1. 1 Chemical Kinetics Chapter 3 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. 2 Chemical Kinetics Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed? Reaction rate is the change in the concentration of a reactant or a product with time (M/s). A B rate = -  [A] tt rate =  [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.

3. 3

4. Meaning and Measurement of Rate

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

6. 6 Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) time red-brown

7. Lets have closer look to the reaction below. Concentration, M Time Products NO or CO 2 Reactants CO or NO 2 . Slope = tg  =Rate Concentration, M Time

8. Meaning and Measurement of Rate Chapter 2

9. Meaning and Measurement of Rate Example 1 and 2 (Page 135) Example 3 Find the rate relationship of reactants and products for the given reaction. N 2 (g) + 3H 2 (g)  2NH 3 (g) Solution

10. Meaning and Measurement of Rate Example 4 In the following decomposition reaction, 2N 2 O 5 → 4NO 2 + O 2 oxygen gas is produced at the average rate of 9.1 × 10 -4 mol · L -1 · s -1. Over the same period, what is the average rate of the following: the production of nitrogen dioxide. the loss of dinitrogen pentoxide. Chapter 2

11. Meaning and Measurement of Rate Solution rate NO 2 production = 4 × (9.1 × 10 -4 mol · L -1 · s -1 ) = 3.6 × 10 -3 mol · L -1 · s -1 rate loss of N 2 O 5 = 2× (9.1 × 10 -4 mol · L -1 · s -1 ) = 1.8 × 10 -3 mol · L -1 · s -1 Chapter 2

12. Meaning and Measurement of Rate Example 5 Consider the following reaction: N 2(g) + 3H 2(g) → 2NH 3(g) If the rate of loss of hydrogen gas is 0.03 mol · L -1 · s -1, what is the rate of production of ammonia? Solution From the balanced equation we see that there are 2 moles NH 3 produced for every 3 moles H 2 used. Thus: rate NH 3 production =2/3 × (0.03 mol · L -1 · s -1 ) = 0.02 mol · L -1 · s -1 Chapter 2

13. Meaning and Measurement of Rate Example 6 0.05 mol of SO 3 gas is produced in 1-L container within 2 minutes according to the reaction below. Find the rate of consumption of SO 2 and O 2 gases in M/s. 2SO 2 (g) + O 2 (g)  2SO 3 (g) Chapter 2

14. Meaning and Measurement of Rate 6.4 Write the rate expressions for the following reactions in terms of the disappearance of the reactants and the appearance of the products: Write the rate expression for the following reaction:

15. 2. Collision Theory Chapter 2

16. 2. Collision Theory Collision theory explains the chemical reactions occur as a result of collisions. For a collision to lead to a reaction; Molecules must be properly oriented, Molecules must have enough energy. Arrhenius proposed that for colliding molecules to lead to a reaction they must be “activated”. The minimum energy which reacting molecules must have for resulting in a reaction is called “activation energy”, Ea. The molecules which have activation energy can form “activated complex” or “transition state”.

17. 2. Collision Theory

18. ∆H PE-RC Diagram for an exothermic reaction

19. 2. Collision Theory PE-RC Diagram for an endothermic reaction ∆H

20. 2. Collision Theory Ea f : Activation energy for forward reaction Ea r : Activation energy for reverse reaction The heat of reaction, ∆H, in terms of activation energies is defined as follows; ∆H = Ea f - Ea r

21. 2. Collision Theory Example 6 If the PE-RC diagram for C + O 2  CO 2 given below; calculate, ∆H, Ea f, and Ea r PE RC CO 2 C + O 2 242 -393 0

22. 2. Collision Theory Solution PE RC CO 2 C + O 2 ∆H 242 -393 0 Ea r Ea f ∆H = Ea f – Ea r = 242- (393 + 242) = - 393 kj

23. 2. Collision Theory Example 7 Following PE-RC diagram for the reaction A 2 + B 2  2AB given below; calculate, ∆H, Ea f, and Ea r PE RC 2AB A 2 + B 2 80 45 55 0

24. 2. Collision Theory Solution PE RC 2AB A 2 + B 2 80 45 55 0 Ea r ∆H Ea f ∆H = Ea f - Ea r = 35 – 25 = 10 kj

25. 2. Collision Theory Example 9 Sketch a potential energy curve that is represented by the following values of ΔH and E a. You may make up appropriate values for the y-axis (potential energy). ΔH = -100 kJ and E a = 20 kJ Is this an endothermic or exothermic reaction?

26. Post Test Consider the reaction 2NO(g) + O 2 (g) 2NO 2 (g) Suppose that at a particular moment during the reaction nitric oxide (NO) is reacting at the rate of 0.06 M/s. (a) At what rate is NO 2 being formed? (b) At what rate is molecular oxygen reacting?

27. Answer the following questions based on the potential energy diagram shown here: a.Endothermic or exothermic reaction? b.Determine the energy of activated complex. c. Determine the Ea, Eaf and Ear. d.Determine the heat of reaction, ΔH, (enthalpy change). Post Test

28. 3.The Rate Law The 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. aA + bB cC + dD Rate = k [A] x [B] y Reaction is xth order in A Reaction is yth order in B Reaction is (x +y)th order overall k is the rate constant, it depends only on temperature and activation energy

29. 3. Rate Law Example 1 Write the possible rate law of the following reactions. a.Cl 2 (g) + H 2 (g) → 2HCl(g) b.Fe(s) + Cl 2 (g) → FeCl 2 (g) c.Fe 2 O 3 (s) + 3H 2 SO 4 (aq) → Fe 2 (SO 4 ) 3 (s) + 3H 2 O(l) d.Ca(s) + 2Ag + (aq) → 2Ca +2 (aq) + Ag(s)

30. 3. Rate Law Example 2 Write the rate law of the following one step reaction, determine the order of reaction in terms of N 2 and H 2, and overall order. N 2 (g) + 3H 2 (g) → 2NH 3 (g)

31. 3. Rate Law Example 3 The rate law for the reaction NH 4 (aq) + NO 2 (aq) N 2 (g) + 2H 2 O(l) is given by rate = k[NH 4 ][NO 2 ]. At 25°C, the rate constant is 3.0 x 10 -4 /M s. Calculate the rate of the reaction at this temperature if [NH 4 ] 0.25 M and [NO 2 ] 0.080 M.

32. 3. Rate Law Example 4 Determine the overall orders of the reactions to which the following rate laws apply: (a)Rate = k[NO 2 ] 2, (b) Rate = k, (c)Rate = k[H 2 ][Br 2 ] 1/2, (d) Rate = k[NO] 2 [O 2 ].

33. F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ][ClO 2 ] 3. Rate Law 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. 1

34. 34 F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k [F 2 ] x [ClO 2 ] y Double [F 2 ] with [ClO 2 ] constant Rate doubles x = 1 Quadruple [ClO 2 ] with [F 2 ] constant Rate quadruples y = 1 rate = k [F 2 ][ClO 2 ]

35. 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 - ]

36. 3. Rate Law Example 5 Consider the reaction A + B products From the following data obtained at a certain temperature, determine the order of the reaction and calculate the rate constant:

37. 3. Rate Law Example 6 Consider the reaction X + Y Z From the following data, obtained at 360 K, (a) determine the order of the reaction, and (b) determine the initial rate of disappearance of X when the concentration of X is 0.30 M and that of Y is 0.40 M.

38. 3. Rate Law Example 7 Consider the reaction A B The rate of the reaction is 1.5 x 10 -2 M/s, when the concentration of A is 0.30 M. Calculate the rate constant if the reaction is (a)first order in A (b)second order in A.

39. Post Test Exp No[A][B]Rate 10.110.2 M/s 20.210.4 M/s 30.120.8 M/s For the reaction A + B  C, find a)Order of the reaction? b) The rate constant (k)? c)The rate, if the concentration of A is 0,3 M and concentration of B is 2 M?

40. Reaction Mechanisms The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions. The sequence of elementary steps that leads to product formation is the reaction mechanism. 2NO (g) + O 2 (g) 2NO 2 (g) N 2 O 2 is detected during the reaction! Elementary step:NO + NO N 2 O 2 Elementary step: N 2 O 2 + O 2 2NO 2 Overall reaction: 2NO + O 2 2NO 2 +

41. 2NO (g) + O 2 (g) 2NO 2 (g) Mechanism:

42. 42 Elementary step:NO + NO N 2 O 2 Elementary step: N 2 O 2 + O 2 2NO 2 Overall reaction: 2NO + O 2 2NO 2 + Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation. An intermediate is always formed in an early elementary step and consumed in a later elementary step. The molecularity of a reaction is the number of molecules reacting in an elementary step. Unimolecular reaction – elementary step with 1 molecule Bimolecular reaction – elementary step with 2 molecules Termolecular reaction – elementary step with 3 molecules

43. The slowest step in a mechanism is the rate-determining step. The slowest step in a mechanism has the highest activation energy value. Therefore the rate law is Rate = k[ Br][H 2 ] For Example, H 2 + Br 2 → 2 HBr Following mechanism has been proposed, 1.stepBr 2 → 2Br (fast) 2.stepBr + H 2 → HBr + H(slow) 3.stepH + Br → HBr (fast)

44. 3. Rate Mechanisms Example 1 Find the rate expression of the following reaction; NH 4 + + HNO 2 → N 2 + 2 H 2 O + H + having following mechanism; 1.Step HNO 2 + H + → H 2 O + NO + (fast) 2.StepNH 4 + → NH 3 + H + (fast) 3.Step NO + + NH 3 → NH 3 NO + (slow) 4. StepNH 3 NO + → H 2 O + H + + N 2 (fast) Rate = k [NH 3 ] [NO + ]

45. 3. Rate Mechanisms PE RC For a reaction PE-RC diagram is given. a.How many elementary steps are there in the reaction? b.Show the activation energies of all steps. c.Which step is the slowest, and rate determining step? d.Is the reaction exothermic or endothermic? Example 2

46. 3. Rate Mechanisms Solution PE RC Ea 1 Ea 2 Ea 3 ∆H

47. 4. Factors Affecting Rate

48. A. Nature of Reactants Generally; If there are many bond must be broken and many bonds must be formed in a reaction then the reaction is slow. The reactions between simple ions are very fast. If the reaction is between many ions and molecules then the reaction is slow. C 3 H 6 O + 4O 2 → 3CO 2 + 3H 2 O(Slow) Ag + (aq) + Cl - (aq) → AgCl(s)(very fast) 2MnO 4 - (aq) + 5SO 3 -2 (aq) + 6H + → 2Mn +2 (aq) + 5SO 4 -2 (aq) + 3H 2 O (fast)

49. 4. Factors Affecting Rate B. Concentration of Reactants Rate of reactions increases when the concentration of reactant is increased. Rate of reactions is directly proportional to the concentration of reactants. For the reaction, 2X (g) +3Y (g) +Z (g) →2K (g) +L (g) the reaction rate increases by a factor of 8 when the volume of the container is halved. Doubling only the concentration of X increases the rate by a factor of 2 and doubling only the concentration of Z has no influence on the reaction rate. Using these data, determine the rate law for the reaction. Example 15 Solutionrate= k[X][Y] 2

50. 4. Factors Affecting Rate B. Concentration of Reactants

51. 4. Factors Affecting Rate B. Concentration of Reactants The following data were recorded to determine the rate equation of the reaction; O 3(g) +NO (g) →O 2(g) +NO 2(g) I. When the concentrations of both O 3 and NO were doubled, the rate increased 4 times. II. When the concentration of NO was held constant but the concentration of O 3 was tripled, the rate was tripled too. A. Find the rate equation for the reaction. B. If both [O 3 ] and [NO] had tripled what would have been the rate? Example 1 Solution A. Rate=k[NO][O 3 ], B. Rate would have increased 9 times.

52. 4. Factors Affecting Rate B. Concentration of Reactants The following data were collected for the reaction, 2X (g) + 3Y (g) + Z (g) → 2K (g) + L (g) Find the rate expression and the value of k. Example 2 Solution Rate =k[X][Y] 2, k = 2 M -2 min -2 Exp No [X] M [Y] M [Z] M Initial Rate (M/min) 1 0.2 0.1 0.1 4.0x10 -3 2 0.2 0.2 0.1 16x10 -3 3 0.2 0.2 0.2 16x10 -3 4 0.1 0.2 0.1 8.0x10 -3

53. 4. Factors Affecting Rate B. Concentration of Reactants The following data are given for the reaction: 2A+B+C → D+2E Find the rate expression, order and the value of rate constant, k. Example 3 Solution Rate =k[A][C], overall order is 1+1=2, k = 0.2 Mmin -1 [A]M[B]M[C]MInitial rate (M/min) 0.10.20.24.0x10 -3 0.10.40.24.0x10 -3 0.20.20.28.0x10 -3 0.30.10.21.2x10 -2 0.10.40.61.2x10 -2

54. 4. Factors Affecting Rate C. Temperature of the System Rates of almost all reactions increase as temperature is increased. It is often stated that rate doubles by increasing 10 o C.

55. 4. Factors Affecting Rate C. Temperature of the System Ea At T o C number of molecules exceeding Ea is a, At (T+20) o C number of molecules exceeding Ea is a+b. The major factor in increasing rate when temperature is increased is increasing effective collisions. No of Molecules ToCToC (T+20) o C Kinetic Energy b a

56. 4. Factors Affecting Rate D. Presence of a Catalyst A catalyst is a substance which increases rate of reaction without being consumed in reactions. It provides an alternate reaction pathway with a lower activation energy.

57. 4. Factors Affecting Rate D. Presence of a Catalyst

58. 4. Factors Affecting Rate D. Presence of a Catalyst Catalyst have the following properties; They remain unchanged after the reaction, Both reverse and forward reactions are affected by catalyst, ∆H remains unchanged, It does not make impossible reactions possible. Depletion of ozone layer is speeded up by CFC gases in which Cl∙ radicals acts as catalyst. O 3 (g) + Cl∙(g) → O 2 (g) + ClO∙(g) ClO∙(g) + O∙(g) → O 2 (g) + Cl∙(g) Example 19

59. 4. Factors Affecting Rate D. Presence of a Catalyst

60. 4. Factors Affecting Rate E. Interacting Area Increasing surface area between colliding particles increases the rate of reactions. Remember the dissolution of powdered sugar and lump of sugar. For the reaction; 2HCl(aq) + CaCO 3 (s) → CaCl 2 (aq) + CO 2 (g) + H 2 O(g) If powdered CaCO 3 is used the reaction will be faster