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1 Chemical Kinetics Chapter 15 An automotive catalytic muffler.

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Presentation on theme: "1 Chemical Kinetics Chapter 15 An automotive catalytic muffler."— Presentation transcript:

1 1 Chemical Kinetics Chapter 15 An automotive catalytic muffler.

2 2 We can use thermodynamics to tell if a reaction is product or reactant favored.We can use thermodynamics to tell if a reaction is product or reactant favored. But this gives us no info on HOW FAST reaction goes from reactants to products.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.KINETICS — the study of REACTION RATES and their relation to the way the reaction proceeds, i.e., its MECHANISM. We can use thermodynamics to tell if a reaction is product or reactant favored.We can use thermodynamics to tell if a reaction is product or reactant favored. But this gives us no info on HOW FAST reaction goes from reactants to products.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.KINETICS — the study of REACTION RATES and their relation to the way the reaction proceeds, i.e., its MECHANISM. Chemical Kinetics Chapter 15

3 3 Thermodynamics or Kinetics Reactions can be one of the following: 1.Not thermodynamically favored (reactant favored) 2.Thermodynamically favored (product favored), but not kinetically favored (slow) 3.Thermodynamically favored (product favored) and kinetically favored (fast)

4 4 Thermodynamics or Kinetics Not thermodynamically favored (reactant favored) Sand (SiO 2 ) will not decompose into Si and O 2

5 5 Thermodynamics or Kinetics Thermodynamically favored (product favored), but not kinetically favored (slow) Diamonds will turn into graphite, but the reaction occurs very slowly

6 6 Thermodynamics or Kinetics Thermodynamic ally favored (product favored) and kinetically favored (fast) Burning of paper in air will turn to ash very quickly

7 7 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 + O 3 ---> BrO + O 2 Step 2:Cl + O 3 ---> ClO + O 2 Step 3:BrO + ClO + light ---> Br + Cl + O 2 NET: 2 O 3 ---> 3 O 2 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 + O 3 ---> BrO + O 2 Step 2:Cl + O 3 ---> ClO + O 2 Step 3:BrO + ClO + light ---> Br + Cl + O 2 NET: 2 O 3 ---> 3 O 2

8 8 Reaction rate = change in concentration of a reactant or product with time.Reaction rate = change in concentration of a reactant or product with time. Know about initial rate, average rate, and instantaneous rate. See Screen 15.2.Know about initial rate, average rate, and instantaneous rate. See Screen Reaction Rates Section 15.1 Reaction Rates Section 15.1

9 9

10 10 Determining a Reaction Rate 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. 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. Dye Conc Time

11 11 Concentrations and physical state of reactants and products (Screens 15.3 and 15.4)Concentrations and physical state of reactants and products (Screens 15.3 and 15.4) Temperature (Screen 15.11)Temperature (Screen 15.11) Catalysts (Screen 15.14)Catalysts (Screen 15.14) Factors Affecting Rates Section 15.2

12 12 ConcentrationsConcentrations Factors Affecting Rates Section 15.2 Rate with 0.3 M HCl Rate with 6.0 M HCl

13 13 Physical state of reactantsPhysical state of reactants Factors Affecting Rates Section 15.2

14 14 Catalysts: catalyzed decomp of H 2 O 2 2 H 2 O 2 --> 2 H 2 O + O 2 2 H 2 O 2 --> 2 H 2 O + O 2 Factors Affecting Rates Section 15.2

15 15 TemperatureTemperature Factors Affecting Rates Section 15.2

16 16 Concentrations and Rates To postulate a reaction mechanism, we study reaction rate andreaction rate and its concentration dependenceits concentration dependence To postulate a reaction mechanism, we study reaction rate andreaction rate and its concentration dependenceits concentration dependence

17 17 Concentrations and Rates Take reaction where Cl - in cisplatin [Pt(NH 3 ) 2 Cl 3 ] is replaced by H 2 O

18 18 Concentrations and Rates Rate of reaction is proportional to [Pt(NH 3 ) 2 Cl 2 ] We express this as a RATE LAW Rate of reaction = k [Pt(NH 3 ) 2 Cl 2 ] where k = rate constant k is independent of conc. but increases with T Rate of reaction is proportional to [Pt(NH 3 ) 2 Cl 2 ] We express this as a RATE LAW Rate of reaction = k [Pt(NH 3 ) 2 Cl 2 ] where k = rate constant k is independent of conc. but increases with T

19 19 Concentrations, Rates, and Rate Laws In general, for a A + b B ---> x X with a catalyst C Rate = k [A] m [B] n [C] p The exponents m, n, and p are the reaction orderare the reaction order can be 0, 1, 2 or fractionscan be 0, 1, 2 or fractions must be determined by experiment!!!must be determined by experiment!!! In general, for a A + b B ---> x X with a catalyst C Rate = k [A] m [B] n [C] p The exponents m, n, and p are the reaction orderare the reaction order can be 0, 1, 2 or fractionscan be 0, 1, 2 or fractions must be determined by experiment!!!must be determined by experiment!!!

20 20 Interpreting Rate Laws Rate = k [A] m [B] n [C] p If m = 1, rxn. is 1st order in AIf m = 1, rxn. is 1st order in A 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. is 2nd order in A. Rate = k [A] 2 Rate = k [A] 2 Doubling [A] increases rate by ? If m = 0, rxn. is zero order.If m = 0, rxn. is zero order. Rate = k [A] 0 Rate = k [A] 0 If [A] doubles, rate ? Rate = k [A] m [B] n [C] p If m = 1, rxn. is 1st order in AIf m = 1, rxn. is 1st order in A 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. is 2nd order in A. Rate = k [A] 2 Rate = k [A] 2 Doubling [A] increases rate by ? If m = 0, rxn. is zero order.If m = 0, rxn. is zero order. Rate = k [A] 0 Rate = k [A] 0 If [A] doubles, rate ?

21 21 Deriving Rate Laws Derive rate law and k for CH 3 CHO(g) ---> CH 4 (g) + CO(g) from experimental data for rate of disappearance of CH 3 CHO Expt. [CH 3 CHO]Disappear of CH 3 CHO (mol/L)(mol/Lsec) Derive rate law and k for CH 3 CHO(g) ---> CH 4 (g) + CO(g) from experimental data for rate of disappearance of CH 3 CHO Expt. [CH 3 CHO]Disappear of CH 3 CHO (mol/L)(mol/Lsec)

22 22 Deriving Rate Laws Rate of rxn = k [CH 3 CHO] 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— mol/Ls = k (0.30 mol/L) 2 k = 2.0 (L / mols) k = 2.0 (L / mols) Using k you can calc. rate at other values of [CH 3 CHO] at same T. Rate of rxn = k [CH 3 CHO] 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— mol/Ls = k (0.30 mol/L) 2 k = 2.0 (L / mols) k = 2.0 (L / mols) Using k you can calc. rate at other values of [CH 3 CHO] at same T.

23 23 Concentration/Time Relations Need to know what conc. of reactant is as function of time. Consider FIRST ORDER REACTIONS For 1st order reactions, the rate law is - ( [A] / time) = k [A] For 1st order reactions, the rate law is - (  [A] /  time) = k [A] Need to know what conc. of reactant is as function of time. Consider FIRST ORDER REACTIONS For 1st order reactions, the rate law is - ( [A] / time) = k [A] For 1st order reactions, the rate law is - (  [A] /  time) = k [A]

24 24 Concentration/Time Relations Integrating - ( [A] / time) = k [A], we get Integrating - (  [A] /  time) = k [A], we get

25 25 Concentration/Time Relations Integrating - ( [A] / time) = k [A], we get Integrating - (  [A] /  time) = k [A], we get [A] / [A] 0 =fraction remaining after time t has elapsed.

26 26 Concentration/Time Relations Integrating - ( [A] / time) = k [A], we get Integrating - (  [A] /  time) = k [A], we get [A] / [A] 0 =fraction remaining after time t has elapsed. Called the integrated first-order rate law.

27 27 Concentration/Time Relations Sucrose decomposes to simpler sugars Rate of disappearance of sucrose = k [sucrose] k = 0.21 hr -1 Initial [sucrose] = M How long to drop 90% (to M)? Sucrose decomposes to simpler sugars Rate of disappearance of sucrose = k [sucrose] k = 0.21 hr -1 Initial [sucrose] = M How long to drop 90% (to M)? Sucrose

28 28 Concentration/Time Relationships Rate of disappear of sucrose = k [sucrose], k = 0.21 hr -1. If initial [sucrose] = M, how long to drop 90% or to M? Use the first order integrated rate law

29 29 Concentration/Time Relationships Rate of disappear of sucrose = k [sucrose], k = 0.21 hr -1. If initial [sucrose] = M, how long to drop 90% or to M? Use the first order integrated rate law

30 30 Concentration/Time Relationships Rate of disappear of sucrose = k [sucrose], k = 0.21 hr -1. If initial [sucrose] = M, how long to drop 90% or to M? Use the first order integrated rate law ln (0.100) = = - (0.21 hr -1 ) time

31 31 Concentration/Time Relationships Rate of disappear of sucrose = k [sucrose], k = 0.21 hr -1. If initial [sucrose] = M, how long to drop 90% or to M? Use the first order integrated rate law ln (0.100) = = - (0.21 hr -1 ) time time = 11 hours

32 32 Using the Integrated Rate Law The integrated rate law suggests a way to tell if a reaction is first order based on experiment. 2 N 2 O 5 (g) ---> 4 NO 2 (g) + O 2 (g) Rate = k [N 2 O 5 ] Time (min)[N 2 O 5 ] 0 (M) ln [N 2 O 5 ]

33 33 Using the Integrated Rate Law 2 N 2 O 5 (g) ---> 4 NO 2 (g) + O 2 (g) Rate = k [N 2 O 5 ] Data of conc. vs. time plot do not fit straight line.

34 34 Using the Integrated Rate Law 2 N 2 O 5 (g) ---> 4 NO 2 (g) + O 2 (g) Rate = k [N 2 O 5 ] Data of conc. vs. time plot do not fit straight line. Plot of ln [N 2 O 5 ] vs. time is a straight line!

35 35 Using the Integrated Rate Law 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) Plot of ln [N 2 O 5 ] vs. time is a straight line! Eqn. for straight line: y = ax + b y = ax + b Plot of ln [N 2 O 5 ] vs. time is a straight line! Eqn. for straight line: y = ax + b y = ax + b

36 36 Graphical Methods for Determining Reaction Order and Rate Constant Zero-order Straight line plot of [A] t vs. t Slope is –k (mol/L*s)

37 37 Graphical Methods for Determining Reaction Order and Rate Constant First-order Straight line plot of ln[A] t vs. t Slope is – k (s -1 )

38 38 Graphical Methods for Determining Reaction Order and Rate Constant Second-order Straight line plot of 1/[A] t vs t Slope is k (L/mol*s)

39 39 Half-Life Section 15.4 and Screen 15.8 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.

40 40 Reaction is 1st order decomposition of H 2 O 2.Reaction is 1st order decomposition of H 2 O 2. Half-LifeHalf-Life

41 41 Half-LifeHalf-Life Reaction after 654 min, 1 half-life.Reaction after 654 min, 1 half-life. 1/2 of the reactant remains.1/2 of the reactant remains.

42 42 Half-LifeHalf-Life Reaction after 1306 min, or 2 half-lives.Reaction after 1306 min, or 2 half-lives. 1/4 of the reactant remains.1/4 of the reactant remains.

43 43 Half-LifeHalf-Life Reaction after 3 half-lives, or 1962 min.Reaction after 3 half-lives, or 1962 min. 1/8 of the reactant remains.1/8 of the reactant remains.

44 44 Half-LifeHalf-Life Reaction after 4 half-lives, or 2616 min.Reaction after 4 half-lives, or 2616 min. 1/16 of the reactant remains.1/16 of the reactant remains.

45 45 Half-Life Section 15.4 and Screen 15.8 Sugar is fermented in a 1st order process (using an enzyme as a catalyst). Sugar is fermented in a 1st order process (using an enzyme as a catalyst). sugar + enzyme --> products sugar + enzyme --> products Rate of disappear of sugar = k[sugar] k = 3.3 x sec -1 What is the half-life of this reaction?

46 46 Half-Life Section 15.4 and Screen 15.8 Solution [A] / [A] 0 = 1/2 when t = t 1/2 Therefore, ln (1/2) = - k t 1/ = - k t 1/2 t 1/2 = / k t 1/2 = / k So, for sugar, t 1/2 = / k = 2100 sec = 35 min Rate = k[sugar] and k = 3.3 x sec -1. What is the half- life of this reaction?

47 47 Half-Life Section 15.4 and Screen 15.8 Solution 2 hr and 20 min = 4 half-lives Half-life Time ElapsedMass Left 1st35 min5.00 g 2nd g 3rd g 4th g Rate = k[sugar] and k = 3.3 x sec -1. Half-life is 35 min. Start with 5.00 g sugar. How much is left after 2 hr and 20 min?

48 48 Half-Life Section 15.4 and Screen 15.8 Solution Radioactive decay is a first order process. Tritium ---> electron + helium Tritium ---> electron + helium 3 H 0 -1 e 3 He 3 H 0 -1 e 3 He If you have 1.50 mg of tritium, how much is left after 49.2 years? t 1/2 = 12.3 years

49 49 Half-Life Section 15.4 and Screen 15.8 Solution ln [A] / [A] 0 = -kt [A] = ?[A] 0 = 1.50 mgt = 49.2 years Need k, so we calc k from: k = / t 1/2 Obtain k = y -1 Now ln [A] / [A] 0 = -kt = - ( y -1 ) (49.2 y) = = Take antilog: [A] / [A] 0 = e = is the fraction remaining! Start with 1.50 mg of tritium, how much is left after 49.2 years? t 1/2 = 12.3 years

50 50 Half-Life Section 15.4 and Screen 15.8 Solution [A] / [A] 0 = is the fraction remaining! Because [A] 0 = 1.50 mg, [A] = mg But notice that 49.2 y = 4.00 half-lives 1.50 mg ---> mg after mg ---> mg after 1 ---> mg after 2 ---> mg after 2 ---> mg after 3 ---> mg after 3 ---> mg after 4 ---> mg after 4 Start with 1.50 mg of tritium, how much is left after 49.2 years? t 1/2 = 12.3 years

51 51 Half-Lives of Radioactive Elements Rate of decay of radioactive isotopes given in terms of 1/2-life. 238 U --> 234 Th + He4.5 x 10 9 y 14 C --> 14 N + beta5730 y 131 I --> 131 Xe + beta8.05 d Element seaborgium 263 Sg 0.9 s

52 52 Darleane Hoffman of UC-Berkeley studies the newest elements. Hoffman is Director of Seaborg Institute. Rec’d ACS Award in Nuclear Chem and the Garvan Medal.


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