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CHM 116: Chapter 14 Reaction Mechanics

by: Alarmel Sira

CHM 116: Chapter 14 Reaction Mechanics CHM 116 Cabirac

Marketplace > Chemistry > CHM 116 Cabirac > CHM 116 Chapter 14 Reaction Mechanics
Alarmel Sira
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These notes cover reaction mechanics and rate laws for elementary reactions
CHM116- general chemistry II
Dr. Cabirac
Class Notes
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This 2 page Class Notes was uploaded by Alarmel Sira on Saturday January 30, 2016. The Class Notes belongs to CHM 116 Cabirac at a university taught by Dr. Cabirac in Spring 2016. Since its upload, it has received 19 views.

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Date Created: 01/30/16
Reaction Mechanics Elementary Reactions:  Elementary Reactions: Individual reactions that makeup an overall reaction  Reaction Mechanism: Sequence of elementary reactions that leads to a product formation  Each reaction had its own activation energy o Every reaction has a ‘hill’ on an energy diagram  Reaction Molecularity: Number of molecules reacting in an elementary step o Unimolecular: 1 reactant molecule o Bimolecular: 2 reactant molecules (could be two of the same molecule) o Termolecular: 3 reactant molecules  Intermediate: Appears in a reaction mechanism, but NOT in the overall reaction o Product that is formed in one reaction, then used up as a reactant in a later reaction Rate Laws:  For elementary reactions ONLY, rate laws can be written using coefficients of reactants as their orders o Ex. For the elementary reaction: A + 2B → C Rate law = k[A][B] 2  Rate law of the slow elementary reaction is the same as the overall rate law if and only if the first step is the slow step o Ex. The rate law for the overall reaction AND the rate-determining (slow) step is Rate law = k[A][B] 2 A + 2B → C slow C + 2D → 2E fast  If the first step is NOT slow, the rate law can be determined by using rate forward rate backwardo find [intermediate] in terms of k ,fk bnd [reactants] o Ex. Rate overallk[C][D] . But because [C] is an intermediate and doesn’t show up in the overall reaction, find [C] in terms of k , k forward backwardnd [reactants]. A + 2B → C fast C + 2D → 2E slow 2 rate forwardk forward][B] = k backward] = rate backward [C] = k forward[B] /k backward So, rate overallk[C][D] = k*(k forward][B] /k backward[D]2 Catalysis:  Homogeneous: Reactants and catalysts are in the same phase (usually liquid) Heterogeneous: Reactants and catalysts are in different phases References Dr. Cabirac’s lectures and notes


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