1. Chapter 14 Chemical Kinetics

2. Kinetics is the study of how fast chemical reactions occur. There are 4 important factors which affect rates of reactions: –reactant concentration, –temperature, –action of catalysts, and –surface area. Goal: to understand chemical reactions at the molecular level. Factors that Affect Reaction Rates

3. Speed of a reaction is measured by the change in concentration with time. For a reaction A  B Suppose A reacts to form B. Let us begin with 1.00 mol A. Reaction Rates

5. –At t = 0 (time zero) there is 1.00 mol A (100 red spheres) and no B present. –At t = 20 min, there is 0.54 mol A and 0.46 mol B. –At t = 40 min, there is 0.30 mol A and 0.70 mol B. –Calculating, Reaction Rates

6. For the reaction A  B there are two ways of measuring rate: –the speed at which the products appear (i.e. change in moles of B per unit time), or –the speed at which the reactants disappear (i.e. the change in moles of A per unit time). Reaction Rates

7. Change of Rate with Time For the reaction A  B there are two ways of Most useful units for rates are to look at molarity. Since volume is constant, molarity and moles are directly proportional. Consider: C 4 H 9 Cl(aq) + H 2 O(l)  C 4 H 9 OH(aq) + HCl(aq) Reaction Rates

9. Change of Rate with Time C 4 H 9 Cl(aq) + H 2 O(l)  C 4 H 9 OH(aq) + HCl(aq) –We can calculate the average rate in terms of the disappearance of C 4 H 9 Cl. –The units for average rate are mol/L·s or M/s. –The average rate decreases with time. –We plot [C 4 H 9 Cl] versus time. –The rate at any instant in time (instantaneous rate) is the slope of the tangent to the curve. –Instantaneous rate is different from average rate. –We usually call the instantaneous rate the rate. Reaction Rates

11. Reaction Rate and Stoichiometry For the reaction C 4 H 9 Cl(aq) + H 2 O(l)  C 4 H 9 OH(aq) + HCl(aq) we know In general for aA + bB  cC + dD Reaction Rates

12. In general rates increase as concentrations increase. NH 4 + (aq) + NO 2 - (aq)  N 2 (g) + 2H 2 O(l) Concentration and Rate

13. For the reaction NH 4 + (aq) + NO 2 - (aq)  N 2 (g) + 2H 2 O(l) we note –as [NH 4 + ] doubles with [NO 2 - ] constant the rate doubles, –as [NO 2 - ] doubles with [NH 4 + ] constant, the rate doubles, –We conclude rate  [NH 4 + ][NO 2 - ]. Rate law: The constant k is the rate constant. Concentration and Rate

14. Exponents in the Rate Law For a general reaction with rate law we say the reaction is mth order in reactant 1 and nth order in reactant 2. The overall order of reaction is m + n + …. A reaction can be zeroth order if m, n, … are zero. Note the values of the exponents (orders) have to be determined experimentally. They are not simply related to stoichiometry. Concentration and Rate

15. Using Initial Rates to Determines Rate Laws A reaction is zero order in a reactant if the change in concentration of that reactant produces no effect. A reaction is first order if doubling the concentration causes the rate to double. A reacting is nth order if doubling the concentration causes an 2 n increase in rate. Note that the rate constant does not depend on concentration. Concentration and Rate

16. First Order Reactions Goal: convert rate law into a convenient equation to give concentrations as a function of time. For a first order reaction, the rate doubles as the concentration of a reactant doubles. The Change of Concentration with Time

17. First Order Reactions A plot of ln[A] t versus t is a straight line with slope -k and intercept ln[A] 0. In the above we use the natural logarithm, ln, which is log to the base e. The Change of Concentration with Time

18. First Order Reactions The Change of Concentration with Time

19. Second Order Reactions For a second order reaction with just one reactant A plot of 1/[A] t versus t is a straight line with slope k and intercept 1/[A] 0 For a second order reaction, a plot of ln[A] t vs. t is not linear. The Change of Concentration with Time

20. Second Order Reactions The Change of Concentration with Time

21. Half-Life Half-life is the time taken for the concentration of a reactant to drop to half its original value. For a first order process, half life, t ½ is the time taken for [A] 0 to reach ½[A] 0. Mathematically, The Change of Concentration with Time

23. Half-Life For a second order reaction, half-life depends in the initial concentration: The Change of Concentration with Time

24. The Collision Model Most reactions speed up as temperature increases. (E.g. food spoils when not refrigerated.) When two light sticks are placed in water: one at room temperature and one in ice, the one at room temperature is brighter than the one in ice. The chemical reaction responsible for chemiluminescence is dependent on temperature: the higher the temperature, the faster the reaction and the brighter the light. Temperature and Rate

25. The Collision Model As temperature increases, the rate increases.

26. The Collision Model Since the rate law has no temperature term in it, the rate constant must depend on temperature. Consider the first order reaction CH 3 NC  CH 3 CN. –As temperature increases from 190  C to 250  C the rate constant increases from 2.52  10 -5 s -1 to 3.16  10 -3 s -1. The temperature effect is quite dramatic. Why? Observations: rates of reactions are affected by concentration and temperature. Temperature and Rate

27. The Collision Model Goal: develop a model that explains why rates of reactions increase as concentration and temperature increases. The collision model: in order for molecules to react they must collide. The greater the number of collisions the faster the rate. The more molecules present, the greater the probability of collision and the faster the rate. Temperature and Rate

28. The Collision Model The higher the temperature, the more energy available to the molecules and the faster the rate. Complication: not all collisions lead to products. In fact, only a small fraction of collisions lead to product. The Orientation Factor In order for reaction to occur the reactant molecules must collide in the correct orientation and with enough energy to form products. Temperature and Rate

29. The Orientation Factor Consider: Cl + NOCl  NO + Cl 2 There are two possible ways that Cl atoms and NOCl molecules can collide; one is effective and one is not. Temperature and Rate

30. The Orientation Factor Temperature and Rate

31. Activation Energy Arrhenius: molecules must posses a minimum amount of energy to react. Why? –In order to form products, bonds must be broken in the reactants. –Bond breakage requires energy. Activation energy, E a, is the minimum energy required to initiate a chemical reaction. Temperature and Rate

32. Activation Energy Consider the rearrangement of methyl isonitrile: –In H 3 C-N  C, the C-N  C bond bends until the C-N bond breaks and the N  C portion is perpendicular to the H 3 C portion. This structure is called the activated complex or transition state. –The energy required for the above twist and break is the activation energy, E a. –Once the C-N bond is broken, the N  C portion can continue to rotate forming a C-C  N bond. Temperature and Rate

34. Activation Energy The change in energy for the reaction is the difference in energy between CH 3 NC and CH 3 CN. The activation energy is the difference in energy between reactants, CH 3 NC and transition state. The rate depends on E a. Notice that if a forward reaction is exothermic (CH 3 NC  CH 3 CN), then the reverse reaction is endothermic (CH 3 CN  CH 3 NC). Temperature and Rate

35. Activation Energy How does a methyl isonitrile molecule gain enough energy to overcome the activation energy barrier? From kinetic molecular theory, we know that as temperature increases, the total kinetic energy increases. We can show the fraction of molecules, f, with energy equal to or greater than E a is where R is the gas constant (8.314 J/mol·K). Temperature and Rate

36. Activation Energy Temperature and Rate

37. The Arrhenius Equation Arrhenius discovered most reaction-rate data obeyed the Arrhenius equation: –k is the rate constant, E a is the activation energy, R is the gas constant (8.314 J/K-mol) and T is the temperature in K. –A is called the frequency factor. –A is a measure of the probability of a favorable collision. –Both A and E a are specific to a given reaction. Temperature and Rate

38. Determining the Activation Energy If we have a lot of data, we can determine E a and A graphically by rearranging the Arrhenius equation: From the above equation, a plot of ln k versus 1/T will have slope of –E a /R and intercept of ln A. Temperature and Rate

40. Determining the Activation Energy If we do not have a lot of data, then we recognize Temperature and Rate

41. The balanced chemical equation provides information about the beginning and end of reaction. The reaction mechanism gives the path of the reaction. Mechanisms provide a very detailed picture of which bonds are broken and formed during the course of a reaction. Elementary Steps Elementary step: any process that occurs in a single step. Reaction Mechanisms

42. Elementary Steps Molecularity: the number of molecules present in an elementary step. –Unimolecular: one molecule in the elementary step, –Bimolecular: two molecules in the elementary step, and –Termolecular: three molecules in the elementary step. It is not common to see termolecular processes (statistically improbable). Reaction Mechanisms

43. Multistep Mechanisms Some reaction proceed through more than one step: NO 2 (g) + NO 2 (g)  NO 3 (g) + NO(g) NO 3 (g) + CO(g)  NO 2 (g) + CO 2 (g) Notice that if we add the above steps, we get the overall reaction: NO 2 (g) + CO(g)  NO(g) + CO 2 (g) Reaction Mechanisms

44. Multistep Mechanisms If a reaction proceeds via several elementary steps, then the elementary steps must add to give the balanced chemical equation. Intermediate: a species which appears in an elementary step which is not a reactant or product. Reaction Mechanisms

45. Rate Laws for Elementary Steps The rate law of an elementary step is determined by its molecularity: –Unimolecular processes are first order, –Bimolecular processes are second order, and –Termolecular processes are third order. Rate Laws for Multistep Mechanisms Rate-determining step: is the slowest of the elementary steps. Reaction Mechanisms

46. Rate Laws for Elementary Steps Reaction Mechanisms

47. Rate Laws for Multistep Mechanisms Therefore, the rate-determining step governs the overall rate law for the reaction. Mechanisms with an Initial Fast Step It is possible for an intermediate to be a reactant. Consider 2NO(g) + Br 2 (g)  2NOBr(g) Reaction Mechanisms

48. Mechanisms with an Initial Fast Step 2NO(g) + Br 2 (g)  2NOBr(g) The experimentally determined rate law is Rate = k[NO] 2 [Br 2 ] Consider the following mechanism Reaction Mechanisms

49. Mechanisms with an Initial Fast Step The rate law is (based on Step 2): Rate = k 2 [NOBr 2 ][NO] The rate law should not depend on the concentration of an intermediate (intermediates are usually unstable). Assume NOBr 2 is unstable, so we express the concentration of NOBr 2 in terms of NOBr and Br 2 assuming there is an equilibrium in step 1 we have Reaction Mechanisms

50. Mechanisms with an Initial Fast Step By definition of equilibrium: Therefore, the overall rate law becomes Note the final rate law is consistent with the experimentally observed rate law. Reaction Mechanisms

51. A catalyst changes the rate of a chemical reaction. There are two types of catalyst: –homogeneous, and –heterogeneous. Chlorine atoms are catalysts for the destruction of ozone. Homogeneous Catalysis The catalyst and reaction is in one phase. Catalysis

52. Catalysis

53. Homogeneous Catalysis Hydrogen peroxide decomposes very slowly: 2H 2 O 2 (aq)  2H 2 O(l) + O 2 (g) In the presence of the bromide ion, the decomposition occurs rapidly: –2Br - (aq) + H 2 O 2 (aq) + 2H + (aq)  Br 2 (aq) + 2H 2 O(l). –Br 2 (aq) is brown. –Br 2 (aq) + H 2 O 2 (aq)  2Br - (aq) + 2H + (aq) + O 2 (g). –Br - is a catalyst because it can be recovered at the end of the reaction. Catalysis

54. Homogeneous Catalysis Generally, catalysts operate by lowering the activation energy for a reaction. Catalysis

55. Catalysis

56. Homogeneous Catalysis Catalysts can operate by increasing the number of effective collisions. That is, from the Arrhenius equation: catalysts increase k be increasing A or decreasing E a. A catalyst may add intermediates to the reaction. Example: In the presence of Br -, Br 2 (aq) is generated as an intermediate in the decomposition of H 2 O 2. Catalysis

57. Homogeneous Catalysis When a catalyst adds an intermediate, the activation energies for both steps must be lower than the activation energy for the uncatalyzed reaction. The catalyst is in a different phase than the reactants and products. Heterogeneous Catalysis Typical example: solid catalyst, gaseous reactants and products (catalytic converters in cars). Most industrial catalysts are heterogeneous. Catalysis

58. Heterogeneous Catalysis First step is adsorption (the binding of reactant molecules to the catalyst surface). Adsorbed species (atoms or ions) are very reactive. Molecules are adsorbed onto active sites on the catalyst surface. Catalysis

59. Catalysis

60. Heterogeneous Catalysis Consider the hydrogenation of ethylene: C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g),  H = -136 kJ/mol. –The reaction is slow in the absence of a catalyst. –In the presence of a metal catalyst (Ni, Pt or Pd) the reaction occurs quickly at room temperature. –First the ethylene and hydrogen molecules are adsorbed onto active sites on the metal surface. –The H-H bond breaks and the H atoms migrate about the metal surface. Catalysis

61. Heterogeneous Catalysis –When an H atom collides with an ethylene molecule on the surface, the C-C  bond breaks and a C-H  bond forms. –When C 2 H 6 forms it desorbs from the surface. –When ethylene and hydrogen are adsorbed onto a surface, less energy is required to break the bonds and the activation energy for the reaction is lowered. Enzymes Enzymes are biological catalysts. Most enzymes are protein molecules with large molecular masses (10,000 to 10 6 amu). Catalysis

62. Enzymes Enzymes have very specific shapes. Most enzymes catalyze very specific reactions. Substrates undergo reaction at the active site of an enzyme. A substrate locks into an enzyme and a fast reaction occurs. The products then move away from the enzyme. Catalysis

63. Enzymes Only substrates that fit into the enzyme lock can be involved in the reaction. If a molecule binds tightly to an enzyme so that another substrate cannot displace it, then the active site is blocked and the catalyst is inhibited (enzyme inhibitors). The number of events (turnover number) catalyzed is large for enzymes (10 3 - 10 7 per second). Catalysis

64. Enzymes Catalysis

65. End of Chapter 14 Chemical Kinetics