Timeline for Dynamic equilibrium in unsaturated solutions
Current License: CC BY-SA 4.0
9 events
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| Oct 14, 2020 at 7:04 | comment | added | Buck Thorn♦ | @adosar My advice: follow the definitions provided in your study material. A chemical change alters chemical properties through a reaction. A chemical reaction is a process that breaks or makes new bonds. But what you call a bond is somewhat arbitrary. We usually mean covalent, metallic or ionic, a strong bond in other words, and not intermolecular interactions such as vdW or H-bonding (with partial covalent/ionic character). In that sense precipitation is a chemical reaction and mixing two liquids is not. But what about ice freezing? | |
| Oct 13, 2020 at 20:00 | comment | added | Antonios Sarikas | "It's better to think of this "reaction" as a phase transition". Strictly speaking is the formation of the solution a chemical reaction? I was searching to see if mixing compounds is considered as a physical or chemical change but there is no clear answer. | |
| Oct 13, 2020 at 19:40 | comment | added | Buck Thorn♦ | @adosar Yes, this should apply generally. It should be true when for cases of limited solubility or miscibility. | |
| Oct 13, 2020 at 17:19 | comment | added | Antonios Sarikas | Does the same happen when we have a binary mixture of two liquids? If we add more and more from the first component then there would be a point where we should expect two phases, one with both components and another with only the first component (like when the salt precipitates)? | |
| Sep 11, 2020 at 17:44 | comment | added | Buck Thorn♦ | @adosar Ksp as written conceals the third "reagent", the solid, which has constant chemical potential. It's better to think of this "reaction" as a phase transition, given that you form a solid with constant chemical potential (at constant p, T). The same would be true of water freezing or boiling at fixed pressure, say. | |
| Sep 10, 2020 at 13:51 | comment | added | Antonios Sarikas | So before the solution becomes saturated there is no solid at all? But this seems contrary to the fact that in any reaction as much as high is the equilibrium constant there is some amount of reactants. | |
| Nov 24, 2019 at 21:46 | history | edited | Buck Thorn♦ | CC BY-SA 4.0 |
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| Nov 24, 2019 at 20:02 | history | edited | Buck Thorn♦ | CC BY-SA 4.0 |
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| Nov 24, 2019 at 18:52 | history | answered | Buck Thorn♦ | CC BY-SA 4.0 |