CHEM 122 1st Edition Lecture 15Outline of Last LectureI. Chemical Kineticsa. Rate of reactionsII. Rate law equationOutline of Current Lecture I. Order of the ReactionII. Exponents in the Rate LawIII. Rate Lawsa. Rate Constant Unitsb. Determining the rate lawc. Properties of logarithmsCurrent LectureOrder of the Reaction: sum of the orders with respect to each reactant.- Based on the exponents in the rate law- Ex.) −¿I¿¿+¿H¿¿¿¿k [ H2Se O3]1¿o First order H2SeO3o Third order in I-o Second order in H+o Overall order= 1+3+2=6These 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.Exponents in the Rate law:- Can be fractional, can have a negative sign- Most often they are small whole numbers- Rate Constant Units:K [ A ]x=Lx−1Mx−1s- Determining the Rate Law:1. Measure the initial rate if the reaction at a particular concentration.2. Then change the reactant concentration to see the effect on the initial rate of reaction.- This allows for the order of each reactant to be determined.- Generally the concentrations are doubled.2N2O5 4NO2 + O2Exp #[N2O5]Initial rate (Lmol∗s¿1 1.0x10^-2 4.8x10^-62 2.0x10^-2 9.6x10^-61. Rate (1)= K[N2O5]x(1)Rate (2)= K[N2O5]x(2) divide each equation putting the rate 2 on top2.K[N2O5]x(2)K[N2O5]x(1)= Rate(2)Rate(1)the K’s cancel out giving you:3.[N2O5]x(2)[N2O5]x(1)= Rate(2)Rate(1)use the properties of exponents 4.[N2O5]❑(2)[N2O5]❑(1)¿¿¿Plug in the numbers for the variables5.2.0 x 10−21.0 x 10−2¿¿9.6 x 10−6(¿)¿x=¿¿4.8 x 10−6Crunch the numbers6.2x=27. X=18. This would result on being the first order with respect to the reactant N2O5Properties of Logarithms:1.[conc ]2x[conc ]1x=([conc]2❑[conc]1❑)x=Rate2Rate1Take log base 10 of each side of the equation2.log([conc]2❑[conc]1❑)x=logRate2Rate1Properties of logs allow you to bring the exponent out to the front3.xlog([conc]2❑[conc]1❑)❑=logRate2Rate1Solve for
View Full Document
Unlocking...