What is the "rate factor" or "second-step rate constant" in the reaction rate equation?

Question

The reaction rate equation can be written as:

$$r=r_0 K \prod_n a_n$$

where the $a_n$s are the activities of each reactant in the reaction equation (in the case where we find the reaction rate from reactants to products, of course. Else the other way around). These activities are multiplied together. (This is often written as concentrations instead, I have seen, which seems to be an idealization.)

$K$ is the equillibrium constant, $K=\exp\left(\frac{-\Delta G}{RT}\right)$.

I am curious as to what exactly the $r_0$ is? In the wikipedia article it is denoted $k_2$ and called the "rate constant for the second step." The second step here refers to the intermediate step at the actived-complex.

So far so good, but what is this quantity at this intermediate step? Is it a reaction rate from the activated-complex to the final product (in that case, how can this be different from the reaction rate from the reactants to this state)?

Answer 1

The implication is that the reaction mechanism for any particular reaction, if written like this:

$$A + B → C$$

might not proceed as one step. Having a $k_2$ indicates that there is a second step to the mechanism, and potentially even several more. $k_{obs}$ is probably a better name for it, because sometimes you see it written as a simplified "one-step" process, such as in Michaelis-Menten Kinetics, when really, it might be $1$ more step, or it might even be $20$, depending on the reaction (I'm exaggerating a bit, but you get the point).

If we take Michaelis-Menten enzyme kinetics as an example:

$$E+S⇌ES→E+P$$

The first step is an equilibrium in which the concentration of the intermediate is assumed to not change because the intermediate is being produced about as fast as it is being consumed. This would be the Steady State Approximation. In other words:

$\frac{d[ES]}{dt} = 0 = k_1[E][S] - k_{-1}[ES] - k_2[ES]$

Eventually...

$[ES] = \frac{k_1}{k_{-1} + k_2}[E][S]$

The second step is where the $ES$ complex does something in such a way that the enzyme unbinds somehow, and the product is made... somehow. Depending on the reactants and products, the mechanism is different.

Anyways, the point is that $k_2$ describes this second step in this type of reaction (which is what you were wondering---why the "second step"?), hence the $2$.

In any case, $k_2$ corresponds to the reaction step(s) that proceed after an intermediate product is formed on the first step for a reaction mechanism.