Equilibrium Constants in Thermo

[skounave]

Many of the equilibrium constants in the thermo database at 25C are very different from literature values. Specifically, the Log K for calcite in thermo.dat is listed as 1.7130, but in all the literature I looked at the log K for calcite is given to be around -8.3

What am I missing?

[Tom Meuzelaar]

Many of the equilibrium constants in the thermo database at 25C are very different from literature values. Specifically, the Log K for calcite in thermo.dat is listed as 1.7130, but in all the literature I looked at the log K for calcite is given to be around -8.3

What am I missing?

Hello:

There is not a log K value for any particular mineral, only an equilibrium constant for a mass action equation that involves a mineral. In this case, the difference stems from how the dissolution reaction for Calcite is written. The log K data you see in the literature reflects the following reaction:

CaCO₃ = CO₃^2- + Ca²⁺

Whereas in thermo.dat, the dissolution for Calcite is written as:

CaCO₃ + H⁺ = HCO₃^- + Ca²⁺

However, if you add the thermo.dat reaction for CO₃^2- (log K = -10.3439) to the previous reaction:

HCO₃^- = CO₃^2- + Ca²⁺

..you'll find that you get both the reaction and equilibrium constant commonly reported in the literature.

I hope that helps,

Tom Meuzelaar

RockWare, Inc.

[skounave]

Thanks, that helps a lot.

Is there someplace that will tell me how the

dissolution for calcite (or others) is written?

[Tom Meuzelaar]

Thanks, that helps a lot.

Is there someplace that will tell me how the

dissolution for calcite (or others) is written?

Sure- in the thermo database, just above the equilibrium constant data, you'll see all of the chemical components involved in the reaction, and their stoichiometric coefficients.

The below image shows what this looks like for Calcite:

Note that a negative symbol in front of the stoichiometric coefficient means the component is on the reactant side (with Calcite), whereas a positive symbol refers to a component on the product side of the reaction.

I hope that helps,

Tom