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Irreversible reactions

  1. Mar 10, 2014 #1
    Why are precipitate and gas evolution reactions irreversible ( why don't the products react once again ) ??
  2. jcsd
  3. Mar 10, 2014 #2
    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.
  4. Mar 10, 2014 #3

    What if the products were to be produced in a closed vessel ?
  5. Mar 10, 2014 #4
    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.
  6. Mar 10, 2014 #5
    That means that there is a back reaction but it's negligible ??
  7. Mar 10, 2014 #6
    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.
  8. Mar 11, 2014 #7


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    Staff: Mentor

    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.
  9. Mar 11, 2014 #8
    Thanks guys, but please I need a simple explanation
  10. Mar 11, 2014 #9


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    Staff: Mentor

    I don't see how to simplify "you are right" further :wink:
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