Why does the rate of a chemical reaction depend on partition functions according to the ratio $Q^\ddagger/Q$? I'm trying to understand isotope effects on the rates of chemical reactions, which, surprisingly, can be very large in some cases. In treatments of this problem, I see a lot of expressions that involve a proportionality of the rate to the following factor:
$$\frac{Q^\ddagger}{Q}$$
Here the $Q$'s are basically partition functions (scaled in a certain way), with $Q$ being for the reactants and $Q^\ddagger$ for the transition state, i.e., the saddle point in the potential energy surface.
I'm a physicist, and I'm finding that the treatments of these topics by chemists discuss a lot of things that I don't care that much about, and don't seem to discuss the physics as much, including the physics behind the occurrence of this factor in the rates. Can anyone give me a physicist's explanation of why this occurs?
I would think that these factors were basically counting the number of classically allowed states available for the process, since a state with a high occupation probability contributes roughly 1 to the partition function. It would then sort of seem to make sense that you'd get a proportionality to $Q^\ddagger$, since each state is available as a pathway. But why should there be an inverse proportionality to $Q$? I don't see why it should matter if there is 1 possible initial state for the reactant or 100. Either way, every reactant molecule is in one of these states and is available to react.
 A: As Sunyam has correctly pointed this out, the factor comes from the transition state theory (TST). It is not easy to find the rigorous derivation of this theory, because there is a lot of handwaving around the field, a book that I can warhmeartedly suggest is Chemical Kinetics and Dynamics by Steinfeld et al. I will try my best to outline some of the main arguments used in the derivation, but I am not sure this comment can do justice here.
So the rigorous transition state theory is based on a number of approximations:


*

*The system can be well described by classical mechanics (I know, it might sound strange, but classical description of nuclear motion works extremely well in chemistry, proper description of nuclear quantum effects is a focus of large current scientific interest.

*The system behaves according to the laws of - classical - statistical mechanics

*There exists a dividing surface in the phase space that can only be crossed from one side, and this crossing corresponds to the chemical reaction in question. 


The derivation usually also assumes that the ensemble describing the system is microcanonical. This is not necessary, but helps a lot with the formulas. 
Based on the approximation (3) above, then we can express the reaction rate as the flux over the transition state. Moreover, since the trajectory of the system from the transition state can only go towards the reactants, then the rate is:
$$ r = \frac{\mathrm{d} N_{TS}}{\mathrm{d}t} = \frac{\mathrm{d} v}{\mathrm{d}t} N $$
In the formula I used the notation $v$ to describe the ratio of the molecules found in the vicinity of the transition state $N_{TS}$ to the total reactant molecules $N$ in the system. Let's assume that the motion through the dividing surface is separable from all the other motions. Then, the kinetic energy of this motion (call this direction $s$) is generally given by:
$$ T_s = \frac{p_s^2}{2 \mu_s} $$
We also have a potential energy $V_0$ that corresponds to the potential energy at the transition state. So the energy that may be distributed along the other modes is $E-V_0-T_s$. 
What is the ratio of the phase space volume of the transition state compared to the whole phase space? If we have $N$ atoms in the system, we generally have $3N$ dimensions in the configuration space, and $6N$ in the phase space. The surface fixes one spatial coordinate, so it is $3N-1$ dimensional in configuration space, and the momentum is also defined, so overall, it is $6N-2$ dimensional. The ratio is simply:
$$v(s) = \frac{\mathrm{d} s \mathrm{d}p_s /h \int_{E_0-T_s-V_0} \mathrm{d}q_2 \mathrm{d} p_2 \mathrm{d}q_3 \mathrm{d}p_3 ... \mathrm{d}q_{3N} \mathrm{d}p_{3N}/h^{3N-1}}{\int_{E} \mathrm{d}q_1 \mathrm{d} p_1 \mathrm{d}q_2 \mathrm{d} p_2 \mathrm{d}q_3 \mathrm{d}p_3 ... \mathrm{d}q_{3N} \mathrm{d}p_{3N}/h^{3N}} $$
But since $\mathrm{d}s = p_s / \mu_s \mathrm{d} t$ we can insert this expression:
$$v(s) = \frac{p_s \mathrm{d}p_s \mathrm{d}t}{ \mu_s  h} \frac{\int_{E-T_s-V_0} \mathrm{d}q_2 \mathrm{d} p_2 \mathrm{d}q_3 \mathrm{d}p_3 ... \mathrm{d}q_{3N} \mathrm{d}p_{3N}/h^{3N-1}}{\int_{E} \mathrm{d}q_1 \mathrm{d} p_1 \mathrm{d}q_2 \mathrm{d} p_2 \mathrm{d}q_3 \mathrm{d}p_3 ... \mathrm{d}q_{3N} \mathrm{d}p_{3N}/h^{3N}} $$
Now the denominator is the state density corresponding to $E$, that is, the classical degeneracy of this energy level. Let's call this $\rho(E)$. The numerator also contains a 'reduced' density of states that only corresponds to the $6N-2$ dimensional phase space region around the transition state. Call this $\rho'(E-T_s-V_0)$. Moving the $\mathrm{d}t$ to the other side we immediately see the rate of the reaction:
$$r = \frac1h \frac{p_s \mathrm{d} p_s}{\mu_s} \frac{\rho'(E-T_s-V_0)}{\rho(E)} N $$
By using the chain rule it's not hard to see that the second term can be written as $\mathrm{d}T_s$. So we have:
$$r = \frac1h \frac{\rho'(E-T_s-V_0)}{\rho(E)} \mathrm{d}T_s N $$
But this rate only corresponds to a fixed $T_s$ value that we've arbitrarily set before. To get the total rate, we need to integrate over all the allowed values:
$$r = \frac1h \frac{N}{\rho(E)} \int_0^{E-V_0} \rho'(E-T_s-V_0) \mathrm{d}T_s  $$
It is common to call the integral the sum of states $N^{TS}(E-V_0)$, which gives you the total number of states that you can have with a given total energy. 
Now we got a nice simple expression for the rate coefficient in terms of the sum of states of the TS and the density of states of the reactants:
$$k(E) =\frac{N^{TS}(E-V_0)}{h \rho(E)}  $$
This expression is the expression for the rate coefficient for a given $E$ energy. To get the temperature-dependent expression, we need to use the Boltzmann-averages for the reactants (also containing the exact same density of states as the expression above) and with the  freedom of setting $N^{TS}(E-V_0)$ = $N^{TS}(E)$ because the number of states below $V_0$ is 0 anyways):
$$ k(T) = \frac{1}{h Q_r} \int_0^{\infty} N^{TS}(E) \exp{(-E/kT)}\mathrm{d}E $$
Now we use the definition from above for the number of states, and change the order of integration once:
$$ k(T) = \frac{1}{h Q_r} \int_0^{\infty} N^{TS}(E) \exp{(-E/k_BT)}\mathrm{d}E  =  \frac{1}{h Q_r} \int_0^{\infty}\exp{(-E/k_BT)} \mathrm{d}E \int_0^{E}\rho(E') \mathrm{d}E' = \frac{1}{h Q_r} \int_0^{\infty} \rho(E') \mathrm{d}E' \int_{E'}^{\infty} \exp{(-E/k_BT)}  \mathrm{d}E $$
Now the second integral is simply ${k_B T}, and the first is, by definition, the partition function of the transition state. Now all we need to do is to shift the energy to a common level and we get the usual expression:
$$ k(T) = \frac{k_b T}{h}\frac{Q_{TS}}{Q_r}\exp{(-\Delta V/k_BT)} $$
I apologize if this post was unnecesarily long, believe me I still feel bad about missing a lot of details, but I had to lose some rigor to keep is somewhat closed. I felt that in order to get a good understanding of the factor, you need to understand where do the sum of states and the density of states come into the play, and the integral change after Boltzmann averaging is also something not so obvious. But if you want to have a concise answer, here you go:
tl;dr: $Q_{TS}/Q_{r}$ approximately gives the ratio of the phase space volumes between the transition state and the reactants, and this ratio has to appear naturally in any rate expression.
