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

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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. - PowerPoint PPT Presentation

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Page 1: General Chemistry Lab

General Chemistry Lab

Rates of Chemical Reactions, I:

The Iodination of AcetoneExperiment 20, p.155

Page 2: General Chemistry Lab

C

O

CH3H3C

acetone

C

O

CH2IH3C+(aq) I2(aq) + H+(aq) + I-(aq)

Rate = k [acetone]m [I2]n [H+]p

Rate = —[I2]

t

We want you to determine the value of m, n, and p.

Abbreviate [acetone] as [A]

Page 3: General Chemistry Lab

C

O

CH3H3C

acetone

C

O

CH2IH3C+(aq) I2(aq) + H+(aq) + I-(aq)

colored colorless

We can follow the course of the reaction by observing color changes.

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.

Rate = k [A]m [I2]n [H+]p

Page 4: General Chemistry Lab

C

O

CH3H3C

acetone

C

O

CH2IH3C+(aq) I2(aq) + H+(aq) + I-(aq)

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+].

Page 5: General Chemistry Lab

C

O

CH3H3C

acetone

C

O

CH2IH3C+(aq) I2(aq) + H+(aq) + I-(aq)

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?

Rate 2 k 2[A]m

Rate 1 k [A]m= = = 2m( )2[A]

[A]

m

Page 6: General Chemistry Lab

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

Page 7: General Chemistry Lab

C

O

CH3H3C

acetone

C

O

CH2IH3C+(aq) I2(aq) + H+(aq) + I-(aq)

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

Rate 2 k 2[H+]p

Rate 1 k [H+]p= = = 2p( )2[H+]

[H+]

p

Page 8: General Chemistry Lab

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

Page 9: General Chemistry Lab

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/