Ch15_Kinetics_1

Ch15_Kinetics_1 - Chemical Kinetics Chemical Kinetics 1...

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Unformatted text preview: Chemical Kinetics Chemical Kinetics 1 Chapter 15 Chapter 15 Chemical Kinetics Chemical Kinetics 2 • But this gives us no info on HOW FAST • But this gives us no info on HOW reaction goes from reactants to products. reaction goes from reactants to products. Br from biomass burning destroys Br from biomass burning stratospheric ozone. (See R.J. Cicerone, stratospheric ozone. (See R.J. Cicerone, Science,, volume 263, page 1243, 1994.) Science volume 263, page 1243, 1994.) Step 1: Br + O3 ---> BrO + O2 Step 1: Br + O3 ---> BrO 2 • KINETICS — the study of REACTION — the study of REACTION RATES and their relation to the way the RATES and reaction proceeds, i.e., its MECHANISM. reaction proceeds, i.e., its MECHANISM MECHANISM. Reaction Rates Reaction Rates NET: NET: 5 Determining a Reaction Rate Determining a Reaction Rate Secttion 15..1 Sec ion 15 1 Screen 15.2 Dye Conc Dye Conc • Reaction rate = change in concentration of a reactant or product with time. • Three “types” of rates –initial rate –average rate –instantaneous rate Step 2: Step 2: Step 3: Step 3: • The reaction mechanism is our goal! • The reaction mechanism is our goal! 4 Blue dye is oxidized Blue dye is oxidized with bleach. with bleach. Its concentration Its concentration decreases with time. decreases with time. The rate — the The rate — the change in dye conc change in dye conc with time — can be with time — can be determined from the determined from the plot. plot. Time Time Page 1 3 The sequence of events at the molecular level that control the speed and outcome of a reaction. • We can use thermodynamics to tell if • We can use thermodynamics to tell if a reaction is product or reactant favored. a reaction is product or reactant favored. An automotive catalytic muffler Reaction Mechanisms Reaction Mechanisms Cl + O3 ---> ClO + O2 Cl + O3 ---> ClO 2 BrO + ClO + llight ---> Br + Cl + O2 BrO ClO ight Cl 2 2 O3 ---> 3 O2 2 O3 ---> 3 O2 Factors Affecting Rates Sectiion 15.2 Sect on 15.2 • Concentrations • and physical state of reactants and products (Screens reactants (Screens 15.3-15.4) • Temperature (Screen 15.11) (Screen • Catalysts (Screen 15.14) (Screen 6 Factors Affecting Rates 7 Factors Affecting Rates Factors Affecting Rates Sectiion 15.2 Sect on 15.2 • Concentrations 8 Rate with 0.3 M HCl 9 Factors Affecting Rates Factors Affecting Rates • Physical state of reactants Catalysts: catalyzed decomp of H2O2 2 H2O2 --> 2 H2O + O2 Rate with 6.0 M HCl 10 Factors Affecting Rates Factors Affecting Rates • Temperature 11 Concentrations and Rates To postulate a reaction mechanism, we study • reaction rate and •reaction and • its concentration its •its concentration dependence Concentrations and Rates Take reaction Take reaction where Cl-- iin where Cl n cisplatin cisplatin [Pt(NH3)2Cl3] [Pt(NH3)2Cl3] is replaced is replaced by H2O by H2O Rate of change of conc of Pt compd = Page 2 Cisplatin Am't of cisplatin reacting (mol/L) elapsed time (t) 12 Concentrations and Rates Cisplatin 13 Cisplatin Rate of change of conc of Pt compd = 14 Am't of cisplatin reacting (mol/L) elapsed time (t) m n p Rate = k [A]m[B]n[C]p • If m = 2, rxn. is 2nd order in A. • If m = 2, rxn. Rate = k [A]2 Rate = k [A]2 The exponents m, n, and p The exponents m, n, and p • are the reaction order are reaction •are Rate of reaction = k [Pt(NH 3)2Cl2] Rate of reaction = k [Pt(NH 3)2Cl2] Doubling [A] increases rate by ________ Doubling [A] increases rate by ________ • If m = 0, rxn. is zero order. • If m = 0, rxn. Rate = k [A]0 Rate = k [A]0 • can be 0, 1, 2 or fractions •can fractions where k = rate constant where rate • must be determined by experiment! •must experiment! k is independent of conc. but increases with T k is independent of conc. Deriving Rate Laws If [A] doubles, rate ________ If [A] doubles, rate ________ 16 Cisplatin Deriving Rate Laws Derive rate law and k for CH3CHO(g) --> CH4(g) + CO(g) from experimental data for rate of disappearance of CH3CHO Rate of rxn = k [CH3CHO]2 Rate of rxn = k [CH3CHO]2 Expt. Expt. [CH3CHO] [CH 3CHO] (mol/L) (mol/L) Now determine the value of k. Use expt. #3 data— Now determine the value of k. Use expt. #3 data— 1 1 2 2 0.10 0.10 0.20 0.20 0.020 0.020 0.081 0.081 3 3 4 4 0.30 0.30 0.40 0.40 0.182 0.182 0.318 0.318 Disappear of CH3CHO Disappear of CH3CHO (mol/L•sec) (mol/L•sec) 15 Rate = k [A]1 Rate = k [A]1 If [A] doubles, then rate goes up by factor of __ If [A] doubles, then rate goes up by factor of __ a A + b B --> x X with a catalyst C with a catalyst C We express this as a RATE LAW We express this as a RATE Interpreting Rate Laws Cisplatin Rate = k [A]m[B]n[C]p Rate = k [A]m[B]n[C]p • If m = 1, rxn. is 1st order in A • If m = 1, rxn. In general, for In general, for Rate of reaction is proportional to [Pt(NH 3)2Cl2] Rate of reaction is proportional to [Pt(NH 3)2Cl2] Cisplatin Concentrations, Rates, & Rate Laws Here the rate goes up by ______ when initial conc. Here the rate goes up by ______ when initial conc. doubles. Therefore, we say this reaction is doubles. Therefore, we say this reaction is _________________ order. _________________ order. 0.182 mol/L•s = k (0.30 mol/L)2 0.182 mol/L•s = k (0.30 mol/L)2 k = 2.0 (L / mol•s) k = 2.0 (L / mol•s) Using k you can calc. rate at other values of [CH3CHO] Using k you can calc. rate at other values of [CH3CHO] at same T. at same T. Page 3 17 Cisplatin Concentration/Time Relations Need to know what conc. of reactant is as Need to know what conc. function of time. Consider FIRST ORDER function of time. Consider FIRST ORDER REACTIONS REACTIONS The rate law is The rate law is [A] = k [ A] time 18 Cisplatin Concentration/Time Relations Integrating - ( [A] // Integrating - ( [A] ln natural logarithm 19 Sucrose decomposes Sucrose decomposes to simpler sugars to simpler sugars Rate of disappearance Rate of disappearance of sucrose of sucrose = k [sucrose] = k [sucrose] k = 0.21 hr-1 k = 0.21 hr-1 time) = k [A], we get time) = k [A], we get [A] = - k t [A]o [A] // [A]00 =fraction remaining after time tt [A] [A] =fraction remaining after time has elapsed. has elapsed. 21 Rate of disappear of sucrose = k [sucrose], k = 0.21 hr-1. If If initial [sucrose] = 0.010 M, how long to drop 90% or to 0.0010 M? Sucrose Use the first order integrated rate law ln Initial [sucrose] = Initial [sucrose] = 0.010 M 0.010 M How long to drop 90% How long to drop 90% (to 0.0010 M)? (to 0.0010 M)? [A] at time = 0 Concentration/Time Relations 20 Concentration/Time Relations 0.0010 M 0.010 M = - (0.21 hr-1) t ln (0.100) = - 2.3 = - (0.21 hr -1) • time time = 11 hours Called the integrated first-order rate law . Called integrated Using the Integrated Rate Law The integrated rate law suggests a way to tell the order based on experiment. 2 N2O5(g) ---> 4 NO 2(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 ln 0 -0.35 -0.70 -1.75 Rate = k [N 2O5] 22 23 Using the Integrated Rate Law Using the Integrated Rate Law 2 N2O5(g) ---> 4 NO 2(g) + O2(g) Rate = k [N 2O5] Plot of lln [[N22O55] vs. time Plot of n N O ] vs. time is a straight line! is a straight line! Eqn. for straight line: Eqn. for straight line: y = ax + b y = ax + b ax l n [N 2 O 5 ] v s . t i m e 0 [N 2 O 5 ] v s . t i m e l n [N 2 O 5 ] v s . t i m e 1 0 ln [N 2O5] = - kt + ln [N2O5]o -2 0 0 -2 0 ti m e 5 Data of conc. vs. time plot do not fit straight line. 0 ti m e 5 Plot of ln [N2 O5] vs. time is a straight line! Page 4 ti m e 5 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) 24 25 26 Half-Life Properties of Reactions Half-Life Sectiion 15..4 & Screen 15..8 Sect on 15 4 & Screen 15 8 Screen 15.7 27 HALF-LIFE is the time it takes for 1/2 a sample is disappear. For 1st order reactions, the concept of HALFLIFE is especially useful. • Reaction is 1st order decomposition of H2O2. 28 29 30 Half-Life Half-Life Half-Life • Reaction after 654 min, 1 half-life. • 1/2 of the reactant remains. • Reaction after 1306 min, or 2 half-lives. • 1/4 of the reactant remains. • Reaction after 3 half-lives, or 1962 min. • 1/8 of the reactant remains. Page 5 31 Half-Life 32 Section 15.4 & Screen 15.8 Half-Life Half-Life Sugar is fermented in a 1st order process (using an enzyme as a catalyst). sugar + enzyme --> products sugar Rate of disappear of sugar = k[sugar] k = 3.3 x 10 -4 sec-1 What is the half-life of this reaction? • Reaction after 4 half-lives, or 2616 min. • 1/16 of the reactant remains. Rate = k[sugar] and k = 3.3 x 10-4 sec-1. What is the halflife of this reaction? Solution [A] / [A]0 = 1/2 when t = t 1/2 Therefore, ln (1/2) = - k • t1/2 - 0.693 = - k • t 1/2 t1/2 So, for sugar, = 0.693 / k t1/2 = 0.693 / k = 2100 sec = Half-Life 34 Half-Life Section 15.4 & Screen 15.8 Section 15.4 & Screen 15.8 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? Solution 2 hr and 20 min = 4 half-lives hr half-lives Half-life Half-life Time Elapsed Mass Left Time 1st 35 min 2.50 g 2nd 70 1.25 g 3rd 105 0.625 g 4th 140 0.313 g Radioactive decay is a first order process. Tritium ---> electron + helium Tritium 3H 0e 3He -1 If you have 1.50 mg of tritium, how much is left after 49.2 years? t 1/2 = 12.3 years Page 6 33 Half-Life Section 15.4 & Screen 15.8 35 35 min 35 Half-Life Section 15.4 & Screen 15.8 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 mg Need k, so we calc k from: k = 0.693 / t 1/2 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 0.0627 fraction 36 Half-Life Section 15.4 & Screen 15.8 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 ! 0.0627 fraction Because [A] 0 = 1.50 mg, [A] = 0.094 mg But notice that 49.2 y = 4.00 half-lives 1.50 mg ---> 0.750 mg after 1 half-life ---> 0.375 mg after 2 ---> 0.188 mg after 3 ---> 0.094 mg after 4 37 38 Half-Lives of Radioactive Elements Rate of decay of radioactive isotopes given in terms of 1/2-life. 238U --> 234Th + He 4.5 x 10 9 y --> 234 14C --> 14N + beta 5730 y --> 14 131I --> 131Xe + beta 8.05 d --> 131 Element 106 - seaborgium 263Sg Sg Page 7 0.9 s 0.9 39 • Darleane Hoffman of UC-Berkeley studies the newest elements. She is Director of Seaborg Institute. • Rec’d ACS Award in Nuclear Chem, the Garvan Medal, and (in 2000) the ACS highest award, the Priestley Medal. ...
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