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Homework Help: Rate Laws

  1. Feb 23, 2017 #1
    1. The problem statement, all variables and given/known data
    An iodine clock kinetics experiment was performed by varying the composition of reactant solutions and measuring resulting rates of reaction (based on the data given, I have determined the rate law to be R = [I-]1[S2O82-]1). Which of these four proposed mechanisms below would be supported by these results?

    Mechanism 1: I-(aq) + I-(aq) ---> I22-(aq) (slow)
    I22-(aq) + S2O82-(aq) ---> I2(aq) + SO42-(aq) (fast)

    Mechanism 2: I-(aq) + S2O82-(aq) ---> [S2O82-·I]3-(aq) (slow)
    [S2O82-·I]3-(aq) + I-(aq) ---> I2(aq) + 2 SO42-(aq) (fast)

    Mechanism 3: I-(aq) + S2O82-(aq) <---> [S2O82-·I]3-(aq) (fast)
    [S2O82-·I]3-(aq) + I-(aq) ---> I2(aq) + 2 SO42-(aq) (fast)

    Mechanism 4: 2I-(aq) + S2O82-(aq) ---> I2(aq) + 2SO42-(aq)

    2. Relevant equations
    Rate law: R = k[A]xy

    3. The attempt at a solution
    Mechanism 1: R = [I-]2. Therefore, this is incorrect.

    Mechanism 2: R = [I-][S2O82-] Therefore, this is correct.

    Mechanism 3: R = [[S2O82-·I]3-][I-]. Therefore, this is incorrect.

    Mechanism 4: R = [I-]2[S2O82-]. Therefore, this is incorrect.

    Answer: Mechanism 2

    Can anyone confirm if I am on the right track? Thanks. All help is very much appreciated.
    Last edited: Feb 23, 2017
  2. jcsd
  3. Feb 23, 2017 #2
    Pardon me for the incorrect rate law equation that I listed.
  4. Feb 23, 2017 #3


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    Homework Helper
    Gold Member

    Yes I'd say you are on the right track. I agree mechanism (2) is consistent with the kinetics.

    However have another look at mechanism 3, which after all is only quantitavely different from 2. You have to work out the rate expression as a function of your experimental variables (which your other three are) not of intermediates. You may find the answer is more nuanced. :oldwink:
    Last edited: Feb 23, 2017
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