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Week 6 Notes

by: Kylie McLaughlin

Week 6 Notes Chem 211

Kylie McLaughlin
General Chemistry II
Dr. James R Horn

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Here's week 6 notes. It covers chapter 16 and the beginning of chapter 17.
General Chemistry II
Dr. James R Horn
Class Notes
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This 3 page Class Notes was uploaded by Kylie McLaughlin on Wednesday October 7, 2015. The Class Notes belongs to Chem 211 at Northern Illinois University taught by Dr. James R Horn in Fall 2015. Since its upload, it has received 27 views. For similar materials see General Chemistry II in Chemistry at Northern Illinois University.


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Date Created: 10/07/15
Week 6 Chapter 16 cont Rate k Aquotm Bquotn k AeAEaRT reates constant in rate law to temperature Ea activation energy minimum energy molecules must have in order to react The Effect of Temperature on Reaction Rate The Arrhenius Equation k AeAEaRT lnk lnA EaRT nk2k1 EaR X 1T2 1T1 k kinetic constant at T Ea is the activation energy R the energy gas constant T kevin temperature A the collision frequency factor sope is equal to the activation energyRgas constant the dependence of possible collisions on the product of reactant concentrations The Importance of Molecular Orientation to have an Effective Collision NO N03 gt 2 N02 A is the frequency factor for A pZ Z the collision frequency p orientation probability factor the nature of the transition state in the reaction between CH3Br and OH CH3Br OH gt CH3OH Br transition state or activated complex Reaction Mechanisms Elementary Step Molecularity Rate Law A gt product unimolecular rate kA 2A gt product bimolecular rate kAquot2 A B gt product bimolecular rate kAB 2A B gt product termolecular rate kAquot2 B The RateDetermining Step of a Reaction Mechanism the overall rate of a reaction is related to the rate of the slowest or rate determining step Correlatino the Mechanism with the Rate Law the elementary steps must add up to the overall equation the elementary steps must be physically reasonable the mechanism must correlate with the observed rate law Catalysts speeds up reaction lowers activation energy things that can take years can happen in seconds each has its own speci c ways of functioning owering the activation energyEa increases the rate constant k and thereby increases the rate of the reaction increases the rate of forward and reverse reactions catayst reaction yields the products more quickly but does not yield more product than the uncatalyzed reaction a catalyst lowers Ea by providing a different mechanism for the reaction through a new lower energy pathway intermediates disappear are not consumed in the reaction there at the beginning and there when it is all over Chapter 17 Equilibrium the Extent of Chemical Reactions Kinetics applies to the speed of a reaction the concentration of product that appears or of reactant that disappears per unit time Equilibrium applies to the extent of a reaction the concentration of product that has appeared after an unlimited time or once no further change occurs at equilibrium rate forward rate reverse a system at equilibrium is dynamic on the molecular level no further net change is observed because changes in one direction are balanced by changes in the other if rate forward rate reverse then k forward reactantsquotm k reverse productsquotn K forward k reverse productsquotn reactantsquotm K the equilibrium constant aso called law of mass action the values of m and n are those of the coef cients in the balanced chemical equation Since this is equilibrium rate of the forward and reverse reactions are equal not the concentrations of reactants and products 0 the Reaction Quotient at any time t the system can be sampled to determine the amounts of reactants calculated in the same manner as K tells us whether the system has come to equilibrium QK or whether the reaction has to proceed further from reactants to products QltK or in the reverse direction from products to reactants QgtK use the molar concentrations of the substances in the reaction symbolized by using for a general reaction aA bB cC dD where a b c and d are the numerical coef cients in the balanced equation Q and K can be calculated as QC CquotC X Dquotd Aquota X Bquotb cCdDaAbB Q 1Qc naA bB cC dD Qc39 Qcquotn for a sequence at equilibrium K overall K1 K2 K3 Expressing Equilibria with Pressure Terms Kc and Kp PV nRT P nV X RT PRT nN m P or m so for 2NOg 02g 2N02g 09 PN02quot2 PN0A2 X 902 QC N02quot2 NOquot2 X 02 Kp Kc RTquotAngas


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