1. 1 PRINCIPLES OF REACTIVITY: CHEMICAL KINETICS CHAPTER 15 1-17 + all bold numbered problems

2. 2 Chapter 15 Overview Calculate each part of the rate law:Calculate each part of the rate law: rate (law) = k [A] m [B] n [C] p Memorize the rate law/decay formula for:Memorize the rate law/decay formula for: –0 order –1 st order –2 nd order Purpose of study rate laws is to find the mechanism of a reaction

3. 3 Chemical Kinetics Chapter 15 An automotive catalytic muffler.

4. 4 CHAPTER OVERVIEW This chapter examines factors that determine the speed of reactions, reaction rates, and the steps that occur during the reaction process, reaction mechanisms.This chapter examines factors that determine the speed of reactions, reaction rates, and the steps that occur during the reaction process, reaction mechanisms. Reaction rates are generally affected by concentration of reactants, temperature, and the presence of a catalyst.Reaction rates are generally affected by concentration of reactants, temperature, and the presence of a catalyst.

5. 5 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

6. 6 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. 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

7. 7 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.Know about initial rate, average rate, and instantaneous rate. Reaction Rates Section 15.1 Reaction Rates Section 15.1

8. 8 Determining a Reaction Rate Blue dye is oxidized with bleach. Its concentration decreases with time. The rate — the change in dye concentration with time — can be determined from the plot. Dye Conc Time

9. 9 RATES OF CHEMICAL REACTIONS In chemical reactions, the rate is usually measured as change in concentration per unit time, -Δ(A)/ Δt, for reactant A.In chemical reactions, the rate is usually measured as change in concentration per unit time, -Δ(A)/ Δt, for reactant A. The rate can also be measured as the rate of appearance of the product.The rate can also be measured as the rate of appearance of the product. Rates are related stoichiometrically.Rates are related stoichiometrically. Reaction rates are always positive numbers.Reaction rates are always positive numbers. In the text, the rate for the decomposition of N 2 O 5 is calculated at several different concentrations.In the text, the rate for the decomposition of N 2 O 5 is calculated at several different concentrations.

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11. 11 RATES OF REACTIONS The average rate can be calculated between any two points.The average rate can be calculated between any two points. The instantaneous rate at any specific time is the slope of the tangent line at that point for the graph of concentration vs. time.The instantaneous rate at any specific time is the slope of the tangent line at that point for the graph of concentration vs. time.

12. 12 Concentrations and physical state of reactants and productsConcentrations and physical state of reactants and products TemperatureTemperature CatalystsCatalysts Factors Affecting Rates Section 15.2

13. 13 ConcentrationsConcentrations Factors Affecting Rates Rate with 0.3 M HCl Rate with 6.0 M HCl

14. 14 Physical state of reactantsPhysical state of reactants Factors Affecting Rates

15. 15 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

16. 16 TemperatureTemperature Factors Affecting Rates

17. 17 In general, the rate of reaction is a function of reactant concentration. This function is determined experimentally, frequently using initial rates. The outcome of the experiment is expressed in a form called the rate equation or rate law.The outcome of the experiment is expressed in a form called the rate equation or rate law. 15.3 EFFECTS OF CONCENTRATION ON REACTION RATE

18. 18 CONCENTRATION AND RATE For the general equation,For the general equation, a A + b B ====> x X with catalyst C a A + b B ====> x X with catalyst C rate = k [A] m [B] n [C] prate = k [A] m [B] n [C] p m, n, and p are frequently positive whole numbers, but they can be negative, fractions, or even zero.m, n, and p are frequently positive whole numbers, but they can be negative, fractions, or even zero. They are determined EXPERIMENTALLYThey are determined EXPERIMENTALLY

19. 19 The Rate Constant, k The rate constant, k, is determined experimentally from the rate data, and has units consistent with the rate equation.The rate constant, k, is determined experimentally from the rate data, and has units consistent with the rate equation. These units include time -1 and concentrationThese units include time -1 and concentration to some power. The magnitude of k is a function of the temperature for the same reaction.The magnitude of k is a function of the temperature for the same reaction. Larger k's mean faster reactions.Larger k's mean faster reactions.

20. 20 The Order of Reaction The total order of the reaction is the sum of the orders for each of the species in the rate law.The total order of the reaction is the sum of the orders for each of the species in the rate law. In the previous example the total order is: m + n + pIn the previous example the total order is: m + n + p As the total order varies, the units on k will vary.As the total order varies, the units on k will vary. For zero order, k has units of M/time.For zero order, k has units of M/time. For first order, k has units of 1/time or time - 1, etc.For first order, k has units of 1/time or time - 1, etc.

21. 21 Rate Laws: What Use There are two uses for the Rate Law:There are two uses for the Rate Law: –1. The ability to find a “future” concentration The we “Integrate” the Rate Law:The we “Integrate” the Rate Law: –2. This gives a time dependent function that tells us “how long” it takes to get to the “future” concentration

22. 22 Determination of the Rate Equation The rate equation is frequently determined using initial rates.The rate equation is frequently determined using initial rates. These experiments measure the rate of the reaction keeping all but one reactant concentration essentially constant and allowing the other to change by only about 1% or 2%.These experiments measure the rate of the reaction keeping all but one reactant concentration essentially constant and allowing the other to change by only about 1% or 2%. The order for this reactant is then determined.The order for this reactant is then determined.

23. 23 Determination of the Rate Equation This process is continued until all the reactant orders have been determined.This process is continued until all the reactant orders have been determined. The order for each reactant is determined algebraically by choosing an appropriate pair of experiments and taking a ratio.The order for each reactant is determined algebraically by choosing an appropriate pair of experiments and taking a ratio. The rate constant, k, is then determined by substitution once all the reaction orders have been calculated.The rate constant, k, is then determined by substitution once all the reaction orders have been calculated. Examples 15.3,4 and Exercises 15.3, 4.Examples 15.3,4 and Exercises 15.3, 4.

24. 24 Concentrations and Rates Take reaction where Cl - in cisplatin [Pt(NH 3 ) 2 Cl 2 ] is replaced by H 2 O Rate of change of conc of Pt compound = Am't of cisplatin reacting (mol/L) elapsed time (t)

25. 25 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 change of conc of Pt compound = Am't of cisplatin reacting (mol/L) elapsed time (t)

26. 26 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!!!

27. 27 Interpreting Rate Laws Rate = k [A] m [B] n [C] p If m = 1, reaction is 1st order in AIf m = 1, reaction 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, reaction is 2nd order in A.If m = 2, reaction is 2nd order in A. Rate = k [A] 2 Rate = k [A] 2 Doubling [A] increases rate by ? If m = 0, reaction is zero order.If m = 0, reaction is zero order. Rate = k [A] 0 Rate = k [A] 0 If [A] doubles, rate ?

28. 28 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) 10.100.020 10.100.020 20.200.081 20.200.081 30.300.182 30.300.182 40.400.318 40.400.318

29. 29 Deriving Rate Laws Rate of rxn = k [CH 3 CHO] 2 Here the rate goes up by ______ when initial concentration doubles. Therefore this reaction is _________________ order. Now determine the value of k. Use expt. #3 data 0.182 mol/Ls = k (0.30 mol/L) 2 k = 2.0 (L / mols) Using k you can calculate the rate at other values of [CH 3 CHO] at same T…”1. Future Conc”

30. 30 Concentration/Time Relations…”2. how long” Need to know that concentration of reactant is as function of time. Consider FIRST ORDER REACTIONS For 1st order reactions, the rate law is rate = - ( Δ [A] / Δ time) = k [A]

31. 31 Concentration/Time Relations Integrating - (  [A] /  time) = k [A], we get [A] / [A] 0 =fraction remaining after time t has elapsed. integrated first-order rate law. Called the integrated first-order rate law. [A] = - k t ln [A] o natural logarithm [A] at time = 0

32. 32 Concentration/Time Relations Sucrose decomposes to simpler sugars Rate of disappearance of sucrose = k [sucrose] k = 0.21 hr -1 Initial [sucrose] = 0.010 M How long to drop 90% ( to 0.0010 M )? Sucrose

33. 33 Concentration/Time Relationships Rate of disappearance 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? ln (0.100) = - 2.3 = - (0.21 hr -1 ) time time = 11 hours Use the first order integrated rate law

34. 34 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 ] o (M) ln [N 2 O 5 ] o 0 1.00 0 1.0 0.705-0.35 2.0 0.497-0.70 5.0 0.173-1.75

35. 35 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!

36. 36 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 ln [N 2 O 5 ] = - kt + ln [N 2 O 5 ] o conc at time t rate const = slope conc at time = 0

37. 37 15.4 Relationships Between Concentration and Time The order for a specific reactant can be determined graphically if the function that produces a linear plot for each order is known.The order for a specific reactant can be determined graphically if the function that produces a linear plot for each order is known. We will derive the algebraic relationship for three orders, zero, first, and second.We will derive the algebraic relationship for three orders, zero, first, and second. For the general equation:For the general equation: A ===> Products rate = -d[A]/dt = k [A] x, x is the order.rate = -d[A]/dt = k [A] x, x is the order.

38. 38 Relationships Between Concentration and Time The linear plot always has time on the horizontal axis and a function of concentration on the vertical axis.The linear plot always has time on the horizontal axis and a function of concentration on the vertical axis. The half-life, t 1/2, is defined as the time required for one-half of the reactant to be consumed:The half-life, t 1/2, is defined as the time required for one-half of the reactant to be consumed: [A] = [A] o /2

39. 39 Zero-OrderZero-Order rate = - d[A]/dt = k [A] 0 = krate = - d[A]/dt = k [A] 0 = k This is the differential form of the rate expression, since it contains a differential.This is the differential form of the rate expression, since it contains a differential. It is also the definition of zero-order.It is also the definition of zero-order. The integrated form for zero-order can be found by separation of variables and integration.The integrated form for zero-order can be found by separation of variables and integration. The result is [A] = - kt + [A] o.The result is [A] = - kt + [A] o. The linear plot is [A] vs. timeThe linear plot is [A] vs. time The half-life is, t 1/2 = [A] o /2k.The half-life is, t 1/2 = [A] o /2k. From the graph, m = - k.From the graph, m = - k.

40. 40 First-OrderFirst-Order rate = -d[A]/dt = k [A] 1rate = -d[A]/dt = k [A] 1 This is the differential form of the rate expression, since it contains a differential.This is the differential form of the rate expression, since it contains a differential. It is also the definition of first-order.It is also the definition of first-order. The integrated form for first-order can be found by separation of variables and integration.The integrated form for first-order can be found by separation of variables and integration. The result is ln[A] = - kt + ln[A] o.The result is ln[A] = - kt + ln[A] o. The linear plot is ln[A] vs. timeThe linear plot is ln[A] vs. time The half-life is, t 1/2 = (ln2)/k.The half-life is, t 1/2 = (ln2)/k. From the graph, m = - k.From the graph, m = - k.

41. 41 Second-OrderSecond-Order rate = - d[A]/dt = k [A] 2rate = - d[A]/dt = k [A] 2 This is the differential form of the rate expression, since it contains a differential.This is the differential form of the rate expression, since it contains a differential. It is also the definition of second-order.It is also the definition of second-order. The integrated form for second-order can be found by separation of variables and integration.The integrated form for second-order can be found by separation of variables and integration. The result is 1/[A] = + kt + 1/[A] o.The result is 1/[A] = + kt + 1/[A] o. The linear plot is 1/[A] vs. timeThe linear plot is 1/[A] vs. time The half-life is, t 1/2 = 1/k[A] o.The half-life is, t 1/2 = 1/k[A] o. From the graph, m = + k.From the graph, m = + k.

42. 42 Summary 0 order: [A] = - kt + [A] o0 order: [A] = - kt + [A] o t 1/2 = [A] o /2k 1 st order: ln[A] = - kt + ln[A] o1 st order: ln[A] = - kt + ln[A] o t 1/2 = (ln1/2)/-k 2 nd order: 1/[A] = + kt + 1/[A] o,2 nd order: 1/[A] = + kt + 1/[A] o, t 1/2 = 1/k[A] o

43. 43 Half-LifeHalf-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.

44. 44 Reaction is 1st order decomposition of H 2 O 2.Reaction is 1st order decomposition of H 2 O 2. Half-LifeHalf-Life

45. 45 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.

46. 46 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.

47. 47 Half-LifeHalf-Life 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 10 -4 sec -1 What is the half-life of this reaction?

48. 48 Half-LifeHalf-Life Solution: ln[A] = - kt + ln[A] o [A] / [A] 0 = 1/2 when t = t 1/2 Therefore, ln (1/2) = - k t 1/2 - 0.693 = - k t 1/2 t 1/2 = 0.693 / k t 1/2 = 0.693 / k So, for sugar, t 1/2 = 0.693 / k = 2100 sec = 35 min Rate = k[sugar] and k = 3.3 x 10 -4 sec -1. What is the half-life of this reaction?

49. 49 Half-LifeHalf-Life Solution 2 hr and 20 min = 4 half-lives Half-life Time ElapsedMass Left 1st35 min2.50 g 2nd701.25 g 3rd1050.625 g 4th1400.313 g 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?

50. 50 Or using the rate law 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 (8400 sec)? ln[A] = - kt + ln[A] o (first order)ln[A] = - kt + ln[A] o (first order)

51. 51 Half-LifeHalf-Life 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 This has been experimentally determined to be a 1 st order reaction.This has been experimentally determined to be a 1 st order reaction.

52. 52 Solution ln [A] / [A] 0 = -kt, however 2 unknown variables [A] = ?[A] 0 = 1.50 mgt = 49.2 years Need k, so we calc k: 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) ln [A] / [A] 0 = - 2.77 ln [A] / [A] 0 = - 2.77 Take antilog: [A] / [A] 0 = e -2.77 = 0.0627 0.0627×1.50mg =.0940mg 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 Half-LifeHalf-Life

53. 53 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 ---> 0.750 mg after 1 1.50 mg ---> 0.750 mg after 1 ---> 0.375 mg after 2 ---> 0.375 mg after 2 ---> 0.188 mg after 3 ---> 0.188 mg after 3 ---> 0.0940 mg after 4 ---> 0.0940 mg after 4 Half-LifeHalf-Life Start with 1.50 mg of tritium, how much is left after 49.2 years? t 1/2 = 12.3 years

54. 54 Half-Lives of Radioactive Elements Rate of decay of radioactive isotopes given in terms of 1/2-life. 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 106 - Seaborgium 263 Sg 0.9 s Try: start with 10g of Sg, see how much is left after an 8 hour work day, assume 1 st order.