# How does cooling increase the selectivity of a reaction producing the kinetic vs the thermodynamic product?

I know that cooling gives a lower ratio of molecules with sufficient energy to overcome the activation barrier, but I'm not sure how this selects for a kinetic or thermodynamic product. I also know that the kinetic product is favoured if there is not enough energy supplied to reach equilibrium, and the opposite for the thermodynamic product. Can cooling select for either product in different circumstances?

First, there is no rule that "the" kinetic product will be different than the thermodynamic product. It depends on the particular reaction you are interested in.

Thus, I think you are really asking "in reactions where the kinetic product is different than the thermodynamic product, what does cooling select for?"

Let's consider two reactions, the first with a reactant R to a thermodynamic product T and a second, separate reaction where R goes to the kinetic product K. The rate constants for these reactions are $k_1$ and $k_2$, respectively.

$$\ce{R ->[k_1] T}$$ $$\ce{R ->[k_2] K}$$

Let's suppose both reactions follow classical transition state theory, which is a very good approximation to many everyday reactions and is a great starting place even for the weird cases that do vary.

TST says (if you assume that the $\kappa$ transmission coefficient is 1):

$k_1 = \frac{k_b T}{h} e^{-\Delta G^{\ddagger}_1/RT }$

$k_2 = \frac{k_b T}{h} e^{-\Delta G^{\ddagger}_2/RT }$

The "kinetic" product is the one with the higher rate constant; this is the definition of kinetic. Doing some algebra,

$$\ln{\frac{k_2}{k_1}} = -\frac{1}{RT}\left(\Delta G^{\ddagger}_2 - \Delta G^{\ddagger}_1 \right)$$

...we see that for $k_2$ to be higher than $k_1$, i.e. for $\ln{\frac{k_2}{k_1}}$ to be greater than 0, then $\Delta G^{\ddagger}_2$ must be smaller than $\Delta G^{\ddagger}_1$ (note the negative sign).

The effect of temperature is still in the equation:

$$\ln{\frac{k_2}{k_1}} = -\frac{1}{RT}\left(\Delta G^{\ddagger}_2 - \Delta G^{\ddagger}_1 \right)$$

The assumption of TST is that $\Delta G^{\ddagger}$ is not a function of temperature. As temperature is lowered, then the $RT$ divisor in the equation gets smaller, which means that the difference $\left(\Delta G^{\ddagger}_2 - \Delta G^{\ddagger}_1 \right)$ affects the ratio of rate constants more. Thus, under the assumptions we have made, cooling will favor the kinetic product more. As temperature is raised, eventually $RT$ becomes much greater than $\left(\Delta G^{\ddagger}_2 - \Delta G^{\ddagger}_1 \right)$, so selectivity for the kinetic product vanishes because $\ln{\frac{k_2}{k_1}} \rightarrow 0$, i.e. $\frac{k_2}{k_1} \rightarrow 1$.

You might ask, how can the ratio go to one; doesn't that mean that we have assumed the equilibrium constant between T and K is also 1? The answer is no, we haven't assumed that, but we have made a number of other assumptions that mean we we won't be able to predict the net rates of T and K formation. The most important one is that we have neglected the reverse reactions $\ce{T -> R}$ and $\ce{K -> R}$; introducing them would give the same qualitative effect of cooling the temperature but would complicate the math considerably (but would let us wind up with net rates of T and K instead of the gross single-directional rates we have calcualted here. Second we have assumed that $\kappa = 1$, which might not always be true.