Question

# How can separation factor (RL), calculated through Langmuir isotherm model, serve as an invaluable factor in the adsorption process?

This factor has been extensively used in the literature, and it was stated that if 0<RL<1, the process is favorable. I believe this inference is not accurate due to the fact that according to the formula, RL=1/(1+C0*k), its value would always be between 0 to 1.
more interestingly, it was mentioned in several papers that If RL= 0 process is irreversible, if RL= 1 it is linear and if RL> 1 it is unfavorable! The values of 0 and 1 can be accepted as asymptotes, but RL>1 ?! It is simply impossible!
It would be great if we could shed light on this problem. It may be the question of several researchers in the future.

Rajith Akalanka Perera
University of Kelaniya
You can use the optimum concentration, which showed maximum removal efficiency.
2 Recommendations

Fernando Vallejos-Burgos
Hi Mojtaba,
I will analyze each case in separate, according to the RL definition you showed [ RL=1/(1+C0*k) ]:
• RL= 0 : This occurs if k is very very large, which means that adsorption is too strong. That is why RL=0 is named irreversible.
• 0 < RL < 1 : This is the standard case when adsorption occurs normally under our conditions. Not so strong, but noticeably occurs and we can notice the shape of the adsorption isotherm. We call this favorable adsorption.
• RL = 1 : This can occur only if k=0. This means that the adsorption isotherm (dependence of adsorbed amount and concentration) is a straight line. So this is linear adsorption.
• RL > 1 : This is not impossible, as you said. We just need a negative k, AND that -1 < C0*k < 0 (check your equation). If denominator of the equation is negative, then RL is negative, and instead of adsorption, we will have desorption. We call this an unfavorable adsorption (that is, desorption).
I hope this is clear!
17 Recommendations

Yogesh C Sharma
Indian Institute of Technology (Banaras Hindu University) Varanasi
Dear Mojtaba,
I have also seen the parameter in several good articles but in my opinion, this is not conclusive. I have not so far seen any system where RL>1 is reported.
Fernando Vallejos-Burgos
Hi Mojtaba,
I will analyze each case in separate, according to the RL definition you showed [ RL=1/(1+C0*k) ]:
• RL= 0 : This occurs if k is very very large, which means that adsorption is too strong. That is why RL=0 is named irreversible.
• 0 < RL < 1 : This is the standard case when adsorption occurs normally under our conditions. Not so strong, but noticeably occurs and we can notice the shape of the adsorption isotherm. We call this favorable adsorption.
• RL = 1 : This can occur only if k=0. This means that the adsorption isotherm (dependence of adsorbed amount and concentration) is a straight line. So this is linear adsorption.
• RL > 1 : This is not impossible, as you said. We just need a negative k, AND that -1 < C0*k < 0 (check your equation). If denominator of the equation is negative, then RL is negative, and instead of adsorption, we will have desorption. We call this an unfavorable adsorption (that is, desorption).
I hope this is clear!
17 Recommendations
Dear Yogesh,
Thank you for your support. In fact, I struggled with this question in my previous works, and even I used this factor in my first ISI paper to show that the adsorption process is favorable.
I am really optimistic that we can reach a concrete solution to this question.
Hi Fernando,
I am really grateful that you participate in my discussions.
As I said in the explanation, the value of 0<=RL<=1 is acceptable, and you described each part separately, and I really appreciate it. But for the RL>1 I couldn't get it.
Do You mean that we could have a negative value for KL? (we definitely know with a negative value for constant rate, the denominator would be less than 1).
In every typical adsorption process, we have a contaminated solution, and we will use fresh adsorbent to remove those contaminations (In laboratory scale). So after the equilibrium time, finally we could only accept RL=1 as an asymptote, which is a sign of no adsorption! let me explain it in another way: There is no adsorbate molecule on the adsorbent that could be released during the process (This is the only condition that we could accept desorption). Can you recommend any papers that had RL>1 with a fresh adsorbent?
Therefore, what I am inclined to mention here is that in adsorption process with the fresh adsorbent, we should not even mention RL>1. However, if the adsorbent is previously used and contains some adsorbate molecules attached to its particles, we can accept the desorption hypothesis and KL lower than zero.
I hope I could convey the main confusing part of this equation.
Fernando Vallejos-Burgos
Hi Mojtaba,
Thanks for following the discussion.
As I said, the only way in that RL>0 is by desorption. Desorption cannot occur if there is nothing already adsorbed (as you correctly state). So, you are right saying that with fresh adsorbent RL>0 cannot occur.
Note: I edited my comment above, to fix the correct condition as -1 < C0*k < 0 instead of simply C0*k < 0.
Dear Fernando,
Now it is much clearer, and both of us agreed on this statement that in adsorption with a fresh adsorbent, RL>1 is not possible.
Thank you for this helpful discussion.
Zaidi Ab Ghani
Universiti Teknologi MARA
Very nice and interesting discussion. Thanks for your sharing, Fernando and Mojtaba.
Pakistan Institute of Engineering and Applied Sciences
nice discussion
Susmita Kamila
Bhubaneswar, Odisha
so far no one has pointed if Rl<0 then what would be the condition
Dear Susmita,
I am of the opinion that the opportunity to discuss this challenging parameter is missing. I personally expect it from researchers to consider this discussion in their future works.
Best regards.
Nur -E- Alam
Bangladesh Council of Scientific and Industrial Research
Dear All,,
i have found a Langmuir equation like this y = 8.454x - 4.365 and R² = 0.815 where the value of k is -4.37. Is it correct?
1 Recommendation
Dear Nur, Did you try different isotherm models such Freundlich, Elovich, etc? Because the coefficient of determination (R2) calculated here is not high enough to suppose that Langmuir is able to correlate the experimental data.
Mehdi Safdarian
Ahvaz Jundishapur University of Medical Sciences
Dear Mojtaba,
I read the discussion between you and Mr. Fernando and I enjoyed it very much.
Your examples are about public adsorption, but in the case of specialized absorbents such as MIPs (molecularly imprinted polymers), this phenomenon may happen. You can refer to this paragraph from the reference below:
" It is well known that template residues are present in the polymer matrix even after exhaustive washing steps and leakage of the template can affect the accuracy of detection in trace analysis.To overcome this bleeding problem the MIP can be synthesized by using a close structural analogue of the analyte as template (also called dummy template) that can capture also the target analyte since very similar cavities are generated"
in acidic condition template residues in the polymer matrix, may will have desorption in equilibrium solution, as a result KL<0 and RL>1.
Best wishes.
1 Recommendation
Dear all brothers, read all the discussion which is very fruitful but i have the same problem can any one tell me how to find out the value of k, can any one give me a brief detail i will be very thankful to you.
Dear Saleem,
We all are happy you found this discussion useful.
I see you have a question about finding K value in Langmuir isotherm. If so, when one obtained isotherm experimental data, he can do the regression to find out which model works better. The one in which the higher (R2) value is found fits the experimental data more accurate. This is how researcher can go for hypothesis and theory behind that model and conclude some results etc.
You can find thousand papers in this area. Should you need more help please don't hesitate to add your comment here.
1 Recommendation
Shahjalal Khandaker
Kyushu University
Could anyone help me regarding the issues:
1). In one adsorption experiment, I got the ∆G (kJ/mol) values as below.
-8521.47 (288K)
-9073.61(298K)
-9624.52(308K)
This actually indicate that the reaction is spontaneous and favorable.
My inquire is what is standard value of ∆G (kJ/mol)?
2). Another inquire, what is the standard value of
∆H (kJ/mol) and ∆S (kJ/mol) in thermodynamics? There is any limit??
Igbinedion University
How does one deal with a situation where the correlation coefficient (R2) is as high as 0.99 yet the Langmuir Isotherm constant K gives negative value? Do we say the model failed to described the data despite the high R2 value or are there other reasons? Please, I will appreciated prompt contributions from all.
1 Recommendation
Danilo Gualberto Zavarize
Instituto Federal de Educação Ciência e Tecnologia do Maranhão (IFMA)
I think that there is a factor which might cause RL to be higher than 1, and it is the agitation speed. In an adsorption system, if you increase agitation speed using the same adsorbent already adsorbed with any amount of adsorbate, there is the possibility that higher speed rates may lead the ion or molecule to unbind from the surface and return to the solution, mainly if it is a physisorption process which is weaker and reversible. It will make the constant to be negative at some point of the process. The effect may be minimal, but I think it is possible. I don't know if there is already, but research regarding this topic would be interesting.
2 Recommendations
Dear Danilo,
According to RL formula, and given the fact that we do not analyse the concentration during the process in isotherm experiments, I believe 0<RL<1.
However I disagreed with your reason, it it does not mean it is incorrect. I would be more than happy to understand the concept you are referring to.
Regards,
Mojtaba
Dear Shahjalal,
You can find the definition of standard value in the text books.
I know there is a range of ∆H for physisorption and chemisorption, indicating the nature of adsorption based on its value. You may find the same for other parameters in the literature.
Sorry couldn't be of help.
Dear Olanrewaju,
I am grateful for your comment.
I personally have never faced a negative value for K in my works. You are recommended to check other models as well.
Please share your new findings in this area to all of us.
Thank you.
Vajiheh Behranvand
Isfahan University of Technology
Dear Fernando,
Ahamed Ashiq
Queen's University
Thank you Mojtaba Hedayati Marzbali for raising this query! Keerthanan Santhirasekaram apt for you!
Keerthanan Santhirasekaram
Bergische Universität Wuppertal
Thanks and separation factor always take value between 0 to 1
Sami Guiza
National Engineering School of Gabes. University of Gabes. Tunisia.
A dimensionless constant called the equilibrium parameter can be used to express the important characteristics of the isotherm as follows:
where KL is the Langmuir constant and C0 is the initial solute concentration . The value of RL shows if the adsorption process is favorable or not as follows:
• RL > 1 : Unfavorable adsorption
• RL = 1 : Linear
• 0 < RL < 1 : Favorable
• RL = 0 irreversible.
Daniel Dittmann
German Environment Agency
thank you very much for this question since I was struggeling with the separation factor, too.
I want to add an enlighting information I found in the book of Eckhard Worch (2012).
In short, RL > 1 is possible because the factor is not necessarily derived from the Langmuir equation but it can always be used in it.
"In this way, any isotherm can be formally reduced to a Langmuir isotherm with a constant separation factor, R*. Such an isotherm transformation can be advantageous because for the case R* = constant, several analytical solutions for breakthrough curve models exist. It has to be noted that the formal application of Equation [...] to other isotherms can lead to separation factors of R* > 1, a range that is originally not covered by the Langmuir isotherm. For the Langmuir isotherm, R* is always lower than 1 [...]."
R* = RL
Thank you for providing us with this valuable information. We are getting some new information now. Much appreciated.
Regards.
Susmita Kamila
Bhubaneswar, Odisha
useful information
Augustus Ilori
Federal University Oye-Ekiti
Dear All,
I am so grateful for your contributions concerning the subject matter particularly the issue of -be RL values. I am currently working on P adsorption in biochar amended soils and I have observed -ve RL values which have been worrisome. But I now understand the cause; most biochars has the tendency of adding P to the soil. Hence, I experienced P-desorption in those soils with exceptions of controls without biochar amendments.
1 Recommendation
Thank you for your support, and also sharing your finding with us.
my contribution is late, hopefully not too late.
Separation factor is about separation (of 2 components).
https://goldbook.iupac.org/terms/view/S05614 gives reasonable definition for chromatography, but it is also defined for liquid-liquid distribution https://goldbook.iupac.org/terms/view/S05615
No idea who first got the idea to 'misuse' it for a one component adsorbate to try to double information already included in adsorption constant.
This is then combined with poor mathematics here in current discussion
' RL= 0 : This occurs if k is very very large' - no it gets small but never 0 , but it might be even be compensated by very small initial concentration, very large k does not mean irreversibility
negative k is not really defined - Langmuir adsorption isotherm is not for desorption.
If I do not miss a point RL provides no additional benefit.
4 Recommendations
Ihsan Habib Dakhil
Al Muthanna University
The value of RL can calculate by the following eq.:
RL=1/(1+C0*k)
Its value should be 0<RL<1.
The value of RL shows if the adsorption process is favorable or not accordining the following values:
• RL > 1 : Unfavorable adsorption
• RL = 1 : Linear
• 0 < RL < 1 : Favorable
• RL = 0 irreversible.
makes no sense to repeat answers of others, even more when RL > 1 'unfavorable adsorption' - is impossible
Moubarek Mebarki
Université des Sciences et de la Technologie d'Oran Mohamed Boudiaf
Ihsan Habib Dakhil
We use different Co (initial conc.) for Langmuir isotherm modeling data i.e. Co=100, Co=300, Co=500 and Co=800. Then which Co would be use for RL ??
1 Recommendation
Ungku Amirul Arif Ungku Abdullah
Universiti Teknologi MARA
Moubarek Mebarki Ihsan Habib Dakhil Have you figure this yet? I also used different initial concentration (Co) for the isotherm study. Which concentration do we pick? Or is it not possible to calculate RL with different Co?
Ihsan Habib Dakhil
Al Muthanna University
You can draw RL versus different initial concentration ( Co ) at optimum conditions of other parameters which selected in your study.
Irum Asif
Fatima Jinnah Women University
Hi, I have a similar problem when I am applying Langmuir Isotherm on different concentrations of dye, and is obtaining RL =2.001 for all the adsorbates showing Langmuir adsorption is unfavourable and R2 value is equal to 0.9975 and 1.00 in some cases. I am using Rl=1/1+(1+ Kl*C0) formula.
Kindly guide in this regard
Irum, read the answers carefully and you should see it is all nonsense.
Asif Ayub
The Islamia University of Bahawalpur
1 Recommendation
Chandra Bhanu Gupt
Indian Institute of Technology Guwahati
It is very interesting discussion regarding the separation factor for the adsorption process and very good topic of research to avoid the confusion on separation factor. But I am still confused why these parameters are used to support the adsorption process.
Bestani Benaouda
I agree with Fernando Vallejos-Burgos
Seema Gupta
University of Delhi
@Dr. Asif, in you tube video calculation of RL is done at Ci =50. Couldnot understand it
1 Recommendation
Ernesto León Alcantar
Autonomous University of Baja California
In the RL equation, Ci represents the highest of the initial concentrations in the experiment
page 6
Rajith Akalanka Perera
University of Kelaniya
You can use the optimum concentration, which showed maximum removal efficiency.
2 Recommendations

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