Slow step vs. Fast step

$K = \frac{k_{forward}}{k_{reverse}}$

jeffreyli1L
Posts: 8
Joined: Wed Sep 21, 2016 2:55 pm

Slow step vs. Fast step

Hello, I have a question about definitions.
What does it mean for a step in a reaction to be the slow step, or the fast step? Is this given, or are we able to figure out which is the slow step based on the reaction? Thank you.

Rachel_Prescott_2M
Posts: 10
Joined: Wed Sep 21, 2016 2:57 pm
Been upvoted: 1 time

Re: Slow step vs. Fast step

The slow step determines the rate of reaction and for this reason it is also called the rate determining step. The slow step serves as a bottleneck of sorts, the rate at which something gets to a bottleneck (fast step) doesn't really matter for the overall rate, but the rate something gets through the bottleneck (slow step) does. This is why the slow step is so important for rates whereas the fast steps are not as important. Generally this information is given I believe but I'm not 100% sure... hope this helped a bit!

AonyaMontoya2B
Posts: 12
Joined: Fri Jun 17, 2016 11:28 am

Re: Slow step vs. Fast step

How do you use the slow, fast and overall reactions to determine the rate law of the reaction?

Winnie Liu 1M
Posts: 11
Joined: Fri Jul 15, 2016 3:00 am

Re: Slow step vs. Fast step

It depends on the mechanism given. The slow step is always the rate determining step. Therefore, all steps after that we can ignore. Since the slow step is an elementary reaction, you can determine the rate law from the stoichiometric chemical equation.

If the mechanism is given as a slow then fast step, we can go ahead and just use the slow step to determine rate law.

If the mechanism is given as a fast then slow step, we would have to use the pre-equilibrium approach. (find K=kforward/kreverse for fast step. Then rearrange and plug into slow step rate law).