## 15.85

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

ClaireHW
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Joined: Fri Sep 29, 2017 7:07 am
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### 15.85

The following rate laws were each derived from an elementary reaction. In each case, write the chemical equation for the reaction, determine its molecularity, and draw a proposed structure for the activated complex.
a. Rate = k[CH3CHO] (products are CH3 and CHO)
b. Rate = k[I]2[Ar] (products are I2 and Ar; the role of Ar is to remove energy as the product forms.)
c. Rate = k[O2][NO] (products are NO2 and O)

I confused as to how to draw a structure for the activated complex? How do you figure out format and order?

Thanks!
Claire Woolson Dis 1K

Cam Bear 2F
Posts: 60
Joined: Thu Jul 27, 2017 3:01 am

### Re: 15.85

Nancy Dinh 2J posted a really good explanation!:

"First you're finding the balanced equation. After that, look at the number of different reactants of the balanced equation. This will determine the molecularity. 1 reactant is unimolecular, 2 is bimolecular, 3 termolecular.

Finally, you draw the activated complex, which will form at the top of the activation barrier and is basically all of the reactants mushed together into one complex because they all collided with each other. Draw dotted lines between places were bonds are formed and broken.

For example, let's look at c:

When O2 collides with NO to form NO2 and O, you can see that the bond in O2 breaks and one of the O's forms a bond with NO to form NO2. Therefore, the activated complex is:

O---(breaking)---O---(forming)---NO"

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