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Rate law expression with solids and liquids

  1. May 12, 2014 #1
    Hello Forum!

    I was wondering:

    In the rate law expressions, I could never find an example with a solid or a liquid in the reactants in my textbook (Chemistry by Zumdahl). I searched Chemistry (Raymond Chang) and Principles of Molecular Chemistry, without any success.

    What happens to solids and liquids who act as reactants in reactions? How do I incorporate them into the rate law? Are they ALWAYS already included in the k constant?
     
  2. jcsd
  3. May 12, 2014 #2

    Qube

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    Gold Member

    Solids and liquids are not incorporated into the K constant because their concentrations are assumed to be constant (which is a pretty good approximation; as you consume either a solid or a liquid, you decrease the total amount but do not generally affect their concentrations).
     
  4. May 13, 2014 #3

    Borek

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    Staff: Mentor

    k (kinetics) or K (equilibrium)?

    For equilibrium presence of the solid is enough - no matter how much of the solid there is, the result (in terms of the final equilibrium position) is always the same.

    For kinetics it is much more complicated, as reaction rate depends on the surface area of the solid. But I have never seen it addressed in the introductory courses.
     
  5. May 13, 2014 #4

    Qube

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    Oops, true, I should have specified that I meant equilibrium Ks only. As Borek says, Ks for kinetics differ.
     
  6. May 13, 2014 #5

    AGNuke

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    Not to mention, the rate law also depends on the reaction mechanism. So, it is impossible to guess on your own unless provided with some incentive.
     
  7. May 13, 2014 #6
    I meant k as in the rate law: for example, rate=k[NO2]^2.
     
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