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Chemistry

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  • $\begingroup$ how would you have got here from the simpler delta G of activation, which is more commonly seen than the difference in the zero point energies? $\endgroup$
    – sweetandtangy
    Commented May 18, 2020 at 12:18
  • $\begingroup$ The $\Delta U$ includes the difference in the minima of the potentials and also accounts for the difference in energy of the lowest vibrational energy level. The $\Delta G$ includes entropy changes, which can be incorporated into the pre-exponential term leaving just the $-\Delta H/RT$ you ask about in the exponential. The thermodynamic approach cannot account for any quantum effects such as zero point energy. $\endgroup$
    – porphyrin
    Commented May 18, 2020 at 12:39