1 / 9

General Chemistry Lab

General Chemistry Lab. Rates of Chemical Reactions, I: The Iodination of Acetone Experiment 20, p.155. ( aq ). I 2 ( aq ). H 3 C. CH 2 I. + H + ( aq ) + I - ( aq ). +. [I 2 ]. Rate = —. acetone.  t. O. O. CH 3. C. C. H 3 C. Rate = k [acetone] m [I 2 ] n [H + ] p.

ballari-taj
Télécharger la présentation

General Chemistry Lab

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. General Chemistry Lab Rates of Chemical Reactions, I: The Iodination of Acetone Experiment 20, p.155

  2. (aq) I2(aq) H3C CH2I + H+(aq) + I-(aq) + [I2] Rate = — acetone t O O CH3 C C H3C Rate = k [acetone]m [I2]n [H+]p We want you to determine the value of m, n, and p. Abbreviate [acetone] as [A]

  3. (aq) I2(aq) H3C CH2I + H+(aq) + I-(aq) + acetone O O CH3 C C H3C colored colorless We can follow the course of the reaction by observing color changes. Rate = k [A]m [I2]n [H+]p The reaction is zero order with respect to I2 (n = 0). So the rate does not depend on [I2] at all, as long as [I2] is not = 0. In the lab, you will make I2 the limiting reagent, present in a large excess of acetone and H+, so the [ ] of these effectively do not change.

  4. (aq) I2(aq) H3C CH2I + H+(aq) + I-(aq) + acetone O O CH3 C C H3C The rate of the reaction is constant during its course under these conditions, so we can vary the rate by changing the initial [A] and [H+].

  5. (aq) I2(aq) H3C CH2I + H+(aq) + I-(aq) + acetone ( ) Rate 2 k 2[A]m Rate 1 k [A]m m 2[A] = 2m O O = = CH3 [A] C C H3C If we double the [A] while holding the [ ] of the other two constant Rate 2 = k 2[A]m [I2]0 [H+]p Rate 1 = k [A]m [I2]0 [H+]p What happened to the other terms in these equations?

  6. So by timing these two reactions, we can discover the order of the reaction with respect to [A] (that is, the value of m).

  7. (aq) I2(aq) H3C CH2I + H+(aq) + I-(aq) + acetone O O CH3 ( ) Rate 2 k 2[H+]p Rate 1 k [H+]p p 2[H+] = 2p C C = = [H+] H3C Next we double the initial [H+] while holding the [ ] of the other two constant Rate 2 = k [A]m [I2]0 2[H+]p Rate 1 = k [A]m [I2]0 [H+]p

  8. So by timing these two reactions, we can discover the order of the reaction with respect to [H+] (that is, the value of p).

  9. Before you arrive at chemistry lab, read and study the introduction to Lab Exercise 20 (p.155). Then read and re-read the “Experimental Procedure” on pp.157-161 until you completely understand what you will do. Next week you will need to do this for Lab Exercise 21: Rates of Chemical Reactions, II. A Clock Reaction (p.165). This ppt file is available on my web sitehttp://www.evangel.edu/Personal/badgers/Web/

More Related