This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 2 is equal to 2 times the rate of the slow step (k 2 [O][O 3 ]), since two molecules of O 2 are formed. Thus, rate of formation of O 2 = 2k 2 [O][O 3 ], but O is an intermediate, solve for O in terms of products and reactants and rate constants. Since the first step is fast and reversible and the second step is slow, the first step is in equilibrium and we can write [O 2 ][O] = k 1 = K 1 or [O] = k 1 [O 3 ] [O 3 ] k1 k1 [O 2 ] substituting: rate = 2k 2 k 1 [O 3 ] 2 k1 [O 2 ] rate = k obs [O 3 ] 2 [O 2 ] What is the order in O 3 ? 2 double O 3 /rate will? multiply by 4 What is the order in O 2 ? 1 double O 2 / multiply by What is the overall order? 1 double both O 3 and O 2 / double...
View
Full
Document
This note was uploaded on 03/02/2012 for the course CHEM 5111 taught by Professor Vogel during the Fall '08 term at MIT.
 Fall '08
 Vogel
 Reaction

Click to edit the document details