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Effects of Crushing a Solid on Equilibrium Position

  1. Jan 10, 2016 #1
    Hi everyone.

    So I have been learning about equilibrium recently in class... and suppose I have the following reaction:

    A(s) + B(aq) <-----> C(aq) + D(g)

    [Note: in parentheses, I indicate the state of my substances]

    So I was told recently that pulverizing my solid "A" would not change the position of equilibrium as nothing will happen.... but I recalled our unit on Kinetics which says that increasing the surface area of my reactants will increase frequency of collisions which will increase the rate of the chemical reaction.

    So, my impression is that pulverizing A would increase the surface area and increase the collisions for the forward reaction. Thus, the forward reaction would speed up, and a shift to the products would occur with [C] and [D] increasing.

    Will the equilibrium shift, or will it not, when I crush a solid, and most importantly, why?

    Thanks for reading and thanks in advance.
     
  2. jcsd
  3. Jan 10, 2016 #2

    Borek

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    Staff: Mentor

    What the backward reaction, will increased surface speed it up too, or not?
     
  4. Jan 10, 2016 #3
    I believe it will not as the solid is only a reactant in the forward reaction and not in the backward one.
     
  5. Jan 10, 2016 #4
    Would you happen to know if this is the case?
     
  6. Jan 10, 2016 #5

    Borek

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    Good point. But product has to be deposited and the deposition speed depends on the amount of available surface. Changing the are influences both the forward and the backward reaction.
     
  7. Jan 10, 2016 #6
    So you are saying that both reactions will proceed in the same speed?

    Wow.

    We did not discuss deposition and deposition speed in class at all... so this is an unfamiliar concept for me.
     
  8. Jan 10, 2016 #7

    Borek

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    No, the CHANGE in both speeds will be identical, and their ratio will remain constant.
     
  9. Jan 10, 2016 #8
    Thanks so much!
     
  10. Jan 11, 2016 #9
    But you are increasing the surface free energy of solid A. Should this not enter into equilibrium calculation?
     
  11. Jan 11, 2016 #10

    Borek

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    Staff: Mentor

    Interesting point. My bet is that if you compare it with everything else, the difference is negligible. Or, in the cases where it matters, when you have a mixture of crystals with any equilibrium allowing transport of the substance between them, smaller ones disappear while the larger ones grow, so the overall surface goes down relatively fast, reducing the possible difference.
     
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