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Solids and Liquids in Hetrogenous Equilibrium

  1. Nov 22, 2011 #1
    We started to learn about chemical equilibrium and equilibrium constants a few weeks ago and something has been bugging me. I don't understand why solids and liquids are not included in the equilibrium constants for some reactions.

    Here's a hypothetical situation with the following reaction

    [tex] A_{(s)} + B_{(g)} \leftrightarrow C_{(g)}[/tex]

    Now the K expression would normally be written as [tex] K = \frac{[C]}{} [/tex]
    with the solid left out since it has constant concentration (density) at a constant temperature.

    At equilibrium the forwards and backwards reaction rate must also be the same. Now from my understanding, heterogeneous reactions only occur at the surface of the solid or liquid, therefore, the rate of the forwards reaction would be affected by the surface area of the "A" solid. If you increase the surface area of "A" by adding more solid or by crushing it into a powdery form wouldn't it increase the rate of the forwards reaction while leaving the backwards reaction rate unaffected and thereby shift the equilibrium to the right? However, according to the K expression changing the surface area of the solid would have no effect on the equilibrium concentrations of either "B" or "C".

    Also I have another related equilibrium question. Would an unsaturated solution be considered to be in dynamic equilibrium? My teacher says it isn't since you can't observe any solid solute in such a condition and all the species in a reaction must be present for a dynamic equilibrium to be established. However, don't microscopic amount of solid solute still exist even in an unsaturated solution?
     
  2. jcsd
  3. Nov 22, 2011 #2
     
  4. Nov 22, 2011 #3

    Ygggdrasil

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    Increasing the surface area of a solid would increase both the forward and backward rate equally. This happens because both the forward and backward reactions must take place on the surface.
     
  5. Nov 23, 2011 #4
    But in the hypothetical reaction I posted there is no solid in the backwards reaction.
     
  6. Nov 24, 2011 #5

    Ygggdrasil

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    Let's say the reaction is crystallization of a salt from solution. Addition of ions from solution to the crystal at the crystal's surface is much faster than nucleating new crystals at a different location in the solution (this is why solutions can be supersaturated). Therefore, the rate of deposition will depend on the surface area of the solid.
     
  7. Nov 25, 2011 #6
    In that situation I would agree that the backwards reaction would depend on the surface area of the solid. However, the solid crystal is essentially placed on both sides of the reaction equation in that situation because in order for the deposition to occur the ion has to come in contact with the crystal's surface. Therefore, I don't believe the logic in that situation will apply to all situations of heterogeneous equilibrium.

    Now let me adjust the hypothetical reaction to be a bit easier to visualize:

    [tex] A_{(s)} + B_{(g)} \leftrightarrow C_{(g)} + D_{(g)} [/tex]

    Notice that I added a second gaseous substance on the right side. Now in this case I don't see how the backwards reaction could possibly change depending on the surface area of the "A" solid since the backwards reaction takes place between two gases.
     
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