- #1
Entanglement
- 438
- 13
Why are precipitate and gas evolution reactions irreversible ( why don't the products react once again ) ??
Reactions: Irreversible Precipitate & Gas Evolution
- Thread starter Entanglement
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- Irreversible Reactions
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Discussion Overview
The discussion centers on the irreversibility of precipitate and gas evolution reactions, exploring why the products do not react again under typical conditions. Participants examine the concepts of solubility, thermodynamics, and equilibrium in relation to these reactions.
Discussion Character
- Exploratory
- Technical explanation
- Debate/contested
Main Points Raised
- Some participants propose that gas evolution reactions are irreversible because the gas has low solubility and escapes into the atmosphere.
- Others argue that precipitate reactions are more complex, involving interactions at the solid-liquid interface, where only the exposed layer of solid interacts with the solvent or solutes.
- A participant suggests that in a closed vessel, thermodynamic principles would apply, potentially leading to an equilibrium state with a mixture of products, reactants, and gases.
- Another participant presents a hypothetical scenario to illustrate how low solubility can affect the equilibrium of a reaction, emphasizing that the presence of a precipitate can limit the concentration of products in solution.
- One participant notes that while there may be a back reaction, it could be negligible due to the low solubility of the precipitate.
- Another participant highlights that in a closed container, these reactions can be viewed as equilibrium processes, using the example of calcium carbonate precipitation and dissolution as a reversible reaction in nature.
- A request for a simpler explanation indicates that some participants may find the technical details challenging to understand.
Areas of Agreement / Disagreement
Participants express differing views on the irreversibility of these reactions, with some asserting that they are irreversible under typical conditions, while others suggest that they can reach equilibrium in closed systems. The discussion remains unresolved regarding the extent of reversibility and the conditions that affect it.
Contextual Notes
Participants reference concepts such as chemical potential, solubility, and thermodynamics, but there are no settled definitions or consensus on how these concepts apply to the irreversibility of reactions discussed.
- #2
Yanick
- 402
- 22
Because they are no longer available for reacting. Assuming the gas has very little solubility in the solution, once it forms it will bubble out and diffuse into the atmosphere.
The precipitate is a bit trickier but not really. When a solution is in contact with a precipitate the interaction between solvent (or whatever solutes may be present) and the solid occurs at the interface of the phases. That is only the exposed layer of solid will ever get to see the solvent or solutes. You can demonstrate this to yourself by dissolving equal masses of sugar in water but using sugar cubes versus confectioner's sugar (or regular powdered sugar, or all three). You'll quickly notice the different behaviors and should be able to relate this back to your situation.
You can also try and read up about an abstract concept called chemical potential, which can be used to, more rigorously, describe such behavior as my answer is just quick, dirty and intuitive.
The precipitate is a bit trickier but not really. When a solution is in contact with a precipitate the interaction between solvent (or whatever solutes may be present) and the solid occurs at the interface of the phases. That is only the exposed layer of solid will ever get to see the solvent or solutes. You can demonstrate this to yourself by dissolving equal masses of sugar in water but using sugar cubes versus confectioner's sugar (or regular powdered sugar, or all three). You'll quickly notice the different behaviors and should be able to relate this back to your situation.
You can also try and read up about an abstract concept called chemical potential, which can be used to, more rigorously, describe such behavior as my answer is just quick, dirty and intuitive.
- #3
Entanglement
- 438
- 13
What if the products were to be produced in a closed vessel ?Yanick said:
- #4
Yanick
- 402
- 22
Then you have to use thermodynamics make some reasonable predictions about the system. If the free energy of the process favors some sort of equilibrium you will have a mixture of products, reactants and gases.
- #5
Entanglement
- 438
- 13
Yanick said:
That means that there is a back reaction but it's negligible ??
- #6
Yanick
- 402
- 22
Not necessarily, it only means that the product is not soluble and is therefore not in the actual solution.
Consider it this way, a completely made up case:
A(aq) + B(aq) ⇔ C(aq) : Keq ~1
C(s) ⇔ C(aq) : Ksp ~ 10-10
The low solubility of C(aq) will force the reaction to proceed very much further than the equilibrium constant may lead you to believe. If we ignore the solubility of C, we would expect [C](aq)/([A](aq)(aq)) ~ 1. However because the solubility is so low the solution never actually gets enough [C](aq) to get to ~1 because as far as the solution is concerned there is very little C(aq) around (it precipitates out and becomes C(s)). Only a thin layer at the solid/solution interface is accessible to A or B, effectively small enough to be considered negligible. So the reaction proceeds to make some more C(aq) which precipitates out. So on and so forth.
Consider it this way, a completely made up case:
A(aq) + B(aq) ⇔ C(aq) : Keq ~1
C(s) ⇔ C(aq) : Ksp ~ 10-10
The low solubility of C(aq) will force the reaction to proceed very much further than the equilibrium constant may lead you to believe. If we ignore the solubility of C, we would expect [C](aq)/([A](aq)(aq)) ~ 1. However because the solubility is so low the solution never actually gets enough [C](aq) to get to ~1 because as far as the solution is concerned there is very little C(aq) around (it precipitates out and becomes C(s)). Only a thin layer at the solid/solution interface is accessible to A or B, effectively small enough to be considered negligible. So the reaction proceeds to make some more C(aq) which precipitates out. So on and so forth.
- #7
Borek
Mentor
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ElmorshedyDr said:
As you seem to be correctly guessing, it is not a whole truth - in a closed container these are just equilibrium reactions, as every other.
Precipitation and dissolution of calcium carbonate in water (in equilibrium with carbon dioxide) are responsible for creating karst formations - and it is the reversibility of the precipitation that is the driving force.
- #8
Entanglement
- 438
- 13
Thanks guys, but please I need a simple explanation
- #9
Borek
Mentor
- 29,180
- 4,612
I don't see how to simplify "you are right" further 
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