Exam_3_review - EXAM 3 REVIEW LBS 172 REACTION MECHANISMS GENERAL-Step by step process of bond making and breaking by which reactants become

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EXAM 3 REVIEW LBS 172 REACTION MECHANISMS GENERAL -Step by step process of bond making and breaking by which reactants become products -Summation of steps must be equal to overall reaction -Example: NO 2 (g) + F 2 (g) Æ FNO 2 + F(g) N O 2 (g) + F(g) Æ FNO 2 (g) Overall: 2NO 2 (g) + F 2 (g) Æ 2FNO 2 (g) TERMS -Elementary reaction: individual step in a mechanism; one chemical event Ex (from above rxn): NO 2 (g) + F 2 (g) Æ FNO 2 + F(g) -Intermediate: species that is formed and subsequently used up in the reaction Ex: F(g) -Molecularity: number of particles colliding in an elementary step 1- unimolecular 2- bimolecular 3- termolecular Ex: NO(g) + F(g) Æ FNO 2 (g) is a bimolecular step -Catalyst: species that is used up by the mechanism and is subsequently reformed RATES FOR ELEMENTARY STEPS -Rate equation of an elementary step is based on reaction stoichiometry -Rate = product of rate constant and the concentrations of each reacting particle (Neither solids nor pure liquid reactants are used in rate equations) For any elementary step: A + B Æ C + D, rate = k[A][B] Ex: Rate for step NO 2 (g) + F 2 (g) Æ FNO 2 + F(g) = k[NO 2 ][F 2 ] RATE EQUATIONS BASED ON REACTION MECHANISMS -The rate of any reaction is limited by its slowest step, the rate of which is often nearly the same as the overall reaction rate -Rate determining step (RDS): the slowest elementary step in a reaction mechanism -The RDS involves the highest total energy of all elementary steps in a given mechanism -The rate of a rate determining step is assumed to be equal to the rate of its overall reaction 1. A + B Æ X + M, Slow, large E a 2. M + A Æ Y, Fast, small E a Overall: 2A + B Æ X + Y Step 1 is RDS, so rate of overall reaction = k step 1 [A][B]
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-In a laboratory setting, proposed mechanisms can be disproved by examining rate dependence on reactants in RDS. Mechanisms cannot, however, be proved correct. -Reaction mechanisms involving an equilibrium step · Common case: a fast equilibrium step followed by a slow step Ex: 1. NO 2 (g) NO(g) + O(g) fast, equilibrium, where k 1 is constant for forward rxn and k -1 is constant for reverse reaction 2. O(g) + NO 2 (g) Æ NO(g) + O 2 (g) slow, k 2 is constant Overall: 2NO 2 (g) Æ 2NO(g) + O 2 (g) -Rate equation may be written as: k 2 [O][NO 2 ] -O is an immeasurable intermediate, and should not appear in the rate equation -Since in equilibrium, k 1 [NO 2 ] = k -1 [NO][O], [O] can be written in terms of measurable species [O] = k 1 [NO 2 ] / (k -1 [NO]) -Rate equation can be written as: k 2 k 1 [NO 2 ] / (k -1 [NO]) [NO 2 ] COLLISION THEORY AND ACTIVATION ENERGY -Collision theory states that for any reaction to occur, three conditions must be met: 1. Reacting molecules must collide 2. Collision must have enough energy to break bonds 3. Molecules must collide in the appropriate orientation -Activation energy (E A ): the minimum amount of energy that must be absorbed by a system to cause it to react -The proportion of molecules in a sample that have energy at or above E
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This note was uploaded on 07/25/2008 for the course LBS 172 taught by Professor Laduca during the Spring '08 term at Michigan State University.

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Exam_3_review - EXAM 3 REVIEW LBS 172 REACTION MECHANISMS GENERAL-Step by step process of bond making and breaking by which reactants become

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