rate determining step order of reaction EK Chem 1101 (242)

[Spiker]

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Does anyone know why EK assume that first order reaction would make it the slowest step?

 

[Vanguard23]

I'm guessing that you're supposed to assume that the dissociation of the single molecule is the slowest step. I don't recall Iodide being a good leaving group, so I suppose that's why.

 

[Spiker]

Well i guess you can say it that way since you have to have a leaving group first which is always slow, however if i didnt know about that leaving group stuff

shouldnt order 1 be faster because it does not require collision of as many particles...

 

[sleepy425]

Does anyone know why EK assume that first order reaction would make it the slowest step?

what are you talking about?? your question doesn't make any sense. tell us exactly what EK says (including the context in which it says this) and we'll be able to help you

 

[johndoe3344]

To OP:

Your question makes no sense. It is not true in general that a first order reaction is the slowest step.

In the specific example, you should know from your OChem that the formation of a carbocation is the rate determining step.

 

[Dr Gerrard]

The slow step however does usually determine the rate of the equation.

You can look at the reaction rate and determine it is first order.

 

[Vanguard23]

Guys, if you look at the problem he's referring to, it will make sense. The problem doesn't explain *why* the first step in the reaction is the rate-determiner. While the question isn't difficult, that's why he didn't go into detail; the question was provided.

 

[Dr Gerrard]

Sorry bro, but everyone does not have that book, and we were just trying to help.