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KU CHEM 135 - Rate Law

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CHEM 135 1st Edition Lecture 9 Outline of Current Lecture I. Average rate examplesII. Rate LawIII. Reaction OrderCurrent LectureExamples:1. In the following chemical reaction, within first 10 sec, [N2O5] dropped from 1M to 0.868M.Find i) Average rate ii) ∆ O2∆ t iii)∆ N O2∆ t2 N2O5-> 4NO2 + O2 Rate= −12∆ [N 2 O5]∆ t= +14[∆ NO 2]∆t= ∆[O2]∆ti) Average Rate= − 12∆[N 2O 5]∆ t= −12[0.868−1][10−0]=−12(−0.0132)= 0.0066 M/sii) Rate= ∆ O2∆ t= 0.0066 M/siii) Rate=14∆ N O2∆t >> 0.0066=14∆ N O2∆t>> 0.0264 m-12. H2O2 can be used as a disinfectant; it decomposes as:2 H2O2>> 2H2O + O2if the rate of appearance of O2 is 0.0014 Ms-1, what is the rate of disappearance of H2O2?Answer: 0.0028 Ms-1These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.3. In the first 10.0 seconds of the reaction, the concentration of I- dropped from 1.000M to 0.868M. a. Calculate the average rate of this reaction in this time intervalb. Determine the rate of change in the concentration H2O2(aq) + 3 I-(aq) + 2H+(aq) >> I3-(aq) + 2H2O (l)a. Rate= -∆[ H2O2]∆ t=-∆I3−¿∆ t=12∆ H2O∆ t∆ H +¿∆ t=¿∆ I −¿∆ t=−12¿13¿Rate= I −¿¿∆ ¿−13¿=−13(0.868−1)10=4.4 ×10−3M /sb. Rate= H +¿¿∆ ¿−12¿H +¿¿∆ ¿4.4 ×10−32=−12¿¿−8.8× 10−3M / sRate Law and Reaction Order:Rate Law: Rate=K[reactant]nK=constant[Reactant]=concentration of reactant n=reaction order*rate law is only with reactants, NO PRODUCTSReaction order tells the relation between the rate and the concentration of the reactant- n is not the coefficient- n is found experimentally- n cannot be predictedA) When n=1, first order reactionxA>>yB- The rate changes by the same factor as the concentrationRate=K[A]K=1/sEx: [A] Rate (M/s)0.1 30 0.2 600.6 180B) When n=2, second order reactionxA>>yB- The rate changes by the square of the factorRate= K[A]2K=M-1S-1C) When n=0, zero order reaction- Rate remains constant


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