I have found how to calculate parameters of pseudo-second order kinetic model, but bad luck with pseudo-first order model. So, how to calculate qe and k1 from the slope and intercept?
If you have nothing to say, then don't say anything! I even can not imagine, how hard it would be for you to spend some extra 15 seconds to write an answer!
P.S. I have read countless publications (biosorption, low cost sorbents), many say how to calculated parameters of pseudo-second order model, but not about the pseudo-first order model.
Thank you for teaching me. While you modified the question the answer is the same: EXPONENTIAL. C/Co = exp(-kt), -kt = 1-Ln(C/Co). Good luck in reading countless publications.
Hi,
It not so difficult:
v=k[A]^x.[B]^y (1) for example.
If x = 1, and the variation of [B]^y is very small ([B] is too high...) --> [B]^y = constant = C
then equation (1) become: v = k[A]^1.constanc C = (k.constant C)[A]^1= k'. [A]^1
and the it becomes first-order
And they call that is pseudo-first order kinetic model
:)
Hope this helps
It would be easy to understand if you mention reaction scheme and how you are following the reaction.
Pseudo first-order kinetic model as given by Lagergren is ln (qe-qt)=lnqe-k1*t. You can calculate only k1 from this equation because qe is known to you from batch adsorption studies. qe is equilibrium adsorption capacity (or adsorption capacity at equilibrium time), which is known to us through experiments. You plot a graph between ln (qe-qt) Vs time, you will get a slope of-k1 and intercept of ln qe, which should match with experimental value.
Thank you! But shouldn't it be
log (q e – q t ) = log (q e ) - (k 1 /0.2303)*t ?
And is this LOG function in Excel or LN? LN - returns the logarithm of a number to a specified base. LOG - returns the logarithm of a number to a specified base.
And it is not always that qe experimental is close to qe calculated...
"Although the correlation coefficient values at high concentration are higher
than 0.90, the experimental q e values do not agree with the calculated ones, obtained from the linear plots (Table 1). This shows that the adsorption of lead onto activated carbon produced from oil palm fibre is not first-order kinetics."
Hi Artis,
It is really simple. The linearised pseudo first order reaction is represented as follows:
Ln (qe - qt) = Ln qe - k1t
where k1 is the pseudo-first order adsorption rate constant and qe and qt are the values of the mass adsorbed per unit mass at equlibrium and at time “t” respectively.
Plot of ln (qe-qt) versus t for an adsorbent would yield a straight line if the adsorption process follows a pseudo-first order kinetic behaviour, and k1 and qe can be calculated from the gradient and the intercept of the line. For many adsorption processes, the pseudo-first order kinetics was found to be suitable for only the initial 20 to 30 minutes of interaction time and not for the whole range of contact times.
The linearised Lagergren second-order kinetics equation may be represented by the following equation :
t/qt = (1/k2qe2) + (1/qe)*t
where k2 is the pseudo-second order rate constant.
The value of k2 depends on the operating conditions, such as, initial pH and solution concentration, temperature, agitation rate etc. It also decreases as the initial solution concentration increases, since it takes a longer time to reach equilibrium.
If the adsorption system follows a pseudo-second order kinetics, then a plot of t/qt versus t would be linear and k2 and qe can be determined from the intercept and gradient of the graph. Check out my papers:
1. Zaynab Aly, Adrien Graulet, Nicholas Scales, Tracey Hanley, “Removal of aluminium from aqueous solutions using PAN-based adsorbents: characterisation, kinetics, equilibrium and thermodynamic studies, Journal of Environmental Science and Pollution Research (2014) 21: 3972-3986. DOI 10.1007/s11356-013-2305-6. 2. Z. Aly and V. Luca, “Uranium extraction from aqueous solution using dried and pyrolyzed tea and coffee wastes”, J. Radioanaly. Nucl. Chem., Volume 295, issue 2, (2013) pages 889-900. DOI: 10.1007/s10967-012-1851-6.
based on your experimental results,plot a graph of log(qe-qt) vs t.
slope= -k/2.303
intercept= log(qe)
from this find qe (this is calulated orntheoretical value). match it with your experimental values (there should be minimum difference between experimental and calculated values) and analyze based on R2 to see if the model fits or not.
Best of luck.
Use least square method:
qt.................... q2t ................ q't...................qt*q't
0.0425 0.0018078 0.0047220 0.0002008 (1)
0.0472 0.0022297 0.0003577 0.0000169 (2)
0.0479 0.0022928 0.0000545 0.0000026 (3)
0.0483 0.0023338 0.0000486 0.0000023 (4)
0.0498 0.0024829 0.0000449 0.0000022 (5)
0.0510 0.0026014 0.0000246 0.0000013 (6)
0.0513 0.0026324 0.0000135 0.0000007 (7)
0.0522 0.0027270 0.0000126 0.0000007 (8)
0.0528 (9)
......................................................................................
0.3902896 0.0191077 0.0052785 0.0002275 (these are sum of above value)
here n=9 , so (n-1)=8
[8 0.3902896; 0.3902896 0.0191077]x=[0.0052785;0.0002275]
solve the above matrix
you will get k1 from here directly=-0.4483, then 0.225
to get qe=0.225/0.4483=.0502
Thank you for all the great comments you have made! I believe this discussion will be useful for many researchers!
@Prithvi Simha; Was there a reason why you carried out the Lagergen model fitting at different concentrations? Is it sufficient to achieve this at a single concentration?
@everyone; I carried out the kinetic fitting using the simple first order kinetic model using the Arrhenius equation and the rate constant was calculated using the Solver in Excel. This rate constant (0.22) was similar to the one (0.18) derived from the Lagergen pseudo first order model. My question is why do we need to use Lagergen model if we can simply use the simple first order kinetic model? I will appreciate your responses.
Thanks
Yes, i confuse about qe same Riry Wirasnita
How does calculate qe? What's time?
Hello there,
The linearised pseudo first order kinetic model is represented as follows:
ln (qe - qt) = lnqe - k1t
where k1 is the pseudo-first order adsorption rate constant and qe and qt are the values of the mass adsorbed per unit mass at equlibrium and at time “t” respectively.
One has to do the experiment at different times and calculate the adsorbed values.
If the maximum contact time is 24 h then it is the equilibrium time - qe.
Plot of ln (qe-qt) versus t for an adsorbent would yield a straight line if the adsorption process follows a pseudo-first order kinetic behaviour.
Using a straight line equation of y = c + mx examining the plot of ln (qe-qt) versus t - k1 can be calculated from the gradient of the line, while the intercept can give you the capacity of the adsorbent at equilibrium ie. qe.
I hope this clears any ambiguity.
Hi,
I don't understand you. If you plot qt versus t, you will get a curve which will show you the maximum adsorption value, qe. Use this value only to calculate qe -qt values for other qt, not the equilibrium one. Of course the equilibrium value qe - qe will be 0.
Thanks Dear all, Could anyone send me a template for all calculation in excel for Pseudo first order kinetics.
Anuj Kumar, I am not sure exactly what you meant by plotting qe-qt vs. time; but as explained in many posts above (I suggest you to read Z.Aly's posts):
Pseudo First Order:
ln (Qe-Qt) = ln Qe -k1t
Y = C -mX
You may find the Qe - Qt by assuming the the last time of your Qt is in equilibrium (hence, the Qe). then you can plot ln (Qe-Qt) vs t which is supposed to produce a straight line (or have a line fitted to the data). The slope (m) will be the rate constant, k, the intersection is the ln Qe (so, Qe = e^(ln Qe) ).
if the 1st order rate equation has some intercept value. So what does it signifies ?
I agree with Selvaraju Sivamani on the calculation of the individual parameters for pseudo first order kinetic model...
My experimental qe values do not agree well with calculated qe values for both pseudo-first order and pseudo-second order adsorption kinetic models. Can I explain applicability of model on the basis of R2 values...
Anyone can provide me any such reference....???
Plot the graph first and get the equation. For example if u want to calculate rate constant for pseudo-first order:
log (qe-qt) versus time
y = m x + c
log (qe-qt) = -(k1/2.303) t + log qe
so, y = log (qe-qt) ; m=-(k1/2.303) ; x=t ; c=log qe
when u plot the graph, please get the equation and compare their values. for example:
y=-0.0087x+1.2034
log (qe-qt) = -(k1/2.303) t + log qe
(k1/2.303)=0.0087, so u will get k1=0.0461
log qe=1.2034, u will get qe=15.97
.: Correct me if i am wrong.
kindly suggest that what is the significance of these parameters and they should be higher or lower. Which condition is favourable.
Add the following formula for pseudofirst order kinetics into the "build new function" option of the origin software.
y=(B*(1-(1/(1+(A*B*x))))), here A and B are the two parameters.
Plot the data and fit with this function to obtain A and B.
A = k2 and B = qe,cal.
What you call the "pseudo-first-order model" is actually an empirical differential equation: dqt/dt=k1(qe-qt), qt(0)=0, whose right-hand side is related to the intuitive concept of linear driving force (LDF). Curiously, this name is also used to its solutions, both for the explicit form with respect to qt (exponential function) and for the implicit form (logarithmic function).
To find the values of k1 and qe, it is best to use the nonlinear regression method with the measure of fit in the form of the mean relative error. I advise against using the graphical method in the ln (qe-qt) and t coordinate system. Regards,
Having the appropriate data, the approximation equation qt=qe(1-exp(-kt)) is used, and using the non-linear regression method, the value of two parameters (qe, k) is determined. However, it must be remembered that they are numerically correlated. Regards,
Himanshu Gupta I´m having the same problem thad you had: my experimental qe values are different to both kind of models. What did you do in this case? Thanks in advance!
Paulina Alferez and Himanshu Gupta
If, in addition to kinetic studies, you perform equilibrium tests (this is usually the case), you know the Ce values (eg. for different initial concentrations of adsorbate, Co) and the corresponding qe values. So these and only these values should be used to determine k1. Why to determine the values of qe again, if they are known from the experiments. Have you ever asked yourself such question? You are also wondering why the value of qeexp can be different from qecal, the answer is very simple - calculation error. This is the first reason and the second reason is that one use exponential or hyperbolic equation transformed to the form, which after indroduction of new variables takes linear form. Recently I wrote a dozen comments to the works posted on the RG, where for the so-called PFO and not only the value of qecal was significantly different from qeexp. They're on my RG side. For example: qeexp = 77.63 and qecal = 5.50 (PFO). Pure nonsense. Regards,
how to calculate the adsorption studies for film by kinetic methods
https://www.khanacademy.org/science/chemistry/chem-kinetics/copy-of-kinetics/v/plotting-data-for-a-first-order-reaction
This link will be useful to find the rate constant for first order reaction. Thank you.
ln (qe-qt)= lnqe - k1*t.
You can calculate only k1 from this equation because qe is known to you from batch adsorption studies. qe is equilibrium adsorption capacity (or adsorption capacity at equilibrium time), which is known to us through experiments. You plot a graph between ln (qe-qt) Vs time, you will get a slope of-k1 .
Pseudo first-order kinetic model as given by Lagergren is ln (qe-qt)=lnqe-k1*t. You can calculate only k1 from this equation because qe is known to you from batch adsorption studies. qe is equilibrium adsorption capacity (or adsorption capacity at equilibrium time), which is known to us through experiments.
Destiny Lover Nf
It's good that you have doubts, because it is impossible. The parameter k2, incorrectly called "pseudo-second-order rate constant", is not a rate constant. It is a lumped parameter that brings together the influence of many different factors.
Themba Dominic Ntuli Do you really want to know or are you just pretending. You will get the q values from batch experiment. If the experiment goes on long enough you will get qe.
Dear Mirostaw Grzesik it was not a pretense. The question still remains, qt and qe bearing in mind that qt is the capacity at a particular time and qe is the capacity at equilibrium
Themba Dominic Ntuli You're right. The concentration denoted q, qt or q(t) is related to any time. It is the current concentration. In contrast, qe refers to t->∞, so it is an equilibrium concentration. I intentionally used the term concentration here, because the adsorption capacity: q=m/M is also concentration.
Aif Ayub shared very useful youtube links for you please follow it as it can be to much good for you during parameters fitting if still feel difficulty then let me know
Well, not everyone has to have a view that matches yours. Links to YouTube are useless because they concern commercial software. Nor do I think it is necessary to disseminate unreliable graphical methods. It's a kind of surreptitious advertising.
Let me add that the basic problem in both demos is the biased data selection, which is to convince viewers that "PSO is the best". It's like reading Ho and McKay's 1998 and 1999 works. I wonder if the authors of both demos based on these works?
M S Islam, Again, again. The equation/function is either linear or non-linear. There is no linear form for any nonlinear equation/function. And PFO and PSO are nonlinear functions.
This video demonstrates how to fit the linear Pseudo first order (PFO) kinetic model in OriginPro easily! https://youtu.be/xPkRj4bcDcc
Dear Artis,
Find the below paper. It may helpful for you.
Article Synthesis of nano-magnetic MnFe2O4 to remove Cr(III) and Cr(...
Kind Regards
Ahmad
what do you mean first and second order pseudo adsorption kinetics model ?
@Siti Ida
in the calculation you made (log (qe-qt) = -(k1/2.303) t + log qe), where did you get the 2.303 from?
Josephine Getz
We usually take the natural logarithm (the ln) of both sides. Instead, she did log and not ln. The difference (I am not sure if this is the term for that) between log and ln is 2.303.
PFO is actually exponential function, so it's best to use the nonlinear method commonly known as nonlinear regresion method. I would not recommend the often used graphical double log plot method here.
Plot a graph of log(qe-qt) vs t.
Slope=-kpseudo1
Intercept= log qe
R2=correlation coefficient
https://www.sciencedirect.com/topics/materials-science/adsorption-kinetics
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