1. Chemical Kinetics The area of chemistry that concerns reaction rates and reaction mechanisms.

2. Outline: Kinetics Reaction Rates How we measure rates. Rate Laws How the rate depends on amounts of reactants. Integrated Rate Laws How to calc amount left or time to reach a given amount. Half-life How long it takes to react 50% of reactants. Arrhenius Equation How rate constant changes with T. Mechanisms Link between rate and molecular scale processes.

3. Factors That Affect Reaction Rates Concentration of Reactants – As the concentration of reactants increases, so does the likelihood that reactant molecules will collide. Temperature – At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy. Catalysts – Speed rxn by changing mechanism.

4. Reaction Rate The change in concentration of a reactant or product per unit of time

5. UNITS FOR RATE Mol/L s Mol/L hr Mol/L day Etc…………………….

6. 2NO 2 (g)  2NO(g) + O 2 (g) Reaction Rates: 2. Can measure appearance of products 1. Can measure disappearance of reactants 3. Are proportional stoichiometrically

7. 2NO 2 (g)  2NO(g) + O 2 (g) Reaction Rates: 4. Are equal to the slope tangent to that point  [NO 2 ] tt 5. Change as the reaction proceeds, if the rate is dependent upon concentration

8. Products and reactants are stoichiometrically related 2 NO 2 → 2 NO + O 2 If the rate of disappearance of NO 2 is 2.0 mol/L s what is the rate of appearance for NO and O 2 ?

9. Page 598 – Problem 17

10. Rate Laws Differential rate laws express (reveal) the relationship between the concentration of reactants and the rate of the reaction. Integrated rate laws express (reveal) the relationship between concentration of reactants and time The differential rate law is usually just called “the rate law.”

11. What is the difference between rate and time?

12. How to write a rate expression R = k [A] x [B] y X and Y represent the order to which the reactant is raised.

13. HOW DO YOU DETERMINE THE ORDER INITIAL RATES METHOD

14. Writing a (differential) Rate Law 2 NO(g) + Cl 2 (g)  2 NOCl(g) Problem - Write the rate law, determine the value of the rate constant, k, and the overall order for the following reaction: Experiment[NO](mol/L) [Cl 2 ] (mol/L)RateMol/L·s 10.2500.250 1.43 x 10 -6 20.5000.250 5.72 x 10 -6 30.2500.500 2.86 x 10 -6 40.5000.500 11.4 x 10 -6

15. Log Rate 3 = (Conc 3 ) log Rate 4 (Conc 4 ) Experiment[NO](mol/L) [Cl 2 ] (mol/L)RateMol/L·s 10.2500.250 1.43 x 10 -6 20.5000.250 5.72 x 10 -6 30.2500.500 2.86 x 10 -6 40.5000.500 11.4 x 10 -6

16. Writing a Rate Law Part 3 – Determine the overall order for the reaction. R = k[NO] 2 [Cl 2 ] Overall order is the sum of the exponents, or orders, of the reactants 2+1 = 3  The reaction is 3 rd order

17. Page 598 – problem 22

18. Page 598 – problem 24

19. Page 599 – problems 26

20. SHORT CUT FOR DETERMINING ORDER If the concentration increases and the rate stays the same – zero order If the concentration doubles and the rate doubles – 1 st order If the concentration doubles and the rate increases by 4 – 2 nd order

21. 2What if you have more than one reactant and one is not held constant? EXPERIMENTInitial [ClO 2 - ] mol/L 2 Initial [F 2 ] mol/LInitial Rate of Increase of [ClO 2 F] 10.0100.102.4 x 10 -3 20.0100.409.6 x 10 -3 30.0200.209.6 x 10 -3 a.Write the rate law for the reaction 2 ClO 2 + → 2 ClO 2 F b.Calculate the numerical value of the rate constant and specify the unit. c.In experiment 2, what is the initial rate of decrease of [F 2 ]. d.Which of the following rate law developed in a? Justify your choice. ClO 2 + F 2 ↔ ClO 2 F 2 fast ClO 2 F 2 → ClO 2 F + F slow ClO 2 + F → ClO 2 F fast or F 2 → 2 F slow 2 ClO 2 + 2 F → ClO 2F fast

22. UNITS FOR K

23. Determining Order with Concentration vs. Time data (the Integrated Rate Law) Zero Order: First Order: Second Order:

24. Rate Laws Summary Zero Order First Order Second Order Rate Law Rate = kRate = k[A]Rate = k[A] 2 Integrated Rate Law [A 0 ] – [A] = ktln[A] = kt [A] 0 1- 1 [A] [A 0 ] = kt Plot the produces a straight line [A] versus tln[A] versus t Relationship of rate constant to slope of straight line Slope = -k Slope = k Half-Life

25. Solving an Integrated Rate Law Time (s) [H 2 O 2 ] (mol/L) 01.00 1200.91 3000.78 6000.59 12000.37 18000.22 24000.13 30000.082 36000.050 Problem: Find the integrated rate law and the value for the rate constant, k A graphing calculator with linear regression analysis greatly simplifies this process!!

26. Time vs. [H 2 O 2 ] Time (s) [H 2 O 2 ] 01.00 1200.91 3000.78 6000.59 12000.37 18000.22 24000.13 30000.082 36000.050 Find 2 points use ∆Y/∆X = A Find 2 more points ∆Y/∆X = appx A The slope of both should be close. Pick the plot that is closest to the same numbers

27. Time vs. ln[H 2 O 2 ] Time (s) ln[H 2 O 2 ] 00 120-0.0943 300-0.2485 600-0.5276 1200-0.9943 1800-1.514 2400-2.04 3000-2.501 3600-2.996 Regression results:

28. Time vs. 1/[H 2 O 2 ] Time (s) 1/[H 2 O 2 ] 01.00 1201.0989 3001.2821 6001.6949 12002.7027 18004.5455 24007.6923 300012.195 360020.000

29. And the winner is… Time vs. ln[H 2 O 2 ] 1. As a result, the reaction is 1 st order 2. The (differential) rate law is: 3. The integrated rate law is: 4. But…what is the rate constant, k ?

30. Finding the Rate Constant, k Method #1: Calculate the slope from the Time vs. ln[H 2 O 2 ] table. Time (s) ln[H 2 O 2 ] 00 120-0.0943 300-0.2485 600-0.5276 1200-0.9943 1800-1.514 2400-2.04 3000-2.501 3600-2.996 Now remember:  k = -slope k = 8.32 x 10 -4 s -1

31. Determining rxn order The decomposition of NO 2 at 300°C is described by the equation NO 2 (g)NO (g) + 1/2 O 2 (g) and yields these data: Time (s)[NO 2 ], M 0.00.01000 50.00.00787 100.00.00649 200.00.00481 300.00.00380

32. Graphing ln [NO 2 ] vs. t yields: Time (s)[NO 2 ], Mln [NO 2 ] 0.00.01000-4.610 50.00.00787-4.845 100.00.00649-5.038 200.00.00481-5.337 300.00.00380-5.573 The plot is not a straight line, so the process is not first- order in [A]. Determining rxn order Does not fit:

33. Second-Order Processes A graph of 1/[NO 2 ] vs. t gives this plot. Time (s)[NO 2 ], M1/[NO 2 ] 0.00.01000100 50.00.00787127 100.00.00649154 200.00.00481208 300.00.00380263 This is a straight line. Therefore, the process is second-order in [NO 2 ].

34. Finding the Rate Constant, k Method #2: Obtain k from the linear regresssion analysis. Now remember:  k = -slope k = 8.35 x 10 -4 s -1 Regression results: y = ax + b a = -8.35 x 10 -4 b = -.005 r 2 = 0.99978 r = -0.9999

35. Reaction Mechanism The reaction mechanism is the series of elementary steps by which a chemical reaction occurs.  The sum of the elementary steps must give the overall balanced equation for the reaction  The mechanism must agree with the experimentally determined rate law

36. Rate-Determining Step In a multi-step reaction, the slowest step is the rate-determining step. It therefore determines the rate of the reaction. The experimental rate law must agree with the rate-determining step

37. Identifying the Rate-Determining Step For the reaction: 2H 2 (g) + 2NO(g)  N 2 (g) + 2H 2 O(g) The experimental rate law is: R = k[NO] 2 [H 2 ] Which step in the reaction mechanism is the rate-determining (slowest) step? Step #1 H 2 (g) + 2NO(g)  N 2 O(g) + H 2 O(g) Step #2 N 2 O(g) + H 2 (g)  N 2 (g) + H 2 O(g) Step #1 agrees with the experimental rate law

38. Identifying Intermediates For the reaction: 2H 2 (g) + 2NO(g)  N 2 (g) + 2H 2 O(g) Which species in the reaction mechanism are intermediates (do not show up in the final, balanced equation?) Step #1 H 2 (g) + 2NO(g)  N 2 O(g) + H 2 O(g) Step #2 N 2 O(g) + H 2 (g)  N 2 (g) + H 2 O(g) 2H 2 (g) + 2NO(g)  N 2 (g) + 2H 2 O(g)  N 2 O(g) is an intermediate

39. Collision Model Key Idea: Molecules must collide to react. However, only a small fraction of collisions produces a reaction. Why?

40. The Collision Model Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation.

41. Activation Energy In other words, there is a minimum amount of energy required for reaction: the activation energy, E a. Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier.

42. Collision Model Collisions must have sufficient energy to produce the reaction (must equal or exceed the activation energy). Colliding particles must be correctly oriented to one another in order to produce a reaction. 1. 2.

43. Factors Affecting Rate Increasing temperature always increases the rate of a reaction.  Particles collide more frequently  Particles collide more energetically Increasing surface area increases the rate of a reaction Increasing Concentration USUALLY increases the rate of a reaction Presence of Catalysts, which lower the activation energy by providing alternate pathways

44. Endothermic Reactions

45. Exothermic Reactions

46. Maxwell–Boltzmann Distributions As the temperature increases, the curve flattens and broadens. Thus at higher temperatures, a larger population of molecules has higher energy.

47. The Arrhenius Equation k = rate constant at temperature T  k = rate constant at temperature T  A = frequency factor  E a = activation energy  R = Gas constant, 8.31451 J/K·mol

48. The Arrhenius Equation, Rearranged  Simplifies solving for E a  -E a / R is the slope when ( 1/T ) is plotted against ln(k)  ln(A) is the y-intercept  Linear regression analysis of a table of (1/T) vs. ln(k) can quickly yield a slope  E a = -R(slope)

49. Catalysis : A substance that speeds up a reaction without being consumed Catalyst: A substance that speeds up a reaction without being consumed : A large molecule (usually a protein) that catalyzes biological reactions. Enzyme: A large molecule (usually a protein) that catalyzes biological reactions. : Present in the same phase as the reacting molecules. Homogeneous catalyst: Present in the same phase as the reacting molecules. : Present in a different phase than the reacting molecules. Heterogeneous catalyst: Present in a different phase than the reacting molecules.

50. Lowering of Activation Energy by a Catalyst

51. Catalysts One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break.

52. Enzymes Enzymes are catalysts in biological systems. The substrate fits into the active site of the enzyme much like a key fits into a lock.

53. Catalysts Increase the Number of Effective Collisions

54. Heterogeneous Catalysis Step #1: Adsorption and activation of the reactants. Carbon monoxide and nitrogen monoxide adsorbed on a platinum surface

55. Heterogeneous Catalysis Step #2: Migration of the adsorbed reactants on the surface. Carbon monoxide and nitrogen monoxide arranged prior to reacting

56. Heterogeneous Catalysis Step #3: Reaction of the adsorbed substances. Carbon dioxide and nitrogen form from previous molecules

57. Heterogeneous Catalysis Step #4: Escape, or desorption, of the products. Carbon dioxide and nitrogen gases escape (desorb) from the platinum surface

58. Slow Initial Step The rate law for this reaction is found experimentally to be Rate = k [NO 2 ] 2 CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. This suggests the reaction occurs in two steps. NO 2 (g) + CO (g)  NO (g) + CO 2 (g)

59. Slow Initial Step A proposed mechanism for this reaction is Step 1: NO 2 + NO 2  NO 3 + NO (slow) Step 2: NO 3 + CO  NO 2 + CO 2 (fast) The NO 3 intermediate is consumed in the second step. As CO is not involved in the slow, rate-determining step, it does not appear in the rate law.

60. Fast Initial Step A proposed mechanism is Step 1 is an equilibrium- it includes the forward and reverse reactions.