1. AP CHEMISTRY CHAPTER 12 KINETICS

2. 2 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

3. Reaction Rate- change in concentration of a reactant or product per unit time. [A] = concentration in mol/L Rate =  [A]  t

4. If the rate expression involves a reactant: Rate = -  [A]  t (negative because [ ] decreases) The above gives the average rate.

6. 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)

8. 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)

9. Differential Rate Law or Rate Equation For the reaction aA+bB + …  gG +hH+ … Rate = k[A] m [B] n … [A] & [B] represent molarities

10. The exponents are positive or negative, integers or fractions. usually positive integers (small whole numbers) k = rate constant -value depends on reaction, temperature and presence of a catalyst -faster the reaction, larger the k value

11. 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.

12. The units of k can be calculated by reaction orders and units of concentration and rate. For example, if rate is in mol/L.s in the above rate law, we can find the units for k as follows: Rate = k so k = mol [A][B] 2 L. s mol mol 2 L L 2 This simplifies to: k = L 2 mol 2. s

13. 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

14. 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.

15. If doubling the initial [ ] of a reactant causes the initial rate to quadruple, the reaction is second order in that reactant.

16. If doubling the initial [ ] of a reactant causes the initial rate to increase 8 times, the reaction is third order in that reactant.

17. 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.

18. Ex. 2A + B  2C [A] [B] Rate. 0.1 0.2 0.10 0.1 0.4 0.20 0.2 0.4 0.80 Rate = k[A] 2 [B] overall reaction order is 3

19. A + 2B  2C [A] [B] Rate 0.1 0.1 0.10 0.2 0.1 0.20 0.4 0.2 0.40 Rate = k[A][B] 0 = k[A] overall reaction order is 1

20. 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

21. A + B  C Exp [A] [B] Rate 1 0.10 0.20 1.0 x 10 -5 2 0.20 0.20 1.0 x 10 -5 3 0.20 0.40 2.0 x 10 -5 Determine the rate law: Rate = k[A] x [B] y Rate = k[A] 0 [B] y Rate = k [B]

22. NH 4 + + NO 2 -  N 2 + 2H 2 Exp [NH 4 + ] [NO 2 - ] Rate(mol/Lmin) 1 0.100 0.005 1.35 x 10 -7 2 0.100 0.010 2.70 x 10 -7 3 0.200 0.010 5.40 x 10 -7 Determine the rate law: Determine the value of k and its units: Rate = k[NH 4 + ] x [NO 2 - ] y Rate = k[NH 4 + ] x [NO 2 - ] 1 Rate = k[NH 4 + ] [NO 2 - ] 1.35 x 10 -7 mol/Lmin = k(0.100)(0.005) k = 2.7 x 10 -4 L/mol. min

23. 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.

24. 1 st order integrated rate law ln [A] 0 = kt [A] t

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

27. 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 = 86400s 1 hr ln[A] 0 = kt ln 1.00 x 10 -2 = 1.16 x 10 -6 (86400) [A] t [A] t [A] t = 9.05 x 10 -3

28. Radioactive decay is first order. 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

29. 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.

30. A Plot of (N 2 O 5 ) Versus Time for the Decomposition Reaction of N 2 O 5

31. 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

32. 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 ln1.00 = 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

33. 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

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

35. Zero Order Rate = k [A] t - [A] o = -kt A plot of [A] versus t produces a straight line with slope -k.

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

37. Reaction Mechanisms 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)

38. molecularity- the number of species that must collide to produce the reaction represented by the elementary step. 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)

39. Rate-determining step -slowest step Lucy clip Elementary Reaction -agrees with the balanced equation Lucy clip

40. Reaction mechanisms must: 1. Add up to the overall balanced equation. 2. Agree with the rate law

41. We can’t prove a mechanism absolutely. We can only come up with a possible mechanism.

42. Ex. Elementary Rxn : NO + N 2 O  NO 2 + N 2 Rate Law: R = k[NO][N 2 O]

43. 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 ]

44. When an intermediate is a reactant in the rate-determining step, the derivation of the rate law is more difficult.

45. Ex. NO 2 + CO  NO + CO 2 k 1 Mechanism: NO 2 + NO 2  NO 3 + NO Both fast w/ k -1 equal rates NO 3 + CO  CO 2 + NO 2 slow

46. Ex. NO 2 + CO  NO + CO 2 k 1 Mechanism: NO 2 + NO 2  NO 3 + NO Both fast w/ k -1 equal rates NO 3 + CO  CO 2 + NO 2 slow Slow step determines rate law (rate-determining step) Rate law: R = k[NO 3 ][CO] But, NO 3 was an intermediate. We must come up with something equal to NO 3 to substitute. k[NO 2 ] 2 = k[NO 3 ][NO] [NO 3 ] = [NO 2 ] 2 [NO] R = k [NO 2 ] 2 [CO] [NO]

47. Ex. Cl 2 + CHCl 3  HCl + CCl 4 Mechanism k 1 Cl 2  2Cl fast k -1 Cl + CHCl 3  HCl + CCl 3 slow Cl + CCl 3  CCl 4 fast Rate Law:

48. Ex. Cl 2 + CHCl 3  HCl + CCl 4 Mechanism k 1 Cl 2  2Cl fast k -1 Cl + CHCl 3  HCl + CCl 3 slow Cl + CCl 3  CCl 4 fast Rate Law: R = k[Cl][CHCl 3 ] Cl is an intermediate k[Cl 2 ] = k[Cl] 2 [Cl 2 ] 1/2 = [Cl] R = k[Cl 2 ] 1/2 [CHCl 3 ]

49. Increasing temperature increases reaction speed. Rate and rate constants often double for every 10 o increase in temperature.

50. Plot Showing the Number of Collisions with a Particular Energy at T 1 and T 2, where T 2  

51. Molecules must collide to react. Only a small portion of collisions produce a reaction.

52. Several Possible Orientations for a Collision Between Two BrNO Molecules

53. Activation energy (E A ) - energy that must be overcome to produce a chemical reaction.

54. 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.

55. 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. Reaction rate is smaller than would be predicted from the number of collisions having sufficient energy to react. This is because of molecular orientations.

56. 2 factors: 1. sufficient energy 2. proper orientation

57. Arrhenius Equation- -can be used to calculate E A ln k = ln A - E A k = rat e constant RT R = 8.314 J/K mol T = Kelvin temp A = frequency factor (constant as temperature changes) E A = activation energy As E A increases, k decreases.

58. The Arrhenius equation describes a line. We can plot 1/T vs ln k and get a straight line whose slope is equal to -E A /R.

59. Plot of In(k) Versus 1/T for the Reaction 2N 2 O 5 (g)    g) + O 2 (g) Slope = -E A /R

60. A variation of the Arrhenius equation can be used to calculate E A or to find k at another temperature if E A is known: ln k 1 = E A 1 - 1 k 2 R T 2 T 1

61. Ex. The rate constant of a first order reaction is 3.46 x 10 -2 s -1 at 298 K. What is the rate constant at 350 K if the activation energy for the reaction is 50.2 kJ/mol? ln k 1 = E A 1 - 1 k 2 R T 2 T 1 ln 3.46 x 10 -2 = 50200 1 - 1 k 2 8.314 350 298 k 2 = 7.02 x 10 -1 s -1

62. Catalysis

63. 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.

66. enzymes- biological catalysts

67. homogeneous catalyst- present in the same phase as the reacting molecules (usually liquid phase)

68. 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.

69. Heterogeneous Catalysis of the Hydrogenation of Ethylene

70. Catalytic Converter