Timeline for Is collision theory applicable only for molecules?
Current License: CC BY-SA 4.0
12 events
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| Apr 27, 2022 at 8:52 | vote | accept | CommunityBot | ||
| Apr 14, 2022 at 5:42 | comment | added | Poutnik | @TomHardy Energy outcome of producing ions depends on how far from each other the are If far enough, it is endothermic. If close enough, it is exothermic. The best if many opposite ions are close enough, forming solid ionic lattice. | |
| Apr 14, 2022 at 4:03 | comment | added | user119245 | @Poutnik "unless ions end bound close enough (Na+Cl−(g)) by coulombic force."- sir, I couldn't understand exactly what you mean by this. Do you mean that large amounts of energy are released (energetically favourable) when Na+ ion and Cl- get closer and get stuck together? | |
| Apr 13, 2022 at 14:26 | review | Close votes | |||
| Apr 22, 2022 at 3:04 | |||||
| Apr 13, 2022 at 14:03 | history | became hot network question | |||
| Apr 13, 2022 at 9:36 | comment | added | Poutnik | Yes it is like that. We would say such a reaction is endothermic, consuming thermal energy. Strictly speaking, I should use thermodynamically favourable via negative Gibbs free energy change, not enthalpy change. ( $\Delta G = \Delta H - T \Delta S$ ), but at these big $H$ changes the entropy term plays just minor role. When energy is released, it is usually dissipated elsewhere, so it cannot be reused to go back. OTOH, if energy is needed, you have to gain it somewhere. | |
| Apr 13, 2022 at 8:55 | comment | added | user119245 | @Poutnik On a related note: how do we determine if a reaction is energetically favorable or not? Do we add the ionization energy of Na (496kJ/mol) and the electron affinity of Cl (-349kJ/mol), and if the answer is positive(147kJ/mol) then we say that the reaction isn't favourable? | |
| Apr 13, 2022 at 8:38 | answer | added | porphyrin | timeline score: 6 | |
| Apr 13, 2022 at 7:55 | comment | added | Poutnik | Note that $\ce{Na(g) + Cl(g) -> Na+(g) + Cl-(g)}$ is energetically unfavourable, unless ions end bound close enough ($\ce{Na+Cl-(g)}$) by coulombic force. Saying that, not speaking about energy needed for formal $\ce{Na(s)}$ atomization and $\ce{Cl2}$ dissociation. | |
| Apr 13, 2022 at 6:48 | history | edited | user119245 | CC BY-SA 4.0 |
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| Apr 13, 2022 at 6:43 | comment | added | Buck Thorn♦ | You may want to explain why the two alternatives you present might or might not agree with collision theory (in particular the second). Collision theory combines a description of the rate at which reactants collide with a model of the fraction of those reactants that have enough kinetic energy to result in a productive collision. The details will change but, provided the assumption that a collision between reactants exceeding the required energy is appropriate, the theory should be able to describe reactions between gases and solids. | |
| Apr 13, 2022 at 6:03 | history | asked | user119245 | CC BY-SA 4.0 |