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Chem 222 notes Week 4

by: Leslie Pike

Chem 222 notes Week 4 Chem 222

Leslie Pike
GPA 3.9

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About this Document

Elementary steps, intermediate molecules, rate laws
College Chemistry 2
Darwin Dahl
Class Notes
Chemistry, chem222, chem 222
25 ?




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This 1 page Class Notes was uploaded by Leslie Pike on Wednesday February 17, 2016. The Class Notes belongs to Chem 222 at Western Kentucky University taught by Darwin Dahl in Spring 2016. Since its upload, it has received 13 views. For similar materials see College Chemistry 2 in Chemistry at Western Kentucky University.


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Date Created: 02/17/16
Chem 222 notes Week 4 Reaction Mechanics: Elementary steps are the steps taking place during the reaction. For example, in the reaction 2NO + O  2NO 2 The presence of N O 2s 2etected during this reaction. Dinitrogen dioxide is produced during an elementary step and consumed during another elementary step; that’s why it doesn’t appear in the equation above. There are two elementary steps to the above reaction: Step 1: NO + NO  N O 2 2 Step 2: N O + O  2NO 2 2 2 2 Notice that dinitrogen dioxide is a product, then it is a reactant. It does not appear in the net reaction. Substances which are formed in an early elementary step and are consumed in a later elementary step are known as intermediates. Catalysts are not generally considered intermediates because they are not produced or consumed by the reaction, generally. Rate laws can be written for intermediates. These do not have to be determined experimentally; instead, the concentration is raised to the power of the balancing coefficient. Using the reaction above gives the results: Rate for first step = k[NO] 2 Rate for second step = k[N O ][O ] 2 2 2 The rate of the slowest elementary reaction should agree with the experimentally determined rate for the overall reaction. A catalyst does not change whether or not a reaction will occur, but it speeds up the rate of the reaction by lowering the activation energy. Given the Arrhenius equation k=Ae -(Ea)/(where k is the rate of the reaction and Ea is the activation energy, a decrease in Ea will result in a corresponding increase in k. This should be obvious from looking at the equation, because e is raised to a negative power.


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