2015 Final 6B

Moderators: Chem_Mod, Chem_Admin

Marissa Petchpradub 2F
Posts: 15
Joined: Wed Sep 21, 2016 2:56 pm

2015 Final 6B

Postby Marissa Petchpradub 2F » Sat Mar 18, 2017 8:06 pm

For this question, I understand that we must use the pre-equilibrium approach to get rid of the intermediate [ES] in the rate law. After doing this, we get k2(k1/k1')[E][S], but [E] is a catalyst and I thought we cannot have catalysts in the rate law as well as intermediates. Can someone please explain why it is okay for the catalyst to appear in the rate law here?

Posts: 27
Joined: Wed Sep 21, 2016 2:56 pm

Re: 2015 Final 6B

Postby Shadi_Keyvani_1D » Sat Mar 18, 2017 9:05 pm

Catalysts are always okay to have in the rate law b/c they are originally present and then reformed but intermediates are not as they are formed and then gone.

Return to “Method of Initial Rates (To Determine n and k)”

Who is online

Users browsing this forum: No registered users and 1 guest