CSU CHEM 113 - Mechanisms with a Fast Initial Step (3 pages)

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Mechanisms with a Fast Initial Step



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Mechanisms with a Fast Initial Step

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Determining the validity of a mechanism.


Lecture number:
7
Pages:
3
Type:
Lecture Note
School:
Colorado State University- Fort Collins
Course:
Chem 113 - General Chemistry II

Unformatted text preview:

CHEM 113 1st Edition Lecture 7 Outline of Last Lecture I Reaction Mechanisms a Elementary steps II Correlating mechanisms with the rate law III The rate determining step of a reaction Outline of Current Lecture II Practicing example problems of reaction mechanisms Current Lecture In class we discussed and solved the following problem The overall reaction 2NO g O2 2NO2 g has an experimental rate law Rate k NO 2 O2 A proposed mechanism is 1 NO g O2 g NO3 fast 2 NO3 g NO g 2NO2 g slow rate determining Let s go through all three criteria to see if this reaction is valid o 1 The elementary steps must add up to the overall balanced equation NO O2 NO3 NO NO3 2NO2 NO O2 NO3 NO NO3 2NO2 2NO O2 2NO2 These workings show that the first criterion is met o 2 The elementary steps must be reasonable These notes represent a detailed interpretation of the professor s lecture GradeBuddy is best used as a supplement to your own notes not as a substitute Both steps are bimolecular so this criterion is met o 3 The mechanism must correlate with the observed rate law To see that this is met we first write rate laws for the elementary steps Note that the first reaction occurs both forwards and in reverse therefore there are two rates Rate 1 forward k1 NO O2 Rate 1 reverse k 1 NO3 k 1 is the rate constant and NO3 is the reactant for the reverse reaction Rate 2 k2 NO3 NO Next we must show that the rate law for the rate determining step step 2 in this case gives the overall rate law How we have it written does not because it contains the intermediate NO3 where an overall rate law only includes the reactants and products We eliminate NO3 from the rate law for step 2 by expressing it in terms of reactants Rate 1 forward Rate 1 reverse or k1 NO O2 k 1 NO3 To express NO3 in terms of reactants we isolate it algebraically NO3 k1 k 1 NO O2 We then substitute this into rate 2 Rate 2 k2 k1 k 1 NO O2 NO Rate 2 k2k1 k 1 NO 2 O2 This is identical to the overall rate law o To summarize we asses the



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