1. Chemical Kinetics Honors Unit 11

2. Chemical Kinetics  Chemical equations do not give us information on how fast a reaction goes from reactants to products.  We can use thermodynamics to tell if a reaction is product – or reactant – favored but how do we know about the speed?

3. Chemical Kinetics  KINETICS = The study of reaction rates and the mechanism (the way the reaction proceeds) *** Only kinetics will tell us how fast the reaction happens!

4. Reaction Rate

5. Rate of Reaction Reaction rate = change in concentration of a reactant or product over time

6. 1) Initial rate = rate at “time zero” 2) Average rate = the rate over a given time interval 3) Instantaneous rate = the slope of the tangent line at a given point Types of Rates

8. Expressing a Rate

9. Collision Theory of Reactants  Reactions occur when molecules collide to exchange or rearrange atoms  Effective collisions occur when molecules have correct energy and orientation

10. Factors Affecting Rates 1. Concentration and physical state of reactants and products o A Rate Law relates the rate of the reaction to the concentration of the reactants 2. Temperature 3. Catalysts o Catalysts are substances that speed up a reaction but are unchanged by the reaction

11. Writing Rate Laws For aA + bB  cC + dD The rate law is: Rate = k[A] m [B] n  k is the rate constant  The exponents must be determined by performing an experiment.  They are NOT derived from the stoichiometry coefficients in an overall chemical equation!!

12. Rate Laws & Orders of Reactions Rate Law for a reaction: Rate = k[A] m [B] n [C] p The exponents m, n, and p:  Are the reaction order  Can be 0, 1, 2, or fractions (may be other whole numbers in fictional examples)  Must be determined by experimentation Overall Order = sum of m, n, and p

13. Interpreting Rate Laws If m = 1 (1 st order) Rate = k [A] 1 If [A] doubles, then the rate doubles (goes up by a factor of 2) If m = 2 (2 nd order) Rate = k [A] 2 If [A] doubles, then rate quadruples (increases rate by a factor of 4) If m = 0 (zero order) Rate = k [A] 0 If [A] doubles, rate does not change! For reaction aA  products Rate = k[A] m

14. Rate Constant, k Relates rate and concentration at a given temperature. General formula for units of k: M (1- overall order) time -1 Overall OrderUnits of k 0M time -1 1Time -1 2M -1 Time -1 3M -2 Time -1

15. Example #1 (part a): Rate Law Problem The initial rate of decomposition of acetaldehyde, CH 3 CHO, was measured at a series of different concentrations and at a constant temperature. CH 3 CHO(g)  CH 4 (g) + CO(g) (a) Using the data, determine the order of the reaction; that is, determine the value of m in the equation Rate = k[CH 3 CHO] m [CH 3 CHO] (mol/L) 0.1620.1950.2730.4100.518 Rate (mol/L*min) 3.154.568.9420.235.2

16. Strategy You are looking at how the concentration affects the rate so compare the two in a proportion! Use the equation: Pick any two points from the given data!

17. Example #1: CH 3 CHO (mol/L)0.1620.1950.2730.4100.518 Rate (mol/L*min) 3.154.568.9420.235.2 CH 3 CHO(g)  CH 4 (g) + CO(g) Rate = k[CH 3 CHO] m

18. Significance of Rate Laws Rate of rxn = k[CH 3 CHO] 2 Here the rate goes up by FOUR when the initial concentration doubles. Therefore, the reaction is SECOND order overall.

19. Example #1 (part b & c): Using the same set of data from part a, and knowing the order of the reaction, determine: b) the value of the rate constant, k (w/ units!) c) the rate of the reaction when [CH 3 CHO] = 0.452 mol/L Strategy:  Use any set of data to find k.  Solve for rate using k, rate order equation, and given concentration.

20. Example #1 (part b & c):

21. Example #2: The data below is for the reaction of nitrogen (II) oxide with hydrogen at 800 o C. 2NO(g) + 2H 2 (g)  N 2 (g) + 2H 2 O(g) (a) Determine the order of the reaction with respect to both reactants and write the rate law (b) calculate the value of the rate constant, and (c) determine the rate of formation of product when [NO]=0.0024 M and [H 2 ]=0.0042 M. Strategy:  Choose two experiments where the concentration of one reactant is constant and the other is changing!  Solve for m and n separately

22. Experiment[NO][H 2 ] 10.00100.00400.12 20.00200.00400.48 30.00300.00401.08 40.00400.00100.48 50.00400.00200.96 60.00400.00301.44 Example #2:

23. Experiment[NO][H 2 ] 10.00100.00400.12 20.00200.00400.48 40.00400.00100.48 50.00400.00200.96 Example #2:

24. Experiment[NO][H 2 ] 10.00100.00400.12 20.00200.00400.48 40.00400.00100.48 50.00400.00200.96 Example #2:

25. Example #3: The initial rate of a reaction A + B  C was measured with the results below. Write the rate law, the value of the rate constant (with units), and the rate of reaction when [A] = 0.050 M and [B] = 0.100 M. Experiment [A] (M)[B] (M)Initial Rate (M/s) 1 0.1 4.0x10 -5 20.10.24.0x10 -5 30.20.116.0x10 -5

26. Example #3:

27. Potential Energy Diagrams Molecules need a minimum amount of energy for a reaction to take place.  Activation energy (E a ) – the minimum amount of energy that the reacting species must possess to undergo a specific reaction Activated complex - a short-lived molecule formed when reactants collide; it can return to reactants or form products.  Formation depends on the activation energy & the correct geometry (orientation)

28. Potential Energy Diagrams

31. Catalyzed Pathway Catalysts lower activation energy!!!

32. Reaction Mechanisms Mechanism – how reactants are converted to products at the molecular level Most reactions DO NOT occur in a single step! They occur as a series of elementary steps (a single step in a reaction).

33. Rate Determining Step Rate determining step – the slowest step in a reaction COCl 2 (g)  COCl (g) + Cl (g)fast Cl (g) + COCl 2 (g)  COCl (g) + Cl 2 (g) slow 2 COCl (g)  2 CO (g) + 2 Cl (g)fast 2 Cl (g)  Cl 2 (g)fast

34. Rate Determining Step COCl 2 (g)  COCl (g) + Cl (g)fast Cl (g) + COCl 2 (g)  COCl (g) + Cl 2 (g) slow 2 COCl (g)  2 CO (g) + 2 Cl (g)fast 2 Cl (g)  Cl 2 (g)fast 2 COCl 2 (g)  2 Cl 2 (g) + 2 CO (g) Adding elementary steps gives the net (or overall) reaction!

35. Intermediates  Intermediates are produced in one elementary step but reacted in another NO (g) + O 3 (g)  NO 2 (g) + O 2 (g) NO 2 (g) + O (g)  NO (g) + O 2 (g) O 3 (g) + O (g)  2 O 2 (g)

36. Catalysts Catalyst – a reactant in an elementary step but unchanged at the end of the reaction  A substance that speeds up the reaction but is not permanently changed by the reaction  Both an original reactant and a final product NO (g) + O 3 (g)  NO 2 (g) + O 2 (g) NO 2 (g) + O (g)  NO (g) + O 2 (g) O 3 (g) + O (g)  2 O 2 (g)

37. Example #4 Cl 2 (g)  2 Cl (g) Fast Cl (g) + CHCl 3 (g)  CCl 3 (g) + HCl (g) Slow CCl 3 (g) + Cl (g)  CCl 4 (g)Fast Identify:  The rate determining step  The overall (net) reaction  The identity of any intermediates  The identity of any catalysts

38. Example #5 H 2 O 2 (aq) + I 1- (aq)  H 2 O(l) + IO 1- (aq) Slow H 2 O 2 (aq) + IO 1- (aq)  H 2 O(l) + O 2 (g) + I 1- (aq) Fast Identify:  The rate determining step  The overall (net) reaction  The identity of any intermediates  The identity of any catalysts

39. Example #6 O 3 (g) + Cl (g)  O 2 (g) + ClO (g) Slow ClO (g) + O (g)  Cl (g) + O 2 (g) Fast Identify:  The rate determining step  The overall (net) reaction  The identity of any intermediates  The identity of any catalysts

40. Chemical Equilibrium UNIT 11 (PART 2) Pb 2+ (aq) + 2 Cl - (aq)  PbCl 2 (s)

41. What is equilibrium? Definition (dictionary.com): a state of rest or balance due to the equal action of opposing forces Chemical Equilibrium: A process where a forward and reverse reaction occur at equal rates ***Not all chemical reactions are reversible!!!

42. General Characteristics of Equilibrium 1.DYNAMIC (in constant motion) 2.REVERSIBLE 3.Can be approached from either direction (reaction can run in the forward direction or the reverse direction) 1.DYNAMIC (in constant motion) 2.REVERSIBLE 3.Can be approached from either direction (reaction can run in the forward direction or the reverse direction) Pink to blue Co(H 2 O) 6 Cl 2  Co(H 2 O) 4 Cl 2 + 2 H 2 O Blue to pink Co(H 2 O) 4 Cl 2 + 2 H 2 O  Co(H 2 O) 6 Cl 2 Video Clip - Cobalt Chloride Complex Equilibrium

43. Characteristics of Dynamic Equilibrium After a period of time, the concentrations of reactants and products are constant. The forward and reverse reactions continue after equilibrium is attained. + Fe 3+ + SCN -  FeSCN 2+

44. Examples of Chemical Equilibria Phase changes such as: H 2 O(s)  H 2 O(liq)

45. Graphing Dynamic Equilibrium Equilibrium achieved when product and reactant concentrations remain constant!!

46. The Equilibrium Expression, K eq K eq = equilibrium constant (for a given T) Brackets "[ ]" = concentration (molarity) "a, b, c, and d" = coefficients from balanced equation The "c" in K c = concentration (K c = a special K eq based on concentration)

47. There are two cases when a species is not shown in the equilibrium expression: #1: SOLIDS – (s) after the formula #2: pure LIQUIDS – (l) after the formula

48. Example #7 Write the equilibrium expression for the oxidation- reduction reaction occurring between iron(III) chloride and tin(II) chloride: 2 FeCl 3 (aq) + SnCl 2 (aq)  2 FeCl 2 (aq) + SnCl 4 (aq)

49. Example #8 Write the equilibrium expression for the replacement of silver ions by copper: Cu (s) + 2 Ag + (aq)  Cu 2+ (aq) + 2 Ag (s)

50. “If a system at equilibrium is stressed, the system tends to shift its equilibrium position to counter the effect of the stress.” Le Chatelier’s Principle

51. How does a “stress” influence equilibrium? The impact of addition of reactants on reaction rate

52. The Seesaw Analogy

53. “Stresses” to a System 1)Changes in amount of species 2)Changes in pressure or volume 3)Changes in temperature 4)Adding an inert substance 5)Adding a catalyst

54. “Stresses” (factors) that can cause changes at equilibrium 1) Changes in amount of species o Add reactant; system shifts to theRIGHT! (produces more products) o Add product; system shifts to theLEFT! (produces more reactants) o Remove reactant; system shifts toLEFT! o Remove product; system shifts toRIGHT!

55. Example #9: Predict the direction of shift based on the following concentration changes for the reaction: CH 4 (g) + 2S 2 (g)  CS 2 (g) + 2H 2 S(g) (A) Some S 2 (g) is added. (B) Some CS 2 (g) is added. (C) Some H 2 S(g) is removed. (D) Some argon gas (an inert gas) is added.

56. “Stresses” (factors) that can cause changes at equilibrium 2) Changes in pressure or volume o If P goes down (same as V goes up), system shifts to increased # of moles of gas o If P goes up (same as V goes down), system shifts to decreased # of moles of gas

57. Example #10: Predict the effect of increasing pressure (decreasing volume) on each of the following reactions. (a) CH 4 (g) + 2S 2 (g)  CS 2 (g) + 2H 2 S(g) (b) H 2 (g) + Br 2 (g)  2HBr(g) (c) CO 2 (g) + C(s)  2CO(g) (d) PCl 5 (g)  PCl 3 (g) + Cl 2 (g)

58. “Stresses” (factors) that can cause changes at equilibrium 3) Changes in temperature o Write heat as a product (exothermic) or reactant (endothermic) o System shifts to get rid of added heat: o System shifts LEFT for exo. reactions as T goes up o System shifts RIGHT for endo. reactions as T goes up

59. Example #11: Predict the effect of increasing temperature on each of the following reactions: (a) CO (g) + 3 H 2 (g)  CH 4 (g) + H 2 O (g)ΔH < 0 (b) CO 2 (g) + C (s)  2 CO (g) ΔH > 0 (c) 4 NH 3 (g) + 5 O 2 (g)  4 NO (g) + 6 H 2 O (g) EXO. (d) 2 H 2 O (g)  2 H 2 (g) + O 2 (g) ENDO.

60. “Stresses” (factors) that can cause changes at equilibrium 4) Adding an inert substance o If a substance is NOT in the reaction (or in the K eq expression) any changes will have NO EFFECT on equilibrium! Ex.) If CO 2 is added to the system below, there would be no effect on the equilibrium. H 2 (g) + Br 2 (g)  2 HBr (g)

61. “Stresses” (factors) that can cause changes at equilibrium 5) Adding a catalyst o What do catalysts do? They increase the rate of the reaction! o Adding a catalyst will not affect equilibrium. It only changes the rate at which you reach equilibrium.

62. Example #12: How can the reaction below be shifted to the right? List all possibilities! CO(g) + 2 H 2 (g) CH 3 OH (g)

63. Example #13: How can the reaction below be shifted to the right? This process (the Haber process) is used in the industry to produce ammonia. List all possibilities! N 2 (g) + 3H 2 (g)  2NH 3 (g) ∆H = negative

64. N 2 (g) + 3 H 2 (g)  2 NH 3 (g) + heat N 2 (g) + 3 H 2 (g)  2 NH 3 (g) + heat K = 3.5 x 10 8 at 298 K K = 3.5 x 10 8 at 298 K Haber-Bosch Process for NH 3