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The qe value calculated earlier (using Ce/qe) is derived from your experimental data, while the qe value, which is theoretical, is determined based on the Langmuir model. Therefore, they differ in terms of being experimental and calculated values, respectively.

Hi there - great question. The R^2 values in the comparison table at around 16:25 are worse than those in the linear graphs, yes. This is because when comparing linearised and non-linear fits, we should compare R^2 values calculated using y-axis data in the original format, with the original units! This means using data expressed as (mg g-1) and not expressed as (Ce/qe) or as ln(qe). if you download the spreadsheet, you can explore the formulae and maths used

Hi there - great question. We want to compare the quality of our optimised model with optimised parameters k and qe with a very poor model. The poor model is a simple average of all y-values (i.e. no model at all!). So we calculate (a) sum of squared differences between experiment and model, and then calculate (b) sum of squared differences between experiment and average of all y-values. And we compare the two. Further explanation can be found elsewhere: www.ncl.ac.uk/webtemplate/ask-assets/external/maths-resources/statistics/regression-and-correlation/coefficient-of-determination-r-squared.html

Definitely! But in this spreadsheet, we show all the mathematical steps, one by one, to improve our understanding of what R^2 is and what it means and how we calculate it :-)

Imagine that qe (mg g-1) = KF (UNITS) * Ce (ug L-1)^(1/n). We can rearrange to get KF (UNITS) = qe (mg g-1) / [Ce (ug L-1)]^(1/n). This is the same as KF (UNITS) = [qe (mg g-1)]*[Ce (ug L-1)]^(-1/n). If we look at the units only, we see that the units for KF will be equal to (mg g-1)*(ug L-1)^(-1/n). Or more generically, the units are (aqueous concentration units)*(adsorbed adsorbate units)*(-1/n). These units are very strange indeed!

That is fine. k1 and k2 have different units, so the comparison of the absolute size of the numbers of k1 and k2 is not a useful comparison to make. For example... pseudo-first order: rate = k1 * (qe-qt) pseudo-second order: rate = k2 * (qe-qt)^2 PFO: rate units = mg g-1 min-1 PSO: rate units = mg g-1 min-1 (the same units for both rate equations) PFO: (qe-qt) units = mg g-1 PSO: (qe-qt)^2 units = mg^2 g-2 PFO: k1 units = min-1 PSO: k2 units = g mg-1 min-1 Different units = comparing the size of the numbers isn't necessarily useful. One example of this is to calculate the initial rates (use the rate equation, but set qt equal to 0). This gives you PFO: rate = k1*(qe-0) PSO: rate = k2*(qe-0)^2 experimentally observed 'rate' is the same for PFO and PSO experimentally observed 'qe' is the same for PFO and PSO However, PFO rate equation uses qe^1, whilst PSO rate equation uses qe^2. So if qe is greater than 1, then k2 has to be smaller than k1 in order for PFO and PSO models to both give the same initial rate. If qe is smaller than 1, then k2 has to be bigger than k1 for PFO and PSO models to both give the same initial rate. Your experiment gives k1 > k2, so my guess is that your qe is greater than 1 (whatever units you are using...). Your k1 values are approximately 10 times bigger than k2. So I guess than your qe is about 10 (with whatever units you are using)... Best wishes

You should get the same answer either way, whether you use natural logs or base 10.

Hi there, the PSO equation you wrote is directly rearranged into the linearised equation used in this video. (To check this, [1.] inverse your equation to give 1/qt=(1+k2*qe*t)/qe^2*k2*t. [2.] Cancel terms to give 1/qt=(1/qe^2*k2*t)+(1/qe). Then [3.] multiple both sides by t to give t/qt=(1/qe^2*k2)+(t/qe). In my spreadsheet, I would have taken the linearised equation, as I am most familiar to working with this, then rearranged to essentially give the original equation (non-linear) that you write in your comment. Hope that helps

Kinetic experiment is usually carried out at a fixed concentration (e.g., 10 ppm), adsorebt dose, pH and temperature as a variation of contact time to get the optimum equilibration time. So, if you wanna use different conecntrations then you need to carry out separate experiment for each concentration such as 10 ppm, 20 ppm etc.

Yes, and you will see some interesting effects, like PSO rate constant k2 decreasing by a factor of x2 when the initial adsorbate concentration is increased x2. We discuss these effects in our paper published in Langmuir journal

pudiste sacar el valor de qt?

Hi there, what is the model you're trying to fit? Is it an adsorption isotherm with 4 fitting parameters? When we increase the number of fitting parameters, we increase the probability that the Solver algorithm fails to find the true best fit, and instead it finds a not-so-good fit. If you can, try to simplify the problem so that you can optimise just two unknowns first. After you know what those two unknowns are, you can use them in the model and calculate the other two unknowns.

@@jaybullen2887 Thank very much for your suggestion. I used Peleg's model (MC = A(aw)^C + B(aw)^D; C<1, D>1) for sorption isotherm.

@@karunratsakulnarmrat7993 I'm not familiar with the Peleg sorption isotherm model. Looking at a paper called "Comparison of the Sorption Properties of Fruit Powder Shampoos Using the BET, GAB, and Peleg Models" - They use a Monte Carlo algorithm to solve the non-linear regression and optimise all four constants. This should avoid the algorithm getting stuck in a local minimum rather than the true minimum. Unfortunately I don't know how to do this using Microsoft Excel's Solver without a lot of manual work (i.e. constrain two parameters, optimise the second two parameters, write down R^2, change the first two parameters, and repeat the optimisation, and generate a whole table of R^2 values then choose the best) - If you solve the problem, then please do share your progress. Good luck!

@@jaybullen2887 Thanks for your information. I recalculate Peleg model same as your way. But this time I change the initial value (previously assigned as "1" for 4 unknows). After I run the solver the results look good. I will try Monte Carlo in order to compare the results. Highly appreciate your time and suggestion.

@@karunratsakulnarmrat7993 Thanks for the update. Glad that you're making progress. You can also make a very nice graph where the x-axis is the value you fix unknown 1 or unknown 2 to be, and the y-axis is the R^2 value after you optimise unknown 3 and unknown 4. You change unknown 1 and 2 to many different values, and write down the R^2 results. Let us know how you get on with the Monte Carlo method and how you implement it.

Hi there, when you say binary adsorption isotherm models, do you mean (a) two different adsorbates, (b) two different adsorbent materials, or (c) two different adsorption mechanisms, e.g. monolayer adsorption followed by multilayer adsorption-surface precipitation?

@@jaybullen2887 Thank you for your reply. I mean adsorption of two different adsorbates. How can I use non-linear method to fit the data to different multiple adsorption models such as extended Langmuir model, modified Langmuir model and so forth?

@@ry8361 Perhaps you could provide the equation that you would like to model? I am not familiar with multi-adsorbate Langmuir adsorption isotherms and I imagine that there are many different approaches. Perhaps contact the author from a recent paper than interests you. Looking at the "Non-modified Langmuir multi-component isotherm" in the following paper: pubs.rsc.org/en/content/articlelanding/2019/ra/c9ra04865k the authors provide the following adsorption isotherm equation: qe,j = Qmax,j * (KL,j * Ce,j) / (1 + Σ,j (KL,j * Ce,j)) where parameter j refers to either adsorbate A, adsorbate B or adsorbate C etc... (It's very hard to write the equation in TH-cam so please refer to the paper!) You could model an adsorption isotherm using this equation in Microsoft Excel, using the Solver function. You would add extra columns since for a binary system you don't just have one independent variable (Ce) and one dependent variable (qe), but you instead have two independent variables (Ce,A and Ce,B) and two dependent variables (qe,A and qe,B). Just like the single-adsorbate system, you would run the Solver and try to improve the R^2 value. The challenge is that you now have 4 fitting parameters instead of 2 (KL,A and KL,B and Qmax,A and Qmax,B). The Solver function is likely to fail to give the best fit, since the algorithm may get stuck at a local minima rather than the universal minimum when trying to minimise the error between the experimental data and the model. To solve this, you could first determine KL,A and Qmax,A by fitting an experimental adsorption isotherm where you only have adsorbate A, and there is zero adsorbate B. Then you can constrain (fix) the values of KL,A and Qmax,A in the binary model to the results from your single-adsorbate experiment, and you will only have to optimise the fitting parameters KL,B and Qmax,B. When you are dealing with complex systems with multiple adsorbates and multiple adsorbents, you may wish to consider using a Surface Complexation Model (SCM) instead. In an SCM, you calculate adsorption equilibrium constants for one adsorbate and one adsorbent at a time, then you combine your experimental parameters with a database of literature parameters to achieve a model that is sensitivity to many factors: not just adsorbate concentration but also pH, ionic strength and competitor ions. I have a short tutorial article about surface complexation modelling here: www.researchgate.net/publication/349947876_Surface_complexation_modelling_Building_tools_to_understand_and_predict_adsorption

De nada - estou feliz que foi util