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Topic: Question about competition (Read 2190 times)
HFlower
Frequent Contributor
Posts: 14
Question about competition
«
on:
21/01/20 00:44 »
With the code below, if Ca adsorbs to the surface in an initial step, can Sr replace it in a subsequent step, even though no direct reaction is given for the exchange? I'm assuming such a reaction would be redundant, but I wanted to be sure. Thanks.
Hfo_sOH + Ca+2 = Hfo_sOHCa+2 -log_k 4.97
Hfo_sOH + Sr+2 = Hfo_sOHSr+2 -log_k 5.01
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dlparkhurst
Global Moderator
Posts: 4034
Re: Question about competition
«
Reply #1 on:
21/01/20 01:38 »
The first species on the right-hand-side of the equation is the one defined by the reaction. If you have two reactions defining the same species, only the last will be used. So, no, you do not need to define an exchange reaction. If you did (and put it last), it would replace the original definition.
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HFlower
Frequent Contributor
Posts: 14
Re: Question about competition
«
Reply #2 on:
21/01/20 12:01 »
Thank you for your clarification, Mr. Parkhurst.
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John Mahoney
Top Contributor
Posts: 73
Re: Question about competition
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Reply #3 on:
21/01/20 16:53 »
I think the question was not about the order of the reactions in a database, but rather Sr+2 replacing Ca+2 on the surface sites. I suggest you answer your own question and try doing it, as a learning exercise. Your question actually included its own answer, so you had the right idea. Load up a surface with Ca+2, SAVE the SURFACE and END the first simulation, then in the next step USE the saved SURFACE and contact it with a different solution that has a high (always relative) concentration of Sr. Some of the Ca should leave the surface and go into solution and the Sr-surface complex should be present on the HFO.
You could also do it in a single step. For example, use the REACTION keyword and add say SrCl2 in various amount. look at the distribution of Ca and Sr surface complexes.
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