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