Timeline for Where does the RT term come from in the derivation for the activation enthalpy from the Eyring equation?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 29, 2020 at 3:38 | vote | accept | sweetandtangy | ||
| May 19, 2020 at 4:37 | vote | accept | sweetandtangy | ||
| May 29, 2020 at 3:38 | |||||
| May 18, 2020 at 12:39 | comment | added | porphyrin | 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. | |
| May 18, 2020 at 12:18 | comment | added | sweetandtangy | 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? | |
| May 18, 2020 at 9:02 | history | answered | porphyrin | CC BY-SA 4.0 |