The author believes that the chemical potential of liquid molecules is dominated by their bonding with the same kind molecules. A liquid drop can be roughly divided as surface molecules and inner molecules. The inner molecules will have an extra pressure ΔP due to the following reasons:

a) With the perception that “bonding potential is negative”, the bonding potential of inner molecules is more negative because of more bonding. That is, their molar chemical potential is lower than that of surface molecules when both are at 1 atm.

b) The molar chemical potential of molecules can be increased by an extra pressure ΔP (similar to a pressed spring).

c) When the liquid drop is at equilibrium, the molar chemical potential of inner molecules must be the same as that of surface molecules. Therefore the inner molecules should have an extra pressure ΔP (in other words, surface molecules will press inward resulting in the extra pressure ΔP).

The smaller is the liquid drop, the less complete is the bonding of surface molecules. Hence the higher is ΔP. The ΔP inside a liquid drop is very essential to the Atomistic 2^nd Law of Thermodynamics. If someone can verify the existence of ΔP (which should be far smaller than 1 atm) and that ΔP increases with decreasing radius of the liquid drop, it certainly will make the Atomistic 2^nd Law of Thermodynamics clearer.

In the book “Physics and Chemistry of Interfaces”, Butt et al. described: the saturated vapor pressure of a liquid drop is higher than that of a flat surface [2] . This phenomenon can be shown graphically in Figure 1 :

The author’s explanation of this phenomena is: the surface of a liquid drop is convex which results in less bonding and a higher molar chemical potential for the surface molecules. Thus a liquid drop will have a higher saturated vapor pressure (compared to a liquid with flat surface). Landau et al. also reported a similar phenomenon that the saturated vapor pressure above a drop of liquid increases with decreasing radius of the drop [3] . All these phenomena can be explained by the differences in liquid curvature at surface.

Similarly, the descending meniscus in a capillary has a convex surface and the surface molecules have a higher molar chemical potential than a flat surface, thus this meniscus also has a higher saturated vapor pressure [3] . The above-mentioned phenomena can be extended further: It may happen that the vapor is supersaturated with respect to the larger droplets but unsaturated with respect to the smaller ones. Then the liquid which evaporates from the smaller drops will condense on the larger ones, which as it were “consume” the small drops [3] .

Conversely, when liquids have concave surfaces, the phenomena on saturated vapor pressure are totally opposite, as also reported in literature [2] - [5] , examples include:

a) The liquid surfaces surrounding gas bubbles are concave, resulting in lower molar chemical potentials and lower saturated vapor pressures inside gas bubbles.

b) The surface of a rising meniscus in capillary is concave, thus it also has a lower saturated vapor pressure than a flat surface. For this reason the vapor generated from flat surface would diffuse to and condense on the meniscus surface (called capillary condensation).

c) The higher does the meniscus rise in capillary, the more is gravitational potential needed for its equilibrium. Hence the lower is the molar chemical potential of the meniscus and the more pronounced is the capillary condensation.

Capillary condensation is also observed in solids with small pores which will act like capillaries. For example, capillary condensation would occur in soil, shales, building materials, man-made foods, thus making these small pores have some condensed liquids [4] - [7] .

Based on the various phenomena mentioned, it can be deduced that the curvature of liquid surface, whether it’s flat, convex or concave, is the main factor determining the saturated vapor pressure. And the larger is the curvature of liquid surface, the more different is the saturated vapor pressure from that of a flat surface. Although this induction complies with the perception that the molar chemical potentials of liquid molecules originate from bonding, more verifications and better descriptions are needed.

When gravitational potential can be neglected, for a liquid drop at room temperature and 1 atm, its molar chemical potential at surface, μ(1)_surface, can be defined as:

$\mu {\left(1\right)}_{\text{surface}}=\mu {\left(1\right)}_{\text{inner}}+\gamma \left(1\right)\cdot A$ (1)

In Equation (1), μ(1)_inner is the molar chemical potential of inner molecules at 1 atm, “γ(1)·A” is the excess molar potential due to the convex surface of liquid drop at 1 atm. Here γ(1) is defined as the excess energy per unit surface area of the liquid drop at 1 atm (the same as traditional definition), “A” is defined as the molar surface area of the liquid drop.

When the inner molecules of liquid drop are at equilibrium with its surface molecules, the necessary and sufficient condition of the Atomistic 2^nd Law of Thermodynamics requires that μ(1)_surface is equal to μ(1 + ΔP)_inner. Here ΔP is defined as the extra pressure of the inner molecules at 1 atm. Based on this condition, the author has derived the following Equation for 1 atm [1] (in its Equation (7)):

$\gamma \left(1\right)\fallingdotseq \left[\Delta P\cdot v\left(1\right)\right]÷A$(2)

In Equation (2), v(1) is the molar volume of liquid at 1 atm, defined as v(1) = M ÷ ρ(1) where M is the molar weight of liquid drop and ρ(1) is its density at 1 atm.

When R is the radius of the liquid drop, its surface area is 4πR² and its volume is (4πR³/3), where π is the circular ratio. The number of moles in this liquid drop is (4πR³/3)·ρ(1) ÷ M. Hence, “A”, defined as the molar surface area of the liquid drop, is equal to the surface area of this liquid drop divided by its moles, i.e.:

$\begin{array}{c}A=4\pi R^2÷\left[\left(4\pi R^3/3\right)\cdot \rho \left(1\right)÷M\right]\\ =\left(3/R\right)\cdot \left(M÷\rho \left(1\right)\right)\\ =\left(3/R\right)\cdot v\left(1\right)\end{array}$(3)

Plug Equation (3) into Equation (2), we’ll get:

$\gamma \left(1\right)\fallingdotseq \left(\Delta P\cdot R\right)/3$(4)

Equation (4) implies：

a) Surface energy γ(1) can be calculated once ΔP and R are obtained. This is different from the conventional way to obtain surface energy γ(1).

b) The value (ΔP·R) of a liquid drop is not necessarily a constant when R changes, neither is its surface energy γ(1).

c) When R is fixed, the surface energy at 1 atm, γ(1), is proportional to the extra pressure ΔP.

Since there is no special condition in the derivation of (Equation (4)), this result should be applicable to all materials. Due to the fact that (a), (b) are different from conventional knowledge, a rigorous verification of (Equation (4)) is necessary.

If Equation (4) is correct, it should contribute considerably to the rationalization of surface energies for materials. If Equation (4) is mistaken, the reasons for being mistaken should be studied for a better understanding. Furthermore, when the liquid drop is under a pressure different from 1 atm, the relation between γ and pressure would be interesting, so is the relation between ΔP and pressure.

In Classic 2^nd Law of Thermodynamics, the molar chemical potential of component i is defined as “the increase in Gibbs free energy due to the addition of one mole of component i, when temperature, pressure and other molecules are fixed”. This definition is relative in nature and it does not mention the origin of molar chemical potential. Since the molar chemical potential of material is the core of the Atomistic 2^nd Law of Thermodynamics when Liu developed this law, a study on the origin of chemical potential is reasonable.

In literature, Cook et al. [8] , Baierlein [9] , Job et al. [10] all tried to study in-depth the molar chemical potentials of materials, but it seemed that they did not have much success. Their comments can be summarized as: “students generally could not perceive the meaning of molar chemical potential, yet it can be applied widely to physical and chemical phenomena”. After searching, it’s also not possible for the author to find the origin of molar chemical potential in Wikipedia, google or DeepSeek.

The author believes, chemical potential is originated from the interaction among molecules of the same kind. In a condensed phase of one component, this interaction is dominated by inter-molecular bonds. According to Wikipedia, these inter-molecular bonds can be categorized as [11] : 1) ionic bond, 2) covalent bond, 3) hydrogen bond, 4) dipole bond, 5) London dispersion forces (commonly called Van der Waal forces).

From the same source, the relative strength of these inter-molecular bonds were differentiated in a table¹: ionic bond is the strongest, covalent bond is the next, while the rest 3 bonds (sometimes all called Van der Waal forces) are the weakest [11] . However, metallic bond is not in this table. From known information, most metals are with relative high melting points (except mercury). Their boiling points are even higher. These facts indicate that the attraction force of the metallic bonds are very strong.

If the molar chemical potentials of condensed phases could be categorized according to the above-mentioned inter-molecular bonds, it should help the progress in the advance of material physics and chemistry. The same is true if extra pressure ΔP and the surface energy γ of a liquid drop could also be categorized based on these inter-molecular bonds.

In the region where a liquid is at equilibrium with its own gas, both phases have the same molar chemical potential. Therefore it’s not difficult to know the molar chemical potential of a gas experimentally. However, if the molar chemical potential of a gas can be expressed as a formula, the applications will be much wider and the convenience much better. For this reason, it’s worth a deep look at the formula for the molar chemical potential of a gas.

All gas molecules are free to move in open space, hence their bonding potentials should be zero. With the perception that chemical potential is originated from the interaction among molecules of the same kind, it can be inferred that the formula of molar chemical potential of a gas should be dominated by its molar concentration. While the well-known phenomena related to gases would serve as the constraints of this formula². After trials, the author came up with the following formula for the molar chemical potential μ_G of a gas G:

${\mu }_G=k_0\cdot \ln \left(k_1{}_G\cdot \left[G\right]\right)$(5)

In Equation (5), [G] is the molar concentration of the gas G; k₀ is a positive constant for every material which only depends on temperature and pressure; k_1G is a constant not only dependent on temperature and pressure, but also dependent on gas G; k_1G·[G] is without unit and $0\le k_{1G}\cdot \left[G\right]<1$.

As the author have said in the conclusions 8 of reference [1] : the values of k₀ and k_1G are yet to be determined. In this communication, the author proposes a procedure to determine the values of k₀ and k_1G, and proposes the methods to verify whether these values are reasonable and consistent.

The following is taking the equilibrium of $aA+bB\leftrightarrow cC+dD$ as an example (a, b, c, d are the moles of gases A, B, C, D respectively). According to Equation (5), μ_A,·μ_B, μ_C, μ_D can be expressed as:

$\begin{array}{l}{\mu }_A=k_0\cdot \ln \left(k_{1A}\cdot \left[A\right]\right),{\mu }_B=k_0\cdot \ln \left(k_{1B}\cdot \left[B\right]\right)\\ {\mu }_C=k_0\cdot \ln \left(k_{1C}\cdot \left[C\right]\right),{\mu }_D=k_0\cdot \ln \left(k_{1D}\cdot \left[D\right]\right)\end{array}$(6)

In Equation (6), [A], [B], [C], [D] are the molar concentrations of gases A, B, C, D respectively.

a) To determine the values of k₀ and k_1A, k_1B, k_1C, k_1D

When $aA+bB\leftrightarrow cC+dD$ spontaneously reaches equilibrium, the total chemical potential of reactants N_i, i.e. ${\displaystyle \sum N_i\cdot {\mu }_i}$, must be equal to the total chemical potential of products N_j, i.e. ${\displaystyle \sum N_j\cdot {\mu }_j}$, hence:

$a\cdot {\mu }_A+b\cdot {\mu }_B=c\cdot {\mu }_C+d\cdot {\mu }_D$(7)

Plug Equation (6) into Equation (7) and after some treatments, we’ll derive (for details, please see the derivation of (Equation (20)) in [1] ):

$\left(k_{1A}^a\cdot k_{1B}^b\right)÷\left(k_{1C}^c\cdot k_{1D}^d\right)=\left({\left[C\right]}^c\cdot {\left[D\right]}^d\right)÷\left({\left[A\right]}^a\cdot {\left[B\right]}^b\right)$(8)

Since Equation (6) can be rewritten as:

$\begin{array}{l}{k}_{1A}=\left[{\text{e}}^{\left({\mu }_A÷k_0\right)}\right]÷\left[A\right],k_{1B}=\left[{\text{e}}^{\left({\mu }_B÷k_0\right)}\right]÷\left[B\right]\\ k_{1C}=\left[{\text{e}}^{\left({\mu }_C÷k_0\right)}\right]÷\left[C\right],k_{1D}=\left[{\text{e}}^{\left({\mu }_D÷k_0\right)}\right]÷\left[D\right]\end{array}$(9)

We can plug Equation (9) into Equation (8) and derive:

$\begin{array}{l}\left({\left\{{\text{e}}^{\left({\mu }_A÷k_0\right)}÷\left[A\right]\right\}}^a\cdot {\left\{\left[{\text{e}}^{\left({\mu }_B÷k_0\right)}\right]÷\left[B\right]\right\}}^b\right)÷\left({\left\{\left[{\text{e}}^{\left({\mu }_C÷k_0\right)}\right]÷\left[C\right]\right\}}^c\cdot {\left\{\left[{\text{e}}^{\left({\mu }_D÷k_0\right)}\right]÷\left[D\right]\right\}}^d\right)\\ =\left({\left[C\right]}^c\cdot {\left[D\right]}^d\right)÷\left({\left[A\right]}^a\cdot {\left[B\right]}^b\right)\end{array}$ (10)

In Equation (10), when the temperature and pressure are fixed, the numeric value of [A], [B], [C], [D], a, b, c, d are known values from experiment; if the reliable values of μ_A, μ_B, μ_C, μ_D are obtainable, the only unknown is k₀. With current computational capability, the numeric value of k₀ can be calculated from Equation (10) without much difficulty. Once the numeric value of k₀ is obtained, the corresponding numeric values of k_1A, k_1B, k_1C, k_1D can be calculated from Equation (6).

b) Verify if k_1A, k_1B, k_1C, k_1D are reasonable

Experimentally, it is known that the equilibrium constant for $aA+bB\leftrightarrow cC+dD$ can be expressed as [12] :

$K_{eq}=\left({\left[C\right]}^c\cdot {\left[D\right]}^d\right)÷\left({\left[A\right]}^a\cdot {\left[B\right]}^b\right)$(11)

Once k_1A, k_1B, k_1C, k_1D are obtained, the value of $\left(k_{1A}^a\cdot k_{1B}^b\right)÷\left(k_{1D}^c\cdot k_{1D}^d\right)$ can be calculated for the left-hand side of Equation (8). Hence under different temperature and pressure, whether or not the values of $\left(k_{1A}^a\cdot k_{1B}^b\right)÷\left(k_{1D}^c\cdot k_{1D}^d\right)$ are equal to $\left({\left[C\right]}^c\cdot {\left[D\right]}^d\right)÷\left({\left[A\right]}^a\cdot {\left[B\right]}^b\right)$³ is a means to verify if k_1A, k_1B, k_1C, k_1D are reasonable.

c) Verify the consistencies of k₀, k_1A, k_1B, k_1C, k_1D

If gas A has chemical reactions other than $aA+bB\leftrightarrow cC+dD$, then k₀ and k_1A can also be obtained from other chemical reactions (the prerequisite is the related molar chemical potentials are reliable). If all the obtained k₀ and k_1A are consistent for all the chemical reactions involving gas A, then Equation (5) is applicable to gas A.

If gases B, C, D have chemical reactions other than $aA+bB\leftrightarrow cC+dD$, then k₀ and k_1B, k_1C, k_1D can also be obtained via other chemical reactions. In these cases, the following should be verified:

1) Whether the k₀ and k_1B, k_1C, k_1D obtained are consistent for all the chemical reactions involving gases B, C, D.

2) Whether the k₀ obtained is the same as that obtained from gas A.

If all k₀, k_1G are reasonable and consistent, it implies that Equation (5) (where k₀, k_1G are from) are also reasonable. Under this condition, whether or not k₀, k_1G can be simplified as formulas is a topic needing study. On the other hand, if k₀, k_1G are not reasonable or not consistent after rigorous verifications, it is likely that Equation (5) is not the correct formula for the molar chemical potential of a gas. In this case, to look into the reasons and to find a better formula are the logical next steps.

Based on the Atomistic 2^nd Law of Thermodynamics and its necessary and sufficient condition, Liu [1] had discussed the spontaneous merging between small and large liquid drops at the two ends of a horizontal tube. Via speculations, this phenomenon was extended to the conditions when the heights of liquid drops are different. These conditions can be shown more clearly in Figures 2(a)-(d) :

Figure 2(a) is the initial state when the small and large liquid drops are connected through a hollow and horizontal tube. Figure 2(b) is in the midway when the small liquid drop is merging toward the large one. Figure 2(c) shows that when the large liquid drop is lifted up with a height h₀ so that the total molar potentials (i.e. the sum of molar chemical potential and molar gravitational potential) of the two drops are equal, therefore neither drop will move. In Fig. 2(d), the large liquid drop is lifted further with a height h₁ (h₁>h₀), so that the large liquid drop is spontaneously merging into the small one.

The phenomena in Figure 2 are based on speculations, and the molar gravitational potentials of liquid drops are involved. If someone could verify Figure 2 experimentally, it would confirm that the Atomistic 2^nd Law of Thermodynamics and its necessary and sufficient condition are applicable to these phenomena.

If the potentials other than chemical potential can be neglected, whether a material can be dissolved in a solvent is determined by that whether or not the total chemical potential of the system can be reduced. If the dissolution of a material will result in a decrease of the total chemical potential, then this material will keep dissolving until saturation is reached or the solid solvent has exhausted (e.g. salt in water).

At saturation the molar chemical potential of the solute is equal to that of the co-existing phases, e.g., the molar chemical potential of the salt dissolved in water is equal to that of the solid salt in water. Another example is that the molar chemical potential of the oxygen dissolved in water is equal to that of the oxygen in air.

If the molar chemical potential of a solute is higher than that of its saturated value due to condition change, then part of the solute will leave the solution. Examples include: salt will crystalize from largely condensed sea water, or the dissolved oxygen in a pond will decrease because of temperature rise. These phenomena certainly comply with the Atomistic 2^nd Law of Thermodynamics or its necessary and sufficient condition.

If the dissolution of a material will result in an increase of total chemical potential in the system, then this material will not dissolve at all, e.g. sand, metals, glass or woods in water. Also, oils are generally insoluble in water. On the other hand, experiments showed that all gases can dissolve in water to certain extents, while the least soluble gas is Helium [13] . Hence, it’s logical to expect that the total chemical potential of the system be reduced when gases dissolve in water.

When a compound A_pB_q dissolves in solvent X via the reaction $A_p{B}_q\leftrightarrow p{A}_{\left(sol.\right)}+q{B}_{\left(sol.\right)}$, it makes sense that its equilibrium condition is similar to the chemical reaction of gases $aA+bB\leftrightarrow cC+dD$. Since the equilibrium condition for $aA+bB\leftrightarrow cC+dD$ is ${\displaystyle \sum N_i\cdot {\mu }_i}={\displaystyle \sum N_j\cdot {\mu }_j}$ (i denotes reactants, j denotes products), it’s logical to expect that the equilibrium condition for $A_p{B}_q\leftrightarrow p{A}_{\left(sol.\right)}+q{B}_{\left(sol.\right)}$ would be:

$\mu \left(A_p{B}_q\right)=p\cdot \mu \left(A_{\left(sol.\right)}\right)+q\cdot \mu \left(B_{\left(sol.\right)}\right)$(12)

In Equation (12), μ(A_pB_q) is the molar chemical potential of the compound A_pB_q, while μ(A_(sol.)) and μ(B_(sol.)) are the molar chemical potentials of components A and B in solution respectively; p and q are the molar numbers of A and B in solution. The concept that “chemical potential is originated from the interaction among the same molecules” should also be applicable to solutions. Hence μ(A_(sol.)) and μ(B_(sol.)) are dominated by molar concentrations [A] and [B], and their formulas are likely:

$\mu \left(A_{\left(sol.\right)}\right)=k_2\cdot \ln \left(k_{3A}\cdot \left[A\right]\right),\mu \left(B_{\left(sol.\right)}\right)=k_2\cdot \ln \left(k_{3B}\cdot \left[B\right]\right)$(13)

Equation (13) is similar to the formula of the molar chemical potential of a gas (but with different constants). In Equation (13), k₂ (with the same unit as μ) is a positive constant common to all solutes in solvent X, and is dependent on temperature and pressure. While k_3A or k_3B (with the unit of concentration⁻¹) not only is dependent on temperature and pressure, but also is dependent on solvent X and solute A or B.

Similar to the formulas of the molar chemical potential of gases, the values of k₂, k_3A and k_3B can be determined and verified via the following procedures:

a) To determine the values of k₂ and k_3A, k_3B

Plug Equation (13) into Equation (12), we’ll get:

$\mu \left(A_p{B}_q\right)=p\cdot k_2\cdot \ln \left(k_{3A}\cdot \left[A\right]\right)+q\cdot k_2\cdot \ln \left(k_{3B}\cdot \left[B\right]\right)$(14)

Now moving the common factor k₂ from right-side to left-side, and moving p, q inside of ln, Equation (14) becomes:

$\mu \left(A_p{B}_q\right)÷k_2=\ln {\left(k_{3A}\cdot \left[A\right]\right)}^p+\ln {\left(k_{3B}\cdot \left[B\right]\right)}^q$(15)

Taking both sides of Equation (15) as the exponent of natural logarithm, Equation (15) becomes:

${\text{e}}^{\mu \left(A_p{B}_q\right)÷k_2}={\left(k_{3A}\cdot \left[A\right]\right)}^p\cdot {\left(k_{3B}\cdot \left[B\right]\right)}^q$(16)

After moving (k_3A)^p and (k_3B)^q to the left-side of Equation (16), we can derive:

${\text{e}}^{\mu \left(A_p{B}_q\right)÷k_2}÷\left[{\left(k_{3A}\right)}^p\cdot {\left(k_{3B}\right)}^q\right]={\left[A\right]}^p\cdot {\left[B\right]}^q$(17)

Now expressing k_3A and k_3B in terms of k₂ in Equation (13):

$k_{3A}={\text{e}}^{\mu \left(A_{\left(sol.\right)}\right)÷k_2}÷\left[A\right],k_{3B}={\text{e}}^{\mu \left(B_{\left(sol.\right)}\right)÷k_2}÷\left[B\right]$(18)

Plugging Equation (18) into the k_3A and k_3B of Equation (17), we’ll derive:

${\text{e}}^{\mu \left(A_p{B}_q\right)÷k_2}÷\left[{\left({\text{e}}^{\mu \left(A_{\left(sol.\right)}\right)÷k_2}÷\left[A\right]\right)}^p\cdot {\left({\text{e}}^{\mu \left(B_{\left(sol.\right)}\right)÷k_2}÷\left[B\right]\right)}^q\right]={\left[A\right]}^p\cdot {\left[B\right]}^q$ (19)

In Equation (19), when the temperature, pressure and solvent are fixed at equilibrium, the numeric values of [A], [B], p, q are known. If the reliable values of μ(A_pB_q), μ(A_(sol.)) and μ(B_(sol.)) are available, then k₂ is the only unknown in Equation (19). Hence it’s not difficult to determine k₂ from Equation (19) under current calculating capability. Once the value of k₂ is determined, the numeric values of the corresponding k_3A and k_3B can be calculated from Equation (13).

b) To verify if k₂, k_3A, k_3B are reasonable

Experimentally the solubility product K_sp of $A_p{B}_q\leftrightarrow p{A}_{\left(sol.\right)}+q{B}_{\left(sol.\right)}$ can be expressed as [14] ：

$K_{sp}={\left[A\right]}^p{\left[B\right]}^q$ (20)

Once the value of k₂, k_3A, k_3B are determined, the value of ${\text{e}}^{\mu \left(A_p{B}_q\right)÷k_2}÷\left[{\left(k_{3A}\right)}^p\cdot {\left(k_{3B}\right)}^q\right]$ for this dissolution can be calculated. At various temperatures and pressures “whether or not the values of ${\text{e}}^{\mu \left(A_p{B}_q\right)÷k_2}÷\left[{\left(k_{3A}\right)}^p\cdot {\left(k_{3B}\right)}^q\right]$ are equal to [A]^p[B]^q (i.e. the experimental K_sp)”⁴ is a means to verify that if k₂, k_3A, k_3B are reasonable.

c) To verify if the values of k₂, k_3A, k_3B are consistent

If component “A” can dissolve in solvent X through other reactions, then k₂ and k_3A can also be obtained through other dissolutions. In this case, all the k₂ and k_3A should be consistent. Otherwise Equation (13) cannot be the right formula for the molar chemical potential of component “A” in solvent X. Similarly for component “B” or other solvent.

If k₂, k_3A, k_3B are reasonable and consistent in all dissolutions, it implies that Equation (13) is also reasonable. Under this condition, whether or not k₂, k_3A and k_3B can be simplified as formulas is a topic needs further study. On the other hand, if k₂, k_3A, k_3B are not reasonable or not consistent in all dissolutions, it is likely that Equation (13) is not the correct formula for the molar chemical potential of solute, and a better formula should be looked for.

When a solid dissolves, its solubility product K_sp is normally defined for large monocrystals [15] . From experiments, it is known that K_sp will become larger when the grain size of this solid decreases [16] [17] . This phenomenon certainly complies with the perception that the molecules of smaller grains are with less complete bonding at surface, thus have a higher molar chemical potential and a larger K_sp.

When two solids with a common ion can both be dissolved in a solvent (e.g. the Cl⁻ in PbCl₂ and NaCl dissolved in water), experiments showed that the solubility products of both solids will basically remain unchanged. Since the solubility of NaCl is much higher than that of PbCl₂ in water, the [Na⁺] and [Cl⁻] will be high and [Pb²⁺] will be very low [18] .

At certain temperature and pressure when a solid is at equilibrium with its liquid, the molar chemical potentials of both phases are equal. Hence it’s not difficult to obtain the molar chemical potential of a solid experimentally.

When there are defects (e.g. dislocations) or residue stresses (e.g. in metal forming, welding or distorted solids) in a solid, theoretically the molar chemical potential of this solid will increase. Similarly, when the crystals of a solid are very small, the molar chemical potential will increase due to largely increased interfaces. Under these conditions, the molar chemical potentials of solids will be higher than they should. Hence it is advisable to minimize interferences as much as possible when measuring the molar chemical potentials of solids.

As to the surface energy of a single-crystal solid, attention should be paid to the direction of the surface plane, because it will make substantial difference. For example, according to the table of Wikipedia [19] , the surface energy is 300 mJ/m² when the surface plane of NaCl is (100). While the surface energy is changed to 400 mJ/m² when the surface plane of NaCl becomes (110). In general, polycrystalline solids are more commonly encountered. Their surface energies can be taken as the average of all planes.

Smaller grains in solids have higher surface energies and higher molar chemical potentials. According to the Atomistic 2^nd Law of Thermodynamics, the higher surface energies will become driving forces for the small grains to grow spontaneously, provided the temperature is high enough so that molecules can move. Examples include sintering processes and Ostwald Ripening in solid solutions [20] - [22] .

When conducting experiments or developing formulas to make the Atomistic 2^nd Law of Thermodynamics clearer, there should be certain constraints to be fulfilled. These reasonable constraints include:

a) To avoid the conditions that the Atomistic 2^nd Law of Thermodynamics is not applicable: As addressed in [1] , the conditions that the Atomistic 2^nd Law of Thermodynamics is not applicable include: 1) when there are kinetic constraints, 2) when the system is not yet at equilibrium, 3) when other driving forces cannot be ignored, 4) when light energy plays an important role.

b) The developed concepts or formulas should comply with the known experimental results: Example include:

1) The bonding conditions at liquid surfaces should comply with their saturated vapor pressures in observations (including capillary condensation);

2) The formulas for the molar potential of a gas should comply with the known phenomena (as described in footnote 2);

3) In the chemical reaction of gases via $aA+bB\leftrightarrow cC+dD$, the calculated K_eq based on k_1A, k_1B, k_1C, k_1D (i.e. $\left(k_{1A}^a\cdot k_{1B}^b\right)÷\left(k_{1C}^c\cdot k_{1D}^d\right)$) should be the same as the experimental K_eq;

4) For a compound dissolved in a solvent via $A_p{B}_q\leftrightarrow p{A}_{\left(sol.\right)}+q{B}_{\left(sol.\right)}$, the calculated K_sp based on μ(A_pB_q) and k₂, k_3A, k_3B (i.e. ${\text{e}}^{\mu \left(A_p{B}_q\right)÷k_2}÷\left[{\left(k_{3A}\right)}^p\cdot {\left(k_{3B}\right)}^q\right]$) should be the same as the experimental K_sp.

c) The constants obtained for the same formula should be consistent: For example, 1) If k₀ and k_1G can be obtained from various chemical reactions, then all the constants k₀ and k_1G should be consistent (as described in Section 2.5). 2) Similarly, if k₂ and k_3A, k_3B can be obtained from various dissolutions in the same solvent, then all the k₂ and k_3A, k_3B should also be consistent (as described in Section 3.2).