CHEM 122 1st Edition Lecture 16Outline of Last LectureI. Order of the ReactionII. Exponents in the Rate LawIII. Rate Lawsa. Rate Constant Unitsb. Determining the rate lawc. Properties of logarithmsOutline of Current Lecture I. Simple First Order Reaction Rate Lawa. Half life of first orderII. Simple Second-Order Reactionsa. Half life of second orderCurrent LectureSimple First Order Reaction Rate Law:Rate=k[A]=−∆ [ A ]∆ tBy using calculus the integrated law is obtained.ln[ A ]t[ A ]o=− kt- Ln: natural logarithm- [A]t: concentration at time t (final)- [A]o: concentration at time 0 (initial)These 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.rearrange this equation to:ln[ A ]t=−kt +ln [ A ]oy = mx + ba plot of ln[A]t vs time gives a straight lineHalf-Life of simple first reactionsHalf Life: the time it takes for the concentration of reactants to decrease to ½ of their initial value.t1 /2=.693khalf- life of simple first reactions only depends on its k value and is independent from the initial concentration.To solve past one half life use the equation:12n[A]on: number of half-lives you need to findSimple second-order reactions: Rate law: rate=k [ A ]2=−∆ [ A]∆ tBy using calculus the integrated rate law is obtained:1[ A ]t=kt+1[ A]oLike the simple first order, if plotted 1/[A]t vs time this will give you a straight line for thegraph.Half-life for a simple second order reaction:t1 /2=1k [ A ]0Half-life for the second order depends on both k and the initial reactant concentration.Finding more than one half-life, each successive half-life is longer than preceding half-life. The time
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