15.53

Moderators: Chem_Mod, Chem_Admin

kanikubari
Posts: 30
Joined: Fri Dec 07, 2018 12:18 am

15.53

Postby kanikubari » Tue Mar 12, 2019 8:18 pm

Why is mechanism C not correct. How does step 1 having a reverse that is both fast and at equilibrium affect this?

A De Castro 14B 2H
Posts: 75
Joined: Fri Sep 28, 2018 12:29 am

Re: 15.53

Postby A De Castro 14B 2H » Tue Mar 12, 2019 9:17 pm

Mechanism C wouldn't agree with the proposed rate law for the same reason that mechanism A wouldn't: the rate law would include CO. Having a reverse reaction in step 1 means that the reactants used to produce NO3 can be used in the rate law. Thus, for (c), rate = k[NO2]2[CO]. But, again, this doesn't agree with the proposed rate law, rate = k[NO2]2.


Return to “Kinetics vs. Thermodynamics Controlling a Reaction”

Who is online

Users browsing this forum: No registered users and 3 guests