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Is activation energy equivalent to Gibbs Free Energy of transition state as related by Eyring equation? $$E_a=\Delta ^\ddagger G \, \, ?$$

Is Gibbs Free Energy of transition state defined by the Gibbs Free Energy of formation? $$\Delta ^\ddagger G :=\Delta _\mathrm{f} G \, \, ?$$ If not, which are the conditions for which such an approximation (second equation) would apply?

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The Arrhenius equation

$$k=A\exp\left(-\frac{E_a}{RT}\right)$$ places all of the $T$-dependence in the exponential factor. The pre-exponential factor is not assumed to be temperature-dependent. By contrast, the Eyring equation

$$k=\frac{\kappa k_\mathrm{B}T}{h}\exp\left(-\frac{\Delta ^\ddagger G^⦵}{RT}\right)$$

which is justified on the basis of statistical-mechanical principles, displays a a temperature-dependence in the pre-exponential factor. So the exponential terms are not expected to be equal, at least theoretically, unless the pre-exponential factor in the Eyring equation becomes T-independent by some coincidental cancellation of terms.

As regards the second equation, I would define the free energy of the transition state more accurately as

$$\Delta ^\ddagger G :=\Delta _\mathrm{f} G(\mathrm{ts})-\Delta _\mathrm{f} G(\mathrm{ini}) $$

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    $\begingroup$ Tech detail: please note that using a circled minus symbol (⊖, \ominus) for the standard state isn't correct. IUPAC suggests either a superscript circle, or a plimsoll symbol. Since it looks like you intended to use plimsoll, you want to stick with a Unicode symbol ⦵ as getting it right with MathJax (or LaTeX, for that matter) is rather tricky. $\endgroup$
    – andselisk ♦
    Commented Jul 6, 2019 at 8:46
  • $\begingroup$ @andselisk That's not the standard state, that's the transition state, and I guess I wanted plimsoll but picked the nearest thing based on detexify, but technically I guess you're right, I didn't check the with IUPAC. I remember the post not so long ago about plimsoll, that was a good one. $\endgroup$
    – Buck Thorn ♦
    Commented Jul 6, 2019 at 17:03
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    $\begingroup$ Either way, a transition state must have a reference point, which is normally the standard state, so there is no contradiction. $Δ^\ddagger G^⦵$ is perfectly fine. Some books and even a Wikipedia article on transition state are using a circled minus — probably due to typographic difficulties in getting the plimsoll symbol right. Here we don't have this problem, hence my edit to comply with IUPAC rules. $\endgroup$
    – andselisk ♦
    Commented Jul 6, 2019 at 17:37

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