A first-order reaction at infinite concentration

Question

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.

Answer 1

When you go to the limit of infinite concentration (obviously you can never get there), a reaction cannot stay first order, because you will always run into some other rate-limiting mechanism thus making the reaction zeroth order in the highly concentrated species.

To give an example, for the binding of 2 molecules if both molecular species are 'infinitely' concentrated, the rate will not be limited by molecules 'finding' each other (as is normally the case for first order reactions), but it will be limited by the finite time binding process itself takes.