## Rate-Determining Step

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

Kathleen Vidanes 1E
Posts: 62
Joined: Fri Sep 29, 2017 7:07 am

### Rate-Determining Step

Sometimes in a reaction mechanism, the 2nd step is the slow step, or rate-determining step. Does this have any significant effect on the overall reaction or the rate law, in contrast from having the 1st step as the slow step? If so, what effects are there?

Ju-Wei Wang 1I
Posts: 31
Joined: Fri Sep 29, 2017 7:05 am

### Re: Rate-Determining Step

The second step being the slow step makes finding the rate law for the overall reaction more difficult because it may contain an intermediate, which is never included in the overall rate law.
Last edited by Ju-Wei Wang 1I on Tue Mar 13, 2018 6:57 pm, edited 1 time in total.

Kathleen Vidanes 1E
Posts: 62
Joined: Fri Sep 29, 2017 7:07 am

### Re: Rate-Determining Step

There was a homework problem that I did (15.53c) in which the 2nd step was the slow step, and it did contain an intermediate (NO3):
Step1 NO2 + NO2 ⇌ NO + NO3 and its reverse (both fast, equilibrium)
Step2 NO3 + CO --> NO2 + CO2 (slow)

However, finding the rate law did not seem to be difficult: Rate = k[NO3][CO]. I found the rate law through the same way that I would have if the 1st step was the slow step -- by multiplying the concentration of the reactants with the rate constant. Was it simply a coincidence? Or is there a different way to calculate the rate law if the 2nd step is the slow step?

Ju-Wei Wang 1I
Posts: 31
Joined: Fri Sep 29, 2017 7:05 am

### Re: Rate-Determining Step

You're correct that the rate law for step 2 is rate = k[NO3][CO], but that is the rate law for that specific step, not for the overall reaction.

Sabrina Fardeheb 2B
Posts: 51
Joined: Sat Jul 22, 2017 3:00 am
Been upvoted: 1 time

### Re: Rate-Determining Step

If the first step is the slow step, then it is significantly easier to solve and determine the rate law as no intermediates appear. However, if the fast step precedes the slow step, then we use the pre-equilibrium approach, which has a more complex effect on the overall reaction as we have to find a concentration that will substitute that of the concentration in order to derive a rate law that will match the observed one.