CHEM 106 1nd Edition Lecture 13Outline of Last Lecture I. Reaction order and Rate constantII. Rate LawIII. Integrated Rate Lawsa. Calculus equationsOutline of Current Lecture II. Rate expression and Rate lawIII. Reaction Half-lifeIV. Rate law from experimental dataCurrent LectureChapter 15: Chemical Kinetics (continued)Rate expression: aA + bB → cC + dDRate= 1a ∆[ A ]∆ t = 1b ∆[B ]∆ t = 1c ∆[C ]∆ t = 1d ∆[ D ]∆ tRate law: mathematical formula that relates how the concentrations of reactants affect the reaction rateZeroth (0) Order First (1) Order Second (2nd) OrderFor A + B → C, generalized rate law is Rate = k [A]m [B]nOverall order is m + nWe determine reaction order by plotting experimental data-For a 1st order reaction, we said ln [A] = ln [A]0 – k t(Integrated 1st order rate law)Plot ln [A] vs t:-if linear, says that reaction is 1st order with regards to [A].Reaction Half-LifeIf the reaction is 1st order: somewhat special situationHalf-life (t12): time it takes for the concentration of the reactant to decrease to one-half its original concentration.T1/2 is related to rate constant, kFor 1st order reaction:ln A¿¿¿[ A ]¿ = - k t at t = t1/2, [A] = [ A ]02ln A¿¿¿[ A ]02¿ = - k t1/2 ln ½ = - k t1/2 t1/2 = ln 2k = 0.693kk = 0.693t12Note: for 1st order reaction, t1/2 & k do not depend on the concentration of the reactantFig. 15.10 pg 719 N2O(g) → N2(g) + ½ O2(g) Moving on to 2nd order reactions:Suppose you plot ln [reactant] vs t & it’s not linear?This tells you its not 1st order. Another possibility is 2nd order.Remember our rate law for 2nd order rate = k [A]2If we integrate with calculus, we get:A¿¿¿1[ A ]=1¿ + k tPlot 1[ A ] vs t linear!Slope = k intercept = A¿¿¿1¿Units for 2nd order rate constant M-1 s -1Half-life for 2nd order reaction:A¿¿¿1[ A ]=1¿ + k t t = t1/2, [A] = [ A ]02A[¿ ¿0]1[ A ]02=1¿ + k t1/2A¿¿¿0¿A¿2¿ = k t1/2t1/2 = 1k [ A0], k = A¿¿t1 /2¿1¿ Half-life for 2nd order reaction does depend on [A]0Probability of reaction ↑ with ↑ [reactants] As [A]0 ↓, t1/2 ↑What about this kind of 2nd order reaction: Rate = k [A] [B]Reaction rate is dependent on both [A] & [B]1st order with regards to each, 2nd order overallHow can we figure out rate law with experimental data?Pseudo 1st orderLet [B]0 >> [A]0, then [B] ≈ [B]0 throughout the entire reactionRate = k [A] [B]0 define k’ = k [B]0Rate = k’ [A]Reaction appears 1st order with regards to [A]Tells us nothing about how the reaction rate depends on [B]!Plot ln [A] vs tSlope = -k’Remember that k’ varies with [B]0Zeroth OrderRate = k [A]0 = kPlot [A] vs t slope = kEx: 15.7 pg 7222ClO → Cl2 + O2Fig. 15.12Given data set,a) What is the rate law?b) What is the rate constant?A + B → CExperiment # [A] (M) [B] (M) Initial Rate (M/s)1 0.100 M 0.100 M 4.0 * 10-52 0.100 M 0.200 M 4.0 * 10-53 0.200 M 0.100 M 16.0 * 10-5a) Rate Law Rate = k [A]m [B]n What are the values of m & n?When [B] changes, initial rate is constant so n = 0 When [A] double, initial rate quadruples so [A]2 or m = 2Rate = k [A]2 [B]0 = k [A]2 What is the rate constant? K = A¿¿¿Rate¿ use data from experiment #1 = 0.100 M¿¿¿4∗1 0−5M /s¿ = 4 * 10-3 M-1 S-1 Can check using data from experiment #3 Rate = k [A]2 = (4 * 10-3 M-1 S-1) (0.200 M)2= 1.6 * 10-4