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Author Topic: Define ion species in Kinetics keyword  (Read 1619 times)

ahmadreza_shojaee

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Define ion species in Kinetics keyword
« on: 10/05/23 15:17 »
Dear all,

Hi
I have a question about the Kinetics keyword.
In the manual it is mentioned that in the formula section of the Kinetics keyword, any charges (+/-) in the formula is ignored. My question is that how can we add ion species in this section? As an example, there is a rate control reaction in which, HSO4- is producing/consuming. How should we enter it in the Kinetics keyword?

Regards,
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dlparkhurst

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Re: Define ion species in Kinetics keyword
« Reply #1 on: 10/05/23 15:42 »
How is it that you can add HSO4- to a solution? Do you have a jar of it that you can pour into the beaker? No, you cannot. Tell me more of the reaction that you want to model.

Think of REACTION as adding or removing a physical entity from the system, or removing one "element" and adding another. The second part can be confusing, but if you have "Doc" (Dissolved organic carbon) as an "element" in your system, then you can model decomposition of organic matter by transferring Doc to CH2O, but no ions are involved in the definition:

Code: [Select]
REACTION
Doc -1 CH2O + 1
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ahmadreza_shojaee

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Re: Define ion species in Kinetics keyword
« Reply #2 on: 10/05/23 15:53 »
Assume we want to model the given reaction using RATE and KINETICS keyword:
Code: [Select]
HCO3- + 4H2 + H+ = CH4 + 3H2O

Code: [Select]
Kinetics
-formula HCO3- -1 H2 -4 H+ -1 CH4 1 H2O 3

I think maybe a possible way is as follows: (but I am not sure)
Code: [Select]
Kinetics
-formula C(4) -1 H(0) -4 H -2 CH4 1 H2O 3
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dlparkhurst

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Re: Define ion species in Kinetics keyword
« Reply #3 on: 10/05/23 16:17 »
PHREEQC will calculate the net change in element concentration from your formula.

Code: [Select]
-formula HCO3- -1 H2 -4 H+ -1 CH4 1 H2O 3

Consider H: you remove H -1 H2 -4 H - 1 and add H4 +1 H2 3 = -1 - 8 - 1 + 4 + 6 = 0

There will be no change in H concentration; C and O also sum to 0. PHREEQC will calculate no reaction. Don't argue, that is simply what the program will do.

If you want to reduce C(4), simply add H2 kinetically.

-formula H2 1

Alternatively, with phreeqc.dat, you could use

-formula CH4 -1 Mtg +1

I don't like that approach because it makes methane whether thermodynamically reasonable or not. The first formulation with H2, lets the system equilibrate among all H, C, O, and other redox elements; for example, it would reduce NO3- before you started to make methane.



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ahmadreza_shojaee

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Re: Define ion species in Kinetics keyword
« Reply #4 on: 10/05/23 16:42 »
Thank you David for your response.

I am using the uncoupled redox approach you mentioned here:

https://phreeqcusers.org/index.php/topic,2023.0.html

Also, in the example 9, this approach is explained for iron oxidation.
The reason is that I want to consider more than one reaction (maybe three/four rate control reactions.).
If you take a look at our previous discussion (above link) about that three reactions in the aqueous phase I mentioned, you can find my desired reactions.

By the way, I think this approach of using RATE and KINETICS would be better, wouldn't be?

What about the below code:
Code: [Select]
-formula C(4) -1 H(0) -4 H(1) -2 C(-4) 1 O(-2) 3

Again, PHREEQC assume no reaction?
When we are talking about elements, do we are referring to elements in Solution_Master_species?
Code: [Select]
SOLUTION_MASTER_SPECIES
H H+ -1.0 H 1.008
O(-2)       H2O 0 0
C CO3-2 2.0 HCO3 12.0111
C(+4) CO3-2 2.0 HCO3
C(-4) CH4 0 CH4

[I am using the uncoupled redox approach by defining new elements database.(in my code C element is defined as Carbonate)]
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dlparkhurst

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Re: Define ion species in Kinetics keyword
« Reply #5 on: 10/05/23 17:11 »
Your reactions need to change the concentrations of elements, not element redox states. Using O and O(-2) will still affect the total O concentrations and will sum to zero.

Again, if you are using the completely uncoupled approach from the post, your reaction would be
Code: [Select]
-formula CH4 -1 Methane +1

You are assuming all the responsibility for making this reaction thermodynamically plausible. It will make methane in the presence of O2, which is not thermodynamically reasonable, unless you add lots of inhibitor factors to make the rate slow in the presence of O(0), N(5), N(3), and every other oxidized valence state. I don't like the approach because it gets so complicated to add the thermodynamic constraints, but its your choice.
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