Berkeley Review: Rate-Determining Step

[chickensoupdr]

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I have a question about the rate-determining step from the Berkeley Review Chem Part II book.

In the Kinetics section, problem #46, it says that adding water would... "decrease the reaction rate." However, water is not a reactant in the rate limiting reaction... so how can it affect the reaction rate??

 

[Fifty 3rds]

I haven't read the question because I'm at work but I'm guessing it has to do with dilution.

Are the reactants liquids or gases? Is it in a pressurized container or just sitting in a beaker?

By adding a third "thing", or a tertium quid, you are decreasing the likelihood that the two correct items will interact and proceed to products. I believe this is a kinetic effect.

If you have two things in a bottle, there's a good chance they collide. Now add a third. Now a fourth. What's the likelihood the two original things will collide?

That's my guess, someone correct me if I'm wrong. ** is this acid-base chemistry?** that could also change what I just wrote.

 

[chickensoupdr]

Well, from what I understand from this problem, the reaction is multistep (the problem tells you which one is the "slow" one, and, therefore, the rate-limiting reaction). Water is not a reactant in the rate-limiting reaction; it is a reactant in one of the "fast" reactions. However, the answer explains how adding water to the reaction would affect the fast reaction, ultimately by pushing the slow step's reaciton backwards... sort of through like a chain reaction. I'm not sure if I really understand it... it seems conflicting.


Thanks for your help!