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The Rate Law in Chemistry

  1. Nov 4, 2006 #1
    "The Rate Law" in Chemistry

    Hey I've just been working on a chemistry lab lately in my chemistry course, and one of the problems on the lab write-up asks to:

    Write the rate law for the following reaction, if it were to occur through a single-step mechanism.
    2I^-(aq) + S2O8^2-(aq) -> 2SO4^2-(aq) + I2(aq)

    Now since it is asking to write it as a single-step mechanism I would assume there is only one step needed.

    Now in the lab it has given us this formula:

    rate = -(delta[S2O8^-2])/(delta t) = -(1/2)(delta [I^-])/(delta t) = (1/2)(delta[SO4^2-])/(delta t) = (delta[I2])/(delta t)

    So could you not just say the only step needed is:
    -(delta[S2O8^-2])/(delta t) = (delta[I2])/(delta t)

    Then for the second part to the question it asks "By examination of the experimentally determined rate law, explain whether this reaction does in fact proceed through a single-step mechanism. Explain."

    Any help on this problem would be great, thanks.
     
  2. jcsd
  3. Nov 5, 2006 #2

    GCT

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    The rate law would be

    rate=k(forward)[I-]^2[S2O8]

    does your experimentally determined rate law concur with this expression?
     
  4. Nov 5, 2006 #3

    GCT

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    The rate law would be

    rate=k(forward)[I-]^2[S2O8]

    does your experimentally determined rate law concur with this expression?
     
  5. Nov 5, 2006 #4
    Ahh I see what you are saying, yeah it's just
    rate=k[S2O8^2-][I^-]^2

    And my rate law does not coincide with this, therefore it cannot be a single-step mechanism. Gotcha! Thanks.
     
    Last edited: Nov 5, 2006
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