Timeline for For homogeneous equilibrium, why are liquids and solids included in the equilibrium constant (when they aren't in heterogeneous equilibria)?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 10, 2016 at 22:16 | vote | accept | K-Feldspar | ||
| Sep 10, 2016 at 6:32 | answer | added | orthocresol | timeline score: 5 | |
| Sep 10, 2016 at 5:16 | comment | added | DHMO | Because liquids are fluid whose concentration still affects the rate of effective collisions. | |
| Sep 10, 2016 at 5:14 | comment | added | K-Feldspar | Thanks for pointing that out. Is there any reason why liquids would be different? | |
| Sep 10, 2016 at 5:12 | comment | added | DHMO | According to the website you cited, "you don't include any term for a solid in the [heterogeneous] equilibrium expression". | |
| Sep 10, 2016 at 5:08 | comment | added | K-Feldspar | Thanks for the response. So in the case where it is say all liquids (or all solids) the rate of reaction would be effected by the liquids and solids since there is nothing else right? Does that mean if we have a reaction which includes both liquids and solids (but not gases or aqueous solutions) then we would include both the liquids and solids in the equilibrium constant, even though it is a heterogeneous equilibrium equation? | |
| Sep 10, 2016 at 5:05 | history | edited | K-Feldspar | CC BY-SA 3.0 |
added 11 characters in body
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| Sep 10, 2016 at 5:04 | comment | added | DHMO | Because experimentally solids and liquids in the first case hardly affect the rate of the reaction. | |
| Sep 10, 2016 at 5:03 | history | asked | K-Feldspar | CC BY-SA 3.0 |