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# RelativeResourceManager.pdf2 - Chapter 22 Reaction...

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45 Chapter 22 Reaction Mechanisms Let’s begin with an example and introduce/review some language. For the stoichiometric reaction: N 2 O 5 + NO 3 NO 2 Step 1: N 2 O 5 NO 2 + NO 3 “unimolecular” Step 2: NO 2 + NO 3 N 2 O 5 bimolecular reverse process Step 3: NO + NO 3 2 NO 2 also bimolecular Rate Laws for these elementary processes Step 1: rate 1 = k 1 [N 2 O 5 ] Step 2: rate 2 = k 2 [NO 2 ] [NO 3 ] Step 3: rate 3 = k 3 [NO] [NO 3 ] The idea is that we have been studying what to do in the laboratory, how to determine rate constants and reaction orders and the like. Now we need to figure out how to devise mechanisms and also how to connect those mechanisms back to rate laws. Thus, figuring out the rate laws for elementary reaction steps is an important first topic. There are three “tricks” we use to make the connection between the mechanism and the rate law. 1. The rate determining step approximation 2. The equilibrium approximation 3. The steady-state approximation Consider the following chemical reaction: A + B C + D A possible mechanism Step 1: A + B X Slow (rate determining) Step 2: X C + D Fast k 1 k 2 k 3 Reaction Mechanism Laboratory Work: Rate Laws, Rate Constants Is Mechanism Consistent k 1 k 2

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46 The molecularity of step 1 is bimolecular since there are two molecules on the left side of the equation. The molecularity of step 2 is unimolecular Step 1: rate 1 = k 1 [A] [B] Step 2: rate 1 = k 1 [X] Remember that the rate law for an elementary reaction step depends ONLY on the reactants, NOT the products. The Rate Determining Step Approximation
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