A first-order reaction at infinite concentration

I have a first-order reaction, for the binding of two molecules, of the form: $\text{k}_\text{f} = \ce{k_1 (M*s)^{-1}}$. What happens at infinite concentration? How is this upperbound typically handled for approximations in chemistry? Do we need to measure additional kinetics parameters in this regime?

If anyone has any specific knowledge, I'm particularly concerned with DNA hybridization, where we have $\text{k}_1 \approx 10^6$. Will we simply converge on the elementary step time for the formation of a base $\text{k}_+ \approx 10^6$ to $10^7$ (Craig et al., 1971; Porschke & Eigen, 1971)? It seems reasonable to me that as $\text{C} \to \infty$ we'd have $\text{k}_\text{f} \to \text{k}_+$, which, I'd guess is the point where the rate limiting step is the time for the molecule to sample different rotamer/etc. configurations. Perhaps it isn't an accident that these orders of magnitude match up?

Also, should we think about the entropy of complex formation in the same way in the limit of infinite concentration? My guess for this would be "yes" since the entropic penalty of holding two molecules together, derived from studies on monoatomic gases, should be concentration independent as long as diffusion/mixing is allowed. In practice, however, we'll probably have a gel at these higher concentrations.

• I would love to try to answer this, but I don't know what you mean by "infinite concentration". Clearly there can be no such thing so you must mean something else. Can you edit your question to show what you mean? – Paul J. Gans Jan 4 '13 at 1:08
• I'd have to agree with Dr. Gans, infinite concentration is an odd concept to consider. – Lighthart Feb 25 '13 at 4:46
• ""I'd have to agree with Dr. Gans, infinite concentration is an odd concept to consider. – Lighthart"" Maybe in a black hole? :=) – Georg Apr 16 '13 at 10:47