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John C. Kotz Paul M. Treichel John Townsend http://academic.cengage.com/kotz Chapter 15 Principles of Reactivity: Chemical Kinetics John C. Kotz • State University of New York, College at Oneonta Important – Read Before Using Slides in Class Instructor: This PowerPoint presentation contains photos and figures from the text, as well as selected animations and videos. For animations and videos to run properly, we recommend that you run this PowerPoint presentation from the PowerLecture disc inserted in your computer. Also, for the mathematical symbols to display properly, you must install the supplied font called “Symb_chm,” supplied as a cross-platform TrueType font in the “Font_for_Lectures” folder in the "Media" folder on this disc. If you prefer to customize the presentation or run it without the PowerLecture disc inserted, the animations and videos will only run properly if you also copy the associated animation and video files for each chapter onto your computer. 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Thank you. 3 Chemical Kinetics Chapter 15 PLAY MOVIE H2O2 decomposition in an insect © 2009 Brooks/Cole - Cengage H2O2 decomposition catalyzed by MnO2 Chemical Kinetics • We can use thermodynamics to tell if a reaction is product- or reactantfavored. • But this gives us no info on HOW FAST reaction goes from reactants to products. • KINETICS — the study of REACTION RATES and their relation to the way the reaction proceeds, i.e., its MECHANISM. • The reaction mechanism is our goal! © 2009 Brooks/Cole - Cengage 4 Reaction Mechanisms The sequence of events at the molecular level that control the speed and outcome of a reaction. Br from biomass burning destroys stratospheric ozone. (See R.J. Cicerone, Science, volume 263, page 1243, 1994.) Step 1: Br + O3 f BrO + O2 Step 2: Cl + O3 f ClO + O2 Step 3: BrO + ClO + light f Br + Cl + O2 NET: 2 O3 f 3 O2 © 2009 Brooks/Cole - Cengage 5 Reaction Rates Section 15.1 • Reaction rate = change in concentration of a reactant or product with time. • Three “types” of rates –initial rate –average rate –instantaneous rate © 2009 Brooks/Cole - Cengage 6 Determining a Reaction Rate Dye Conc PLAY MOVIE See Chemistry Now , Chapter 15 Time © 2009 Brooks/Cole - Cengage Blue dye is oxidized with bleach. Its concentration decreases with time. The rate — the change in dye conc with time — can be determined from the plot. 7 Determining a Reaction Rate See Active Figure 15.2 © 2009 Brooks/Cole - Cengage 8 9 Factors Affecting Rates • Concentrations • and physical state of reactants and products • Temperature • Catalysts © 2009 Brooks/Cole - Cengage Concentrations & Rates Section 15.3 Mg(s) + 2 HCl(aq) f MgCl2(aq) + H2(g) 0.3 M HCl 6 M HCl PLAY MOVIE © 2009 Brooks/Cole - Cengage 10 11 Factors Affecting Rates • Physical state of reactants PLAY MOVIE © 2009 Brooks/Cole - Cengage 12 Factors Affecting Rates Catalysts: catalyzed decomp of H2O2 2 H2O2 f 2 H2O + O2 PLAY MOVIE © 2009 Brooks/Cole - Cengage 13 Catalysts 1. CO2(g) e CO2 (aq) See Page 702 2. CO2 (aq) + H2O(liq) e H2CO3(aq) 3. H2CO3(aq) e H+(aq) + HCO3–(aq) • Adding trace of NaOH uses up H+. Equilibrium shifts to produce more H2CO3. • Enzyme in blood (above) speeds up reactions 1 and 2 © 2009 Brooks/Cole - Cengage 14 Factors Affecting Rates • Temperature Bleach at 54 ˚C Bleach at 22 ˚C PLAY MOVIE © 2009 Brooks/Cole - Cengage Iodine Clock Reaction 1. Iodide is oxidized to iodine H2O2 + 2 I- + 2 H+ f 2 H2O + I2 2. I2 reduced to I- with vitamin C I2 + C6H8O6 f C6H6O6 + 2 H+ + 2 I- When all vitamin C is depleted, the I2 interacts with starch to give a blue complex. © 2009 Brooks/Cole - Cengage 15 Iodine Clock Reaction © 2009 Brooks/Cole - Cengage 16 17 Concentrations and Rates To postulate a reaction mechanism, we study • reaction rate and • its concentration dependence © 2009 Brooks/Cole - Cengage Concentrations and Rates Take reaction where Cl- in cisplatin [Pt(NH3)2Cl3] is replaced by H2O Rate of change of conc of Pt compd Am't of cisplatin reacting (mol/L) = elapsed time (t) © 2009 Brooks/Cole - Cengage 18 Concentrations & Rates Rate of change of conc of Pt compd Am't of cisplatin reacting (mol/L) = elapsed time (t) Rate of reaction is proportional to [Pt(NH3)2Cl2] We express this as a RATE LAW Rate of reaction = k [Pt(NH3)2Cl2] where k = rate constant k is independent of conc. but increases with T © 2009 Brooks/Cole - Cengage 19 Concentrations, Rates, & Rate Laws In general, for a A + b B f x X with a catalyst C Rate = k [A]m[B]n[C]p The exponents m, n, and p • are the reaction order • can be 0, 1, 2 or fractions • must be determined by experiment! © 2009 Brooks/Cole - Cengage 20 Interpreting Rate Laws Rate = k [A]m[B]n[C]p • If m = 1, rxn. is 1st order in A Rate = k [A]1 If [A] doubles, then rate goes up by factor of __ • If m = 2, rxn. is 2nd order in A. Rate = k [A]2 Doubling [A] increases rate by ________ • If m = 0, rxn. is zero order. Rate = k [A]0 If [A] doubles, rate ________ © 2009 Brooks/Cole - Cengage 21 Deriving Rate Laws Derive rate law and k for CH3CHO(g) f CH4(g) + CO(g) from experimental data for rate of disappearance of CH3CHO Expt. [CH3CHO] (mol/L) 1 0.10 0.020 2 0.20 0.081 3 0.30 0.182 4 0.40 0.318 © 2009 Brooks/Cole - Cengage Disappear of CH3CHO (mol/L•sec) 22 Deriving Rate Laws Rate of rxn = k [CH3CHO]2 Here the rate goes up by ______ when initial conc. doubles. Therefore, we say this reaction is _________________ order. Now determine the value of k. Use expt. #3 data— 0.182 mol/L•s = k (0.30 mol/L)2 k = 2.0 (L / mol•s) Using k you can calc. rate at other values of [CH3CHO] at same T. © 2009 Brooks/Cole - Cengage 23 Concentration/Time Relations What is concentration of reactant as function of time? Consider FIRST ORDER REACTIONS The rate law is Ę[A] Rate = = k [A] Ętime © 2009 Brooks/Cole - Cengage 24 Concentration/Time Relations Integrating - (∆ [A] / ∆ time) = k [A], we get ln is natural logarithm [A] ln = - kt [A]0 [A] at time = 0 [A] / [A]0 =fraction remaining after time t has elapsed. Called the integrated first-order rate law. © 2009 Brooks/Cole - Cengage 25 Concentration/Time Relations 26 Sucrose decomposes to simpler sugars Rate of disappearance of sucrose = k [sucrose] If k = 0.21 hr-1 and [sucrose] = 0.010 M How long to drop 90% (to 0.0010 M)? Glucose © 2009 Brooks/Cole - Cengage Concentration/Time Relations Rate of disappear of sucrose = k [sucrose], k = 0.21 hr-1. If initial [sucrose] = 0.010 M, how long to drop 90% or to 0.0010 M? Use the first order integrated rate law 0.0010 -1 ln = (0.21 h )t 0.010 ln (0.100) = - 2.3 = - (0.21 hr-1)(time) time = 11 hours © 2009 Brooks/Cole - Cengage 27 Using the Integrated Rate Law The integrated rate law suggests a way to tell the order based on experiment. 2 N2O5(g) f 4 NO2(g) + O2(g) Time (min) 0 1.0 2.0 5.0 [N2O5]0 (M) 1.00 0.705 0.497 0.173 ln [N2O5]0 0 -0.35 -0.70 -1.75 Rate = k [N2O5] © 2009 Brooks/Cole - Cengage 28 Using the Integrated Rate Law 2 N2O5(g) f 4 NO2(g) + O2(g) Rate = k [N2O5] Data of conc. vs. time plot do not fit straight line. © 2009 Brooks/Cole - Cengage Plot of ln [N2O5] vs. time is a straight line! 29 Using the Integrated Rate Law Plot of ln [N2O5] vs. time is a straight line! Eqn. for straight line: y = mx + b ln [N 2O5] = - kt + ln [N 2O5]o conc at time t rate const = slope conc at time = 0 All 1st order reactions have straight line plot for ln [A] vs. time. (2nd order gives straight line for plot of 1/[A] vs. time) © 2009 Brooks/Cole - Cengage 30 31 Properties of Reactions © 2009 Brooks/Cole - Cengage 32 Half-Life HALF-LIFE is the time it takes for 1/2 a sample is disappear. For 1st order reactions, the concept of HALF-LIFE is especially useful. See Active Figure 15.9 © 2009 Brooks/Cole - Cengage 33 Half-Life • Reaction is 1st order decomposition of H2O2. © 2009 Brooks/Cole - Cengage 34 Half-Life • Reaction after 1 half-life. • 1/2 of the reactant has been consumed and 1/2 remains. © 2009 Brooks/Cole - Cengage 35 Half-Life • After 2 half-lives 1/4 of the reactant remains. © 2009 Brooks/Cole - Cengage 36 Half-Life • A 3 half-lives 1/8 of the reactant remains. © 2009 Brooks/Cole - Cengage 37 Half-Life • After 4 half-lives 1/16 of the reactant remains. © 2009 Brooks/Cole - Cengage Half-Life Sugar is fermented in a 1st order process (using an enzyme as a catalyst). sugar + enzyme f products Rate of disappear of sugar = k[sugar] k = 3.3 x 10-4 sec-1 What is the half-life of this reaction? © 2009 Brooks/Cole - Cengage 38 Half-Life Rate = k[sugar] and k = 3.3 x 10-4 sec-1. What is the halflife of this reaction? Solution [A] / [A]0 = fraction remaining when t = t1/2 then fraction remaining = _________ Therefore, ln (1/2) = - k · t1/2 - 0.693 = - k · t1/2 t1/2 = 0.693 / k So, for sugar, t1/2 = 0.693 / k = 2100 sec = 35 © 2009 Brooks/Cole - Cengage min 39 Half-Life Rate = k[sugar] and k = 3.3 x 10-4 sec-1. Half-life is 35 min. Start with 5.00 g sugar. How much is left after 2 hr and 20 min (140 min)? Solution 2 hr and 20 min = 4 half-lives Half-life Time Elapsed Mass Left 1st 35 min 2.50 g 2nd 70 1.25 g 3rd 105 0.625 g 4th 140 0.313 g © 2009 Brooks/Cole - Cengage 40 Half-Life Radioactive decay is a first order process. Tritium f electron + helium 3H 0 e 3He -1 t1/2 = 12.3 years If you have 1.50 mg of tritium, how much is left after 49.2 years? © 2009 Brooks/Cole - Cengage 41 Half-Life Start with 1.50 mg of tritium, how much is left after 49.2 years? t1/2 = 12.3 years Solution ln [A] / [A]0 = -kt [A] = ? [A]0 = 1.50 mg t = 49.2 y Need k, so we calc k from: k = 0.693 / t1/2 Obtain k = 0.0564 y-1 Now ln [A] / [A]0 = -kt = - (0.0564 y-1)(49.2 y) = - 2.77 Take antilog: [A] / [A]0 = e-2.77 = 0.0627 0.0627 = fraction remaining © 2009 Brooks/Cole - Cengage 42 Half-Life Start with 1.50 mg of tritium, how much is left after 49.2 years? t1/2 = 12.3 years Solution [A] / [A]0 = 0.0627 0.0627 is the fraction remaining! Because [A]0 = 1.50 mg, [A] = 0.094 mg But notice that 49.2 y = 4.00 half-lives 1.50 mg f 0.750 mg after 1 half-life f 0.375 mg after 2 f 0.188 mg after 3 f 0.094 mg after 4 © 2009 Brooks/Cole - Cengage 43 Half-Lives of Radioactive Elements Rate of decay of radioactive isotopes given in terms of 1/2-life. 238U f 234Th + He 4.5 x 109 y 14C f 14N + beta 5730 y 131I f 131Xe + beta 8.05 d Element 106 - seaborgium - 263Sg 0.9 s © 2009 Brooks/Cole - Cengage 44 MECHANISMS A Microscopic View of Reactions Mechanism: how reactants are converted to products at the molecular level. RATE LAW f MECHANISM experiment f theory PLAY MOVIE © 2009 Brooks/Cole - Cengage 45 Activation Energy Molecules need a minimum amount of energy to react. Visualized as an energy barrier - activation energy, Ea. PLAY MOVIE Reaction coordinate diagram © 2009 Brooks/Cole - Cengage 46 MECHANISMS & Activation Energy Conversion of cis to trans-2-butene requires twisting around the C=C bond. Rate = k [trans-2-butene] © 2009 Brooks/Cole - Cengage 47 MECHANISMS Cis Transition state 48 Trans Activation energy barrier © 2009 Brooks/Cole - Cengage MECHANISMS 49 Energy involved in conversion of trans to cis butene energy Activated Complex +262 kJ cis -266 kJ 4 kJ/mol trans © 2009 Brooks/Cole - Cengage Mechanisms • Reaction passes thru a TRANSITION STATE where there is an activated complex that has sufficient energy to become a product. ACTIVATION ENERGY, Ea = energy req’d to form activated complex. Here Ea = 262 kJ/mol © 2009 Brooks/Cole - Cengage 50 MECHANISMS Also note that trans-butene is MORE STABLE than cis-butene by about 4 kJ/mol. Therefore, cis f trans is EXOTHERMIC This is the connection between thermodynamics and kinetics. © 2009 Brooks/Cole - Cengage 51 Effect of Temperature • Reactions generally occur slower at lower T. In ice at 0 oC Room temperature PLAY MOVIE Iodine clock reaction. See Chemistry Now, Ch. 15 H2O2 + 2 I- + 2 H+ f 2 H2O + I2 PLAY MOVIE © 2009 Brooks/Cole - Cengage 52 Activation Energy and Temperature Reactions are faster at higher T because a larger fraction of reactant molecules have enough energy to convert to product molecules. In general, differences in activation energy cause reactions to vary from fast to slow. © 2009 Brooks/Cole - Cengage 53 Mechanisms 1. Why is trans-butene e cis-butene reaction observed to be 1st order? As [trans] doubles, number of molecules with enough E also doubles. 2. Why is the trans e cis reaction faster at higher temperature? Fraction of molecules with sufficient activation energy increases with T. © 2009 Brooks/Cole - Cengage 54 More About Activation Energy Arrhenius equation — Rate constant k Ae Frequency factor Temp (K) -E a / RT Activation 8.31 x 10-3 kJ/K•mol energy Frequency factor related to frequency of collisions with correct geometry. Ea 1 ln k = - ( )( ) + ln A R T © 2009 Brooks/Cole - Cengage 55 Plot ln k vs. 1/T f straight line. slope = -Ea/R More on Mechanisms A bimolecular reaction Reaction of trans-butene f cis-butene is UNIMOLECULAR - only one reactant is involved. BIMOLECULAR — two different molecules must collide f products PLAY MOVIE Exo- or endothermic? © 2009 Brooks/Cole - Cengage 56 Collision Theory Reactions require (a) activation energy and (b) correct geometry. O3(g) + NO(g) f O2(g) + NO2(g) 1. Activation energy PLAY MOVIE © 2009 Brooks/Cole - Cengage 2. Activation energy and geometry PLAY MOVIE 57 Mechanisms O3 + NO reaction occurs in a single ELEMENTARY step. Most others involve a sequence of elementary steps. Adding elementary steps gives NET reaction. PLAY MOVIE © 2009 Brooks/Cole - Cengage 58 Mechanisms Most rxns. involve a sequence of elementary steps. 2 I- + H2O2 + 2 H+ f I2 + 2 H2O Rate = k [I-] [H2O2] NOTE 1. Rate law comes from experiment 2. Order and stoichiometric coefficients not necessarily the same! 3. Rate law reflects all chemistry down to and including the slowest step in multistep reaction. © 2009 Brooks/Cole - Cengage 59 Mechanisms Most rxns. involve a sequence of elementary steps. 2 I- + H 2 O 2 + 2 H + f I2 + 2 H 2 O Rate = k [I-] [H2O2] Proposed Mechanism Step 1 — slow HOOH + I- f HOI + OH- Step 2 — fast HOI + I- f I2 + OH- Step 3 — fast 2 OH- + 2 H+ f 2 H2O Rate of the reaction controlled by slow step — RATE DETERMINING STEP, rds. Rate can be no faster than rds! © 2009 Brooks/Cole - Cengage 60 Mechanisms 2 I- + H2O2 + 2 H+ f I2 + 2 H2O Rate = k [I-] [H2O2] Step 1 — slow HOOH + I- f HOI + OH-Step 2 — fast HOI + I- f I2 + OHStep 3 — fast 2 OH- + 2 H+ f 2 H2O Elementary Step 1 is bimolecular and involves I- and HOOH. Therefore, this predicts the rate law should be Rate [I-] [H2O2] — as observed!! The species HOI and OH- are reaction intermediates. © 2009 Brooks/Cole - Cengage 61 Rate Laws and Mechanisms 62 NO2 + CO reaction: Rate = k[NO2]2 PLAY MOVIE Two possible mechanisms Two steps: step 1 PLAY MOVIE Single step PLAY MOVIE Two steps: step 2 © 2009 Brooks/Cole - Cengage Ozone Decomposition over Antarctica 2 O3 (g) f 3 O2 (g) © 2009 Brooks/Cole - Cengage 63 64 Ozone Decomposition Mechanism 2 O3 (g) f 3 O2 (g) Proposed mechanism Step 1: fast, equilibrium O3 (g) e O2 (g) + O (g) Step 2: slow O3 (g) + O (g) f 2 O2 (g) Rate = k © 2009 Brooks/Cole - Cengage [O3 ]2 [O2 ] CATALYSIS 65 Catalysts speed up reactions by altering the mechanism to lower the activation energy barrier. Dr. James Cusumano, Catalytica Inc. See Chemistry Now, Ch 15 PLAY MOVIE What is a catalyst? PLAY MOVIE Catalysts and the environment PLAY MOVIE © 2009 Brooks/Cole - Cengage Catalysts and society CATALYSIS In auto exhaust systems — Pt, NiO 2 CO + O2 f 2 CO2 2 NO f N2 + O2 PLAY MOVIE © 2009 Brooks/Cole - Cengage 66 CATALYSIS 2. Polymers: H2C=CH2 ---> polyethylene 3. Acetic acid: CH3OH + CO f CH3CO2H 4. Enzymes — biological catalysts © 2009 Brooks/Cole - Cengage 67 CATALYSIS Catalysis and activation energy MnO2 catalyzes decomposition of H2O2 2 H2O2 f 2 H2O + O2 PLAY MOVIE Uncatalyzed reaction Catalyzed reaction © 2009 Brooks/Cole - Cengage 68 Iodine-Catalyzed Isomerization of cis-2-Butene See Figure 15.15 © 2009 Brooks/Cole - Cengage 69 Iodine-Catalyzed Isomerization of cis-2-Butene © 2009 Brooks/Cole - Cengage 70