ExamKrackers Chemistry Lecture 2 Exam Passage III

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MedGrl@2022

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Does anyone have a good understanding of the passage and questions and answers of ExamKrackers Chemistry Lecture 2 Exam Passage III?

I am confused as to why Reaction 2 is the fastest reaction and not the slowest. 😕 In addition, how can Equation 1 be used to determine the rate of Reaction 1?

If reaction 1 is the slow step shouldn't the rate equation be [S2O8]^3=rate?

I understand that 1/2[S2O8]=[I3] but there is also the excess I3 that is used to form the blue-black I2 complex.

How is this rate law the same as described in chapter 2? Chapter 2 has the initial rate law as being rate=k[reactants] of the slowest step. Time was not a factor. I3 is a product of reaction 1 and a reactant of reaction 2. Therefore, doesn't the equation tell us that reaction 2 is the slowest step?

Any helpful advice on how to properly understand this passage would be great.
 
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