CHEM 135 1st Edition Lecture 11Outline of Last Lecture I. Rate vs. reaction concentrationII. Complex rate law and reaction ordera. Overall reaction orderb. ExamplesIII. Integrated Rate Law Outline of Current Lecture I. Integrated Rate Lawa. Graphsb. Equationsc. ExamplesII. Half lifea. Equationsb. ExamplesCurrent LectureIntegrated Rate LawZero order graph- [A] on y-axis and time on x-axis, slope= -kFirst order graph- ln[A] on y-axis and time on x-axis, slope= -kSecond order graph- 1/[A] on y-axis and time on x-axis, slope= kEquations:Zero order- [A]= -kt + [A]initialThese 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.First order- ln[A]= -kt + ln[A]initialSecond order- 1/[A]= kt + 1/[A]initialExamples:The reaction is first order in cyclopropane and has a measured rate constant of 3.36 x 10-5s-1 at 720K. If the initial cyclopropane concentration is 0.0445 M, what will be the cyclopropane concentration after 235.0 min?Given:k= 3.36 x 10-5s-1[A]0=0.0445 M[A]t=?t= 235min= 14100 secondsln[A]= -kt + ln[A]oln[A]= -(3.36 x 10-5)(14100) + ln(0.0445)ln[A]= -3.585[A]=e-3.585[A]= 0.0277 MHalf LifeTime it takes for the concentration of the reactant to fall to one-half its original valueZero order reaction- t1/2=[Ainitial]/2kFirst order reaction- t1/2=0.693/kSecond order reaction- t1/2= 1/k[A0]Ex:A substance decays with second order kinetics. What is the half-life given a rate constant that is 0.37 M-1s-1 and an initial concentration of 0.75 M?t1/2= 1/k[A0]= 1/2 (0.37 x .75)=3.6sCarbon-14 has a half-life of 5720 years and this is a first order reaction. If a piece of wood has converted 25% of the carbon-14, then how old is it?T1/2= 5720 years75% remains[A]= [0.75 A0]find k  t1/2= 0.693/k5720=0.693/kk=1.21 x 10-4ln[A]/ [A0]= -ktln[0.75]/ [A0]= -1.21 x 10-4t-0.287= (-1.21x10-4)ttime= 2372