Geochemist's Workbench Support Forum

# How does the built-in rate law work?

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[admin notice: the below is from the former GWB users group email distribution list. This message was originally posted 11/3/2005]

Posted by: Megan Knight

Does anyone know how the built-in, zero-order rate law in GWB

version 6.0 prevents concentrations from going negative when the rate

constant is sufficiently high? I'm trying to write some scripts that

first mimic, then go beyond the default rate law, but I first need to

understand what's actually going on in the default. Thanks!

Posted by: Craig Bethke

Hi Megan,

The principle of detailed balancing holds that at equilibrium the forward and reverse rates of a reaction must be equal, so that the net reaction rate is zero. The principle appears as the familiar (1-Q/K) term (or, more precisely, a [1-(Q/K)^n]^m term) in the built-in rate law.

This term goes to zero as the reaction proceeds towards equilibrium, which of course is the point at which the activity product Q equals the equilibrium constant K. Even a zero-order rate law, then, requires that a reaction stop before any of its reactants are driven to negative mass.

For example, the equilibrium point of a reaction

A -> B

occurs at a positive (although perhaps very small) activity of A, no matter how unstable A is relative to B. So non-negativity is implicit in the form of the rate law, rather than explicit in the GWB coding.

To check this out, you might set a "far from equilibrium" zero-order law (i.e., one without the 1-Q/K business). Type the law into the "Custom rate law" field and run the model. The software will, if you set a sufficiently long time span, integrate the law past the point of equilibrium to the point where the mass of the limiting species goes negative, where it will get stuck.

As a side note, all this is very nice as long as the limiting reactant is an aqueous species. But how about a mineral species that dissolves completely before the fluid becomes saturated: a grain of salt in a glass of water? In the heterogeneous case, since 1-Q/K never vanishes, why doesn't mineral mass go negative? The answer is that as the mineral dissolves, its surface area approaches zero and hence so does reaction rate.

Again, the built-in law won't drive mass negative. And, again, you are free to set (incorrectly) a law that will -- for example, by omitting surface area term -- in which case the model will get stuck at the point of complete dissolution.

Hope this helps,

Craig

P.S. I think you'll find an example script for a kinetic redox reaction on page 51 of the GWB 6.0 Reaction Modeling Guide useful; the corresponding C function is on page 55. These should help you get started.

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