1. Chemical Kinetics

2. Kinetics Chemical kinetics is the branch of chemistry concerned with the rate of chemical reactions and the mechanism by which chemical reactions occur. Presented by Mark Langella, PWISTA.com

3. Collision Theory  In order for two particles to react chemically, they must collide. Not only must they collide, but it must be an “effective collision.” That is, they must have the correct amount of energy and collide with the proper orientation in space.  Any factor which increases the likelihood that they will collide will increase the rate of the chemical reaction. Presented by Mark Langella, PWISTA.com

4. FACTORS WHICH AFFECT THE RATE OF A CHEMICAL REACTION (Bonds Must Break)  Surface Area/ Contact Area (Opportunity for collisions)  Concentration ( Increase Frequency)  Temperature ( Increase Frequency)  Catalyst ( Effective Collisions)  Nature of Reactants ( Effective collisions) Presented by Mark Langella, PWISTA.com

5. Surface Area /Contact Area  Sugar Cube vs. Packet of Sugar  Steel vs Steel Wool  Two Sided Candle  Coffee Creamer  Alka-tablets Presented by Mark Langella, PWISTA.com

6. Concentration  Sudsy Kinetics  Magnesium in Various Concentrated Acids  Which Reactant has the greater concentration effect ( that’s another story) Presented by Mark Langella, PWISTA.com

7. Which of the following hydrogen peroxide and sodium iodide mixtures will give the greatest rate of reaction? Presented by Mark Langella, PWISTA.com

8. Temperature  Spoiling of Fruit  Alka Seltzer Hot and Cold  Light Stick Kinetics Presented by Mark Langella, PWISTA.com

9. Temp vs. Reaction rate  The effect of temperature on the rate of a chemical reaction can best be explained in terms of the kinetic molecular theory. The higher the temperature, the faster molecules move. The faster they more, the faster they collide and therefore, react at a faster rate. Presented by Mark Langella, PWISTA.com

10. Every ten degrees rate doubles Presented by Mark Langella, PWISTA.com

11. Arrhenius Equation - Collisions must have enough energy to produce the reaction (must equal or exceed the activation energy). - Orientation of reactants must allow formation of new bonds. Presented by Mark Langella, PWISTA.com

12. Temperature

13. Plot showing the number of collisions with a particular energy at T 1 & T 2, where T 2 > T 1 -- Boltzman Distribution. Presented by Mark Langella, PWISTA.com

14. Arrhenius Equation (continued)  k = rate constant  A = frequency factor  E a = activation energy  T = temperature  R = gas constant Presented by Mark Langella, PWISTA.com

15. Arrhenius Equation  If the natural logarithm of each side of the Arrhenius Equation is taken, the following equation results:  ln(k) = -E a /R (1/T) + ln(A)  y = mx + b   m = -E a /R when ln(k) is plotted versus 1/T. Presented by Mark Langella, PWISTA.com

17. THE RELATIONSHIP BETWEEN k AND T Presented by Mark Langella, PWISTA.com

18. Since the second form of the equation follows the form, y = mx + b, then a graph of ln k vs 1/T should give a straight line. The activation energy of the reaction may be calculated from the slope of the ln k vs 1/T plot. Presented by Mark Langella, PWISTA.com

19. If you have data of two points Presented by Mark Langella, PWISTA.com

24. Final Notes on Kinetics As T increases, so does k. As T increases, so does k. E a &  E are independent of T. E a &  E are independent of T. A catalyst lowers E a and increases the rate of both k f & k r. A catalyst lowers E a and increases the rate of both k f & k r. Presented by Mark Langella, PWISTA.com

25. Light Stick Kinetics  Bend a light stick to break the glass tube inside mixing the hydrogen peroxide with the phenyl oxalate ester. Repeat this several times to eliminate any large sections of glass and expedite mixing. Place the light stick aside for two to three hours. Your instructor may have activated and aged the light sticks for you. Presented by Mark Langella, PWISTA.com

26.  Prepare the data collection device by slipping the one-hole #00 stopper prepared for this experiment over the temperature probe. It might be helpful to lubricate the hole in the stopper with silicone oil or spray. Cut the top from an aged and activated light stick with a sharp knife or scissors. Divide its contents equally with another lab group into two small test tubes. Presented by Mark Langella, PWISTA.com

27.  Place the temperature probe into the mixture. Adjust the position of the temperature probe so that it is centered in the solution but just visible when you sight into the light sensor hole. It must not extend too deep as to interfere with the light sensor readings. Clamp the foam block onto the ring stand with a large clamp. Gently force the light sensor into the hole prepared for it on the side of the foam block. It should be deep enough to fit securely but not touch the test tube. Presented by Mark Langella, PWISTA.com

28.  Place the switch on the light sensor to the 600 lux sensitivity. Plug the light sensor into Channel one (CH1) of a powered Lab Pro and the temperature probe into Channel two (CH2).  Connect the LabPro to your computer by means of the appropriate cable. Double click the Logger Pro icon on your computer. It should auto-ID both sensors. If not, check your connections and try again. If your instructor has a file setup for this experiment, open it now. Presented by Mark Langella, PWISTA.com

29.  Remove the test tube and temperature probe from the foam block, placing it into a beaker of hot (45 - 50°C) water. Using the temperature probe as a handle, stir the water with the test tube observing the temperature until it remains more or less constant. Remove the test tube from the water, dry the outside, and insert it into the foam block.  When everything is ready, click on the green “COLLECT” icon on the icon toolbar, and observe the data. The experiment will run for fifteen minutes. Presented by Mark Langella, PWISTA.com

30.  When the measurement is complete, disassemble the apparatus, returning all items to their designated areas. Dispose of the spent light stick solution as indicated by your instructor.  Clean both the temperature probe and the test tube with a brush using soap. Presented by Mark Langella, PWISTA.com

31.  1 a) Click on the label, Lumination (lux) on the vertical axis of your graph. Select the natural log of the light intensity, “ln Illumination,” for the vertical axis. Click on the label, “Time (s),” on the horizontal axis of your graph. Select “Inverse Temperature.” Auto scale your graph. Your graph now displays the ln intensity vs 1/temperature data necessary to find the activation energy. Presented by Mark Langella, PWISTA.com

32.  1 (b) Drag your curser across the most linear portion of your log I vs. 1/T plot. The linear section should be highlighted by a dark gray rectangle. Select the linear fit icon in the icon toolbar. The best fit solution will be displayed. Record these values.  1 (c) Calculate the activation energy for the light stick reaction from the slope found in Question #1b above. Express your answer in kilojoules per mole (kJ/mol). Presented by Mark Langella, PWISTA.com

33. 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. Presented by Mark Langella, PWISTA.com

34. Energy plots for a catalyzed and an uncatalyzed pathway for an endothermic reaction. Presented by Mark Langella, PWISTA.com

35. Effect of a catalyst on the number of reaction-producing collisions. A greater fraction of collisions are effective for the catalyzed reaction. Presented by Mark Langella, PWISTA.com

36. Heterogeneous Catalysis  1.Adsorption and activation of the reactants.  2.Migration of the adsorbed reactants on the surface.  3.Reaction of the adsorbed substances.  4.Escape, or desorption, of the products. Steps: Oscillating Flask Demo Presented by Mark Langella, PWISTA.com

37. Homogeneous Catalysis  Catalyst is in the same phase as the reacting molecules.  NO (g) + 1/2O 2(g) ----> NO 2(g)  NO 2(g) ----> NO (g) + O (g)  O 2(g) + O (g) ----> O 3(g)  3/2 O 2(g) ----> O 3(g)  What is the catalyst in this reaction?  What are the intermediates? Presented by Mark Langella, PWISTA.com

38. Nature of Reactant  Whoosh Bottle  Closed vials with 2 cm of White phenolphthalein in NaCl added to calcium Hydroxide White phenolphthalein in NaCl added to calcium Hydroxide Ferric Chloride added to Sodium Acetate Ferric Chloride added to Sodium Acetate Presented by Mark Langella, PWISTA.com

39. Spontaneity tendency for a reaction to occur. tendency for a reaction to occur. does not mean that the reaction will be fast! does not mean that the reaction will be fast! Diamonds will spontaneously change into graphite, but the process is so slow that it is not detectable. Diamonds will spontaneously change into graphite, but the process is so slow that it is not detectable. Presented by Mark Langella, PWISTA.com

40. Reaction Mechanism the steps by which a chemical process occurs. the steps by which a chemical process occurs. allows us to find ways to facilitate reactions. allows us to find ways to facilitate reactions. can be changed by the use of a catalyst. can be changed by the use of a catalyst. Presented by Mark Langella, PWISTA.com

41. Reaction Rate  Change in concentration (conc) of a reactant or product per unit time. Presented by Mark Langella, PWISTA.com

43. Reaction Rate  It is customary to work with positive reaction rates, so a negative sign is used in some cases to make the rate positive.  Rates determined over a period of time are called average rates.  Instantaneous rate equals the negative slope of the tangent line. Presented by Mark Langella, PWISTA.com

44. The concentrations of nitrogen dioxide, nitric oxide, oxygen plotted versus time. Presented by Mark Langella, PWISTA.com

45. Representation of the reaction of 2 NO 2(g) ---> 2 NO (g) + O 2(g). a) t = 0 b) & c) with increased time, NO 2 is changed into NO and O 2. Presented by Mark Langella, PWISTA.com

46. Rate Laws (differential)  Rate = k[NO 2 ] n  k = rate constant  n = rate order Presented by Mark Langella, PWISTA.com

47. Types of Rate Laws  Differential Rate Law: expresses how rate depends on concentration.  Integrated Rate Law: expresses how concentration depends on time. Presented by Mark Langella, PWISTA.com

48. Rate Laws Summary Differential rate law -- rate of a reaction depends on concentration. Differential rate law -- rate of a reaction depends on concentration. Integrated rate law -- concentration depends on time. Integrated rate law -- concentration depends on time. Rate laws normally only involve concentrations of reactants. Rate laws normally only involve concentrations of reactants. Experimental determination of either rate law is sufficient. Experimental determination of either rate law is sufficient. Experimental convenience dictates which rate law is determined experimentally. Experimental convenience dictates which rate law is determined experimentally. Rate law for a reaction often indicates reaction mechanism. Rate law for a reaction often indicates reaction mechanism. Presented by Mark Langella, PWISTA.com

49. Plot of the concentration of N 2 O 5 as a function of time for the reaction 2N 2 O 5(soln) ---> 4NO 2(soln) + O 2(g). Rate at 0.90 M is twice the rate at 0.45 M. Presented by Mark Langella, PWISTA.com

50. Method of Initial Rates  Initial Rate: the “instantaneous rate” just after the reaction begins.  The initial rate is determined in several experiments using different initial concentrations. Presented by Mark Langella, PWISTA.com

51. Overall Reaction Order  Sum of the order of each component in the rate law.  rate = k[H 2 SeO 3 ][H + ] 2 [I  ] 3  The overall reaction order is 1 + 2 + 3 = 6. Presented by Mark Langella, PWISTA.com

52. First-Order Rate Law  Integrated first-order rate law is  ln[A] =  kt + ln[A] o  y = mx + b  If a reaction is first-order, a plot of ln[A] versus time is a straight line. For aA  Products in a 1st-order reaction, Presented by Mark Langella, PWISTA.com

53. Half-Life of a First-Order Reaction  t 1/2 = half-life of the reaction  k = rate constant  For a first-order reaction, the half-life does not depend on concentration. Presented by Mark Langella, PWISTA.com

54. Plot of [N 2 O 5 ] versus time for the decomposition reaction of N 2 O 5. Note that the half-life for a 1st order reaction is constant. Presented by Mark Langella, PWISTA.com

55. Second-Order Rate Law  For aA  products in a second-order reaction,  Integrated rate law is y = mx + b If a reaction is second-order, a plot of 1/[A] versus time is a straight line. Presented by Mark Langella, PWISTA.com

56. Half-Life of a Second-Order Reaction  t 1/2 = half-life of the reaction  k = rate constant  A o = initial concentration of A  The half-life is dependent upon the initial concentration. Presented by Mark Langella, PWISTA.com

57. Zero-Order Rate Law  For aA---> products in zero-order reaction,  Rate = k[A] o = k  The integrated rate law is  [A] = -kt + [A] 0  y = mx + b  If a reaction is zero-order, the plot of [A] versus time is a straight line.  Example -- surface of a solid catalyst cannot hold a greater concentration of reactant. Presented by Mark Langella, PWISTA.com

58. KNOW THIS TABLE!!!! Presented by Mark Langella, PWISTA.com

59. A Summary  1.Simplification: Conditions are set such that only forward reaction is important.  2.Two types: differential rate law  integrated rate law  3.Which type? Depends on the type of data collected - differential and integrated forms can be interconverted. Presented by Mark Langella, PWISTA.com

60. A Summary (continued)  4.Most common: method of initial rates.  5.Concentration v. time: used to determine integrated rate law, often graphically.  6.For several reactants: choose conditions under which only one reactant varies significantly (pseudo first-order conditions). Presented by Mark Langella, PWISTA.com

61. Reaction Mechanism - The series of steps by which a chemical reaction occurs. - A chemical equation does not tell us how reactants become products - it is a summary of the overall process. Presented by Mark Langella, PWISTA.com

62. Reaction Mechanism (continued)  The reaction  has many steps in the reaction mechanism. Presented by Mark Langella, PWISTA.com

63. Often Used Terms  Intermediate: formed in one step and used up in a subsequent step and so is never seen as a product.  Molecularity: the number of species that must collide to produce the reaction indicated by that step.  Elementary Step: A reaction whose rate law can be written from its molecularity.  uni, bi and termolecular Presented by Mark Langella, PWISTA.com

65. Reaction Mechanism Requirements  1.The sum of the elementary steps must give the overall balanced equation for the reaction.  2.The mechanism must agree with the experimentally determined rate law. Presented by Mark Langella, PWISTA.com

66. Rate-Determining Step  In a multistep reaction, it is the slowest step. It therefore determines the rate of reaction. Presented by Mark Langella, PWISTA.com

67. Fast Equilibrium Reaction  When the 1st step is not the slow step, but a fast equilibrium, the rate can be determined as follows:  2A + B ---> C 2nd order in B, 1st order in A.  Step 1 A + B D (Fast equilibrium)  Step 2 D + B ----> E Slow  Step 3 E + A ----> C + B Fast  A + B D  D + B ---> E  E + A ---> C + B  2 A + B ---> C 1st requirement that elementary steps equal overall reaction is met. Presented by Mark Langella, PWISTA.com

68. Fast Equilibrium Reaction (continued)  k f [A][B] = k r [D] (fast)  [D] = k f /k r ([A][B])  Rate = k 2 [D][B] (slow)  Substitute fast equation in terms of [D] into slow reaction.  Rate = k 2 k f /k r ([A][B][B])  Rate = k[A][B] 2  2nd requirement is also met, this is, then, a possible mechanism. Presented by Mark Langella, PWISTA.com