Slide 1Slide 2Slide 3Reactants & Products over TimeSlide 5Slide 6Slide 7Slide 8Slide 9Practice ProblemSlide 11Slide 12Slide 13Slide 14Rate LawsSlide 16Slide 17Slide 18Slide 19Slide 20Slide 21Slide 22Slide 23Slide 24Slide 25Slide 26Slide 27Slide 28Slide 29Slide 30Slide 31Slide 32Slide 33Slide 34ExampleExample ContinuedSlide 37Slide 38Slide 39Slide 40Slide 41Slide 42Slide 43K + CH3I KI + •CH3Slide 45Slide 46Slide 47Slide 48Slide 49Slide 50Slide 51Slide 52Slide 53Slide 54Slide 55Slide 56Slide 57Slide 58Slide 59Slide 60Slide 61Chemical KineticsChapter 13Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.Chemical KineticsThermodynamics – does a reaction take place?Kinetics – how fast does a reaction proceed?Reaction rate is the change in the concentration of a reactant or a product with time (M/s).A Brate = -[A]trate = [B]t[A] = change in concentration of A over time period t[B] = change in concentration of B over time period tBecause [A] decreases with time, [A] is negative. 13.1A B13.1rate = -[A]trate = [B]ttimeReactants & Products over TimeBr2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g)time393 nmlightDetector[Br2]  Absorption13.1Br2 (aq) + HCOOH (aq) 2Br– (aq) + 2H+ (aq) + CO2 (g)average rate = –[Br2]t= –[Br2]final – [Br2]initialtfinal - tinitialslope oftangentslope oftangentslope oftangentinstantaneous rate = rate for specific instance in time13.1rate  [Br2]rate = k [Br2]k = rate[Br2]13.1= rate constant= 3.50 x 10–3 s–1Reaction Rates and Stoichiometry13.12A BTwo moles of A disappear for each mole of B that is formed.rate = [B]trate = –[A]t12aA + bB cC + dDrate = –[A]t1a= –[B]t1b=[C]t1c=[D]t1dWrite the rate expression for the following reaction:CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)rate = –[CH4]t= –[O2]t12=[H2O]t12=[CO2]t13.1Practice ProblemGiven the following reaction: 8 MnO4 − + 14 H+ + 5 S2O3 2− → 8 Mn 2+ + 7 H2O + 10 SO4 2 −a) Express the general rate of reaction in terms of each of the reactants and productsb) If the rate of appearance of H2O is 0.022 M/s, what is the rate of disappearance of S2O3 2−The Rate Law13.2The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers.aA + bB cC + dDRate = k [A]x[B]yreaction is xth order in Areaction is yth order in Breaction is (x + y)th order overallF2 (g) + 2ClO2 (g) 2FClO2 (g)Determine x and y in the rate law Rate = k [F2]x[ClO2]yDouble [F2] with [ClO2] held constant:The rate doublesTherefore, x = 1Quadruple [ClO2] with [F2] held constant:The rate quadruplesTherefore, y = 1The rate law isRate = k [F2]1[ClO2]113.2F2 (g) + 2ClO2 (g) 2FClO2 (g)rate = k [F2][ClO2]Rate Laws•Rate laws are always determined experimentally.•Reaction order is always defined in terms of reactant (not product) concentrations.•The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.113.2Determine the rate law and calculate the rate constant for the following reaction from the following data:S2O82– (aq) + 3I– (aq) 2SO42– (aq) + I3– (aq)Experiment[S2O82 – ][I – ]Initial Rate (M/s)1 0.08 0.0342.2 x 10–42 0.08 0.0171.1 x 10–43 0.16 0.0172.2 x 10–4rate = k [S2O82–]x[I–]yDouble [I–], rate doubles (experiment 1 & 2)y = 1Double [S2O82–], rate doubles (experiment 2 & 3)x = 1k = rate[S2O82–][I–]=2.2 x 10–4 M/s(0.08 M)(0.034 M) = 0.08/M•s13.2rate = k [S2O82–][I–]Rate LawsExample: Determine the rate law for the following reaction given the data below.H2O2 (aq) + 3 I- (aq) + 2H+ (aq) I3- (aq) + H2O (l)[H2O2] [I-] [H+] InitialExpt # (M) (M) (M) Rate (M /s)1 0.010 0.010 0.00050 1.15 x 10-62 0.020 0.010 0.00050 2.30 x 10-63 0.010 0.020 0.00050 2.30 x 10-64 0.010 0.010 0.00100 1.15 x 10-6Rate Laws•Rate = k[H2O2]m[I-]n[H+]p•Compare Expts. 1 and 2:•Rate 2 = 2.30 x 10-6 M/s = 2.00•Rate 1 = 1.15 x 10-6 M/s•Rate 2 = k (0.020 M)m(0.010 M)n(.00050 M)p = 2.00 •Rate 1 k (0.010 M)m(0.010 M)n(.00050 M)p2.00 =(0.020)m = 2.0m = 2.00 (0.010)m2.0n = 2.00 only if n = 1Rate = k[H2O2]1[I-]1[H+]pRate Laws•Rate = k[H2O2]1[I-]n[H+]p•Compare Expts. 1 and 3:•Rate 3 = 2.30 x 10-6 M/s = 2.00•Rate 1 = 1.15 x 10-6 M/s•Rate 3 = k (0.010 M)1(0.020 M)n(.00050 M)p = 2.00 •Rate 1 k (0.010 M)1(0.010 M)n(.00050 M)p•2.00 = (0.020)n = 2.0n = 2.00 (0.010)n2.0n = 2.00 only if n = 1Rate = k[H2O2]1[I-]1[H+]pRate Laws•Rate = k[H2O2]1[I-]1[H+]p•Compare Expts. 1 and 4:•Rate 4 = 1.15 x 10-6 M/s = 1.00•Rate 1 = 1.15 x 10-6 M/s•Rate 4 = k (0.010 M)1(0.010 M)1(.00100 M)p = 1.00 •Rate 1 k (0.010 M)1(0.010 M)1(.00050 M)p•1.00 = (0.00100)p = 2.0p = 1.00 (0.00050)p2p = 1.00 only if p = 0Rate = k[H2O2]1[I-]1[H+]0Rate = k[H2O2] [I-]Practice Problem•A chemical reaction A + 2 B + C  D + E was studied by the initial rate method with the following results:Rate M s -1[A] M [B] M [C] M1.32x10-32.0x10-21.8x10-21.0x10-25.28x10-34.0x10-21.8x10-21.0x10-21.58x10-24.0x10-25.4x10-21.0x10-25.94x10-43.0x10-21.8x10-22.0x10-3 (a) Determine the experimental rate law. You must explain how you determined this (b) Calculate the numerical value of the rate constant (with units).First-Order Reactions13.3A productrate = -[A]trate = k [A]k = rate[A]= 1/s or s-1M/sM=[A]t= k [A]–[A] is the concentration of A at any time t[A]0 is the concentration of A at time t=0[A] = [A]0exp(–kt)ln[A] = ln[A]0 – kt13.32N2O5 4NO2 (g) + O2 (g)k = 5.714 X 10–4 s–1The reaction 2A B is first order in A with a rate constant of 2.8 x 10–2 s–1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ?ln[A]t = ln[A]0 – ktkt = ln[A]0 – ln[A]t =ln[A]0 – ln[A]k= 66 s[A]0 = 0.88 M[A]t = 0.14 Mln[A]0[A]k=ln0.88 M0.14 M2.8 x 10–2 s–1=13.3Half-Life of First-Order Reactions13.3The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration.t½ = t when [A] = [A]0/2ln[A]0[A]0/2k=t½ln2k=0.693k=What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10–4 s–1?t½ln2k=0.6935.7 x 10–4 s–1= = 1200 s = 20 minutesHow do you know decomposition is first order?units of k (s-1)Half-Life