1. AP 3/21 ◦ TURN IN THE EQUATION WRITING HW ◦ Today we will start the Kinetics unit and move through this unit by next Friday. ◦ You will need the Kinetics packet from the side table, a calul and a clean piece of paper. ◦ You have a test on Wednesday over thermo. The papers you received before spring break are due the day you test. ◦ WE have a lot to cover before May 2 nd. Please be prepared to put in work until then. ◦ Tuesday will be the first review day for the AP test. WE will cover basic stoichiometry stuff. The review will start at 3 and end around 3:45.

2. INTRO KINETICS ACTIVITY ◦ On the paper provided: ◦ List the FIVE factors that affect the reaction rate ◦ Draw a picture depicting each factor ◦ Give a brief description of each ◦ You are allowed to use your phone to help. If you printed off the NMSI packet, that will also be EXTREMELY useful. This one activity could take care of multiple questions on the AP exam. Please make sure you know these factors and are able to describe them with “AP words”

3. AP 3/22 ◦ Today you will need the two packets from the side table, kinetics notes, a calculator, something to write with, and a piece of paper to answer a few warm-up questions. ◦ You will learn how to determine simple rate laws. Monday, we will practice the more difficult rate laws. ◦ Thermo test tomorrow. Study the ppts on my website, NMSI’s notes and practice problems. There will be vocabulary on the test, so make sure you know what everything means. The test will be in TWO parts and will look very similar to the actual AP test. You are able to use a calculator and the AP formula chart you are GIVEN (NOT YOUR OWN). ◦ HW DUE MONDAY: simple rate law packet you picked up today AND WATCH THE FIRST PART OF NMSI (pages 1-3)

4. Warm-UP ◦ 1. What is a catalyst? ◦ 2. Describe what a catalyst such as KMnO 4 does when added to hydrogen peroxide. ◦ 3. Describe how successful collisions occur in order to cause a reaction. USE THE AP WORDS. (look at NMSI on your phone)

5. AP CHEMISTRY CHAPTER 12 Although we can know a lot of things about a chemical reaction, including the reactants and expected products, the stoichiometry and the energy, we must understand the rate of the reaction to truly grasp what is occurring. Although graphite can be compressed into a diamond, the reaction is so slow that we cannot view this in our lifetime. Therefore, we must choose to examine only those reactions that occur at a reasonable rate.

6. 6 Chemical Kinetics ◦ Thermodynamics tells us if a reaction can occur ◦ Kinetics tells us how quickly the reaction occurs ◦ some reactions that are thermodynamically feasible are kinetically so slow as to be imperceptible

7. 12.1 Reaction Rates ◦ Reaction Rate- change in concentration of a reactant or product per unit time. ◦ We will write the concentration in mol/L as [A] where “A” is the substance ◦ If the rate expression involves a reactant: ◦ Rate = -  [A] (negative because [ ] decreases)  t ◦ The above gives the average rate.

8. Instantaneous Rates ◦ To get an instantaneous rate, we can compute the slope of a line tangent to the curve at that point. ◦ Rate = -(slope of the tangent line)

9. We Use “Initial Rates” ◦ The rate of a reaction is not constant but changes with time because concentrations change with time. ◦ We will only work with reaction rates that are “initial rates” (reverse reaction is negligible)

10. Examining Rates of Reactions ◦ This is the decay of nitrogen dioxide. ◦ 2NO 2  2NO + O 2 ◦ There are some key things to note here: ◦ As NO 2 decreases in concentration, the concentration of each product increases. ◦ The concentration decrease of NO 2 is equal to the concentration increase of NO, but is twice that of O 2. This is reflected by the mol ratio.

11. AP 3/30 ◦ Today we will review the differential rate law and then go over the graphs for the integrated rate law. ◦ You will need: the paper from the side table, a calculator, the “simple rate law” worksheet, and maybe your kinetics notes. ◦ You will have a quiz on Friday over everything we have covered so far, including the factors that affect a reaction rate. ◦ HW for the weekend: REACTION MECHANISMS worksheet. We have not covered this, so it will force you to watch NMSI. I will go over this concept on Monday. You need to try to work through the problems.

12. 12.2 Rate Laws: An Introduction ◦ Rate laws can express how quickly a reactant is consumed or the rate at which a product is produced. In this chapter, however, we will focus on the consumption of reactants. ◦ We will make the assumption that only the forward reaction is occurring; this means that calculations of the reverse reaction do not have to be considered here.

13. 12.2 Rate Laws: An Introduction ◦ Differential Rate Law or Rate Equation- a rate law that expresses how the rate depends on concentration. ◦ For the reaction aA + bB + …  gG + hH + … ◦ Rate = k[A] m [B] n … ◦ [A] & [B] represent molarities

14. Rate = k[A] m [B] n ◦ The exponents are positive or negative, integers or fractions. ◦ Usually positive integers (small whole numbers) ◦ k = rate constant ◦ The value of k depends on the type of reaction, temperature and presence of a catalyst ◦ The faster the reaction, larger the k value

15. The Order of Reaction ◦ The exponents determine the order of the reactants. The sum of the exponents is the order of the reaction. ◦ R = k[A][B] 2 is first order in A, second order in B and third order overall. ◦ The units of k can be calculated by reaction orders and units of concentration and rate.

16. Finding Units for the Rate Constant ◦ For example, if rate is in mol/L. s in the above rate law, we can find the units for k as follows:

17. The “shortcut” to determining units of k is as follows: k = L x /mol x. time. ◦ The value of x will be one less than the order of the reaction. ◦ If the reaction is 3rd order, k = L 2 /mol 2. time. ◦ If the reaction is 2nd order, k = L/mol. time ◦ If the reaction is 1st order, k = 1/time

18. 12.3 Determining the Form of the Rate Law ◦ Determining Differential Rate Laws from Experimental Data ◦ If doubling the initial [ ] of a reactant causes the initial rate to double, the reaction is first order in that reactant. ◦ If doubling the initial [ ] of a reactant causes the initial rate to quadruple, the reaction is second order in that reactant. ◦ If doubling the initial [ ] of a reactant causes the initial rate to increase 8 times, the reaction is third order in that reactant. ◦ If doubling the initial [ ] of a reactant does not change the initial rate, the reaction is zero order in that reactant and that reactant is removed from the rate law.

19. The reaction is ____ order in terms of A.The reaction is ____ order in terms of A. The reaction is ____ order in terms of B.The reaction is ____ order in terms of B. The rate law is ____________________.The rate law is ____________________. The reaction is ____ order overall.The reaction is ____ order overall. Rate = k[A] 2 [B] 2 nd 1 st 3 rd Ex. 2A + B  2C [A][B]Rate 0.400.200.10 0.40 0.20 0.800.400.80

20. The reaction is ____ order in terms of NH 4 +.The reaction is ____ order in terms of NH 4 +. The reaction is ____ order in terms of NO 2 -.The reaction is ____ order in terms of NO 2 -. The rate law is ____________________.The rate law is ____________________. The reaction is ____ order overall.The reaction is ____ order overall. Rate = k[A][B] 1 st 2 nd Ex. NH 4 + + NO 2 -  N 2 + 2H 2 O [ NH 4 + ][ NO 2 - ]Rate 0.1000.00501.35 x 10 -7 0.1000.0102.70 x 10 -7 0.2000.0105.40 x 10 -7

21. Pre-AP: I AM OUT TODAY. Please read the following for what to do today ◦ Finish the Ideal Gas Law worksheet. ◦ You can use your phone to help you.

22. AP 3/28 ◦ DO NOT TURN IN YOUR HOMEWORK. Today we will continue practicing the stuff we learned Friday. ◦ I am out today. You will need to click through these ppts to help you answer the rate law questions. We will go over them better tomorrow. Please be good and try to answer these. I put practice problems on the ppts with answers and explanations. ◦ IMPORTANT: Tomorrow is STAAR testing. I still have ALL my classes BUT you will need to GO THROUGH SILVER TO GET TO MY CLASS. YOU WILL NEED TO GO UP THE STAIRS AND DOWN THE BACK STAIRS BY THE COMPUTER LABS. I will send out a remind101 in the morning to help you remember. ◦ WE are going to try to finish up Kinetics this week and spend two weeks on Equilibrium, one week on Electrochem, and one week on Acids/Bases…AND THEN WE ARE DONE!

23. AP 3/29 ◦ DO NOT TURN IN YOUR HOMEWORK. Today we will continue practicing the stuff we learned Friday WITH ME. Then we will go over the graphs for the integrated rate law. ◦ You will have no homework tonight but I HIGHLY SUGGEST WATCHING NMSI BEFORE FRIDAY. ◦ You will need the simple rate law practice. The “not so simple” can be thrown away OR you can look at the key I made. The chance of AP putting one of those on your test is low. ◦ There is no afterschool review this week due to STAAR. I will possibly do TWO DAYS NEXT week (Monday and Tuesday) ◦ WE are going to try to finish up Kinetics this week and spend two weeks on Equilibrium, one week on Electrochem, and one week on Acids/Bases…AND THEN WE ARE DONE!

24. The topic we were covering on Friday was called “Determining Differential Rate Laws from Experimental Data”. Answer the following questions related to this. 1. What are the units for the rate constant (baby k) if the order of the reaction is 1 st, 2 nd, 3 rd order. 2. The rate law we are on relates what two things? 3. Explain in AP words the following: ◦ [A] 0 ◦ [B] 1 ◦ [A] 1 [B] 2 4. What is the total order of the reactions above?

25. A  B [A] Rate 0.05 3 x 10 -4 0.10 1.2 x 10 -3 0.20 4.8 x 10 -3 Determine the rate law: Rate = k[A] x Rate = k[A] 2 2 2 4 4 Because 2 2 =4, A is 2 nd order

26. A + B  C Exp [A] [B] Rate 1 0.10 0.20 1.0×10  5 2 0.20 0.20 1.0×10  5 3 0.20 0.40 2.0×10  5 Determine the rate law: Rate = k[A] x [B] y Rate = k[A] 0 [B] y Rate = k [B] 1 2 1 1 2 2 Because 2 1 =2, B is 1 st order Because 2 0 =1, A is 0 order

27. AP 3/31 Today we will review more differential rate law problems and then go over the graphs for the integrated rate law. You will need: a calculator, the “simple rate law” worksheet, and maybe your kinetics notes. You will have a quiz on Friday over everything we have covered so far, including the factors that affect a reaction rate. HW for the weekend: REACTION MECHANISMS worksheet. We have not covered this, so it will force you to watch NMSI. I will go over this concept on Monday. You need to try to work through the problems.

28. A + 2B  2C Ex. [A] [B] Rate 1 0.1 0.1 0.10 2 0.2 0.1 0.20 3 0.4 0.2 0.40 Determine the rate law: Rate = k[A] x [B] y Rate = k[A] 1 [B] y Rate = k[A] 1 [B] 0 Rate = k[A] 212 2 2 2 Because 2 1 =2, A is 1 st order No control in this example. Because doubling A causes the rate to double, doubling B must have had no effect. B is zero order.

29. Experiment[A][B]Initial Rate mol L -1 s -1 110.12.015.00 219.83.9980.0 340.22.0080.0 Determine the rate law: Rate = k[A] x [B] y 4 2 =16 so [A] is 2 nd order 1416 Rate = k[A] 2 [B] y 2 216 Because A is 2 nd order, doubling A will cause the rate to quadruple. When both A and B were doubled, the rate was 16 times greater. This means that B is also 2 nd order. Rate = k[A] 2 [B] 2 Determine the value of k and its units. 5.00 mol L -1 s -1 = k(10.1mol/L) 2 (2.01mol/L) 2 k = 0.0121 L 3 mol -3 s -1

30. #1 From Simple Rate Law Experiment[A][B]Initial rate 1325 23420 36440

31. #2 From Simple Rate Law Experiment[A][B]Initial rate 11123 24 3416

32. 5Br - (aq) + BrO 3 - (aq) + 6H + (aq)  3Br 2 (l) + 3H 2 O 3. In a study of the kinetics of the reaction represented above, the following data were obtained at 298K. ExperimentInitial [Br - ] (mol/L) Initial [BrO 3 - ] (mol/L) Initial [H + ] (mol/L) Rate of Disappearance of BrO 3 - (mol/L. s) 10.001000.005000.1002.50 x 10 -4 20.002000.005000.1005.00 x 10 -4 30.001000.007500.1003.75 x 10 -4 40.001000.015000.2003.00 x 10 -3 a)From the data given above, determine the order of the reaction for each reactant listed below. Show your reasoning. a)Br - b)BrO 3 - c)H + 1 st order because doubling [Br - ] doubles the rate if the other reactants are held constant. 1 st order because 1.5x [BrO 3 - ] =1.5x the rate if the other reactants are held constant. 2 nd order because doubling [H + ] quadruples the rate if the other reactants are held constant. AP Practice Problem: This is in the back of your notes!

33. ExperimentInitial [Br - ] (mol/L) Initial [BrO 3 - ] (mol/L) Initial [H + ] (mol/L) Rate of Disappearance of BrO 3 - (mol/L. s) 10.001000.005000.1002.50 x 10 -4 20.002000.005000.1005.00 x 10 -4 30.001000.007500.1003.75 x 10 -4 40.001000.015000.2003.00 x 10 -3 a)Write the rate law for the overall reaction b)Determine the value of the specific rate constant for the reaction at 298K. Include the correct units. Rate = k [Br - ][BrO 3 - ][H + ] 2 2.50 x 10 -4 mol/L. s=k[0.00100M][0.00500M][0.100M] 2 k = 5000 L 3 /mol 3. s

34. 12.4 The Integrated Rate Law ◦ Integrated Rate Law- expresses how the concentration of the reactant depends on time ◦ Instead of changing initial concentrations and using multiple experiments, one experiment is done and concentration changes over time are measured.

35. A Note… ◦ The rate law we choose to determine by experiment often depends on what types of data are easiest to collect. If we can conveniently measure how the rate changes as the concentrations are changed, we can readily determine the differential rate law. On the other hand, if it is more convenient to measure eth concentration as a function of time, we can determine the form of the integrated rate law.

36. 1 st order integrated rate law ln [A] 0 = kt [A] t “Naughty Katy”

37. A plot of ln[A] vs t always gives a straight line for a 1 st order reaction. The slope = -k.

38. Ex. At 400 o C, the 1 st order conversion of cyclopropane into propylene has a rate constant of 1.16 x 10 -6 s -1. If the initial concentration of cyclopropane is 1.00 x 10 -2 mol/L at 400 o C, what will its concentration be 24.0 hrs after the reaction begins? 24 hrs 3600s = 86,400s ln [A] 0 = kt 1 hr [A] t ln 1.00 x 10 -2 mol/L = 1.16 x 10 -6 s -1 (86,400s) [A] t [A] t = 9.05 x 10 -3

39. Second order integrated rate law Rate = k[A] 2 1 - 1 = kt [A] t [A] 0 A plot of 1/[A] t versus t produces a straight line with slope k. Differential rate law

40. (a) A Plot of In(C 4 H 6 ) Versus t (b) A Plot of 1/(C 4 H 6 ) Versus t

41. A Plot of (A) Versus t for a Zero-Order Reaction

42. Zero Order Rate = k [A] t  [A] o =  kt A plot of [A] versus t produces a straight line with slope -k. Differential rate law This equation is not found on the AP formula sheet and will not have to be calculated. Make sure that you understand the graphing.

43. An easy way to keep the graphs straight is to remember CLR=0,1,2 C=concentration, L = ln concentration, R=reciprical of concentration If the concentration vs time is straight, it is C (0 order). If natural log of concentration vs time is straight it is L=1 st order. If reciprical of concentration vs time is straight it is R=2 nd order. CLR 0 1 2

44. PRACTICE: What order is this reaction?

46. Rate Laws: A Summary

47. AP 4/6 ◦ Pick up the equilibrium packet from the side table ◦ You will need that packet, your kinetics notes, a blank piece of paper, something to write with, and possibly a highlighter. ◦ WE will finish up Kinetics today and introduce equilibrium. ◦ You will answer a free response question over kinetics and then we will grade your FRQ from the test quickly. ◦ HW tonight: Watch NMSI for equilibrium. You need to watch her until you see something called the reaction quotient (stop at exercise 7). Also, you need to google the iodine clock reaction and write a TWO paragraph description over this reaction and how it relates to kinetics.

48. Practice Reaction Mechanism: Determine the rate law for the net reaction based on the given mechanism. What is the molecularity of the rate determining step? What is the overall order of the reaction H 2 (g) + ICl(g) → HClI(g) + H(g) (Slow) H(g) + ICl(g) → HCl(g) + I(g) (Fast) HClI(g) → HCl(g) + I(g) (Fast) I(g) + I(g) → I 2 (g) (Fast)

49. Practice Reaction Mechanism : Determine the rate law for the net reaction based on the given mechanism. What is the molecularity of the rate determining step? What is the overall order of the reaction Step 1: HCOOH + H+  HCOOH 2 + (fast) Step 2: HCOOH 2 +  H 2 O + HCO+ (slow) Step 3: HCO+  CO + H+ (fast)

50. 21.2 The Kinetics of Radioactive Decay ◦ Every radioactive sample has a decay rate, or amount of particles that will decay over time. ◦ This is a first-order process ◦ Rate = -(ΔN/Δt) = kN

51. Half Life ◦ Half-life (t 1/2 ) is the length of time required for the concentration of a reactant to decrease to half of its initial value. ◦ t 1/2 = 0.693/k

52. A fast reaction with a short t 1/2 has a large k. A slow reaction with a long t 1/2 has a small k. A Plot of (N 2 O 5 ) Versus Time for the Decomposition Reaction of N 2 O 5

53. Example: The decomposition of SO 2 Cl and Cl 2 is a first order reaction with k = 2.2 x 10 - 5 s -1 at 320 o C. Determine the half-life of this reaction. t 1/2 = 0.693/k t 1/2 = 0.693/(2.2 x 10 -5 s -1 ) t 1/2 = 31500 s or 525 min or 8.75 hours

54. Example: The decomposition of N 2 O 5 dissolved in CCl 4 is a first order reaction. The chemical change is: 2N 2 O 5  4NO 2 + O 2 At 45 o C the reaction was begun with an initial N 2 O 5 concentration of 1.00 mol/L. After 3.00 hours the N 2 O 5 concentration had decreased to 1.21 x 10 -3 mol/L. What is the half-life of N 2 O 5 expressed in minutes at 45 o C? ln [A] 0 = kt [A] t [A] t ln 1.00 = k(180 min) 1.21 x 10 -3 k = 0.0373 min -1 t 1/2 = 0.693/k t 1/2 = 0.693/0.0373 min -1 = 18.6 min

55. More Advanced Half-life Problems ◦ Ex. The radioactive decay of thallium-206 to lead-206 has a half-life of 4.20 min. Starting with 5.00 x 10 22 atoms of thallium-206, calculate the number of such atoms left after 42.0 min. t 1/2 = 0.693/k 4.20 min = 0.693/k k = 0.165 min -1 ln(N 0 /N t ) = kt ln(5.00 x 10 22 /N t ) = 0.165(42.0) N t = 4.89 x 10 19 atoms

56. A freshly isolated sample of 90 Y was found to have an activity of 9.8 x 10 5 disintegrations per minute at 1:00 PM on December 3, 1995. At 2:15 PM on December 17, 1995, its activity was redetermined and found to be 2.6 x 10 4 disintegrations per minute. Calculate the half-life of 90 Y. time = 14 days, 1 ¼ hours = 337.25 hours ln(N 0 /N t ) = kt ln (9.8 x 10 5 /2.6 x 10 4 ) = k(337.25 hr) k = 0.011 hr -1 t 1/2 = 0.693/k t 1/2 = 0.693/0.011 hr -1 t 1/2 = 63 hours

57. Second order integrated rate law (aka Differential Rate Law) ◦ Rate = k[A] 2 1 - 1 = kt [A] t [A] 0 A plot of 1/[A] t versus t produces a straight line with slope k.

58. (a) A Plot of In(C 4 H 6 ) Versus t (b) A Plot of 1/(C 4 H 6 ) Versus t

59. A Plot of (A) Versus t for a Zero-Order Reaction Zero Order Rate Law ◦ Rate = k ◦ [A] t - [A] o = -kt ◦ A plot of [A] versus t produces a straight line with slope -k.

60. Rate Laws: A Summary

61. AP 4/4 ◦ Today you will help me grade the quiz you took Friday and the multiple choice portion of your test. We will also start reaction mechanisms and review what we have done so far. ◦ You will need a piece of paper, a marker, your Kinetics notes, and possibly the reaction mechanisms HW (not due today) ◦ We will finish up kinetics tomorrow and start equilibrium Wednesday. A ◦ Afterschool tutoring tomorrow we will THERMODYNAMICS. WE will go over the questions on your test and talk about test taking strategies for this topic. ◦ You have ONE MONTH before the AP test and we still have a lot to do. IF YOU HAVE NOT STARTED STUDYING, YOU NEED TO DO SO STARTING NOW. IT WILL BE ALMOST IMPOSSIBLE TO PASS THAT TEST WITHOUT SELF DISCIPLINE AND GOOD STUDY HABITS FOR THE NEXT 4 WEEKS.

62. AP 4/5 ◦ Please pick up the papers from the side table ◦ If you turned in your reaction mechanisms paper, please pick that up from the your periods box of graded papers. ◦ If you did not turn in the six question assignment from yesterday, it is due now. We will go over that at the beginning of class. ◦ Today we will discuss reaction mechanisms (12.5) and a model for chemical kinetics (12.6). Tomorrow, we will go over a few half-life problems (just in case they show up on the AP test). We will also go over catalysis tomorrow. Then we will start equilibrium. You have a possible FRQ/multiple choice quiz on Friday. ◦ Today after school, we will go over the thermo test. We should be done by 3:45. The review starts at 3.

63. Answer the following on a clean piece of paper. ◦ 1. What is the integrated rate law? How does this law differ from the differential rate law? ◦ 2. Draw and label a graph for zero, first, and second order rates. ◦ 3. Define reaction mechanisms, molecularity, and intermediates. ◦ 4. Define, describe, and give an example of a unimolecular, bimolecular, and termolecular step. (THE EXAMPLE MUST BE OF A REACTION THAT HAS REACTANTS AND PRODUCTS OF REAL COMPOUNDS) ◦ 5. What is the rate determining step in a series of reactions? ◦ 6. What is the physical significance of rate expressions for elementary steps?

64. 12.5 Reaction Mechanisms ◦ Reaction mechanisms diagram a series of steps in which a chemical reaction occurs. ◦ Intermediate- a species that is neither a product nor a reactant in the overall equation Is used up in a subsequent step Elementary step- a reaction whose rate law can be written from its molecularity (balanced equation) Molecularity- the number of species that must collide to produce the reaction represented by the elementary step.

66. Reaction Mechanisms ◦ Unimolecular step- a reaction step involving only one molecule. ◦ Bimolecular step- a reaction step involving the collision of two molecules (Rate law always 2 nd order)

67. Rate-determining step -slowest step

68. Elementary Reaction -agrees with the balanced equation Reaction mechanisms must: 1. Add up to the overall balanced equation. 2. Agree with the rate law We can’t prove a mechanism absolutely. We can only come up with a possible mechanism.

69. Ex. Elementary Rxn (LIKE NMSI #10): NO + N 2 O  NO 2 + N 2 Rate Law: Rate = k[NO][N 2 O] Ex. The reaction 2NO 2 (g) + F 2 (g)  2NO 2 F is thought to proceed via the following two-step mechanism: NO 2 + F 2  NO 2 F + F slow F + NO 2  NO 2 F fast Rate law for the reaction: Rate = k [NO 2 ][F 2 ]

70. Example NMSI #10

71. Example: #3 from “Practice Problems - Chemical Kinetics” The mechanism of a reaction is shown below. a) What is the overall reaction? b) Which compounds are intermediates? c) Predict the rate law based on this mechanism. d) What is the overall order of the reaction? HOOH + I ¯  HOI + OH ¯ (slow) HOI + I ¯  I 2 + OH ¯ (fast) 2OH ¯ + 2H 3 O +  4 H 2 O(fast)

72. The iodine clock reaction uses a solution of iodate ion (for to which an acidified solution of sodium bisulfite is added.) Iodide ion is generated by the following slow reaction between the iodate and bisulfite: IO 3 − (aq) + 3HSO 3 − (aq) → I − (aq) + 3HSO 4 − (aq) This is the rate determining step. The iodate in excess will oxidize the iodide generated above to form iodine: IO 3 − (aq) + 5I − (aq) + 6H + (aq) → 3I 2 + 3H 2 O (l) However, the iodine is reduced immediately back to iodide by the bisulfite: I 2 (aq) + HSO 3 − (aq) + H 2 O (l) → 2I − (aq) + HSO 4 − (aq) + 2H + (aq) When the bisulfite is fully consumed, the iodine will survive (i.e., no reduction by the bisulfite) to form the dark blue complex with starch. http://jchemed.chem.wisc.edu/JCESoft/CCA/CCA3/MVHTM/CLOCKRX/CLOCK1.HTM We will vary the iodate concentration tomorrow in our experiment!

73. Plot Showing the Number of Collisions with a Particular Energy at T 1 and T 2, where T 2   12.6 A Model for Chemical Kinetics ◦ Molecules must collide to react. Only a small portion of collisions produce a reaction. ◦ Increasing temperature increases reaction speed. Rate and rate constants often double for every 10 o increase in temperature. ◦ Activation energy (E A )- energy that must be overcome to produce a chemical reaction.

74. Rate of reaction depends on E A, not  E.  E has no effect on rate of reaction. ◦ The higher the E A, the slower the reaction at a given temperature. ◦ Molecules and collisions have varying energies. As temperature increases, more collisions will have sufficient energy to overcome the activation energy. As temperature doubles, the fraction of effective collisions increases exponentially.

75. Several Possible Orientations for a Collision Between Two NOBr Molecules Reaction rate is smaller than would be predicted from the number of collisions having sufficient energy to react. This is because of molecular orientations. ◦ Two factors determine if a reaction will occur or not: ◦ 1. Sufficient energy ◦ 2. Proper molecular orientation

76. Catalysis catalyst- substance that speeds up a reaction without being consumed. -produces a new reaction pathway with a lower activation energy -A catalyst lowers the E A for both the forward and reverse reaction.

78. Enzymes are biological catalysts.

79. Types of Inorganic Catalysts ◦ homogeneous catalyst- present in the same phase as the reacting molecules (usually liquid phase) ◦ heterogeneous catalyst- exists in a different phase ◦ Usually involves gaseous reactants being adsorbed on the surface of a solid catalyst (such as a car’s catalytic converter) ◦ Absorption involves penetration. Heterogeneous Catalysis of the Hydrogenation of Ethylene

80. Catalytic Converter