1. Chemistry 102(60) Summer 2001 n Instructor: Dr. Upali Siriwardane n e-mail:upali@chem.latech n Office: CTH 311 Phone 257-4941 n Office Hours: n 8:30-10:30 a.m., M, W, Tu,Th, F (Test 1): Chapter 12 (Test 2): Chapter 13. (Test 3): Chapter 14 (Test 4): Chapter 15. (Test 5): Chapter 17.

2. Chapter 13. Rates of Reactions Chemical Kinetics n the branch of chemistry dealing with the rates of chemical reactions. n Using chemical kinetics we can find time to complete a reaction, the effect of temperature on the rate n Effect of other substances (catalysts or inhibitors) on the reactions

3. How do you measure rates n Rates are related to the time it required to decay reactants or form products. n The rate reaction = change in concentration of reactants/products per unit time

4. An example reaction Gas buret Constant temperature bath

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

6. Rate n a A --> b B n based on reactants rate = -(1/a)  [A]/  t n Based on products rate = +(1/b)  [B]/  t  [A]= [A] f - [A] I Change in A  t= t f - t i Change in t

7. Rate of Reaction n 2 N 2 O 5 (g) -----> 4 NO 2 (g) + O 2 (g) based on reactants rate = -(1/2)  [N 2 O 5 ]/  t n Based on products rate = +(1/4)  [NO 2 ]/  t rate = +(1/1)  [O 2 ]/  t n

8. Rate of Appearance & disappearance n 2 N 2 O 5 (g) -----> 4 NO 2 (g) + O 2 (g) Disappearance is based on reactants rate = -(  [N 2 O 5 ]/  t n Appearance is based on products rate =  [NO 2 ]/  t rate =  [O 2 ]/  t n converting rates of Appearance.. rate = (  [NO 2 ]/  t = - 4/2  [N 2 O 5 ]/  t  [O 2 ]/  t = - 1/2  [N 2 O 5 ]/  t

9. Chemical Kinetics Definitions and Concepts n a) rate law n b) rate constant n c) order n d) differential rate law n c) integral rate law n

10. n Every chemical reaction has a Rate Law n The rate law is an expression that relates the rate of a chemical reaction to a constant (rate constant-k) and concentration of reactants raised to a power. n The power of a concentration is called the order with respect to the particular reactant. Rate Law

11. n E.g. A + B -----> C rate  [A] l [B] m n rate = k [A] l [B] m ; k = rate constant n [A] = concentration of A n [B] = concentration of B n l = order with respect to A n m = order with respect to B n l & m have nothing to do with stoichiometric coefficients

12. Rate Constant n E.g. A + B -----> C rate  [A] l [B] m n rate = k [A] l [B] m ; n k = rate constant n proportionality constant of the rate law

13. Rate Law n E.g. n 2 N 2 O 5 (g) -----> 4 NO 2 (g) + O 2 (g) rate  [N 2 O 5 ] 1 n rate = k [N 2 O 5 ] 1 ;k = rate constant n [N 2 O 5 ] = concentration of N 2 O 5 n 1 = order with respect to N 2 O 5 n Rate and the order are obtained by experiments

14. Order n The power of the concentrations is the order with respect to the reactant. n E.g. A + B -----> C n rate = k [A] 1 [B] 2 n The order of the reaction with respect to A is one (1). n The order of the reaction with respect to B is two (2). n Overall order of a chemical reaction is equal to the sum of all orders(3).

15. Finding rate laws n Method of initial rates. n 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 n The change in rate is used to determine the order for that specific reactant. The process is repeated for each reactant.

16. How do you find order? n A + B -----> C n rate = k [A] l [B] m ; n Hold concentration of other reactants constant n If [A] doubled, rate doubled n -1st order, [2A] 1 = 2 1 x [A] 1, 2 1 = 2 n b) If [A] doubled, rate quadrupled n -2nd order, [2A] 2 = 2 2 x [A] 2, 2 2 = 4 n c) If [A] doubled, rate increased 8 times -3rd order, [2A] 3 = 2 3 x [A] 3, 2 3 = 8 n

17. Method of Initial Rates n A + B ----> C n The rate law : rate = k [A] x [B] y n [A],mol/L [B],mol/L rate, mol/LS n 4.6 x 10 -4 3.1 x 10 -5 2.08 x 10 -3 n 4.6 x 10 -4 6.2 x 10 -5 4.16 x 10 -3 n 9.2 x 10 -4 6.2 x 10 -5 1.664 x 10 -2

18. Order wrt A n 1.664 x 10 -2 k (9.2 x 10 -4 ) x (6.2 x 10 -5 ) y n --------------- = -------------------------------------- n 4.16 x 10 -3 k (4.6 x 10 -4 ) x (6.2 x 10 -5 ) y n n 1.664 x 10 -2 (9.2 x 10 -4 ) x n --------------- = -------------- n 4.16 x 10 -3 (4.6 x 10 -4 ) x n 4 = (9.2 x 10 -4 )/(4.6 x 10 -4 ) x = 2 x n 4 = 2 x n x = 2

19. Order wrt B n 4.16 x 10 -3 k (4.6 x 10 -4 )x (6.2 x 10 -5 ) y n ----------- = ------------------------------ n 2.08 x 10 -3 k (4.6 x 10 -4 )x (3.1 x 10 -5 ) y n n 4.16 x 10 -3 (6.2 x 10-5) y n ----------- = ------------- n 2.08 x 10 -3 (3.1 x 10-5) y n 2 = (6.2 x 10-5/3.1 x 10-5)y = 2 y n 2 = 2 y n y = 1 n The rate law : rate = k [A] 2 [B] 1

20. Units of the Rate Constant n 1 first order: k = ─── = s -1 n s n L second order k = ─── n mol s

21. n Rate Law Differential Rate Law Integral Rate rate  k [A] 0  [A]/  t =k ; ([A] 0 =1)[A] f -[A] i = -kt n rate  k [A] 1  [A]/  t = k [A] ln [A] o /[A] t = kt rate  k [A] 2  [A]/  t = k [A] 2 1/ [A] f = kt + 1/[A] i n

22. Differential Rate Law n Normal form Differential form n zero order rate = k [A] 0 rate = -  [A]/  t n = k ( [A] 0 =1) n first order rate law rate = k [A] 1 rate = -  [A]/  t n = k [A] 1

23. Integral Rate Law n Differential form Integral form n zero order rate = -  [A]/  t = [A] f -[A] i = -kt n = k ( [A] 0 =1) n first order rate law rate = -  [A]/  t ln [A] o /[A] t = kt n = k [A] 1

24. Finding rate laws Graphical method. R ate 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

25. 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 ]

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

27. First Order Reactions n A ----> B n Differential rate law  [A] - ───── =k  [A]  t [A] t [A] 0 ln ─── = - k t   or ln ─── = k t [A] 0  [A] t

28. Half-life form of 1st order t 2 is defined as time for [A] 0, the initial concentration to decay half the original value ie 1/2 x [A] 0 =  [A] t. 

29. t 2 equation  0.693 =k t 2 n 0.693 t 2 = ---- n k

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

32. 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 ]

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

34. Effective collision n Reactants must have sufficient energy and the proper orientation for a collision to result in a reaction.

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

36. What are the factors that affect rates of chemical reactions? n a) Temperature n b) Concentration n c) Catalysts n d) Particle size of solid reactants

37. 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

38. Potential Energy Curves n Exothermic Reactions n Endothermic Reactions n Effect of catalysts n Effect of temperature

39. Examples of reaction profiles Exothermic reaction Endothermic reaction

40. Examples of reaction profiles High activation energy Low heat of reaction Low activation energy High heat of reaction

41. Arrhenius Equation n Rate constant (k) n k = A e- E a /RT n n A = frequency factor n E a = Activation energy n R = gas constant n T = Kelvintemperature

42. Rate and temperature n 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)

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

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

45. Calculation of E a n k = A e- E a /RT n ln k = ln A - E a /RT n log k = log A - E a / 2.303 RT n using two set of values n log k 1 = log A - E a / 2.303 RT 1 n log k 2 = log A - E a / 2.303 RT 2 n log k 1 - log k 2 = - E a / 2.303 RT 2 + E a / 2.303 RT 1 n log k 1 / k 2 = E a / 2.303 R[ 1/T 1 - 1/T 2 ]

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

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

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

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

50. Reaction Mechanisms n 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. n A possible reaction mechanism might be: n Step one n Step oneNO 2 + F 2 NO 2 F + F n Step two n Step twoNO 2 + F NO 2 F n Overall n Overall 2NO 2 + F 2 2NO 2 F

51. Reaction Mechanism n Elementary Reactions: n NO 2 + F 2 --> NO 2 F + F (slow) n F + NO 2 --> NO 2 F (fast) n Molecularity? Of Elementary Reactions n unimolecular, bimolecular, termolecular?

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

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

54. Catalysis n Catalyst A 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. n Enzymes n Enzymes are biological catalysts. n Inhibitor A substance that decreases the rate of reaction.

55. Catalysts Lowers E a

56. 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.

57. Heterogeneous catalysis

58. Enzymes n 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).

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

60. The active site n 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

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