Timeline for Why only structures corresponding to valleys on PES are taken into account?
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| Jun 11, 2020 at 10:20 | history | edited | CommunityBot |
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| Sep 30, 2017 at 19:12 | answer | added | jheindel | timeline score: 1 | |
| Sep 29, 2017 at 2:42 | comment | added | Martin - マーチン♦ | When you are using the Gibbs energy, then the vibrational modes within the minimum are already considered, and the nuclei arrangement at angles slightly off the minimum correspond to vibrational distortions. | |
| Sep 28, 2017 at 11:03 | answer | added | user213305 | timeline score: 1 | |
| Sep 27, 2017 at 16:06 | comment | added | Tyberius♦ | @HernanMiraola by analogy, think of a two state system: you are in an airplane flying at a constant height or you are on a level stretch on the ground. You can pass between the two states and briefly be at a point that doesn't coincide with those two states. But you will rapidly collapse to one or the other. If you jump out of the plane, there is zero probability that you hang in the air somewhere in between the ground and the plane. The same is true if you jump from the ground. The space in between the states is not occupiable in this sense. | |
| Sep 26, 2017 at 23:28 | history | tweeted | twitter.com/StackChemistry/status/912821305543741440 | ||
| Sep 26, 2017 at 22:35 | comment | added | user43021 | @Tyberius it is an occupiable state of the system, why not? It is a position of nucleus.. | |
| Sep 26, 2017 at 22:17 | comment | added | Tyberius♦ | I think this is a good question because I would like to see a rigorous explanation of this. However, my intuition from the previous iteration of this question remains the same that the issue in how we define a state. I think the argument at the end of your post is invalid because it implicitly assumes that $150^{\circ}$ is an occupiable state of the system. The only way it would be at that configuration is a brief fluctuation of around the bottom of the $180^{\circ}$ well or en route to the $60^{\circ}$ well. | |
| Sep 26, 2017 at 21:25 | history | edited | user43021 | CC BY-SA 3.0 |
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| Sep 26, 2017 at 19:20 | history | edited | user43021 | CC BY-SA 3.0 |
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| Sep 26, 2017 at 16:46 | history | edited | user43021 | CC BY-SA 3.0 |
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| Sep 26, 2017 at 16:30 | history | edited | user43021 | CC BY-SA 3.0 |
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| Sep 26, 2017 at 16:25 | history | edited | user43021 | CC BY-SA 3.0 |
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| Sep 26, 2017 at 16:24 | comment | added | user43021 | No, I mean, why estimate mean gibbs free energy as the sum of gibbs free energy of only most stable conformers?@a-cyclohexane-molecule. Thanks for the comments.. | |
| Sep 26, 2017 at 16:22 | comment | added | a-cyclohexane-molecule | Are you only considering global minima, then? In your question statement, you mention local minima, and I'm not sure to which you refer in the last part of your post. | |
| Sep 26, 2017 at 16:01 | history | edited | user43021 | CC BY-SA 3.0 |
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| Sep 26, 2017 at 15:54 | history | edited | user43021 | CC BY-SA 3.0 |
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| Sep 26, 2017 at 15:48 | history | edited | user43021 | CC BY-SA 3.0 |
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| Sep 26, 2017 at 15:43 | comment | added | user43021 | No, the last part of the post tries to explain that point.@a-cyclohexane-molecule | |
| Sep 26, 2017 at 15:42 | history | edited | user43021 | CC BY-SA 3.0 |
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| Sep 26, 2017 at 15:41 | comment | added | a-cyclohexane-molecule | It could be that the unstable conformers are so low a fraction of the species as to be negligible. | |
| Sep 26, 2017 at 15:39 | history | edited | user43021 | CC BY-SA 3.0 |
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| Sep 26, 2017 at 15:33 | history | edited | user43021 | CC BY-SA 3.0 |
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| Sep 26, 2017 at 15:29 | answer | added | obackhouse | timeline score: 0 | |
| Sep 26, 2017 at 15:29 | history | edited | user43021 | CC BY-SA 3.0 |
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| Sep 26, 2017 at 14:58 | history | asked | user43021 | CC BY-SA 3.0 |