Rate of adsorption/desorption in the kmcos example.

[Michele]

Dear kmcos experts,

I have a question regarding the rate constan for the adsorption/desorption of molecule in the example provided in the kmcos package. The example can be found in the link below:

https://github.com/kmcos/intro2kmcos/blob/master/task_material/COoxRuO2__build.py

In this example the rate of adsorption is:

p_COgas*bar*A/2/sqrt(2*pi*umass*m_CO/beta)

which contain the partial pressure p_COgas and the rate of desorption is:

p_COgas*bar*A/2/sqrt(2*pi*umass*m_CO/beta)*exp(beta*(E_CO_cus-GibbsGas_COgas)*eV)

So from my understand the rate constant the adsorption k_ads is:

p_COgas*bar*A/2/sqrt(2*pi*umass*m_CO/beta)

and the reversed reaction is calculated from K_std = k_ads/k_des which lead to the appearance of exponential term *exp(beta*(E_CO_cus-GibbsGas_COgas)*eV) in the rate of desorption. As a result both adsorption and desorption contain the partial pressure in their rate expression.

However, when I check this paper (https://pubs.acs.org/doi/suppl/10.1021/acs.jctc.9b01249/suppl_file/ct9b01249_si_001.pdf).

The rate and rate constant is calculated using the quations below. 

The rate constant k_ads does not contain the partial pressure. So if I use the equation K_std = k_ads/k_des to calculate the reversed reaction. Then the rate constant of the desorption k_des should not contain the partial pressure. As a result, the expression of reaction rate for desorption should not contain the partial pressure (which is different with the example).

I really appreciate if someone can explain to me the difference.

Michele,

Bests

[Aditya "Ashi" Savara]

This is a good question, and is something I wanted to change in the documentation of kmcos. If you see any examples of it in documentation and examples, maybe it will be good to point out.

By definition, a rate constant in physical chemistry is like this: rate=k[A][B] , for example, which is like rate = k*P*Theta

In kmcos, what is often called a rate constant is actually a transition frequency. In spatial KMC, if the species configurations conditions are met (something similar to concentration) then the event occurs.   However, in the current version of kmcos, the gas phase species are not accounted for as species with positions -- so the pressure cannot be included in the "conditions" for the transition frequency.  Instead, the pressure is included as a term in the transition frequency directly.

Sometimes when people write kmc papers (and maybe in some places of the kmcos documentation / examples) the "transition frequency" becomes referred to as a rate constant. I prefer to use the term transition frequency to distinguish these.

It looks like the current wikipedia entry uses the word "transition rate": https://en.wikipedia.org/wiki/Kinetic_Monte_Carlo

And as you can see from the wikipedia references, that is not a new term. Perhaps kmcos (and perhaps I) should consistently use the phrase "transition rate" for the kmc constant.

[Michele]

Dear Aditya,

Thank you a lot for your explaination.

Lets me take an example to see if I understand your answer correctly.

If the adsorption process is (*) + A <-> *A and k_ads and k_des is the real rate constant for the adsorption (forward) and desorption (backward), respectively.

The rate of forward reaction (rate_f=k_f[*][A]) is rate_ads = k_ads*P_A*Theta1. Since the gas phase species are not accounted for as species with positions and  the pressure cannot be included in the "conditions", the rate constant in kmcos is expressed as k_ads_kmcos = k_ads*P_A. As a result it contains partial pressure.

Now consider the reversed reaction. The rate of this reaction (rate_b=k_b[*A]) is rate_des = k_des*Theta2. In this case, the rate constant in kmcos should be expressed as k_des_kmcos = k_des ?. If I use the K_std = k_ads/k_des relation, I can rewrite it as k_des_kmcos = k_ads/K_std.

In that case, I am a little bit confused why it has the partial pressure in the equation of the desorption process.

Bests

Michele

[Aditya "Ashi" Savara]

[I thought I sent this email already, but it is not appearing in the online history so I am sending it again, and editing some of what I wrote.]

We should be a bit careful with our wording on this issue.

(1)

You wrote:

"Since the gas phase species are not accounted for as species with positions and  the pressure cannot be included in the "conditions", the rate constant in kmcos is expressed as k_ads_kmcos = k_ads*P_A. As a result it contains partial pressure."

What you are saying is correct, though it's not the wording I would use. As you probably noticed,  during adding of processes in kmcos we have an argument named "rate constant" . That argument corresponds to the kmc transition frequency when the conditions are met. My personal preference is to call the kmc variable a transition frequency, and to only use the word "rate constant" for the physical chemistry concept. The physical chemistry concept of rate constants has concentrations for all reactants.  So I would prefer if the kmcos argument name becomes changed, which will avoid confusion for people in the future. Your equation and understanding seem to be correct. 

(2)

"Now consider the reversed reaction. The rate of this reaction (rate_b=k_b[*A]) is rate_des = k_des*Theta2. In this case, the rate constant in kmcos should be expressed as k_des_kmcos = k_des ?"

This is correct under certain circumstances: if the transition frequency (when the conditions are met) are the same as the rate constant, then yes, they will be equal. For a first order desorption process, they will be the same.

For second order process, k_des can be written in absolute units (like molecules/m^2) or in relative units (like theta).  If using relative units, then k_des_kmcos=k_des even for second order process.

(3)

"If I use the K_std = k_ads/k_des relation, I can rewrite it as k_des_kmcos = k_ads/K_std."

For the third equation, I might have to be a bit careful. K_std means the thermodynamic rate constant. That means that K_std= e^-Delta(G/RT) = [Products]*standard-state/[reactants]*standard-state  and also means that k_ads and k_des must be written with the correct units to support this equation. So if you write everything out, and find that the units are all working correctly then you can probably make k_des_kmcos = k_ads/K_std.  But it depends on which units you use for the various terms.

One way I can suggest to improve your understanding (and to make a kmcos example or add an exercise for a kmcos example) is to calculate Langmuir adsorption for simple adsorption desorption. If you create a simple adsorption desorption case for a single site type, then you can calculate with pencil and paper (or excel) which pressure will be required to reach coverage of 0.75 and 0.50, then you can check if you get that coverage when putting the kmcos simulation at that pressure.

I think it would be great if we had a "MyFirstAdsorptionEquilibrium" example in kmcos, and if it explained some of these issues. It would also be good to show what happens when somebody makes a mistake with the units.

My initial impression is that the pressure should not normally appear in the desorption rate constant even with the approach we discussed. However, in the example you gave, we see this:

         rate_constant='p_COgas*bar*A/2/sqrt(2*pi*umass*m_CO/beta)*exp(beta*(E_CO_cus-GibbsGas_COgas)*eV)')

Here rate_constant is the kmcos rate constant (not the physical chemistry one). Also, we see that the exponential has this term: E_CO_cus-GibbsGas_COgas .   I am not sure if the GibbsGas term may have the pressure in it (as noted in some other kmcos thread https://groups.google.com/g/kmcos-users/c/VH5aqxkoqnY/m/tdCMbmexAAAJ )

So I would suggest to first try to understand simple adsorption desorption case with single site available for adsorption (without using GibbsGas feature). Then, if you want to, you can make a second version of your example where you add in the GibbsGas feature to see if pressure needs to be included in k_des_kmcos or if it is a mistake to include pressure there.

[Nam Tran]

Hi Michele,

I think some of your question has been answered by admin in my thread.

https://groups.google.com/g/kmcos-users/c/VH5aqxkoqnY

Cheers

Nam