I am studying the effect of diffusion on liquid chemical reactions. The model is very simple in that it assumes that striations or slabs of fixed thickness of one reactant are next to the other. Typical widths (L) of the striations or slabs are on the order of microns. However, the larger the value of L the greater the effect of diffusion on the overall reaction. For the reaction schemes presented below, this means that more S is produced assuming that A is originally present in a slight stoichiometric excess to B.
Originally, the chemical reaction being studied was as follows:
The rate constants at 25 C are $k_1$ = 12,000 and $k_2$ = 2.07 both in $m^3/(mol*s)$. If diffusion does not affect this reaction scheme, then the amount of S produced is very, very small (<0.1%). The first reaction is considered to be virtually instantaneous. Thus when comparing the "characteristic" reaction time to the "characteristic" diffusion time of the reactants, the following relationship was used:
$$M = \frac{k_2c_{bo}L^2}{D}$$
where $k_2$ is the reaction rate constant of the second, relatively slow reaction, $c_{bo}$ is the initial concentration of B, L is the striation or slab thickness and D is the diffusion coefficient. This has worked fine for analyzing the data.
Now, I want to update to the following reaction scheme:
where the rate constants are as follows: The rate constants at 25 C are $k_{1o}$ = 920, $k_{1p}$ = 12240, $k_{2o}$ = 1.84 and $k_{2p}$ = 22 all in $m^3/(mol*s)$.
My question is: can M be rewritten in terms of this new reaction scheme? Let's assume that the reaction of A with B by both mechanisms (ortho and para) is still fast enough and that the reaction of R with B is still the rate limiting step for the reaction (or is this now a bad assumption?).
It looks like the 2nd pathway of the reaction of A with B is an order of magnitude faster than the first (12,240 versus 920). The 2nd pathway then reacts with B with a rate constant $k_{2o}$ = 1.84. So is it reasonable to just replace $k_2$ in the expression for M above with $k_{2o}$ for the 4 step reaction scheme? Or do I need to use some kind of average of $k_{2o}$ and $k_{2p}$, which doesn't seem quite right. Or something else.......
Here is some extra information. Kinetic studies on the two reaction schemes under conditions where diffusion is negligible, but A and B are present in the same amounts initially, yield the following results (quantities are in mole/$m^3$):
Two reaction scheme: R = 0.0333; S = 0.0004452
Four reaction scheme: p-R = 0.03095; o-R = 0.003144; S = 0.000731.
So for the four reaction scheme, the amount of S produced is higher.
Again, I am trying to redefine M based on the new 4 step reaction scheme.
