Rate Laws

  • Thread starter cvc121
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  • #1
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1. Homework Statement
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. Homework 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:

Answers and Replies

  • #2
61
1
Pardon me for the incorrect rate law equation that I listed.
 
  • #3
epenguin
Homework Helper
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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:
 
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