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

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

3. 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

4. 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

5. 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.”

6. 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

8. Writing a Rate Law Part 1 – Determine the values for the exponents in the rate law: 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 1.14 x 10 -5 In experiment 1 and 2, [Cl 2 ] is constant while [NO] doubles. Rate = k[NO] x [Cl 2 ] y The rate quadruples, so the reaction is second order with respect to [NO] In experiment 2 and 4, [NO] is constant while [Cl 2 ] doubles. The rate doubles, so the reaction is first order with respect to [Cl 2 ] 12

9. Writing a Rate Law Part 2 – Determine the value for k, the rate constant, by using any set of experimental data: Experiment[NO](mol/L) [Cl 2 ] (mol/L)RateMol/L·s 10.2500.250 1.43 x 10 -6 Rate = k[NO] 2 [Cl 2 ]

10. Writing a Rate Law Part 3 – Determine the overall order for the reaction. Rate = 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

11. Complete problem 33…now!

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

13. 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!!

14. 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 y = ax + b a = -2.64 x 10 -4 b = 0.841 r 2 = 0.8891 r = -0.9429 Regression results:

15. 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: y = ax + b a = -8.35 x 10 -4 b = -.005 r 2 = 0.99978 r = -0.9999

16. 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 y = ax + b a = 0.00460 b = -0.847 r 2 = 0.8723 r = 0.9340 Regression results:

17. 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 ?

18. 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

19. 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

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

25. Complete problem 40…now!

27. Green food coloring reacts with bleach. After 4.00 min the remaining [FC]=0.0300M. What is the initial concentration of [FC] 0 ?

29. What’s your k value ? Highest Rate Wins!!!!

30. Complete problem 48

31. Complete problem 50

32. Complete problem 52

33. Complete problem 54

34. 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

35. 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

36. 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: Rate = 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

37. 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. Complete problem 61…soon!

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

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

48. 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

49. 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)

50. Complete problem 71

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

53. Lowering of Activation Energy by a Catalyst

54. Catalysts Increase the Number of Effective Collisions

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

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

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

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