Is enthalpy of activation the same as activation energy? Is there a purely mathematical way to find that value?

Question

I asked about the effect of pressure on a reaction, and since then I learned a lot more. One of the things I learned was the Eyring equation:

$$\kappa = \frac{\kappa_B T}{h} e^\frac{\Delta^\ddagger S°}{R} e^\frac{- \Delta^\ddagger H°}{R}$$

where:

$\begin{multline} \begin{split} \kappa &= \text{reaction rate} \\ \kappa_B &= \text{Boltzmann constant} \\ T &= \text{temperature} \\ h &= \text{Planck's constant} \\ R &= \text{gas constant} \\ \Delta^\ddagger S° &= \text{entropy of activation} \\ \Delta^\ddagger H° &= \text{enthalpy of activation} \\ \end{split} \end{multline}$

My questions:

1. Is the enthalpy of activation the same thing as the activation energy?

2. Is there a purely mathematical way to determine both $\Delta^\ddagger S°$ and $\Delta^\ddagger H°$? (without having a laboratory to use to observe the reaction)

Answer 1

(1) Enthalpy of activation is a component of activation energy. The other component is temperature * entropy of activation. Enthropy controls "heat difference" while entropy controls "measure of disorder difference".

$\Delta$G = $\Delta$H - T*$\Delta$S. This applies to activation energy too.

(2) DFT functionals rely on 4 parameters per atom make predictions. So, if you give 4*5=20 parameters, then you can in theory predict behaviour of any compound, build with only C, H, N, O, S atoms. This covers most of biochemistry, excluding nucleic acids (you would need additional 4 parameters to characterize P).

These 4 parameters are known for most of the atoms (excluding some elements, going after uranium). But those four parameters were experimentally measured by fitting behavior of modeled molecules to real molecules.

Should we count it as "purely theoretically predict?"

Also there are purely ab initio methods like Hartree-Fock with even less input required. You would still need some input. In math you can build many different plausible worlds. For example purely "newtonian" vs "newtonian + quantum mechanics" words. experiment is the only way to tell, which model is correct.