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

Kim
Posts: 33
Joined: Wed Sep 21, 2016 3:00 pm
Been upvoted: 1 time

page 73 in the course reader discusses the pre equilibrium approach
(my question is under the pic)

I just don't understand where the 1/2d[NO2]/dt came from and why that equals k2[N2O2][O2]

Matt Goff 1F
Posts: 10
Joined: Wed Nov 18, 2015 3:00 am

We are trying to see if the reaction mechanism proposed accurately leads us to the given observed rate of k[NO]^2[O2].

1/2d[NO2]/dt comes from looking at the balanced reaction on the first line of this page and writing out the rate in terms of the formation of NO2. You do that the same way it's done on pg 59 of the course reader. So that expression represents the total reaction rate. This is equal to k2[N2O2][O2] because the second intermediate step (STEP 2 on the page) was proposed to be the slow step, which means it is rate determining step. The rate expression for the slow step should equal the expression for the observed rate, so you can set

1/2d[NO2]/dt=k2[N2O2][O2]

The rest of the page shows how the proposed mechanism is confirmed as the expression you end with is the same (just with a more specific breakdown of the constants) as the observed rate. Hope this helps!