1. Chemical Kinetics

2. Kinetics The study of reaction rates. Spontaneous reactions are reactions that will happen - but we can’t tell how fast. (Spontaneity implies nothing about speed) For example: Diamond will spontaneously turn to graphite – so slow it is not detectable. Reaction mechanism- the steps by which a reaction takes place.

3. Reaction Rate Defined as the change in the concentration of a reactant or product per unit time. Rate=Conc.of A at t 2 -Conc. of A at t 1 t 2 – t 1 t 2 – t 1 Rate = [A] D t

4. For this reaction: N 2 + 3 H 2  2 NH 3 As the reaction progresses the concentration H 2 goes down CONCENTRATIONCONCENTRATION TIME [H2]

5. l As the reaction progresses the concentration N 2 goes down 1/3 as fast [H 2 ] [H2] [N2] CONCENTRATIONCONCENTRATION TIME

6. l As the reaction progresses the concentration NH 3 goes up. [H 2 ] [N 2 ] [NH 3 ]

7. Calculating Rates Average rates are taken over long intervals Instantaneous rates are determined by finding the slope of a line tangent to the curve at any given point because the rate can change over time

8. Factors that Affect Reaction Rates Nature of the reactants Concentration of the reactants Temperature Presence of a catalyst

9. Defining Rate We can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product. In our example: N 2 + 3H 2  2NH 3 -[N 2 ] = -3[H 2 ] = 2[NH 3 ] t t t

10. Rate Laws Reactions are reversible. As products accumulate they can begin to turn back into reactants. Initially, the rate will depend on only the amount of reactants present. Therefore, reactions must be studied at a point soon after the reactants are mixed. This is called the Initial rate method.

11. Rate Laws The concentration of the products do not appear in the rate law because this is an initial rate. The order must be determined experimentally, and can’t be obtained from the equation

12. 2 NO 2  2 NO + O 2 The rate will only depend on the concentration of the reactants. Rate = k[NO 2 ] n This is called a rate law expression. k is called the rate constant. n is the “order” of the reactant -usually a positive integer and must be determined experimentally Overall reaction order is the sum of the orders for each individual reactant.

13. Types of Rate Laws Differential Rate law - describes how rate depends on concentration (often simply called rate law) Integrated Rate Law - describes how concentration depends on time. For each type of differential rate law there is an integrated rate law and vice versa. Rate laws can help us better understand reaction mechanisms.

14. Determining the form of a rate law (Method of Initial Rates) The initial rate is the instantaneous rate right after the reaction begins (t=0) Eliminates the effect of the reverse reaction. The initial concentrations of the reactants are varied. The reaction rate is determined for each trial. See the example problem on p. 536

15. A + 2B  C Trial Initial [A] Initial [B] Initial Rate of formation of [C] 1 0.010 M 1.5 x 10 -6 M/s 2 0.010 M 0.020 M 3.0 x 10 -6 M/s 3 0.020 M 0.010 M 6.0 x 10 -6 M/s

16. The rate law for the reaction is 1.Rate = k [A][B] 2.Rate = k [A] 2 [B] 3.Rate = k [A][B] 2 4.Rate = k [A] 2 [B] 2 1234567891011121314151617181920 21222324252627282930

17. A + B + C  products The rate law is determined to be rate = k[A][B] 2 What happens to the reaction rate when we make the following changes to the concentrations?

18. The concentration of A is doubled, while B and C remain unchanged. 1.The rate remains the same. 2.The rate doubles. 3.The rate is cut in half. 4.The rate is quadrupled. 5.The rate is increased by a factor of 8. :10 1234567891011121314151617181920 21222324252627282930

19. The concentration of B is doubled while A and C remain unchanged. 1.The rate remains the same. 2.The rate doubles. 3.The rate is cut in half. 4.The rate quadruples. 5.The rate increases by a factor of 8. :10 1234567891011121314151617181920 21222324252627282930

20. The concentration of C is doubled while A and B remain unchanged. 1.The rate remains the same. 2.The rate is doubled. 3.The rate is cut in half. 4.The rate is quadrupled. 5.The rate is increased by a factor of 8. :10 1234567891011121314151617181920 21222324252627282930

21. The concentration of all three reactants are doubled simultaneously. 1.The rate remains the same. 2.The rate is doubled. 3.The rate is cut in half. 4.The rate is quadrupled. 5.The rate increases by a factor of 8. 10 123456789 11121314151617181920 21222324252627282930

22. The concentration of A is cut in half while B and C are doubled. 1.The rate remains the same. 2.The rate is doubled. 3.The rate is cut in half. 4.The rate is quadrupled. 5.The rate increases by a factor of 8. :10 1234567891011121314151617181920 21222324252627282930

23. Which of the following is the rate law for a first order reaction? 1.Rate = k[A] 2.Rate = k[A] 2 3.Rate = k[A][B] 4.Rate = k[A] 2 [B] 5.Rate = k[A] 2 [B] 2 :10 1234567891011121314151617181920 21222324252627282930

24. Which is a rate law that is second order with respect to B? 1. Rate = k[A] 2.Rate = k[A] 2 3.Rate = k[A][B] 4.Rate = k[A] 2 [B] 5.Rate = k[A] 2 [B] 2 :10 1234567891011121314151617181920 21222324252627282930

25. Which is the rate law for a third order reaction? 1.Rate = k[A] 2.Rate = k[A] 2 3.Rate = k[A][B] 4.Rate = k[A] 2 [B] 5.Rate = k[A] 2 [B] 2 :10 1234567891011121314151617181920 21222324252627282930

26. Which rate law represents a reaction in which the initial rate will increase by a factor of eight when [A] and [B] are doubled? 1.Rate = k[A] 2.Rate = k[B] 2 3.Rate = k[A][B] 4.Rate = k[A] 2 [B] 5.Rate = k[A] 2 [B] 2 :10 1234567891011121314151617181920 21222324252627282930

27. Which rate law represents a reaction in which the rate will increase by a factor of two when [A] and [B] are doubled? 1.Rate = k[A] 2.Rate = k[B] 2 3.Rate = k[A][B] 4.Rate = k[A] 2 [B] 5.Rate = k[A] 2 [B] 2 :10 1234567891011121314151617181920 21222324252627282930

28. Which rate law represents a reaction in which the rate will not change when [A] is doubled and [B] is held constant? 1.Rate = k[A] 2.Rate = k[B] 2 3.Rate = k[A][B] 4.Rate = k[A] 2 [B] 5.Rate = k[A] 2 [B] 2 :10 1234567891011121314151617181920 21222324252627282930

29. A reaction occurs according to the following rate law: rate = k[A]. If the temperature of the reaction chamber were increased, which of the following would be true? 1.The rate of reaction and the rate constant will increase. 2.The rate of reaction and the rate constant will not change. 3.The rate of reaction will increase and the rate constant will decrease. 4.The rate of reaction will increase and the rate constant will not change. 5.The rate of reaction will not change and the rate constant will increase. :10 1234567891011121314151617181920 21222324252627282930

30. Collision Theory of Reaction Rates For a reaction to occur, particles must collide. Increasing the concentration or reactants results in a greater number of collisions and therefore, a faster rate. Not all collisions are effective collisions.

31. Effective Collisions For a collision to be effective: 1) particles must possess a necessary minimum energy to break bonds and form new ones (activation energy) 2) particles must have proper orientation at the time of the collision. (see page 589)

32. Factors Affecting Rate of Rxn Increasing temperature – ALWAYS increases rate of reaction –Particles collide more frequently and energetically Increasing surface area – increases rate of reaction Increasing concentration – USUALLY increases rate of reaction Presence of Catalysts – lowers activation energy by providing alternate pathways

33. Reaction Mechanism The step by step pathway by which a reaction occurs is called its mechanism. The individual steps are called elementary steps. The rate law for an elementary step can be written from its molecularity. Molecularity is defined as the number of species that must collide to produce the reaction indicated by that step.

34. Examples of Elementary Steps Elementary Step Molecularity Rate Law A  products Unimolecular Rate = k[A] A + A  products Bimolecular Rate =k[A] 2 A + B  products Bimolecular Rate =k[A][B] A + A + B  products Termolecular Rate=k[A] 2 [B] A + B + C  products TermolecularRate=k[A][B][C]

35. Reaction Mechanism (continued) In most mechanisms, one step is slower than the others. A reaction can never occur faster than its slowest step. The slow step is the rate-determining step.

36. Determining Possible Reaction Mechanisms The balanced equation for the overall reaction is equal to the sum of all of the individual steps. The rate law expression matches the coefficients of the rate determining step.

37. Example NO 2 + CO  NO + CO 2 rate = k[NO 2 ] 2 One proposed mechanism: NO 2 + NO 2  N 2 O 4 (slow) NO 2 + NO 2  N 2 O 4 (slow) N 2 O 4 + CO  NO + CO 2 + NO 2 (fast) N 2 O 4 + CO  NO + CO 2 + NO 2 (fast) NO 2 + CO  NO + CO 2 NO 2 + CO  NO + CO 2 N 2 O 4 is an intermediate (forms in one step and is consumed in a later step) This proposed mechanism is consistent with the experimentally determined rate law.

38. Another possible mechanism NO 2 + NO 2  NO 3 + NO (slow) NO 3 + CO  NO 2 + CO 2 (fast) NO 3 + CO  NO 2 + CO 2 (fast) NO 2 + CO  NO + CO 2 NO 2 + CO  NO + CO 2 In order to be a possible mechanism the following two criteria must be met: 1) The individual steps must add up to the overall reaction 2) The mechanism is consistent with the experimentally determined rate law expression

39. Equilibrium Step An equilibrium step is a step in which a product can rapidly re-form the reactants to reach equilibrium. The concentration of products is equal to the concentration of the reactants at equilibrium. An equilibrium step is indicated by a double arrow. ( ) Substitutions can be made in the rate law expression between the reactants and the products.

40. Example 2NO + Br 2  2NOBr 2NO + Br 2  2NOBr rate = k[NO] 2 [Br 2 ] rate = k[NO] 2 [Br 2 ] Proposed mechanism: NO + Br 2 NOBr 2 (fast) NO + Br 2 NOBr 2 (fast) NOBr 2 + NO  2NOBr (slow) NOBr 2 + NO  2NOBr (slow) 2NO + Br 2  2NOBr 2NO + Br 2  2NOBr Rate = k[NOBr 2 ][NO] can be substituted with Rate =k[NO][Br 2 ][NO] or Rate=k[NO] 2 [Br 2 ]

41. A multi step reaction takes place by the following mechanism: A + B  C + D A + C  D + E Which of the species shown is an intermediate in the reaction? 1.A 2.B 3.C 4.D 5.E :10 1234567891011121314151617181920 21222324252627282930

42. Which of the following is the rate law for the predicted mechanism for the production of NO 2 ? NO + O 2 NO 3 (fast) NO 3 + NO  2NO 2 (slow) 1.Rate = k[NO][O 2 ] 2.Rate = k[NO 3 ][NO] 3.Rate = k[NO] 2 [O 2 ] 4.Rate = k[NO 2 ] 2 :10 1234567891011121314151617181920 21222324252627282930

43. Homework problems to be completed by tomorrow Chapter 12: problems 45, 47, 53, 68

44. Integrated Rate Laws Integrated Rate Laws express the reactant concentration as a function of time (instead of rate as a function of the reactant concentration). We will only be considering reactions involving a single reactant.

45. First-Order Rate Laws Consider the following reaction: 2N 2 O 5  4NO 2 + O 2 The rate law (differential) is determined to be Rate = k[N 2 O 5 ] This means that if the concentration of N 2 O 5 is doubled, the rate of production of the products is also doubled.

46. Integrated First-Order Rate Law The previous differential rate law can be integrated and as a result, written in the following form: ln[N 2 O 5 ] = -kt + ln[N 2 O 5 ] 0 Where: ln indicates the natural log t is the time [N 2 O 5 ] is the concentration at time “t” [N 2 O 5 ] o is the initial concentration

47. Integrated First-Order Rate Laws All integrated first order rate laws take the following general form: ln[A] = -kt + ln[A] o Items to note: 1) The equation shows how the concentration of A depends on time 1) The equation shows how the concentration of A depends on time

48. Integrated First-Order Rate Laws ln[A] = -kt + ln[A] o 2) The above equation is of the form y=mx + b, where a plot of y vs. x is a straight line with slope m and intercept b. In the above rate law: In the above rate law: y=ln[A] x=t m=-k b=ln[A] o As a result, plotting ln[A] vs. time always gives a straight line. (This fact is often used to test whether a reaction is 1 st order or not) As a result, plotting ln[A] vs. time always gives a straight line. (This fact is often used to test whether a reaction is 1 st order or not) 3) The integrated rate law for a 1 st order reaction can also be written as: ln([A] o /[A]) = kt

49. Half-Life The time required for a reactant to reach half of its original concentration is called the half-life of a reactant and is designated with the symbol t 1/2. The general equation for the half-life of a first order reaction is t 1/2 =0.693/k (Half-life does not depend on concentration)

50. Second-Order Rate Laws The integrated second order rate law has the form: 1/[A] = kt + 1/[A] o Note: 1) A plot of 1/[A] vs. time produces a straight line with a slope=k. 2) The half-life equation for a 2 nd order reaction is: t 1/2 = 1/k[A] o

51. Zero-Order Rate Laws The rate law for a zero order reaction is: rate = k The integrated rate law for a zero order reaction is [A] = -kt + [A] o Note: 1) A plot of [A] vs. time results is a straight line 2) the half life equation is: t 1/2 = [A] 0 /2k

52. Example Problem p. 569-571 Complete homework problems 32,34,39-42, 44, 46, 48, and 89.