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Unformatted text preview: Chemical Kinetics Chemical Kinetics
Chapter 15 Chapter 15 1 Chemical Kinetics Chemical Kinetics
• 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. • But this gives us no info on HOW FAST • But this gives us no info on reaction goes from reactants to products. reaction goes from reactants to products. 2 Reaction Mechanisms Reaction Mechanisms
The sequence of events at the molecular level that control the speed and outcome of a reaction.
Br from biomass burning destroys Br from 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 > 2 Step 2: Step 2: Step 3: Step 3: NET: NET: Cl + O3 > ClO + O2 Cl + O3 > ClO 2 BrO + ClO + light > Br + Cl + O2 BrO 2 2 O3 > 3 O2 2 O3 > 3 O2 3 • KINETICS — the study of REACTION — the study of RATES and their relation to the way the RATES reaction proceeds, i.e., its MECHANISM. reaction proceeds, i.e., its MECHANISM. • The reaction mechanism is our goal! • The reaction mechanism is our goal!
An automotive catalytic muffler Reaction Rates Reaction Rates
Section 15.1 Section 15.1 4 5 Determining a Reaction Rate Determining a Reaction Rate
Screen 15.2 Factors Affecting Rates
Section 15.2 Section 15.2 6 • Reaction rate = change in concentration of a reactant or product with time.
• Three “types” of rates –initial rate –average rate –instantaneous rate
Dye Conc Dye Conc 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. • Concentrations • and physical state of reactants and products (Screens (Screens
15.315.4) • Temperature (Screen 15.11) (Screen • Catalysts (Screen 15.14) (Screen Time Time Page 1 Factors Affecting Rates
Section 15.2 Section 15.2 7 8 9 Factors Affecting Rates Factors Affecting Rates
• Physical state of reactants Factors Affecting Rates Factors Affecting Rates
Catalysts: catalyzed decomp of H2O2 2 H2O2 > 2 H2O + O2 > • Concentrations Rate with 0.3 M HCl Rate with 6.0 M HCl 10 11 Factors Affecting Rates Factors Affecting Rates
• Temperature Concentrations and Rates
To postulate a reaction mechanism, we study Concentrations and Rates
Take reaction Take reaction where Cl in in where cisplatin cisplatin [Pt(NH3)2Cl3] [Pt(NH3)2Cl3] is replaced is replaced by H2O by H2O
Cisplatin 12 • reaction rate and •reaction • its concentration •its concentration dependence Rate of change of conc of Pt compd = Am't of cisplatin reacting (mol/L) elapsed time (t) Page 2 Cisplatin Concentrations and Rates
Rate of change of conc of Pt compd = Am't of cisplatin reacting (mol/L) elapsed time (t) 13
Cisplatin Concentrations, Rates, & Rate Laws 14 Cisplatin Interpreting Rate Laws 15 In general, for In general, for 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. 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 __ • If m = 2, rxn. is 2nd order in A. • If m = 2, rxn. Rate = k [A]2 Rate = k [A]2 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 If [A] doubles, rate ________ If [A] doubles, rate ________ a A + b B > x X with a catalyst C with a catalyst C
m n p Rate = k [A]m[B]n[C]p Rate of reaction is proportional to [Pt(NH 3)2Cl2] Rate of reaction is proportional to [Pt(NH 3)2Cl2] We express this as a RATE LAW We express this as a RATE Rate of reaction = k [Pt(NH 3)2Cl2] Rate of reaction = k [Pt(NH 3)2Cl2] where k = rate constant rate where k is independent of conc. but increases with T k is independent of conc. The exponents m, n, and p The exponents m, n, and p • are the reaction order •are • can be 0, 1, 2 or fractions •can fractions • must be determined by experiment! •must experiment! Cisplatin Deriving Rate Laws 16 Cisplatin Deriving Rate Laws 17 Cisplatin Concentration/Time Relations 18 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 Here the rate goes up by ______ when initial conc. Here the rate goes up by ______ when initial conc. conc. doubles. Therefore, we say this reaction is doubles. Therefore, we say this reaction is _________________ order. _________________ order. Now determine the value of k. Use expt. #3 data— Now determine the value of k. Use expt. #3 data— expt. 0.182 mol/L•s = k (0.30 mol/L)2 0.182 mol/L•s = k (0.30 mol/L) 2 mol/L•s mol/L) Expt. Expt. Expt. 1 1 2 2 3 3 4 4 [CH3CHO] [CH3CHO] (mol/L) mol/L) (mol/L) 0.10 0.10 0.20 0.20 0.30 0.30 0.40 0.40 Disappear of CH3CHO Disappear of CH3CHO (mol/L•sec) mol/L•sec) (mol/L•sec) 0.020 0.020 0.081 0.081 0.182 0.182 0.318 0.318 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 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] calc. at same T. at same T. [A] = k [ A] time Page 3 Cisplatin Concentration/Time Relations
Integrating  ( [A] // Integrating  ( [A] time) = k [A], we get time) = k [A], we get 19 Concentration/Time Relations
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 hr1 k = 0.21 hr1 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)? 20 Concentration/Time Relations 21 Sucrose Rate of disappear of sucrose = k [sucrose], k = 0.21 hr1. If If initial [sucrose] = 0.010 M, how long to drop 90% or to 0.0010 M? Use the first order integrated rate law ln natural logarithm [A] =  k t [A]o
[A] at time = 0 ln 0.0010 M 0.010 M =  (0.21 hr1) t ln (0.100) =  2.3 =  (0.21 hr 1) • time [A] // [A]00 =fraction remaining after time tt [A] [A] =fraction remaining after time =fraction has elapsed. has elapsed. Called the integrated firstorder rate law . time = 11 hours 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) (M) 1.00 0.705 0.497 0.173 ln [N2O5]0 0 0.35 0.70 1.75 22 Using the Integrated Rate Law
2 N2O5(g) > 4 NO 2(g) + O2(g) Rate = k [N 2O5] 23 Using the Integrated Rate Law
0 24 l n [N2 O5 ] v s . t i m e 1 [N2 O5 ] v s . t i m e 0 l n [N2 O5 ] v s . t i m e Plot of ln [N22O55] vs. time Plot of ln [N O ] vs. time vs. is a straight line! is a straight line! Eqn. for straight line: Eqn. for straight line: Eqn. y = ax + b y = ax + b
ln [N 2O5] =  kt + ln [N2O5]o 2 0
0 0 ti m e 5 2 0 ti m e 5 ti m e 5 conc at time t rate const = slope conc at time = 0 Rate = k [N 2O5] Data of conc. vs. conc. vs. time plot do not fit straight line. Plot of ln [N2 O5] vs. vs. time is a straight line! All 1st order reactions have straight line plot for ln [A] vs. time. vs. (2nd order gives straight line for plot of 1/[A] vs. time) vs. Page 4 25 26 27 Properties of Reactions
Screen 15.7 Section 15.4 & Screen 15.8 Section 15.4 & Screen 15.8 HalfLife HalfLife
• Reaction is 1st order decomposition of H2O2. HALFLIFE is the time is it takes for 1/2 a sample is disappear. For 1st order reactions, the concept of HALFLIFE is especially useful. 28 29 30 HalfLife
• Reaction after 654 min, 1 halflife. • 1/2 of the reactant remains. HalfLife
• Reaction after 1306 min, or 2 halflives. • 1/4 of the reactant remains. HalfLife
• Reaction after 3 halflives, or 1962 min. • 1/8 of the reactant remains. Page 5 31 HalfLife HalfLife
• Reaction after 4 halflives, or 2616 min. • 1/16 of the reactant remains. Section 15.4 & Screen 15.8 HalfLife 32 Section 15.4 & Screen 15.8 HalfLife 33 Sugar is fermented in a 1st order process (using Sugar an enzyme as a catalyst). sugar + enzyme > products Rate of disappear of sugar = k[sugar] k = 3.3 x 10 4 sec1 sec What is the halflife of this reaction? Rate = k[sugar] and k = 3.3 x 104 sec1. What is the halfsec life of this reaction? Solution [A] / [A]0 = 1/2 when t = t 1/2 1/2 Therefore, ln (1/2) =  k • t1/2  0.693 =  k • t 1/2 t1/2 So, for sugar, = 0.693 / k 0.693
35 min 35 t1/2 = 0.693 / k = 2100 sec = 0.693 Section 15.4 & Screen 15.8 HalfLife 34 Section 15.4 & Screen 15.8 HalfLife 35 Section 15.4 & Screen 15.8 HalfLife 36 Rate = k[sugar] and k = 3.3 x 10 4 sec1. Halflife sec 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 halflives Halflife 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 Radioactive decay is a first order process. Tritium > electron + helium 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 12.3 Start with 1.50 mg of tritium, how much is left after 49.2 years? t1/2 = 12.3 years 12.3 Solution ln [A] / [A]0 = kt [A] = ? [A]0 = 1.50 mg t = 49.2 mg 1.50 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 y1) • (49.2 y) =  2.77 Take antilog: [A] / [A]0 = e2.77 = 0.0627 antilog: 0.0627 0.0627 = fraction remaining Page 6 Section 15.4 & Screen 15.8 HalfLife 37 38 39 HalfLives of Radioactive Elements
Rate of decay of radioactive isotopes given in terms of 1/2life. 238U > 234Th + He 234 4.5 x 10 9 y 14C > 14N + beta 14 5730 y 131I 131 > 131Xe + beta 8.05 d Element 106  seaborgium 263Sg 0.9 s Start with 1.50 mg of tritium, how much is left after 49.2 years? t1/2 = 12.3 years 12.3 Solution [A] / [A]0 = 0.0627 0.0627 0.0627 is the fraction remaining! remaining! Because [A] 0 = 1.50 mg, [A] = 0.094 mg 1.50 But notice that 49.2 y = 4.00 halflives 1.50 mg > 0.750 mg after 1 halflife > 0.375 mg after 2 > 0.188 mg after 3 > 0.094 mg after 4 • Darleane Hoffman of UCBerkeley 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. Page 7 ...
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This note was uploaded on 01/11/2011 for the course ENGINEERIN MAE 107 taught by Professor Pozikrizdis during the Fall '08 term at San Diego.
 Fall '08
 POZIKRIZDIS
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