Hi Salim,

You are absolutely right; the equation is dimensionally inconsistent. This is because it is incomplete. I suggest you check, for example, "Zeolites and Clay Minerals as Sorbents and Molecular Sieves", R. M. Barrer, 1978, AP, pp. 104-105.

First of all, it is important to keep in mind that you are trying to calculate the standard Gibbs energy of adsorption, ∆Gº. The Gibbs energy of adsorption is the same in both phases (since they're at equilibrium), so ∆G=0. On the other hand, typically ∆Gº≠0. The value of ∆Gº depends on the standard states chosen. The correct van't Hoff-like equation is:

K=exp(-∆Gº/RT)

where K is a non-dimensional equilibrium constant. Since ∆Gº is dependent of the standard states, so is K. However, Kc, if defined as qe/Ce, is not. The relation between these two variables is:

K=Kc.Ceº/qeº=Kc/Kcº

where Ceº and qeº refer to the concentrations at the standard states chosen. Thus, the complete form of the equation you are looking for is:

∆Gº=-RT ln(Kc/Kcº)

To be absolutely clear, ∆Gº depends on the two independent standard states chosen for the system. This choice is completely arbitrary, and it is entirely up to you to decide which these should be. By calculating ∆Gº as you mentioned, you are, in effect, choosing a standard state where Kcº=1. The units of Kcº will be the same as those of Kc. For qe in mol.g-1 and Ce in mol.L-1, you get Kcº = 1 L.g-1.

To compare ∆Gº between different adsorption systems, the reference states must be the same (or, at least, have the same Kcº). So, if you wish to compare your result to those you mentioned, you must use the same formula (although now you know there is a Kcº "hidden" in the equation). Naturally, you may, for some reason, wish to choose different standard states, but the ∆Gº obtained will not be comparable to those other results.

Hope this is helps.

Rui Afonso

thank you,

Seelig's paper is helpful, he multplied Kc in equation 1  by The factor 55.5 which is

the molar concentration of water (mol.l-1). 

Hi Salim,

You are absolutely right; the equation is dimensionally inconsistent. This is because it is incomplete. I suggest you check, for example, "Zeolites and Clay Minerals as Sorbents and Molecular Sieves", R. M. Barrer, 1978, AP, pp. 104-105.

First of all, it is important to keep in mind that you are trying to calculate the standard Gibbs energy of adsorption, ∆Gº. The Gibbs energy of adsorption is the same in both phases (since they're at equilibrium), so ∆G=0. On the other hand, typically ∆Gº≠0. The value of ∆Gº depends on the standard states chosen. The correct van't Hoff-like equation is:

K=exp(-∆Gº/RT)

where K is a non-dimensional equilibrium constant. Since ∆Gº is dependent of the standard states, so is K. However, Kc, if defined as qe/Ce, is not. The relation between these two variables is:

K=Kc.Ceº/qeº=Kc/Kcº

where Ceº and qeº refer to the concentrations at the standard states chosen. Thus, the complete form of the equation you are looking for is:

∆Gº=-RT ln(Kc/Kcº)

To be absolutely clear, ∆Gº depends on the two independent standard states chosen for the system. This choice is completely arbitrary, and it is entirely up to you to decide which these should be. By calculating ∆Gº as you mentioned, you are, in effect, choosing a standard state where Kcº=1. The units of Kcº will be the same as those of Kc. For qe in mol.g-1 and Ce in mol.L-1, you get Kcº = 1 L.g-1.

To compare ∆Gº between different adsorption systems, the reference states must be the same (or, at least, have the same Kcº). So, if you wish to compare your result to those you mentioned, you must use the same formula (although now you know there is a Kcº "hidden" in the equation). Naturally, you may, for some reason, wish to choose different standard states, but the ∆Gº obtained will not be comparable to those other results.

Hope this is helps.

Rui Afonso

thank you Rui,

Thank you Paul

I read your paper which is very interesting paper (especially the adsorption capacity which is very high), but in your supplementary materials I do not understand one thing (if we have many points of qe and Ce in each isotherm at constant temperature, it mean that we have many Kc for each temperature but in reality we need just one Kc for each temperature to plot ln(Kc) vs (1/T).

I hope that you understand my question.

Salim

Dear Paul

thank you it is well easy and clear.

but the plot of Kc=qe/Ce vs qe is not linear, i prefer using the plot of ln(Kc) vs qe which give a strghit line and the intercept with ln(Kc) axis will give us directly the value of  ln(Kc) at this temperature.

Salim.

 Hallo, Can somebody explain me what is correct way to calculate the equilibrium constant to calculate free Gibbs energy G=-RT ln(K), please. In the discussion above you suggest that K=qe/Ce, however even in the reference materials some of you suggested it is not so clear, at least for me. Majority used the above equation, however e.g.

Foo, K. Y., & Hameed, B. H. (2012). Adsorption characteristics of industrial solid waste derived activated carbon prepared by microwave heating for methylene blue. Fuel processing technology, 99, 103-109 - used Langmuir constant instead of Equilibrium const.; and

 Xiaoli Yuan, W.X., Juan An, Jianguo Yin, Xuejiao Zhou, andWenqiang Yang Kinetic and Thermodynamic Studies on the Phosphate Adsorption Removal by Dolomite Mineral. Journal of Chemistry ID 853105, 1-8 (2015) - used Freundlish constant.

I doubt that I will receive similar results applying constants from L or F and  K=qe/Ce.

Dear Doctor,

In  the literature several methods are used for the determination of the thermodynamic parameters for adsorption systems (liquid-solid). The common point to all of these methods is the calculation of  the standard free energy change of the  adsorption at equilibrium DG°. In all cases DG° is calculated by this equation  DG°=-RT ln(K) with K is  thermodynamic equilibrium Constant. The most cited  methods for the calculation of the constant K are:

1- K= KL 

The constant K  is assumed to be equal  to the constant  KL  of Langmuir isothermin mol.l-1  .

Ho, Y. S., & McKay, G. (1999). Pseudo-second order model for sorption processes. Process Biochemistry, 34(5), 451-465.

2- K=qe/Ce

Saleem, M. M. R. J., Afzal, M., Qadeer, R., & Hanif, J. (1992). Selective adsorption of uranium on activated charcoal from electrolytic aqueous solutions.Separation science and technology, 27(2), 239-253.

3- K=Cad/Ce 

with ;

Cad : Equilibrium concentration of the adsorbate in the solid phase (mg.l-1)

Ce : Residual equilibrium concentration (mg.l-1)

Namasivayam, C., & Ranganathan, K. (1993). Waste Fe (III)/Cr (III) hydroxide as adsorbent for the removal of Cr (VI) from aqueous solution and chromium plating industry wastewater. Environmental Pollution, 82(3), 255-261.

4- n = Ln (K) and DG°=-RT.n 

with n is the freundlich isotherm constant.

Bell, J. P., & Tsezos, M. (1987). Removal of hazardous organic pollutants by biomass adsorption. Journal (Water Pollution Control Federation), 191-198.

5- K=as/ae 

Biggar, J. W., & Cheung, M. W. (1973). Adsorption of picloram (4-amino-3, 5, 6-trichloropicolinic acid) on Panoche, Ephrata, and Palouse soils: a thermodynamic approach to the adsorption mechanism. Soil Science Society of America Journal, 37(6), 863-868.

in my Phd work i compared between the results obtained by this methods and in overall i found that a certain homogeneity exist between the obtained different values from this different methods.

Dear Salim Bousba ; Dear Hai Tran 

    The equilibrium is between the milligrams of adsorbate on the surface divided by the milligrams of adsorbate in the solution. Therefore, Qe should multiply by the weight of adsorbent and Ce should multiply with the volume of solution. All units will be removed and K will be without units

Exp: Qe=10mg/gm,   Ce=2mg/ml   , Volume =20 ml,   wight of adsorbate=0.25gm

Keq=Qe*Wt/Ce*Volume

you can see all units should be removed.

Unfortunately, Many papers calculate the wrong Keq, i.e., with units.

Please, refere to 

www.ijppsjournal.com/Vol6Issue4/9132.pdf

Dear  Hai Tran

Thank you, i downloaded your paper and I’m very happy because your conclusion confirm my own results obtained in my phD thesis witch reveal that thermodynamics parameters estimated by various methods had the same sign but in my phD work i found that all results (by various methods) had the same order of magnitude and conduct  to the same thermodynamics interpretation for (DG, DH and DS).

congratulations 

how to calculate Ce in above formula ?

 Ce is the concentration of the solution (adsorbate), Ce is determinate by experimental measurement (not calculated).

Please can any one explain how to calculate, Kc in

∆G = –RT lnKc for gas phase adsorption such as CO2?

Dear All,

I think the equilibrium constant is a dimensionless parameter. Let us consider the equilibrium A = B.  The equilibrium constant is defined as K = a(A)/a(B) where a(A) is the activity of A and a(B) is the activity of B.  The activities are defined as a(A) =g(A)[m(A)/m^0(A)] and a(B) =[g(B)[m(B)/m^0(B)] where m(A) and m(B) are the molal or molar concentrations and g(X) is the activity coefficient for the chosen concentration scale, where X = A or B.  This is so, because the activity coefficient does not have units and neither does the activity.  The quantity m(X)^0 is the concentration in the standard state, which is assigned a value of 1.  For dilute solutions, g(A) and g(B) can be set equal to 1, in which case a(A) =[m(A)/m^0(A)] and a(B) =[m(B)/m^0(B)], so that the equilibrium constant becomes K= [m(A)/m^0(A)]/[m(B)/m^0(B)].Thus K has no units.

Kind regards,

Zahed.

Dear all,

How about the thermodynamic calculation for the gas phase adsorption? How to obtain the value of Kc?

Thanks.

Thank you Dr. Hai Tran

Your article is very helpful.

@Hussein Alhakeim Sir, I love your answer and the concept. However, I want to ask you that is this your way of solving the problem or is there any reference research article in which this method has been used.

Hello experts,

I have gone through your quires.i am facing the same.i hope you can help me out.

I have only one Isotherm Langmuir constant for different conc(5 ppm,10 ppm,15 ppm,20 ppm) at same temp.

for thermodynamic study I have Different Temperature(288,298.308,318,328 k)points for one concentration(10 ppm).so i have only one KL values from .so how to plot ln K vs 1/T ?

or can i use K=Qe/Ce to find out different K value to plot the graph..?

The following articles introduce the most correct manner for calculation of thermodynamic parameters and contain valuable explanations about this subject.

Journal of Molecular Liquids Volume 273, January 2019, Pages 425-434 Journal of Molecular Liquids Volume 280, 15 April 2019, Pages 298-300

Plot a graph of Kc(dimensionless) vs 1/T. You can then get ur change in enthalpy and entropy from equation of the plot. You can then get your Gibbs free energy using dG=dH-TdS or dG=-RTlnKc

You have to multiply the value in L/g by atomic mass of adsorbate the by 55.55mol/L. But if your kd is in L/mol multiply directly by 55.55. The concentration of pure water is 55.5mol/L the unit will cancel out the unit of Kd (L/mol) after being converted from L/g to L/mol. Hope this helps if not you can still contact me please. Cheers

The so-called "entropy " doesn't exist at all.

During the process of deriving the so-called entropy, in fact, ΔQ/T can not be turned into dQ/T. That is, the so-called "entropy " doesn't exist at all.

The so-called entropy was such a concept that was derived by mistake in history.

It is well known that calculus has a definition,

any theory should follow the same principle of calculus; thermodynamics, of course, is no exception, for there's no other calculus at all, this is common sense.

Based on the definition of calculus, we know:

to the definite integral ∫T f(T)dQ, only when Q=F(T), ∫T f(T)dQ=∫T f(T)dF(T) is meaningful.

As long as Q is not a single-valued function of T, namely, Q=F( T, X, …), then,

∫T f(T)dQ=∫T f(T)dF(T, X, …) is meaningless.

1) Now, on the one hand, we all know that Q is not a single-valued function of T, this alone is enough to determine that the definite integral ∫T f(T)dQ=∫T 1/TdQ is meaningless.

2) On the other hand, In fact, Q=f(P, V, T), then

∫T 1/TdQ = ∫T 1/Tdf(T, V, P)= ∫T dF(T, V, P) is certainly meaningless. ( in ∫T , T is subscript ).

We know that dQ/T is used for the definite integral ∫T 1/TdQ, while ∫T 1/TdQ is meaningless, so, ΔQ/T can not be turned into dQ/T at all.

that is, the so-called "entropy " doesn't exist at all.

.

.

.

Why did the wrong "entropy" appear ?

In summary , this was due to the following two reasons:

1) Physically, people didn't know Q=f(P, V, T).

2) Mathematically, people didn't know AΔB couldn‘t become AdB directely .

If people knew any one of them, the mistake of entropy would not happen in history.

Please read my paper and those answers of the questions related to my paper in my Projects.

https://www.researchgate.net/publication/230554936_Entropy_A_concept_that_is_not_a_physical_quantity

I think you can find your answer in this publication:

Article Is the Free Energy Change of Adsorption Correctly Calculated?

The conclusion is:

You can use the Langmuir constant (K) if you have very low equil° conc°

Or even Kc=Cads/Ceq in similar case (where Cads is the "concentration of adsorbed species" )...

But the Kd=qads/Ceq seems not to be good (for the reasons you gave above)...

Best

Very useful discussion

This paper will be helpful.