasize the point that when one discusses something that depends on Gibbs free energy, such as solubility, it is important to fully specify the thermodynamic states of all reactants and products. Let us take this one step further. The topic of this thread is the calculation of solubility from the free energy of solvation. This proposition is true, provided that the problem is properly posed. However, to properly pose the problem it is necessary to specify the full thermodynamic state of the solute both before and after the solvation process.

For example, the free energy of solvation for a molecule in which the final state is at a 1 molar concentration is different from the free energy of solvation in which the final state is at 0.001 molar. This is true, even if the enthalpy of solvation is the same in each case. The reason is that the entropy (strictly speaking, the partial molar entropy) of the compound is different in the two cases.

Similarly, the Gibbs free energy of solvation is different if the initial state is a gas-phase compound at a pressure of 1 Bar than if the initial state of the gas-phase compound is at 0.001 torr, and this in turn is different from the case in which the initial state is a crystalline solid.

In the case of solubility the problem must posed in the following terms: Given an initial state in which the solute is in the crystalline form of interest, what is the Gibbs free energy for the transfer of the compound from the crystal to the solution? In other words, the starting point must be the correct relevant phase, i.e. the analyte in the correct relevant crystalline form.

This means, for example, that the free energy for the transfer from the gas phase to the solution phase is not what is needed. This quantity can be useful, but only if one can also determine the free energy for the transfer of the analyte from the crystal to the gas phase, and even then one must be careful to match the pressure (or number density) of the gas phase for the two parts of the calculation.

This is still not a complete discussion, but I will break for now.

In some early posts in this discussion I used the terms “standard state” and "thermodynamic activity", but I didn’t say what they mean or why they are used. Let me now steer the discussion in the direction of standard states and the concept of thermodynamic activity, which we can just call "activity" for short.

So far the discussion has been presented in terms of fugacities, quantities which are closely related to pressure. In fact, for some calculations pressures can be substituted in place of fugacities, depending on the conditions of the process and the accuracy desired. For example, the fugacity of pure nitrogen gas at 1 Bar pressure is 0.9996, and given the fact that in free energy calculations the fugacity enters in the form of a logarithmic function, using pressure in place of fugacity would be quite accurate in this case. At lower pressures the errors of using pressure in place of fugacity would be even smaller.

One problem with using fugacities for most practical purposes is that the vapor pressures of many substances are too small to measure. This would be good in one way, because one could use pressure in place of fugacity without any significant loss of accuracy, but if one can’t measure the vapor pressure in equilibrium with some component of a phase one would not know what numbers for pressure to substitute into the thermodynamic equations. Hence, there would be no practical way to use fugacities to do free energy calculations or measurements.

Standard states and activities are a way to circumvent this problem. When thermodynamic problems are cast in the form of standard states there is, in most cases, no need to know the actual values of fugacities. One can then cast the thermodynamic problems in terms of standard states and activities. Activity is a kind of relative fugacity defined as being relative to the fugacity of a standard state.

As an aside, please do not confuse “standard state” with “standard temperature and pressure” as sometimes used in discussions of gases. The terms do not mean the same thing and bear little relationship to each other.

I am going to take a bit of an indirect route in going from fugacities to standard states and activities. Let us start with the equation I presented earlier, which is the equation for the Gibbs free energy for the process of compressing or expanding a gas under isothermal conditions. This was designated as equation (2) in an earlier post. It was also designated as equation (3).

Delta(G)=RTln(f2/f1) (3)

I am going to re-write this as

G(2)-G(1)=RTln(f2/f1) (4)

where G(1) is the Gibbs free energy of the gas at the initial pressure and G(2) is the Gibbs free energy of the gas at the final pressure. As before, R is the gas constant, T is the absolute temperature, f1 and f2 are the fugacities of the gases at the initial and final pressures respectively, and ln() is the natural logarithm function.

Let us define a one of the fugacities, f1, as the fugacity of an as-yet undefined reference state and the corresponding G(1) as the Gibbs free energy of the same gas at a fugacity of f1. We will symbolize these as f’ and G’. I am avoiding the use of conventional superscripts for standards states at this point, and I am also avoiding using the term “standard state” to refer to this reference state. Please just trust me on this.

The Gibbs free energy for compression or expansion of a gas from the as-yet unspecified reference state to a state of arbitrary pressure is

G-G’=RTln(f/f’) (5)

Note that I have substituted G in place of G(2) and G’ in place of G(1). Similarly, f takes the place of f2 and f’ takes the place of f1. One can re-write equation (5) as

G=G’+RTln(f/f’)

I am going to terminate this message now so it is not too long, and will continue the discussion in a new message.

I forgot to number the last equation in my last post. I re-write the equation here, along with an equation number:

G=G’+RTln(f/f’) (6)

Let us now consider the Gibbs free energy for the transfer of a component from one condensed phase to another. Let the fugacity of a gas in equilibrium with the initial phase be denoted by f’ and the fugacity of a gas in equilibrium with the final phase be denoted by f.

The Gibbs free energy for the transfer of the component from the initial phase to a gas of fugacity f’ is zero. The reason is that f’ is the fugacity of a gas in equilibrium with same material existing as a component of the condensed phase, and the Gibbs free energy is zero for any process occurring under equilibrium conditions.

Next, compress (or expand) the gas from a fugacity of f’ to a fugacity of f. The Gibbs free energy of that process is given by equation 5, which I write again as

G-G’=RTln(f/f’) (5)

Next, transfer the component from the gas phase at fugacity f to a condensed phase in equilibrium with the gas at fugacity f. The Gibbs free energy of that process is zero.

Adding up the Gibbs free energies of the individual steps, we have the Gibbs free energy for the transfer of the material between two condensed phases:

G-G’=RTln(f/f’) (5)

As an important note regarding terminology, based on the above discussion we can say that the fugacity of the component under consideration in the initial condensed phase is f’ and the fugacity of the same component in the final condensed phase is f. In other words, the fugacity of a component in a condensed phase is the same as the fugacity of the same component as it would exist in a gas in equilibrium with the same condensed phase.

Now let us shift gears and focus the discussion on reference states. The discussion above is framed in terms of a gas-phase reference state, and the fugacity of that state is denoted as f’. It is also possible to define reference states that are not gases. However, let me start discussing reference states by focusing on gases.

There is a particular reference state for gas-phase substances that is of particular interest. It goes by the name of “standard state”. It is possible to define the standard state in a number of non-equivalent ways. For example, one could pick a state corresponding to a real gas at a fugacity of f=1. Note that I did not specify the units. The units could be torr, i.e. a standard state of 1 torr, or bar, i.e. a standard state of 1 bar, or atmospheres, i.e. a standard state of 1 atmosphere, or newtons per square meter, i.e. a standard state of 1 newton per square meter, etc. Thus, when one specifies a standard state it is necessary to specify the relevant units.

Until a few decades ago it was customary to choose a standard of one atmosphere. However, the current recommendation is to use units of bar, so the standard state is at a fugacity of one bar. Those two choices differ by about 1.3%, and this difference can show us as errors in thermodynamic calculations if one is careless about keeping track of which standard state is being used.

Two paragraphs ago I said that one could choose a standard state as being a real gas at a unit pressure, e.g. 1 bar. While this is a possible choice, it is virtually never the choice used in practice. By mutual agreement in the thermodynamic community one does not use the real gas as the standard state but rather one uses certain fictitious state. The fictitious state is that of an ideal gas at unit fugacity. Since pressure equals fugacity for ideal gases the standard state of a gas is the ideal gas at unit pressure, such as 1 bar.

We now confront a conceptual difficulty that is often a stumbling block for people trying to understand standard states. The problem is that real gases are not ideal, so it seems an oxymoron to say that the standard state of a real gas is an ideal gas at unit fugacity. Thus, one has to add one more concept to the definition of standard state, i.e. it is a state obtained by extrapolating the properties of the real gas from zero pressure (where the gas is ideal) to one bar, using the properties of an ideal gas as the extrapolating function.

It is worth noting that both the internal energy and enthalpy of the standard state thus defined are the same as the internal energy and enthalpy of the gas at zero pressure, but the entropy and Gibbs free energy of the standard state are not the same as the entropy and Gibbs free energy of the gas at zero pressure. In fact, the entropy of a gas at zero pressure is infinite, and the Gibbs free energy of a gas at zero pressure is negative infinity. Thus, one should never say, as is sometimes done, that the standard state of a gas is the state of zero pressure.

In terms of notation, whenever a thermodynamic function of a substance in its standard state is discussed it is customary to use a superscript that looks like a degree sign. However, I don’t have that symbol on my keyboard.

I will interrupt the discussion now. When I come back I will discuss standard states for substances that are components of condensed phases, and this will bring the concept of “activity” into play.

Free energy of solvation is only part of the solubility process. Would the solute be a solid you will need its enthalpy of fusion as well as the mixing enthalpy in any case.

Javad,

Sorry for the delay in responding. It will take a few more posts, and I hope at that point all will become clear. It may take a few days before i can get back on this task, so please be patient

I am ready to resume the discussion of standard states. Recall that the discussion left off in the discussion of the standard state of a gas. Recall that the standard state was defined as the ideal gas at unit pressure (normally one chooses one bar). This presents a bit of a quandary, because real gases are not ideal. However, given enough thermodynamic information on the real gas it is possible to calculate the Gibbs free energy for the process of taking the gas from a real, physically realizable state to the standard state.

What is needed is needed is the Gibbs free energy for expanding the gas from the state in which we are interested, i.e. a state of some specified pressure, to a state of very low pressure. What do we mean by low pressure? We simply mean a state where the fugacity is (for all practical purposes) equal to the pressure. I will just call this pressure “the very low pressure” for sake of discussion. Let us symbolize this by the following expression:

Delta(G)compression_real=experimentally determined number

Then we calculate the Gibbs free energy of compressing an ideal gas from the very low pressure back to a pressure of 1 bar, assuming that we are defining the standard state in terms of bar. Recall that the Gibbs free energy for compression of a gas is given by equation (2):

Delta(G)=RTln(f2/f1) (2)

As a reminder, the (2) in this expression denotes the equation number, not multiplication by a factor of 2.

However, for an ideal gas the fugacity equals the pressure, so equation (2) for the process of compressing the gas to a fugacity of 1 is given by equation (6):

Delta(G)=RTln(1/Pvl) (6)

Where Pvl denotes the very low pressure discussed above.

We now add the the Gibbs free energies of the two processes:

Delta(G)=Delta(G)compression_real+RTln(1/Pvl)=G(standard state)-G(real gas) (7)

where G(standard state) is the Gibbs free energy of the gas in the standard state and G(real gas) is the Gibbs free energy of the corresponding real gas at the pressure we are interested in.

We would usually use a symbol of G with a superscript that looks like a degree sign to indicate G(standard state) and we would just use G to denote G(real gas).

Note that we don’t actually know either G(standard state) or G(real gas) in absolute terms, but we can know the difference between the two using the procedure outlined above.

Well, it's been a few days since my last post on this thread. In fact, it's been almost a year. Sorry for the delay, and I hope that at least someone is out there to continue reading about this topic.

Previously I talked about the standard state of a gas and the free energy of converting a real gas to a gas at the standard state. In descriptive terms this is done by breaking the process down into two parts. The first part is to expand a gas along a pathway that corresponds to its real physical properties. The end point of the expansion is a very low pressure, by which I mean a pressure where the fugacity can be taken as essentially equal to the pressure.

The Gibbs free energy of this first process is

Delta_G = G_realgas_low_pressure - G_realgas_high_pressure

= RTln(f_realgas_low_pressure/f_realgas_high_pressure)

I hope you don’t mind that I have shifted the notation to a more descriptive form.

Keeping in mind that we have set up the calculation on the low pressure side so that the pressure is low enough that it behaves as an ideal gas, and the pressure equals the fugacity of the gas, we can re-write delta G for the first process as

Delta_G=RTln(P_realgas_low_pressure/f_realgas_high_pressure)

= RTln(P_idealgas_low_pressure/f_realgas_high_pressure)

Note that I have substituted P_idealgas_low_pressure for P_realgas_low_pressure because the pressure is low enough to treat the gas as ideal.

The second part is to compress the gas along an ideal gas curve. Of course, a real gas does not follow the ideal gas law as pressure increases, so what we are defining here is a fictitious state, i.e. the state in which our material behaves as an ideal gas. We are going to take the gas up to a pressure of 1 bar. Note that because the gas is ideal the pressure equals the fugacity. We will define the end point of this compression as the ideal gas at 1 bar. The Gibbs free energy of this process is

Delta_G = RTln(f_idealgas_low_pressure/f_idealgas_1_bar)

= RTln(f_idealgas_low_pressure/1_bar)

= RTln(P_idealgas_low_pressure/1_bar)

Now let us combine the first process (the expansion of a real gas to very low pressure) and the second process (the compression of an ideal gas to 1 bar). The adding the Gibbs free energies, and performing a little bit of algebra, we end up with the following for the combined process.

Delta_G = G_idealgas_1_bar – G_realgas_high_pressure

= RTln(1_bar/f_realgas_high_pressure)

This can be re-arranged as follows:

G_realgas_high_pressure = G_idealgas_1_bar + RTln(f_realgas_high_pressure/1_bar)

We give G_idealgas_1_bar a special name. We call it the standard state, and we define the standard state as the ideal gas at unit pressure, usually using units of bar for pressure. Conventional notation is to designate it as G with a superscript that looks like a degree sign. I will just call it G(standard_state). Also, the designation G_realgas_high_pressure can refer to any pressure, so let’s just call it G, and similarly let’s just call f_realgas_high_pressure as f. We can now write

G = G(standard_state) + RTln(f/1_bar)

We also give the quantity f/1_bar a name. We call it the activity. It is the fugacity of the gas divided by the fugacity of the standard state:

G = G(standard_state) + RTln(a)

where a is the activity.

So far we have only treated gases. However, it turns out that the definition of activity is the same for any material, provided we define activity as the fugacity divided by the fugacity of the standard state. The difference is that the standard state is not an ideal gas at unit pressure. More later.

Now let us consider solutions. Most of what we discussed before applies also to solutions, but with a few of additional concepts and subtleties thrown in.

We start by discussing Henry’s law, which has been around for more than 200 years.

“At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid."

Actually, Henry’s law is not a law of nature because no solute obeys it perfectly, but it represents a limiting case, i.e. as far as is known all solutes obey Henry’s law in the limit of infinite dilution.  It thus holds a position similar to the ideal gas law, and in fact in the regime where Henry’s law is valid (the limit of infinite dilution) the vapor pressure also approaches zero, which are the conditions for the validity of the ideal gas law for the solute vapor.

Now we approach the first subtlety. Henry’s law is generally expressed in terms of pressure, but this is actually not the most useful form for the underlying concept. A more useful statement of Henry’s law would be that the fugacity of a gas is proportional to the concentration of the gas dissolved in a solution, provided the gas is in equilibrium with a solution. In other words, conceptually we just substitute fugacity in place of pressure.

Of course, in the regime where Henry’s law is truly valid (in the limit of infinite dilution) the ideal gas law is also valid for the solute gas that is in equilibrium with the solution (in the limit of zero pressure), hence the fugacity and the pressure are the same. Furthermore, for most substances the partial pressure of the solute is so low that we can just use pressure without a significant loss of accuracy, but to be rigorously correct in what follows it is more useful to frame the discussion in terms of fugacity.

Let us consider the process of transferring a solute in a solution to the gas phase. We assume the process occurs at constant temperature and pressure. At equilibrium the Gibbs free energy of the process is zero. There’s no surprise here, because this is always true for a process occurring under equilibrium conditions at constant pressure and temperature.

Now consider the following process, broken down into three steps. 1) Transfer a solute of finite concentration from solution to vapor. Delta_G=0. 2) Expand the vapor to a very low pressure. Delta_G = RTln(f_low/f_high), where f_low is the fugacity at low pressure and f_high is the fugacity in equilibrium with the solution. 3) Transfer the low-fugacity vapor to a solution whose concentration is such that its fugacity equals f_low. Delta_G = 0. The total free energy for this three-step process is:

Delta_G = RTln(f_low/f_high)

Now assume that the concentration of the low-concentration solution is low enough that the solute follows Henry’s law. (I’m referring here to the slightly modified form of Henry’s law framed in terms of fugacity rather than pressure.) Let us consider another three-step process with the starting point being the same as the end point of the previous three-step process. However, the end point is different. Instead of being a real solution at an arbitrary concentration, it is a fictitious solution that follows Henry’s law, and the end point is unit concentration: 1) Transfer the solute to the gas phase, with the gas phase being at equilibrium. Delta_G = 0. 2) Compress the gas from f_low to f_high_unit_Henry. Here we define f_high_unit_Henry to be the fugacity of the fictitious state of unit concentration (e.g. 1 molal, 1 molar, unit mole fraction, or whatever). Delta_G = RTln(f_high_unit_Henry/f_low). Adding the free energies of these three steps we have

Delta_G = RTln(f_high_unit_Henry/f_low)

Now add all of the processes together (6 steps in all). The Gibbs free energy is

Delta_G = RTln(f_high_unit_Henry/f_low)+ RTln(f_low/f_high)

Or

Delta_G = RTln(f_high_unit_Henry/f_high)

Now, f_high_unit_Henry is based on the fictitious solution that obeys Henry’s law, so we can make the following substitution.

F_high_unit_Henry = kC = k*1molar

So

Delta_G = RTln(k*1molar/f_high)

Here I have assumed we are working in molar concentration units. If other units are in use, such as molal, then it would be

Delta_G = RTln(k*1molal/f_high)

or whatever, but for sake of illustration let us continue to work in molar units.

Now, if the real solution were to obey Henry’s law we would have

f_high = k*C_high

Where C_high is the concentration of the high-concentration solution. Keep in mind that this is a real solution, not a fictional one. However, because it is a real solution this last expression doesn’t work, so we introduce a parameter gamma that makes the following expression work:

f_high = gamma*k*C_high

This equation defines gamma, and gamma is called the activity coefficient.

We can now write the equation

Delta_G = RTln(k*1molal/f_high)

as

Delta_G = RTln(k*1molal/ (gamma*k*C_high)) = -RTln(gamma*k*C_high /1molal)

We can now re-write this as

G_1_Henry - G_real_high = -RTln(gamma* C_high /1molal)

Where G_1_Henry is the Gibbs free energy of the solute in the idealized solution at 1 molal that obeys Henry’s law, and G_real_high is the Gibbs free energy of the solute in the real solution at arbitrary concentration. To be rigorously correct, when I say something like “Gibbs free energy of the solute in solution” I mean “partial molar Gibbs free energy of the solute in solution.” We can talk about what “partial molar” means later.

Now, the state corresponding to G_1_Henry is an idealized state based on a fictitious material, the one obeying Henry’s law at all concentrations, and further specify that the concentration is at unit concentrations. We usually symbolize it by putting a superscript on it that looks like a degree sign. We also usually write G with a bar over it to specify “partial molar”. I will call it G_standard_state instead, mainly because I can’t use the fonts I want. We also give gamma* C_high /1molal a special name. We call it the activity, and we symbolize it as a. In addition, since G_real_high can refer to any concentration, we just symbolize it as G with a bar over the top. However, I don’t have the font for a bar over the top of the G, so I will just call it G. We therefore have (after substitutions and rearrangements)

G = G_standard_state + RTln(a)

We can also write

G = G_standard_state_state + RT(gamma*C/1molar)

or, playing a little a little fast and loose with the dimensionality (as is customary in this case)

G = G_standard_state + RT(gamma*C)

I say “fast and loose” because the gamma*C has dimensions of concentration, and you can’t take the logarithm of a dimensioned number. Strictly speaking, the form gamma*C/1molar is the correct form, but you usually see it written as gamma*C. Lewis and Randall discuss this issue somewhat in their book on thermodynamics.

This completes the treatment of activity for solutes. Later I will discuss the activity for pure solids.

I promised to deal with the activity of solids. Here is where things become both more interesting and less interesting.

The activity of a solid is the fugacity of the vapor that would be in equilibrium with the a given solid divided by the fugacity of the standard state. Most solids have a very low vapor pressure, so we can usually safely substitute pressures in place of fugacities, and I will sometimes use that conceptual simplfication, with the proviso that it will not always apply.

The standard state of a solid is defined as the most stable form of the solid at a specified temperature and at unit total pressure. In the old days the pressure was specified as 1 atmosphere. The current convention is to specify 1 bar. The two pressures are almost the same, and as a practical matter the ~1% difference does not make a noticeable difference for solids because their vapor pressures vary only very weakly with total applied pressure.

So here is the boring part. Solids are often encountered in their most stable forms. Therefore if we take the vapor pressure of a solid in its most stable form, and divide it by the vapor pressure of the standard state, which is defined as the solid in its most stable form, we end up with a value of 1. In other words, solids often have an activity of 1.

Here's where it becomes more interesting. If we apply additional pressure to a solid its vapor pressure increases. (I won't explain why, but it is something that can be calculated with thermodynamics, if you know the appropriate physical properties.) Therefore the activity of a solid increases with pressure.

OK, here's where it gets boring again. The change in vapor pressure of a solid when you increased the total pressure applied to the solid is small, so the activity normally doesn't change much when you increase the pressure.

Here's where it gets interesting again. If you apply very high pressures then you can get a noticeable change in activity. It is usually a small but not quite negligible difference, which brings it back to the almost boring category.

Ah, but things get interesting again when you consider that many solids can exist in several different forms. A classic example is carbon, which can exist as graphite, diamond, and perhaps other forms. In the case of carbon, graphite is the most stable form under ordinary conditions, so it defines the standard state. Diamond is a less stable form under ordinary conditions, which implies that the vapor pressure of diamond is greater than the vapor pressure of carbon. The thermodynamic activity of graphite is 1 because it is the vapor pressure of graphite divided by the vapor pressure of the standard state, which is also graphite, but the thermodynamic activity of diamond is greater than 1 because its vapor pressure is higher than the vapor pressure of the standard state, which is graphite.

If you have a drug that could exist in several different crystalline forms you would have the same situation as with carbon, i.e. the different crystalline forms would have different activities. The most stable form would have an activity of 1, and the activities of the other forms would be >1.

One of the implications of this is that two different crystalline forms of a drug will have different solubilities.It all ties back to the fact that the different crystalline forms have different vapor pressures, or more rigorously speaking, different fugacities.. I won't go through all of the thermodynamic relationships to show this in detail, but from what has been written above it should be more or less obvious. Basically, If a solid is in equilibrium with a liquid then the fugacity of the solute in the liquid is the same as the fugacity of the same material in the solid. Therefore, if two different crystalline forms of a solid have different vapor fugacities (loosely speaking, different vapor pressures) then the corresponding solutions with which they may be equilibrated will also have different fugacities with respect to the solutes. Since the vapor pressure of a solute always increases with the concentration of the solute, two different crystalline forms of a material must have two different solubilities.

By now you are probably saying "but wait, I can't even measure the vapor pressure of a non-volatile solid, so how can any of this be relevant?" Well, in terms of the conceptual explanation and deriving the thermodynamic relationships it doesn't matter if you can measure the vapor pressure or not. The thermodynamic analysis still applies. Furthermore, when it comes to determining activities they are seldom determined by direct methods, i.e. by measuring vapor pressures. The fundamental reason is that activities are based on ratios, and as long as you can determine the ratio by some method it doesn't matter whether you can measure the numerator and denominator separately. Hence, we often rely on indirect methods, such as the Gibbs-Duhem relationship, electrochemical measurents, or other methods.

Hi Alan,

thanks for your efforts in explaining the activity of solids, and what it actually means. However, I do not fully agree with the sentence:

"The thermodynamic activity of graphite is 1 because it is the vapor pressure of graphite divided by the vapor pressure of the standard state, which is also graphite, but the thermodynamic activity of diamond is greater than 1 because its vapor pressure is higher than the vapor pressure of the standard state, which is graphite."

Where Graphite and Diamond differ, is in their standard potential, given that the DG° of the reaction is non-zero. Moreover, I guess you cannot talk, in a thermodynamic sense, of the vapour pressure of diamond, when diamond should - thermodynamically speaking - even exist.

Finally, the standard chemical potential is defined as the potential of the pure phase at a given temperature T and at a pressure p°. So, by the definition itself, the activity of a pure solid should be one. Moreover, the change in pressure (which will have an effect on the vapour pressure) will change the standard potential rather than the activity.

My two cents for today. Many greets,

Leo

One more comment. Things become even more interesting and controversial when one deals with ions. For more than 80 years the prevailing view has been that activities of ions (single ion activities) are thermodynamically undefineable quantities. The view has been that one can deal with positive/negative ion combinations having zero net charge using something called mean ionic activity coefficients, but that the whole concept of activity breaks down when applied separately to an ion and a counter ion.

A counter ion is an ion in a solution of opposite charge. The concentration of a counter ion and an ion are such as to give electrical balance to the solution. For example, you never see a solution of just sodium ion. There is always an ion of opposite charge present, such as chloride.

The concept that single ion activities are not measurable (or not even defineable) creates an embarrassing situation for chemists because pH is defined as the negative of the logarithm of the hydrogen ion activity, particularly since, as chemists, we routinely measure pH on a frequent basis.

The idea that you can't determine single ion activities (and by implication that you can't determine single ion activity coefficients) goes back at least to some papers published by Guggenheim in the late 1920s.

However, it turns out that it is possible to define single ion activities in a way that is thermodynamically rigorous, and furthermore the definition leads to a method of determining single ion activities using thermodynamically reversible measurements. I just had a paper accepted for publication that gives a thermodynamic proof that one can define and measure single ion activities. The paper will appear in the journal ChemPhysChem sometime in the near future. Watch for the paper. I think you will find it interesting reading.