1. 18 - 1 Chemical Kinetics Rates of Reactions Rate and Concentration Finding Rate Laws First-Order Reactions Rate and Temperature Theory of Reaction Rates Reaction Mechanisms Catalysts Steady-State Approximations

2. 18 - 2 Rates of reactions Rate of a chemical reaction. The change in the quantity of a reactant or product that takes place in a period of time. rate = = concentration later - concentration earlier time later - time earlier  [ ]  t

3. 18 - 3 Rates of reactions To study rates of reaction, you must: Identify the reactants and products. Carry out the reaction. Measure the concentrations of one of the reactants or products at known intervals. You must have a way to measure at least one of the species involved. Continuous monitoring methods should be used whenever possible.

4. 18 - 4 An example reaction Decomposition of N 2 O 5. Dinitrogen pentoxide is known to decompose completely by the following reaction. 2N 2 O 5 (g) 2N 2 O 4 (g) + O 2 (g) This reaction can be conducted in an inert solvent like carbon tetrachloride. When N 2 O 5 decomposes, N 2 O 4 remains in solution and O 2 escapes and can be measured.

5. 18 - 5 An example reaction We can easily measure the oxygen as dinitrogen pentoxide decomposes. Temperature must be maintained to within +0.01 o C. The reaction flask must be shaken to keep oxygen from forming a supersaturated solution. It is found that the reaction initially occurs very rapidly but gradually slows down.

6. 18 - 6 An example reaction Gas buret Constant temperature bath

7. 18 - 7 An example reaction Time (s)Volume STP O 2, mL 0 3001.15 6002.18 9003.11 12003.95 18005.36 24006.50 30007.42 42008.75 54009.62 6600 10.17 7800 10.53 Here are the results for our experiment.

8. 18 - 8 An example reaction Volume, mL O 2 Time, s The rate of O 2 production slows down with time. The rate of O 2 production slows down with time.

9. 18 - 9 Average rates We can calculate the average rate of oxygen formation during any time interval as: Average rate of O 2 formation =  V O 2  t Time, sRate O 2 * 0 3000.0038 6000.0034 9000.0031 12000.0028 18000.0024 24000.0019 30000.0015 * The rates shown here have units of mL O 2 at STP /s. Note how the rate decreases with time.

10. 18 - 10 Instantaneous rates We know that the rate of our reaction is constantly changing with time. Instantaneous rate The rate of reaction at any point in time. It can be found by taking the tangent of our earlier plot. Initial rate of reaction The rate of formation at time zero when the reactants are initially mixed. Volume, mL O 2 Time, s

11. 18 - 11 An example reaction, again Since we know what the stoichiometry for our reaction is, we can calculate the concentration of N 2 O 5 during our reaction. 2N 2 O 5 (g) 2N 2 O 4 (g) + O 2 (g) For each mole of O 2 produced, two moles of N 2 O 5 will have decomposed. The rate of reaction will be: rate of reaction = = -  [O 2 ]  t  [N 2 O 5 ]  t 1212

12. 18 - 12 An example reaction, again Volume, mL O 2 [N 2 O 5 ] Time, s

13. 18 - 13 General reaction rates For the general reaction: a A + b B +... e E + f F +... The rate of reaction can be expressed as: Rate === - 1a1a  [A]  t - 1b1b  [B]  t 1e1e  [E]  t 1f1f  [F]  t

14. 18 - 14 Rate and concentration We can develop a quantitative relationship between instantaneous rates and concentration. By drawing tangents along the curve for [N 2 O 5 ], we can measure the following rates of reaction. Rate of N 2 O 5 Time, s decomposition, m/s 0 7.29 x 10 5 300 6.46 x 10 5 600 5.80 x 10 5 900 5.21 x 10 5 1200 4.69 x 10 5 1800 3.79 x 10 5 2400 3.04 x 10 5 3000 2.44 x 10 5 4200 1.59 x 10 5 5400 1.03 x 10 5

15. 18 - 15 Rate and concentration The earlier data indicated that the rate is directly proportional to concentration. rate = k [ N 2 O 5 ] We can verify this by calculating the value for k for the various rates we measured. k, sec -1 Time,s (rate/[N 2 O 5 ]) 03.65 x 10 -4 3003.59 x 10 -4 6003.60 x 10 -4 9003.62 x 10 -4 12003.61 x 10 -4 18003.64 x 10 -4 24003.62 x 10 -4 30003.59 x 10 -4 42003.61 x 10 -4 54003.68 x 10 -4

16. 18 - 16 Rate and concentration For the general reaction: a A + b B +... e E + f F +... the rate expression will often have the form of: rate = k [A] x [B] y... krate constant x, yorder for A & B, respectively x + yorder for the reaction Note: Note: The order is NOT the same as the coefficients for the balanced reaction.

17. 18 - 17 Finding rate laws Method of initial rates. The order for each reactant is found by: Changing the initial concentration of that reactant. Holding all other initial concentrations and conditions constant. Measuring the initial rates of reaction The change in rate is used to determine the order for that specific reactant. The process is repeated for each reactant.

18. 18 - 18 N 2 O 5 example The following data was obtained for the decomposition of N 2 O 5. Experiment[N 2 O 5 ]Initial rate, M/s 1 0.100 3.62 x 10 -5 2 0.200 7.29 x 10 -5 We know that the rate expression is: rate = k [N 2 O 5 ] x Our goal is to determine what x (the order) is.

19. 18 - 19 N 2 O 5 example For exp. 27.29 x 10 -5 M/s = k (0.200 M) x For exp. 13.62 x 10 -5 M/s = k (0.100 M) x We can now divide the equation for experiment two by the one for experiment one. 7.29 x 10 -5 M/s k (0.200 M) x 3.62 x 10 -5 M/s k (0.100 M) x which give2.01 = (2.00) x andx = 1 (first order reaction) =

20. 18 - 20 A more complex example The initial rate of reaction was obtained for the following reaction under the conditions listed. A + B + C... Exp. [A] [B] [C] Initial rate, M/s 1 0.030 0.010 0.0501.7 x 10 -8 2 0.060 0.010 0.050 6.8 x 10 -8 3 0.030 0.020 0.0504.9 x 10 -8 4 0.030 0.010 0.1001.7 x 10 -8 With this series of experiments, the concentration of a single reactant is doubled in concentration for experiments 2-4, compared to experiment one.

21. 18 - 21 A more complex example While this type of problem is more time consuming, its not any more difficult than the previous example. Order for A Use experiments one and two. 6.8 x 10 -8 M/s(0.060 M) x 1.7 x 10 -8 M/s (0.030 M) x 4.0 = (2.0) x By inspection, x = 2 =

22. 18 - 22 A more complex example Order for B Use experiments one and three. 4.9 x 10 -8 M/s(0.010 M) y 1.7 x 10 -8 M/s (0.020 M) y 2.9 = (2.0) y The order is not obvious by inspection. You must take the logarithm of both sides and solve for y. ln 2.9 = y ln 2.0 y = 1.54 or 3232

23. 18 - 23 A more complex example Order for C Use experiments one and four. Experiment [C] Initial Rate 10.0501.7 x 10 -8 40.1001.7 x 10 -8 Here the rate did not change when [C] was doubled. This is an example of a zero order reaction. z = 0

24. 18 - 24 A more complex example We can now write the overall rate law. rate = [A] 2 [B] 3/2 [C] 0 or since [C] has no effect on the rate: rate = [A] 2 [B] 3/2 The overall order for the reaction is: x + y + z = 2 + 3/2 + 0 = 3 1/2

25. 18 - 25 Finding rate laws Graphical method. Using integrated rate laws, one can produce straight line plots. The order for a reactant if verified if the data fits the plot. Rate integrated Graph Slope Order law rate law vs. time 0 rate = k [A] t = -kt + [A] 0 [A] t -k 1 rate = k[A] ln[A] t = -kt + ln[A] 0 ln[A] t -k 2 rate=k[A] 2 = kt + k 1 [A] 0 1 [A] t 1 [A] t

26. 18 - 26 Finding rate laws 0 order plot 1st order plot 2nd order plot As you can see from these plots of the N 2 O 5 data, only a first order plot results in a straight line. As you can see from these plots of the N 2 O 5 data, only a first order plot results in a straight line. Time (s) [N 2 O 5 ] 1/[N 2 O 5 ] ln[N 2 O 5 ]

27. 18 - 27 First order reactions Reactions that are first order with respect to a reactant are of great importance. Describe how many drugs pass into the blood stream or used by the body. Often useful in geochemistry Radioactive decay Half-life (t 1/2 ) The time required for one-half of the quantity of reactant originally present to react.

28. 18 - 28 Half-life From our N 2 O 5 data, we can see that it takes about 1900 seconds for the concentration to be reduced in half. It takes another 1900 seconds to reduce the concentration in half again. From our N 2 O 5 data, we can see that it takes about 1900 seconds for the concentration to be reduced in half. It takes another 1900 seconds to reduce the concentration in half again. Time (s) [N 2 O 5 ]

29. 18 - 29 Half-life The half-life and the rate constant are related. t 1/2 = t 1/2 = Half-life can be used to calculate the first order rate constant. For our N 2 O 5 example, the reaction took 1900 seconds to react half way so: k == = 3.65 x 10 -4 s -1 0.693k 0.693 t 1/2 0.693 1900 s

30. 18 - 30 Rate and temperature Reaction rates are temperature dependent. Here are rate constants for N 2 O 5 decomposition at various temperatures. T, o C k x 10 4, s -1 20 0.235 25 0.469 30 0.933 35 1.82 40 3.62 45 6.29 k x 10 4 (s -1 ) Temperature ( o C)

31. 18 - 31 Rate and temperature Arrhenius equation The relationship between rate constant and temperature is mathematically described by the Arrhenius equation. k = A e Aconstant E a activation energy Ttemperature, Kelvin Rgas law constant -E a / RT

32. 18 - 32 Rate and temperature An alternate form of the Arrhenius equation is: ln k = + ln A If ln k is plotted against 1/T, a straight line of slope -E a /RT is obtained. Activation energy - E a The energy that molecules must have in order to react. ( ) 1T1T EaREaR -

33. 18 - 33 Calculation of E a from N 2 O 5 data ln k T -1

34. 18 - 34 Theories of reaction rates Collision theory Based on kinetic-molecular theory. It assumes that reactants must collide for a reaction to occur. They must hit with sufficient energy and with the proper orientation so as to break the original bonds and form new ones. As temperature is increased, the average kinetic energy increases - so will the rate. As concentration increases, the number of collisions will also increase, also increasing the rate.

35. 18 - 35 Effective collision Reactants must have sufficient energy and the proper orientation for a collision to result in a reaction.

36. 18 - 36 Temperature and E a As the temperature is increased, a higher fraction of molecules will have a kinetic energy that is greater that the activation energy. T 1 < T 2 < T 3 Kinetic energy Fraction of molecules Having kinetic energy T3T3 T2T2 T1T1 EaEa

37. 18 - 37 Transition state theory As reactants collide, they initially form an activated complex. The activated complex is in the transition state. It lasts for approximately 10-100 fs. It can then form products or reactants. Once products are formed, it is much harder to return to the transition state, for exothermic reactions. Reaction profiles can be used to show this process.

38. 18 - 38 Effective collision Activated Complex Activated Complex A temporary state where bonds are in the process of breaking and forming.

39. 18 - 39 This type of plot shows the energy changes during a reaction. This type of plot shows the energy changes during a reaction. Reaction profile HH activation energy Potential Energy Reaction coordinate

40. 18 - 40 Examples of reaction profiles Exothermic reaction Endothermic reaction

41. 18 - 41 Examples of reaction profiles High activation energy Low heat of reaction Low activation energy High heat of reaction

42. 18 - 42 Reaction mechanisms A detailed molecular-level picture of how a reaction might take place. activated complex = bonds in the process of breaking or being formed

43. 18 - 43 Reaction mechanisms Elementary process Each step in a mechanism.Molecularity The number of particles that come together to form the activated complex in an elementary process. 1 - unimolecular 2 - bimolecular 3 - termolecular

44. 18 - 44 Reaction mechanisms For elementary processes, the exponents for each species in the rate law are the same as the coefficients in the equation for the step. For our earlier example, the rate law is: rate = k [NO] [O 3 ]

45. 18 - 45 Reaction mechanisms In general, the rate law gives the composition of the activated complex. The power of a species in the rate law is the same as the number of particles of the species in the activated complex. If the exponents in the rate law are not the same as the coefficients of the equation for the reaction, the overall reaction must consist of more than one step. Lets look at N 2 O 5 - again!

46. 18 - 46 Reaction mechanisms Earlier we found that for: 2N 2 O 5 2N 2 O 4 + O 2 The rate law was: rate = k [N 2 O 5 ] According to the equation, it should be second order but the data shows it to be first order. The reaction must involve more than one step.

47. 18 - 47 Reaction Mechanisms Consider the following reaction. 2NO 2 (g) + F 2 (g) 2NO 2 F (g) If the reaction took place in a single step the rate law would be: rate = k [NO 2 ] 2 [F 2 ] However, the experimentally observed rate law is: rate = k [NO 2 ] [F 2 ]

48. 18 - 48 Reaction Mechanisms Since the observed rate law is not the same as if the reaction took place in a single step, we know two things. More than one step must be involved The activated complex must be produced from two species. A possible reaction mechanism might be: Step one Step oneNO 2 + F 2 NO 2 F + F Step two Step twoNO 2 + F NO 2 F Overall Overall2NO 2 + F 2 2NO 2 F

49. 18 - 49 Reaction Mechanisms Rate-determining step. slow When a reaction occurs in a series of steps, with one slow step, it is the slow step that determines the overall rate. Step one Step oneNO 2 + F 2 NO 2 F + F Expected to be slow. It involves breaking an F-F bond. Step two Step twoNO 2 + F NO 2 F Expected to be fast. A fluorine atom is very reactive.

50. 18 - 50 Reaction Mechanisms Since step one is slow, we can expect this step to determine the overall rate of the reaction. NO 2 + F 2 NO 2 F + F This would give a rate expression of: rate = k 1 [NO 2 ] [ F 2 ] This agrees with the experimentally observed results.

51. 18 - 51 Catalysis Catalyst CatalystA substance that changes the rate of a reaction without being consumed in the reaction. Provides an easier way to react. Lower activation energy. Still make the same products. Enzymes Enzymes are biological catalysts. Inhibitor InhibitorA substance that decreases the rate of reaction.

52. 18 - 52 Catalysis Types of catalysts Homogeneous Homogeneous - same phase Catalyst is uniformly distributed throughout the reaction mixture Example - I - in peroxide. Heterogeneous Heterogeneous - different phase Catalyst is usually a solid and the reactants are gases or liquids Example - Automobile catalytic converter.

53. 18 - 53 Heterogeneous catalysis

54. 18 - 54 Enzymes Biological catalysts Typically are very large proteins. Permit reactions to ‘go’ at conditions that the body can tolerate. Can process millions of molecules every second. substrates Are very specific - react with one or only a few types of molecules (substrates).

55. 18 - 55 Classification of enzymes Based on type of reaction Oxireductase Oxireductasecatalyze a redox reaction Transferase Transferasetransfer a functional group Hydrolase Hydrolasecause hydrolysis reactions Lyase Lyasebreak C-O, C-C or C-N bonds Isomerases Isomerasesrearrange functional groups Ligase Ligasejoin two molecules

56. 18 - 56 The active site Enzymes are typically HUGE proteins, yet only a small part is actually involved in the reaction. The active site has two basic components. catalytic site binding site Model of trios-phosphate-isomerase Model of trios-phosphate-isomerase

57. 18 - 57 Characteristics of enzyme active sites Catalytic site Where the reaction actually occurs. Binding site Area that holds substrate in proper place. Enzymes uses weak, non-covalent interactions to hold the substrate in place based on R groups of amino acids. Shape is complementary to the substrate and determines the specificity of the enzyme. Sites are pockets or clefts on the enzyme surface.

58. 18 - 58 Characteristics of enzyme active sites Catalytic site Binding site Substrate Enzyme

59. 18 - 59 Enzyme-substrate reaction