1. Chemical Kinetics Chapter 13 Dr. Ali Bumajdad

2. Chapter 13 Topics Rate of a Reaction Reaction Rates and Stoichiometry The Rate Law Relationship between Reactant Concentration and Time 1) 1)First-order reactions 2) 2)Second-ordered reactions 3) 3)Zero-ordered reactions The Effect of Activation Energy and Temperature on Rate Collision Theory The Arhenius Equation Finding Ea from graph and from equation Catalysis Chemical Kinetics Dr. Ali Bumajdad

3. Chemical Kinetics = Chemistry and Time Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed? - Why it is important: To answer questions like: 1) 1)How quickly a medicine is able to work? 2) 2)How fast the depletion of Ozone? 3) 3)How rapidly food spoils? 4) 4)How fast steel rust?

4. Reaction rate is the change in the concentration of a reactant or a product with time (M/s). A B rate = - [A][A] tt rate = [B][B] tt  [A] = change in concentration of A over time period  t  [B] = change in concentration of B over time period  t Because [A] decreases with time,  [A] is negative. Rate of a reaction

5. A B rate = -  [A] tt rate = [B][B] tt time

6. Kinetics can be studied by following a change in property as a function of time This property could be: 1) UV absorption 2) Pressure 3) Conductivity 4) Concentration 5) Gas volume

7. Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) time 393 nm light Detector  [Br 2 ]   Absorption

8. Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) slope of tangent slope of tangent slope of tangent average rate = -  [Br 2 ] tt = - [Br 2 ] final – [Br 2 ] initial t final - t initial instantaneous rate = rate for specific instance in time (slope of a tangent)

9. rate  [Br 2 ] rate = k [Br 2 ] k = rate [Br 2 ] = rate constant = 3.50 x 10 -3 s -1

10. 2H 2 O 2 (aq) 2H 2 O (l) + O 2 (g) PV = nRT P = RT = [O 2 ]RT n V [O 2 ] = P RT 1 rate =  [O 2 ] tt RT 1 PP tt = measure  P over time

12. Reaction Rates and Stoichiometry 2A B Two moles of A disappear and one mole of B that is formed. =  [B] tt rate = -  [A] tt 1 2 aA + bB cC + dD In this case, the rate will be different if we consider the disappearance of A than if we consider the formation of B In order to make it the same we should divide by the coefficient In general for: rate = -  [A] tt 1 a = -  [B] tt 1 b =  [C] tt 1 c =  [D] tt 1 d (1)

13. Q) Write the rate expression for the following reaction: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) rate = -  [CH 4 ] tt = -  [O 2 ] tt 1 2 =  [H 2 O] tt 1 2 =  [CO 2 ] tt Sa. Ex. a) How is the rate of disappearance of ozone related to the rate of appearance of oxygen in the following equation: 2O 3 (g)3O 2 (g) b) If at certain time the rate of appearance of O 2,  [O 2 ]/  t = 6.0  10 -5 M/s what is the value of the rate of disappearance of O 3, -  [O 3 ]/  t at the same time. 4.0  10 -5 M/s

16. The Rate Law Rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers that can be determined by experiment only. Rate Law is different than the rate of reaction aA + bB cC + dD reaction is xth order in A reaction is yth order in B reaction is (x +y)th order overall rate = -  [A] tt 1 a Rate = k [A] x [B] y (2) Rate = k [A] x [B] y Reactant not product

17. F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ] x [ClO 2 ] y When Double [F 2 ] with [ClO 2 ] constant  x = 1 Quadruple [ClO 2 ] with [F 2 ] constant  y = 1  rate = k [F 2 ][ClO 2 ] Rate doubles Rate quadruples

18. F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ][ClO 2 ] Rate Laws 1 Rate laws are always determined experimentally. Reaction order is always defined in terms of reactant (not product) concentrations. The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation.

19. Q) Determine the rate law and calculate the rate constant for the following reaction from the following data: S 2 O 8 2- (aq) + 3I - (aq) 2SO 4 2- (aq) + I 3 - (aq) Experiment [S 2 O 8 2- ][I - ] Initial Rate (M/s) 10.080.0342.2 x 10 -4 20.080.0171.1 x 10 -4 30.160.0172.2 x 10 -4 rate = k [S 2 O 8 2- ] x [I - ] y Double [I - ], rate doubles (experiment 1 & 2) y = 1 Double [S 2 O 8 2- ], rate doubles (experiment 2 & 3) x = 1 k = rate [S 2 O 8 2- ][I - ] = 2.2 x 10 -4 M/s (0.08 M)(0.034 M) = 0.08/M s rate = k [S 2 O 8 2- ][I - ]

21. First-Order Reactions For reaction of the type A product rate = -  [A] tt and rate = k [A] [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0 Relationship between Reactant Concentration and Timeand  [A] tt = k - [A]  [A] tt = k [A] -  Reactions with first overall order Using Calculus ln[A] = ln[A] 0 - kt (3a) (3b) [A] = [A] 0 exp(-kt)

22. 2N 2 O 5 4NO 2 (g) + O 2 (g) k = rate [A] = 1/s or s -1 M/sM/s M = Unit of first order rate constant ln[A] = ln[A] 0 - kt (3a) Y = b + mX Where m = slope b = intercept

23. Q) The reaction 2A B is first order in A with a rate constant of 2.8 x 10 -2 s -1 at 80 0 C. How long will it take for A to decrease from 0.88 M to 0.14 M ? kt = ln[A] 0 – ln[A] t = ln[A] 0 – ln[A] k = 66 s [A] 0 = 0.88 M [A] = 0.14 M ln [A] 0 [A] k = ln 0.88 M 0.14 M 2.8 x 10 -2 s -1 = ln[A] = ln[A] 0 - kt (3a)

24. half-life, t ½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. t ½ = t when [A] = [A] 0 /2 ln [A] 0 [A] 0 /2 k = t½t½ ln2 k = 0.693 k = k = ln2 k t 1/2 = (4) Independent of initial concentration First-Order Reactions 1) It take the same time for the concentration of reactant to decrease from 0.1M to 0.05M as it does from 0.05M to 0.025M 2) If you know t 1/2 you know k

25. A product First-order reaction # of half-lives [A] = [A] 0 /n 1 2 3 4 2 4 8 16

26. Q) What is the half-life of N 2 O 5 if it decomposes with a rate constant of 5.7 x 10 -4 s -1 ? t½t½ ln2 k = 0.693 5.7 x 10 -4 s -1 = = 1200 s = 20 minutes How do you know decomposition is first order? units of k (s -1 )

27. For reactions of the type A product rate = -  [A] tt and rate = k [A] 2  [A] tt = k [A] 2 and - [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0 t ½ = t when [A] = [A] 0 /2 Second-Order ReactionsReactions with second overall order k = Unit of second order rate constant rate [A] 2 = 1/Ms M/sM/s M2M2 = Using Calculus where 1 [A] = 1 [A] 0 + kt (5) t ½ = 1 k[A] 0 (6)

28. A product rate = -  [A] tt rate = k [A] 0 = k k = rate [A] 0 = M/s  [A] tt = k - [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0 t ½ = t when [A] = [A] 0 /2 Zero-Order Reactions[A] = [A] 0 - kt(7) t ½ = [A] 0 2k2k(8)

29. Summary of the Kinetics for reactions of type aA  products

30. Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions OrderRate Law Concentration-Time Equation Half-Life 0 1 2 rate = k rate = k [A] rate = k [A] 2 ln[A] = ln[A] 0 - kt 1 [A] = 1 [A] 0 + kt [A] = [A] 0 - kt t½t½ ln2 k = t ½ = [A] 0 2k2k t ½ = 1 k[A] 0 k Unit s -1 M -1 s -1 M s -1

31. ln[A] = ln[A] 0 - kt (3a) ln [A] = - kt (3c) [A] 0 (a) (b,c)

32. Plot of ln [A] versus t should be linear for first order reaction Plot of ln [A] versus t should be linear for first order reaction Since [A]  P (PV=nRT), then lnP =-kt +ln P 0 Since [A]  P (PV=nRT), then lnP =-kt +ln P 0

33. 0.693 k = ln2 k t 1/2 = (4)

34. 1 [A] = 1 [A] 0 + kt (5) t ½ = 1 k[A] 0 (6)

35. - is the minimum amount of energy required to initiate a chemical reaction The Effect of Activation Energy (E a ) and Temperature on Rate -Usually rate increase by increasing T - Rate increase by decreasing E a - Unit: J/mol - Unit: K

36. Collision Theory 1) How reaction Occurs? By collisions between molecules or atoms 2) Does all collisions results in a reactions? No, a) The collision should be strong enough (more than the activation energy or the reaction) b) The collision should be with the right orientation 3) What factor affect the speed (rate) of the reactions? a) a)Collision frequency (A) b) b)Temperature c) c)Activation energy (E a )

37. Orientation effect

38. Exothermic ReactionEndothermic Reaction A + B AB C + D + +

39. E a is the activation energy (J/mol) R is the gas constant (8.314 J/Kmol) T is the absolute temperature (K) A is the frequency and orientation factor (Arrhenius equation) The Arhenius Equation k = A exp( -E a / RT ) (9) lnk = - EaEa R 1 T + lnA (10) Y = m X + b k  T k  1/E a

40. lnk = - EaEa R 1 T + lnA Y = m X + b 1) Finding E a from graph

41. lnk 1 = EaEa R 1 T1T1 lnA - lnk 2 = EaEa R 1 T2T2 lnA - 2) Finding E a from Equation Subtracting 2 from 1 (1) (2) 1 T2T2 lnk 1 - lnk 2 = EaEa R 1 T1T1 - EaEa R + 1 T2T2 EaEa R 1 T1T1 - 1 T2T2 ln k 1 = EaEa R 1 T1T1 - k 2 (11)

42. Sa. Ex.

43. lnk = - EaEa R 1 T + lnA (10) Y = m X + b

44. 1 T2T2 ln k 1 = EaEa R 1 T1T1 - k 2 (11)

45. Catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. k = A exp( -E a / RT )EaEa k rate catalyzed > rate uncatalyzed E a < E a ‘ UncatalyzedCatalyzed Catalysis

46. In heterogeneous catalysis, the reactants and the catalysts are in different phases. In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid. Haber synthesis of ammonia Ostwald process for the production of nitric acid Catalytic converters Acid catalysis Base catalysis

47. N 2 (g) + 3H 2 (g) 2NH 3 (g) Fe/Al 2 O 3 /K 2 O catalyst Haber Process

48. Ostwald Process Hot Pt wire over NH 3 solution Pt-Rh catalysts used in Ostwald process 4NH 3 (g) + 5O 2 (g) 4NO (g) + 6H 2 O (g) Pt catalyst 2NO (g) + O 2 (g) 2NO 2 (g) 2NO 2 (g) + H 2 O (l) HNO 2 (aq) + HNO 3 (aq)

49. Catalytic Converters CO + Unburned Hydrocarbons + O 2 CO 2 + H 2 O catalytic converter 2NO + 2NO 2 2N 2 + 3O 2 catalytic converter

50. Enzyme Catalysis: very selective biological catalyst

51. uncatalyzed enzyme catalyzed rate =  [P] tt rate = k [ES]