Q:

Relationship between magnitudes of forward and reverse kinetic rate constants

Consider the reversible unimolecular reaction:
$$\ce{A <=>[k_1][k_2] B}$$
We know that the forward reaction is often considerably more thermodynamically favourable than the reverse reaction, and therefore the relationship $k_1 \gg k_2$ holds between the rate constants. The rate constants are in the same units, and so it is possible to write this relationship. A similar example would be a reaction that is bimolecular on both sides, $\ce{A + B <=> C + D}$, etc.
However, consider this reversible reaction:
$$\ce{A + B <=>[k_1][k_2] C}$$
Assume the reaction proceeds at rate $k_1 [\ce{A}][\ce{B}]$ in the forward direction and $k_2 [\ce{C}]$ in reverse. If the overall reaction rate has units $M s^{-1}$, then $k_1$ necessarily has units $M^{-1} s^{-1}$ and $k_2$ has units $s^{-1}$.
My question is: For this second case, we can no longer impose that $k_1 \gg k_2$, due to difference in units, but can we say anything about their relationship? Does setting $k_1$ to some value constrain the choice of $k_2$ in any way?

A:

A chemical reaction may be a very complicated thing, involving many different paths and elementary reactions. Basically it always leads to the formation of an equilibrium, where the lowest possible energy will be the most favoured. Under certain conditions more than one state will be populated and hence there will be enough potential energy to interconvert molecules in certain states. We usually refer to that as an dynamic equilibrium.
Changes under equilibrium conditions will most likely be elementary reactions (or an infinitely complicated chain of these.)
The population of the various states seems to be static at equilibrium, since the overall change of enthalpy is zero. In a dynamic picture that only means, that forward and backward reaction cancel each other. In that sense they are the same reaction in different directions. Therefore they can be described as one trajectory on the potential energy surface. (If there is a chemical reaction involving many different elementary reactions and the equilibrium still occurs, then the mathematical description will be much more complicated as all elementary reactions will be correlated. However, the main conclusions should stay the same.)
\begin{aligned}
\Delta G_r &= \Delta G_r^\circ + \mathcal{R}T\cdot \ln K = 0\\
\Delta G_r^\circ &= − \mathcal{R}T \cdot \ln K
\end{aligned}
The equilibrium constant may be derived from the mass-action law and should be defined as a product of activities (for the standard state). For $\ce{A + B <=> C}$ this will lead to
$$K^\circ= \frac{a(\ce{C})}{a(\ce{A})\cdot a(\ce{B})}$$
The activities are proportional to concentrations (in first approximation) and are unit less ($c^\circ$ being the standard concentration $1 \:\mathrm{mol/L}$)
$$a=\gamma\frac{c}{c^{\circ}}$$
For reasonable dilutions ($c\to0\:\mathrm{mol/L}$) one can assume that the activity coefficient becomes one $\gamma\approx1$ and therefore rewrite the equilibrium constant with concentrations.
$$K^\circ= \frac{c(\ce{C})/c^\circ}{c(\ce{A})\cdot c(\ce{B})/(c^\circ)^2}$$
At equilibrium the forward and backward reaction are coupled and therefore have to reflect the equilibrium constant.
$$K^\circ= \frac{k_f}{k_b} = \frac{c_{\text{eq}}(\ce{C})/c^\circ}{c_{\text{eq}}(\ce{A})\cdot c_{\text{eq}}(\ce{B})/(c^\circ)^2}$$
As one can see, the constants also need to have the same unit. As they are at equilibrium, they are the same reaction with different populated states. 
If you are moving away from the equilibrium, thing will start to get a little bit more messy. On reaction will be faster than the other and rates will not be comparable any more.

