Open System – Exchange of CO2 and O2

The Thermodynamic System and Its Environment

The aqueous solution (together with its mineral assembly) represents our ‘thermodynamic system’. Three types of coupling to the environment (e.g., the atmosphere) are distinguished:

  Exchange of Energy Exchange of Matter
Isolated system no no
Closed system yes no
Open system yes yes

In hydrochemistry, equilibrium calculations are usually perfomed for the closed system. Additionally, aqion affords open-system calculations for the exchange of CO2 and/or O2:

•  CO2 exchange by presetting CO2 partial pressure (or pCO2)
•  O2 exchange by presetting the redox potential (or pe value)

In both cases just as much CO2 or O2 is added to or removed from the water until the presetted pCO2 or pe value is attained. The supply of CO2 and O2 in the environment is unlimited.

What is the aim of this procedure?

Using the ‘open CO2 system’ we are able to attain chemical equilibrium of the aqueous solution with the CO2 in the atmosphere. On the other hand, by exchanging O2 the initial water can be oxidized (pe ≥ 6) or reduced (pe ≤ 0), i.e. we can change the redox equilibrium of the aqueous solution (and trigger the precipitation of specific minerals).

Settings. The value of pCO2 and/or pe is set either in the input window or – in case of reactions (button Reac) – in the corresponding setup panel (button setup). These are the two options of the two calculation pathway.

The Distinction of the Open-System Calculations

Open-system calculations cannot be mimicked or completely replaced by the reaction module which simply adds chemicals to the water. This is because the exchange of CO2 and/or O2 implies both addition and removal of chemicals:

aqion_offenes_vs_geschlossenes_System

Two examples should demonstrate the idea behind the open-system calculations.

Example 1: Equilibrium with Atmospheric CO2

Given a 10 mM CaCl2 solution we perform the calculations in two different ways: the first based on pCO2, the second based on CO2 addition.

Way 1. To generate the 10 mM CaCl2 solution: button New, button Reac, then enter 10 mmol/L CaCl2. To bring the water into CO2(gas) equilibrium click on Setup and activate the checkbox “Open CO2 System” (the default value pCO2=3.408 represents the partial pressure of CO2 in the atmosphere). Then button Start.

The calculated solution has pH = 5.59 and DIC = 0.0161 mM.1

(By click on button next you get the output table, and by a second click on next you get all data that characterizes the carbonate system – among them the calculated pCO2 value. Evidently, in this case pCO2 is exactly 3.408.)

Way 2. The calculation above tells us that 0.0161 mM DIC is required to attain the equilibrium state. This knowledge allows us to generate the equilibrium solution by adding two reactants to pure water:

CaCl2 10 mmol/L
CO2 0.0161 mmol/L

If you perform this simple reaction calculation using Reac (but now without pCO2) the same water composition is obtained: pH = 5.59 and DIC = 0.0161 mM.1

Note that we get the correct result because we know the dosage of CO2 beforehand. Otherwise you should trial and error until the desired pCO2 value (displayed in the carbonate-system output panel) is attained. Thus, the second way is rather impractical.

Example 2: Redox Equilibrium at pe=10 (Oxidation)

A solution of 10 mM FeCl2 has pH 5.88 and pe -1.84, which manifests a water in a strongly reducing state. The aim is to enhance the pe value to 10 (oxidation). Again as in example 1 we solve this problem in two ways.

Way 1. Button New, button Reac, then enter 10 mmol/L FeCl2. To set the pe value at 10 click on Setup and activate the checkbox “Open Redox System”. Run the calculation with button Start.

The obtained oxidized water has pH = 3.67 and pe = 10.2

In the subsequent output table you find in the raw “O2 exch” the amount of O2 added to the water: 0.145 mM. (This information will be used in the second calculation below.)

Way 2. We repeat the above calculation by adding two reactants to pure water:

FeCl2 10 mmol/L
O2 0.145 mmol/L

Performing this reaction calculation using Reac (but now without setting the pe value) the same oxidized water is obtained: pH = 3.67 and pe = 10.2

As mentioned in the first example above this approach is impractical because it requires the knowledge of the O2-dosage beforehand.

Footnotes

  1. By click on Details you will get more information. 2

  2. This refers to Output 1 (whereas Output 2 is ignored to simplify the discussion). 2

[last modified: 2014-12-02]