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Quick question on iodination of acetone.

  1. Aug 25, 2008 #1
    In class today we did a lab with acetone (4M), iodine (.0050M), and HCl (1.0M). The lab was one kinetics and on determining the rate constant, rates, and orders. Which I've done fine, but I'm stuck on this question: Why is the concentration of I2 so much less then the other reactants?

    Any help would be appreciated, and it's due tomorow =]


    Rate=2.1 (Ms)^(-1) x 10^(-5) x (Acet) x (H+)

    Second order reaction.

    First order with respect to Acet and H+, zero order with respect to I2

    Just to prove I've done the work =]
  2. jcsd
  3. Aug 26, 2008 #2
    the rate of reaction is already slow because of the acetone and H+ reacting. if you use low concentrations of them, the reaction would be even slower.

    and since iodine is zero order, using a very low concentration does not affect the rate. probably you have been given low concentration because of economic reasons or maybe to reduce health hazards when pipetting iodine, because it is toxic.
  4. Aug 26, 2008 #3
    Ahhh thanks, that makes sense.
  5. Aug 29, 2008 #4


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    You followed the rate of reaction by monitoring the loss of yellow (iodine) color in solution. Concentrated solutions of iodine would not likely give you linearity with respect to concentration (Beer's law).
  6. Aug 29, 2008 #5
    i guess this makes more sense!!!
  7. Feb 7, 2011 #6
    Iodine is the main determinant in the reaction. So, the reaction will continue until it's gone. Therefore, increasing the concentration of Iodine would cause the reaction to go on much longer and usually a reaction that only takes a few minutes gives you much easier and more accurate numbers to deal with.
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