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1 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Today… Turn in: –Nothing Our Plan:

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Presentation on theme: "1 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Today… Turn in: –Nothing Our Plan:"— Presentation transcript:

1 1 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Today… Turn in: –Nothing Our Plan: –Test Results –Videos/Notes –Investigation 10 Pre-Lab Homework (Write in Planner): –Be prepared for Investigation 10 next class and have report started.

2 2 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Introduction to Chemical Kinetics What is Kinetics? I’ll let Hank explain… EBc Stop at 3:20

3 3 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Kinetics in Action… Clock Reactions… wDuiYhttp://www.youtube.com/watch?v=BqeWpy wDuiY

4 4 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Unit Preview… Some Reactions Are fast: –Acid/base neutralization –Sodium + water –PPT reactions Are slow: –Aluminum oxidation –Iron oxidizing –Plastic bottle decomposing

5 5 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen And some depend… On temperature: fireflies On conditions: Iron oxide (humid or desert) On enzymes: many processes in our body On a catalyst: 2CO + 2NO --> 2CO 2 + N 2 (automobile pollution and catalytic converters)

6 6 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Chemical kinetics is the study of: –the rates of chemical reactions –factors that affect these rates –the mechanisms by which reactions occur Reaction rates vary greatly – some are very fast (burning, precipitation) and some are very slow (rusting, disintegration of a plastic bottle in sunlight). Chemical Kinetics: A Preview

7 7 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Variables in Reaction Rates Concentrations of reactants: Reaction rates generally increase as the concentrations of the reactants are increased. Temperature: Reaction rates generally increase rapidly as the temperature is increased. Surface area: For reactions that occur on a surface rather than in solution, the rate increases as the surface area is increased. Catalysts: Catalysts speed up reactions and inhibitors slow them down.

8 8 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Think of it like this… chemical-reactions-and-get-a-datehttp://ed.ted.com/lessons/how-to-speed-up- chemical-reactions-and-get-a-date

9 9 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Theories of Chemical Kinetics: Collision Theory Before atoms, molecules, or ions can react, they must first collide. An effective collision between two molecules puts enough energy into key bonds to break them. The activation energy (E a ) is the minimum energy that must be supplied by collisions for a reaction to occur. A certain fraction of all molecules in a sample will have the necessary activation energy to react; that fraction increases with increasing temperature. The spatial orientations of the colliding species may also determine whether a collision is effective.

10 10 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Distribution of Kinetic Energies At higher temperature (red), more molecules have the necessary activation energy.

11 11 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Key Point Orientation of molecules at the time of their collision will determine whether they react or not!

12 12 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Analogy - Car Crash Example Energy and orientation of cars during a car crash can establish the change that occurs to the cars.

13 13 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Importance of Orientation One hydrogen atom can approach another from any direction … … and reaction will still occur; the spherical symmetry of the atoms means that orientation does not matter. Effective collision; the I atom can bond to the C atom to form CH 3 I Ineffective collision; orientation is important in this reaction.

14 14 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Transition State Theory The configuration of the atoms of the colliding species at the time of the collision is called the transition state. The transitory species having this configuration is called the activated complex. A reaction profile shows potential energy plotted as a function of a parameter called the progress of the reaction. Reactant molecules must have enough energy to surmount the energy “hill” separating products from reactants.

15 15 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Activated complex Species formed as a result of collisions between energetic molecules that is an intermediate between the reactants and the products of a reaction. Once formed the activated complex dissociates either into products or back to the reactants

16 16 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Reaction Profile A graphical representation of a chemical reaction in terms of the energies of the reactants, activated complexes and products

17 17 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Reaction Profile

18 18 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Exothermic

19 19 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Endothermic

20 20 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen CO (g) + NO 2 (g)  CO 2 (g) + NO (g) A Reaction Profile

21 21 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen An Analogy for Reaction Profiles and Activation Energy

22 22 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Stop! Investigation 10 - We’re going to test the effect of different variables on the rate of reaction next class. Complete steps 1 – 6 on the lab handout and be prepared to conduct the lab next class.

23 23 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Today… Turn in: –Nothing Our Plan: –Investigation 10 Homework (Write in Planner): –Lab Report Due Friday

24 24 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Today… Turn in: –Lab Report (rubric on top) Our Plan: –Notes & Practice Homework (Write in Planner): –Work on the Ch. 13 Homework

25 25 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen The Meaning of Rate The rate of a reaction is the change in concentration of a product per unit of time (rate of formation of product). Rate is also viewed as the negative of the change in concentration of a reactant per unit of time (rate of disappearance of reactant). The rate of reaction often has the units of moles per liter per unit time (mol∙L –1 ∙s –1 or M∙s –1 )

26 26 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen If the rate of consumption of H 2 O 2 is 4.6 M/h, then … … the rate of formation of H 2 O must also be 4.6 M/h, and … … the rate of formation of O 2 is 2.3 M/h

27 27 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen 2 H 2 O 2 --> 2 H 2 O + O 2 … mol H 2 O 2 reacted in 60 s. Rate = mol H 2 O 2 /L 60 s = M H 2 O 2 s –1 1 L g O 2 ( mole) produced in 60 s means …

28 28 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Example 13.1 Consider the hypothetical reaction A + 2B --> 3C + 2D Suppose that at one point in the reaction, [A] = M and 125 s later [A] = M. During this time period, what is the average (a) rate of reaction expressed in M∙s –1 and (b) rate of formation of C, expressed in M∙min –1.

29 29 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Try It Out! EX 13.1 A: Consider the hypothetical reaction 2A + B → 2C + D Suppose that at some point during the reaction [D] = M and that 2.55 min (that is 2 min 33 sec) later [D] = M. a)What is the average rate of reaction during this time period, expressed in M min -1 ? b)What is the rate of formation of C expressed in M s -1 ?

30 30 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Rate of Reaction Expressed as the negative of the Slope of a Tangent Line Average rate (green dotted line) Initial Rate ( blue solid line) Instantaneous Rate (red line) Question: Over what time interval are the instantaneous rates greater than the average rate measured for the 600 sec period?

31 31 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Average vs. Instantaneous Rate Instantaneous rate is the slope of the tangent to the curve at a particular time. We often are interested in the initial instantane- ous rate; for the initial concentrations of reactants and products are known at this time.

32 32 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Example 13.2 Use data from Table 13.1 and/or Figure 13.5 to (a)determine the initial rate of reaction and (b)calculate [H 2 O 2 ] at t = 30 s.

33 33 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Try It Out EX 13.2 A: From Figure 13.5, a)Determine the instantaneous rate of reaction at t = 300 s. b)Use the result of a) to calculate a value of [H 2 O 2 ] at t = 310 s.

34 34 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen The Rate Law of a Chemical Reaction The rate law for a chemical reaction relates the rate of reaction to the concentrations of reactants. The exponents (m, n, p…) are determined by experiment. Exponents are not derived from the coefficients in the balanced chemical equation, though in some instances the exponents and the coefficients may be the same. The value of an exponent in a rate law is the order of the reaction with respect to the reactant in question. The proportionality constant, k, is the rate constant. aA + bB + cC …→ products rate = k[A] n [B] m [C] p …

35 35 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Rate Law Examples Rate = k[A] 1 = k[A]Reaction is first order in A Rate = k[A] 3 Reaction is third order in A Rate = k[A] 2 Reaction is second order in A If we triple the concentration of A in a second-order reaction, the rate increases by a factor of ________.

36 36 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Effect of Concentration on Rate Order of RxnConcentration Change Effect on Rate 0Double/Triple…Nothing (2 0 ) 1DoubleDouble (2 1 ) 1TripleTriple (3 1 ) 2DoubleQuadruple (2 2 ) 2Triple8x (3 2 ) 2Quadruple16x (4 2 ) 3Double9x (2 3 ) 3Triple27x (3 3 )

37 37 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen More About the Rate Constant k The rate of a reaction is the change in concentration with time, whereas the rate constant is the proportionality constant relating reaction rate to the concentrations of reactants. The rate constant remains constant throughout a reaction, regardless of the initial concentrations of the reactants. The rate and the rate constant have the same numerical values and units only in zero-order reactions. For reaction orders other than zero, the rate and rate constant are numerically equal only when the concentrations of all reactants are 1 M. Even then, their units are different.

38 38 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen To find the overall order of a reaction… Add the orders for each compound. Example: –rate = [A] 2 [B] 1 is 3 rd order overall –How about rate = [A] 0 [B] 1 [C] 1 ?

39 39 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Units of k (p. 536) Overall Reaction Order Units of k ZeroM s -1 Firsts -1 SecondM -1 s -1 ThirdM -2 s -1

40 40 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Method of Initial Rates The method of initial rates is a method of establishing the rate law for a reaction—finding the values of the exponents in the rate law, and the value of k. A series of experiments is performed in which the initial concentration of one reactant is varied. Concentrations of the other reactants are held constant. When we double the concentration of a reactant A, if: –there is no effect on the rate, the reaction is zero-order in A. –the rate doubles, the reaction is first-order in A. –the rate quadruples, the reaction is second-order in A. –the rate increases eight times, the reaction is third-order in A.

41 41 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen The concentration of NO was held the same in Experiments 1 and 2 … … while the concentration of Cl 2 in Experiment 2 is twice that of Experiment 1. The rate in Experiment 2 is twice that in Experiment 1, so the reaction must be first order in Cl 2. Which two experiments are used to find the order of the reaction in NO? How do we find the value of k after obtaining the order of the reaction in NO and in Cl 2 ?

42 42 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Example 13.3 For the reaction 2 NO(g) + Cl 2 (g) → 2 NOCl(g) described in the text and in Table 13.2, (a) what is the initial rate for a hypothetical Experiment 4, which has [NO] = M and [Cl 2 ] = M? (b) What is the value of k for the reaction?

43 43 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen So, when looking at data….. Zero order reaction: initial rate is unaffected (2 0 ) 1 st order reaction: double the concentration, doubles the rate (2 1 =2) 2 nd Order Reaction: Initial rate increases fourfold (2 2 =4) 3 rd Order Reaction: Initial rate increases eightfold (2 3 =8)

44 44 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Try It Out Below is some rate data for the hypothetical reaction, 2A + B --> C. What is the rate law for this reaction? Experiment[A] 0 [B] 0 Rate (M/s) 12.0 M1.0 M M M1.0 M0.100

45 45 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Let’s Take a Break With your partner, Complete Part 1 of the Partner Review

46 46 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Sum of exponents is equal to zero

47 47 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Rate of Reaction= k[A] 0 The concentration time graph is a straight line with a negative slope Zero Order

48 48 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Zero Order Rate of the reaction: –IS equal to k –remains constant throughout the reaction –is the negative of the slope of the line when graph Molarity vs. time (see pg. 544)

49 49 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Rate is independent of initial concentration A Zero-Order Reaction rate = k[A] 0 = k

50 50 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Zero Order Integrated rate equation: [A] t = -kt + [A] 0 y = mx + b

51 51 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Zero Order (cont.) y = [A] t = conc of A at some time x = t = time b = [A] 0 m = -k (m, the slope of the straight line)

52 52 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen First-Order Reactions In a first-order reaction, the exponent in the rate law is 1. Rate = k[A] 1 = k[A] Look! It’s an equation for a straight line! ln [A] t – ln [A] 0 = –kt ln [A] t = –kt + ln [A] 0 At times, it is convenient to replace molarities in an integrated rate law by quantities that are proportional to concentration. The integrated rate law describes the concentration of a reactant as a function of time. For a first-order process: ln [A] t [A] 0 = –kt

53 53 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Rate of reaction= k[A] 1 Integrated rate equation ln[A] t = -kt + ln[A] 0 y = mx + b

54 54 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen 1 st Order Easy test for a first order reaction is to plot the natural log of reactant conc. vs. time and see if the graph is linear.

55 55 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Decomposition of H 2 O 2

56 56 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Example 13.4 For the first-order decomposition of H 2 O 2 (aq), given k = 3.66 x 10 –3 s –1 and [H 2 O 2 ] 0 = M, determine (a) the time at which [H 2 O 2 ] = M and (b) [H 2 O 2 ] after 225 s.

57 57 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Another Example H initially at a conc of 2.32 M, is allowed to decompose. What will the [H ] be 1200 s later? Use k = 7.3 x s -1 for this first order decomposition.

58 58 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Try It Out EX 13.4 A: The decomposition of nitramide, NH 2 NO 2, is a first order reaction: NH 2 NO 2 (aq) → H 2 O (l) + N 2 O (g) The rate law is rate = k[NH 2 NO 2 ], with k = 5.62 x 10-3 min-1 at 15C. Starting with M NH 2 NO 2, a)At what time will [NH 2 NO 2 ] = M b)What is [NH 2 NO 2 ] after 6.00 hours?

59 59 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Half-life Time required for ½ of the reactant to be consumed Equation: t 1/2 = ln2 = k k

60 60 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Example 13.5 Use data from Figure 13.7 (p. 542) to evaluate the a)half-life and b)Rate constant for the first order decomposition of N 2 O 5 at 67 C: N 2 O 5 (g) → 2NO 2 (g) + ½ O 2 (g)

61 61 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Example 13.5 A Use the result of Example 13.5 to determine a)The time required to reduce the quantity of N 2 O 5 to 1/16 of its initial value and b)The mass of N 2 O 5 remaining after a 4.80 g sample of N 2 O 5 has decomposed for 10.0 min.

62 62 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Sum of exponents = 2

63 63 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Second-Order Reactions A reaction that is second order in a reactant has a rate law in which the exponent for that reactant is 2. Rate = k[A] 2 The integrated rate law has the form: The half-life of a second-order reaction depends on the initial concentration as well as on the rate constant k: 1 1 –––– = kt + –––– [A] t [A] 0 What do we plot vs. time to get a straight line? 1 t ½ = ––––– k[A] 0

64 64 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Example 13.7 The second-order decomposition of HI(g) at 700 K is represented in Figure HI(g) → ½ H 2 (g) + ½ I 2 (g) Rate = k[HI] 2 What are the: (a) rate constant and (b) half-life of the decomposition of 1.00 M HI(g) at 700 K?

65 65 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Try It Out EX 13.7 A: If, in the second order reaction A → products, it takes 55 s for the concentration of reactant A to fall to 0.40 M from an initial concentration of 0.80 M, what is the rate constant k for the reaction?

66 66 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Summary of Kinetic Data

67 67 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Short Cut!! [A] vs time (straight line = Zero order) ln [A] vs. time ( straight line = first order) 1/[A] vs. time (straight line = second order)

68 68 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Try it in your note packet! Time, min[A]ln[A]1/[A]

69 69 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Crash Course Review 3VEBchttp://www.youtube.com/watch?v=7qOFtL 3VEBc 3:10 – 5:30

70 70 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Let’s Take a Break Complete Part 2 of the Partner Review.

71 71 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Today… Turn in: –Get out a piece of notebook paper Our Plan: –Scavenger Hunt Review –Investigation 11 Pre-Lab Homework (Write in Planner): –Prepare Lab Report –Homework Problems (Due 2/12)

72 72 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Pre Lab Complete steps 1 – 5 on the handout and have your formal lab report started. Be prepared to experiment next class.

73 73 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Today… Turn in: –Nothing Our Plan: –Investigation 11 Homework (Write in Planner): –Lab Report due Next Class (2/10)

74 74 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Today… Turn in: –Lab Report (rubric on top) Our Plan: –Review Activity –Elephant Toothpaste Demo –Notes – Catalysts & Rate Determining Steps –Finish Homework Problems Homework (Write in Planner): –Homework Problems

75 75 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Effect of Temperature on the Rates of Reactions In 1889, Svante Arrhenius proposed the following expression for the effect of temperature on the rate constant, k: k = Ae – E a /RT The constant A, called the frequency factor, is an expression of collision frequency and orientation; it represents the number of collisions per unit time that are capable of leading to reaction. The term e – E a /RT represents the fraction of molecular collisions sufficiently energetic to produce a reaction.

76 76 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen I like this equation better… Look at Eq on p. 551 of the text!

77 77 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Example 13.9 Estimate a value of k at 375 K for the decomposition of dinitrogen pentoxide illustrated in Figure 13.15, given that k = 2.5 x 10 –3 s –1 at 332 K.

78 78 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Try it Out EX 13.9 B: Di-tert-butyl peroxide (DTBP) is used as a catalyst in the manufacture of polymers. In the gaseous state, DTBP decomposes to acetone and ethane by a first- order reaction. (C 4 H 9 ) 2 O 2 (g) → 2(CH 3 ) 2 CO (g) + C 2 H 6 (g) The half-life of DTBP is 17.5 h at 125C and 1.67 h at 145C. What is the activation energy, E a, of the decomposition reaction?

79 79 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Reaction Mechanisms Analogy: a banana split is made by steps in sequence: slice banana; three scoops ice cream; chocolate sauce; strawberries; pineapple; whipped cream; end with cherry. A chemical reaction occurs according to a reaction mechanism—a series of collisions or dissociations— that lead from initial reactants to the final products. Like making a banana split, three molecules will not collide simultaneously very often, so steps of a reaction mechanism involve only one or two reactants at a time.

80 80 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Reaction Mechanisms An elementary reaction represents, at the molecular level, a single step in the progress of the overall reaction. A proposed mechanism must: –account for the experimentally determined rate law. –be consistent with the stoichiometry of the overall or net reaction.

81 81 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Molecularity The molecularity of an elementary reaction refers to the number of free atoms, ions, or molecules that collide or dissociate in that step. Termolecular processes are unusual, for the same reason that three basketballs shot at the same time are unlikely to collide at the same instant …

82 82 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen The Rate-Determining Step The rate-determining step is the crucial step in establishing the rate of the overall reaction. It is usually the slowest step. Some two-step mechanisms have a slow first step followed by a fast second step, while others have a fast reversible first step followed by a slow second step. Slow Fast Mechanism for 2 NO + O 2 --> 2 NO 2

83 83 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen An Example Given the reaction: 2A + 2B → C + Drate = k[A] 2 [B] Could take place by the following three-step mechanism: I.A + A ↔ X (fast) II.X + B → C + Y (slow) III.Y + B → D (fast)

84 84 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Intermediates X & Y are called intermediates because they appear in the mechanism, but they cancel out of the balanced equation. They are products from one reaction and then a reactant in the next.

85 85 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen An Example The steps of a reaction mechanism must add up to equal the balanced equation with all intermediates cancelling out. Let’s try it with our example. I.A + A ↔ X (fast) II.X + B → C + Y (slow) III.Y + B → D (fast)

86 86 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Rate-determining Step As in any process where many steps are involved, the speed of the whole process can’t go faster than the speed of the slowest step in the process. The slowest step of a reaction is the rate- determining step. Because the slowest step is the most important step in determining the rate of a reaction, the slowest step and the steps leading up to it are used to see if the mechanism is consistent with the rate law for the overall reaction.

87 87 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen An Example Let’s look at our reaction again and show that it is consistent with the rate law: 2A + 2B → C + Drate = k[A] 2 [B] I.A + A ↔ X (fast) II.X + B → C + Y (slow) III.Y + B → D (fast)

88 88 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen An Example (p. 199 Cracking AP Chemistry Exam) 1.Write rate law for slow step. 2.X is an intermediate, we need to eliminate it from the rate law. 3.Solve for [X] in terms of [A]. 4.Substitute for [X] in our second step. 5.Now we have the rate law.

89 89 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Fast step: I 2 2 I Slow step: 2 I + H 2 2 HI k1k1 k –1 k2k2 Example For the reaction H 2 (g) + I 2 (g) --> 2 HI(g), a proposed mechanism is below. What is the net equation for the overall reaction, and what is the order of the reaction according to this mechanism?

90 90 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen EX A The decomposition of nitrosyl choride, 2NOCl (g) → 2NO (g) + Cl 2 (g) Is a first-order reaction. Propose a mechanism for this reaction consisting of one fast step and one slow step.

91 91 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Catalyst A catalyst provides an alternative reaction pathway of lower activation energy Participates in a chemical reaction w/o undergoing permanent change Speeds up without being consumed in the reaction. It is neither a reactant nor product in a reaction.

92 92 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Catalysis – how does it work In general, a catalyst works by changing the mechanism of a chemical reaction. Often the catalyst is consumed in one step of the mechanism, but is regenerated in another step. The pathway of a catalyzed reaction has a lower activation energy than that of an uncatalyzed reaction, so more molecules at a fixed temperature have the necessary activation energy.

93 93 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Effect of Catalyst on Reaction Profile and Activation Energy A catalyst lowers the activation energy, making it easier for the reactants to “climb the energy hill” and form the products.

94 94 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen A Kinetics Pick Up Line…

95 95 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Homogeneous Catalysis Ozone decomposition catalyzed by chlorine atoms has a much lower activation energy and proceeds much more rapidly than the uncatalyzed reaction

96 96 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Heterogeneous Catalysis Many reactions are catalyzed by the surfaces of appropriate solids. A good catalyst provides a higher frequency of effective collisions. Four steps in heterogeneous catalysis: –Reactant molecules are adsorbed. –Reactant molecules diffuse along the surface. –Reactant molecules react to form product molecules. –Product molecules are desorbed (released from the surface).

97 97 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Heterogeneous Catalysis Hydrogen is adsorbed onto the surface of a nickel catalyst. A C=C approaches … … and is adsorbed. Hydrogen atoms attach to the carbon atoms, and the molecule is desorbed.

98 98 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen A Surface-Catalyzed Reaction

99 99 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Enzyme Catalysis Enzymes are high-molecular-mass proteins that usually catalyze one specific reaction, or a set of similar reactions. The reactant substance, called the substrate (S), attaches itself to an area on the enzyme (E) called the active site, to form an enzyme-substrate complex (ES). The enzyme–substrate complex decomposes to form products (P), and the enzyme is regenerated.

100 100 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Factors Influencing Enzyme Activity The rates of enzyme-catalyzed reactions are influenced by factors such as concentration of the substrate, concentration of the enzyme, acidity of the medium, and temperature.

101 101 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Mercury Poisoning: An Example of Enzyme Inhibition When Hg reacts with an enzyme … … the Hg binds to sulfur atoms … … changing the shape of the active site, so that it no longer “fits” the substrate.

102 102 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Homogeneous Catalysis Step 1:A + catalyst → intermediate + C Step 2: B + intermediate → D + catalyst Net equation: A + B → C + D

103 103 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Identifying Catalysts and intermediate species O 3 + Cl∙ --> ClO∙ + O 2 ClO· + O∙ --> Cl∙ + O 2 Catalyst? Intermediate? Net Equation?

104 104 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Crash Course Review 3VEBchttp://www.youtube.com/watch?v=7qOFtL 3VEBc 7:15 - end

105 105 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Finish the HW All homework problems are due on Wednesday!

106 106 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Today… Before Class: –Mark Homework Questions on the Board Our Plan: –Homework Questions/Check HW –Worksheet Race –Study Guide Homework (Write in Planner): –Test Next Class –Breakfast Club 6 am on Friday

107 107 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Study Guide Changes 7 – do 39 a only 11 – do 25 a and b only

108 108 General Chemistry 4 th edition, Hill, Petrucci, McCreary, Perry Hall © 2005 Prentice Hall © 2005 Chapter Thirteen Today… Turn in: –Nothing Our Plan: –Study Guide Questions –Test Homework (Write in Planner): –Complete the POGIL (Day 1 ONLY)


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