1. Vanessa N. Prasad-Permaul CHM 1046 Valencia Community College

2. Introduction 1. Chemical kinetics is the study of reaction rates 2. For a chemical reaction to be useful it must occur at a reasonable rate 3. It is important to be able to control the rate of reaction 4. Factors that influence rate a) Concentration of reactants (molarity) b) Nature of reaction, process by which the reaction takes place c) Temperature d) Reaction mechanism (rate determining step) e) Catalyst

3. Reaction rate Reaction rate- Positive quantity that expresses how the conc. of a reactant or product changes with time 2N 2 O 5(g)  4NO 2(g) + O 2(g) 1. [ ] = concentration, molarity, mole/L 2. [N 2 O 5 ] decreases with time 3. [NO 2 ] increases with time 4. [O 2 ] increases with time 5. Because of coefficients the concentration of reactant and products does not change at the same rate 6. When 2 moles of N 2 O 5 is decomposed, 4 moles of NO 2 and 1 mole of O 2 is produced

4. 2N 2 O 5(g)  4NO 2(g) + O 2(g) -  [N 2 O 5 ] =  [NO 2 ] =  [O 2 ] 2 ½ -  [N 2 O 5 ] because it’s concentration decreases, other positive because they increase Rate of reaction can be defined by dividing by the change in time,  t rate = -  [N 2 O 5 ] =  [NO 2 ] =  [O 2 ]  t 2  t ½  t

5. Reaction Rates Generic Formula: aA + bB  cC + dD rate = -  [A] = -  [B] =  [D] =  [C] a  t b  t d  t c  t

6. EXAMPLE 13.1: CONSIDER THE REACTION OF NO 2 WITH F 2 TO GIVE NO 2 F 2NO 2 (g) + F 2 (g) 2NO 2 F(g) HOW IS THE RATE OF FORMATION RELATED TO THE RATE OF REACTION OF F 2 ? NO 2 F =  [NO 2 F] F 2 = -  [F 2 ]  t  t  [NO 2 F] = -  [F 2 ] 2  t  t

7. EXERCISE 13.1: CONSIDER THE REACTION OF NO 2 WITH F 2 TO GIVE NO 2 F 2NO 2 (g) + F 2 (g) 2NO 2 F(g) HOW IS THE RATE OF FORMATION RELATED TO THE RATE OF REACTION OF NO 2 ?

8. EXAMPLE 13.2: CALCULATE THE AVERAGE RATE OF DECOMPOSITION OF N 2 O 5 N 2 O 5(g) 2NO 2(g) + 1/2O 2(g) N 2 O 5 = -  [N 2 O 5 ]  t -(0.93x10 -2 M – 1.24x10 -2 M) = -0.31x10 -2 M = 5.2x10 -6 M/s 1200s – 600s 600s TIME[N2O5] 600s1.24 x10 -2 M 1200s0.93x10 -2 M

9. EXERCISE 13.2: IODIDE ION IS OXIDIZED BY HYPOCHLORITE ION IN BASIC SOLUTION. CALCULATE THE AVERAGE RATE OF REACTION OF I - DURING THIS TIME INTERVAL: I - (aq) + ClO - (aq) Cl - (aq) + IO - (aq) TIME[I - ] 2.00s0.00169M 8.00s0.00101M

10. Reaction Rate and Concentration The higher the conc. of starting reactant the more rapidly a reaction takes place 1. Reactions occur as the result of collisions between reactant molecules 2. The higher the concentration of molecules, the greater the # of collisions in unit time and a faster reaction 3. As the reactants are consumed the concentration decreases, collisions decrease, reaction rate decreases 4. Reaction rate decreases with time and eventually = 0, all reactants consumed 5. Instantaneous rate, rate at a particular time 6. Initial rate at t = 0

11. Rate Expression and Rate Constant Rate expression / rate law: rate = k[A] where k = rate constant, varies w/ nature and temp. [A] = concentration of A Rate law: an equation that related the rate of a reaction to the concentration of reactants (and catalyst) raised to various powers

12. Order of rxn involving a single reactant General equationrate expression A  products rate = k[A] m where m=order of reaction m=0 zero order m=1 first order m=2 second order m, can’t be deduced from the coefficient of the balanced equation. Must be determined experimentally!

13. Order of rxn involving a single reactant Rate of decomposition of species A measured at 2 different conc., 1 & 2 rate 2 = k[A 2 ] m rate 1 = k[A 1 ] m By dividing we can solve for m, to find the order of the reaction Rate 2 = [A 2 ] m Rate 1 [A 1 ] m (Rate 2 /Rate 1 ) = ([A 2 ]/[A 1 ]) m

14. EXAMPLE 13.3: BROMIDE ION IS OXIDIZED BY BROMATE ION IN ACIDIC SOLUTION 5Br - (aq) + BrO 3 - (aq) + 6H + (aq) 3Br 2(aq) + 3H 2 O (l) Rate = k[Br - ][BrO 3 - ][H + ] 2 WHAT IS THE ORDER OF REACTION WITH RESPECT TO EACH REACTANT SPECIES? WHAT IS THE OVERALL ORDER OF REACTION? Br - IS FIRST ORDER BrO 3 - IS FIRST ORDER H + IS SECOND ORDER THE REACTION IS FOURTH ORDER OVERALL (1+1+2)=4

15. EXERCISE 13.3: WHAT ARE THE REACTIO ORDERD WITH RESPECT TO EACH REACTANT SPECIES FOR THE FOLLOWING REACTION? WHAT IS THE OVERALL REACTION? NO 2(g) + CO (g) NO (g) + CO 2 (g) RATE = k[NO 2 ] 2

16. Example 1 CH 3 CHO (g)  CH 4(g) + CO (g) [CH 3 CHO].10 M.20 M.30 M.40 M Rate (mol/L s).085.34.761.4 Using the given data determine the reaction order

17. Example 1 Rate 2 = [A 2 ] m Rate 1 [A 1 ] m 4 = 2 m log 4m = 2 log 2 second order rate = k[CH 3 CHO] 2 once the order of the rxn is known, rate constant can be calculated, let’s calculate the rate constant, k

18. Example 1 rate = k[CH 3 CHO] 2 rate =.085 mol/L s conc =.10 mol/L k = rate = 0.085 mol/L s = 8.5 L/mol s [CH 3 CHO] 2 (0.10 mol/L) 2 now we can calc. the rate at any concentration, let’s try.55 M

19. Example 1 rate = k[CH 3 CHO] 2 rate = 8.5 L/mol s [.55] 2 = 2.6 mol/ L s Rate when [CH 3 CHO] =.55 M is 2.6 mol/L s

20. Order of rxn with more than 1 reactant aA + bB  prod rate exp: rate = k [A] m [B] n m = order of rxn with respect to A n = order of rxn with respect to B Overall order of the rxn is the sum, m + n

21. Key  When more than 1 reactant is involved the order can be determined by holding the concentration of 1 reactant constant while varying the other reactant. From the measured rates you can calculate the order of the rxn with respect to the varying reactant

22. Example 2 (CH 3 ) 3 CBr + OH -  (CH 3 ) 3 COH + Br - Exp. 1Exp. 2Exp. 3Exp. 4Exp. 5 [(CH 3 ) 3 CBr]0.51.01.51.0 [OH - ]0.05 0.100.20 Rate (mol/L s)0.0050.010.0150.01 Find the order of the reaction with respect to both reactants, write the rate expression, and find the overall order of the reaction

23. Reactant concentration and time Rate expressionrate = k[A] Shows how the rate of decomposition of A changes with concentration More important to know the relation between concentration and time Using calculus: Integrated rate equations relating react conc. to time

24. For the following rate law, what is the overall order of the reaction? Rate = k [A] 2 [B] 1. 1 2. 2 3. 3 4. 4

25. Zero Order Zero order:A  Products rate = k [A] 0 – [A] = kt t½ = [A] 0 2k m = 0: zero order - rate is independent of the concentration of reactant. Doubling the concentration has no effect on rate.

26. First Order First Order:A  Products rate = k[A] ln [A] o /[A] = kt t½ =.693/k [A] o = original conc. of A [A] = Conc. of A at time, t k = first order rate constant ln = natural logarithm m = 1: first order - rate is directly proportional to the concentration of the reactant. Doubling the concentration increases the rate by a factor of 2.

27. Second Order Second order: A  Products rate = k[A] 2 1 – 1 = kt t½ = 1 [A] [A] 0 k[A] 0 m = 2: second order - the rate is to the square of the concentration of the reactant. Doubling the concentration increases the rate by a factor of 4.

28. EXAMPLE 13.4: IODIDE ION IS OXIDIZED IN ACIDIC SOLUTION TO TRIIODIDE ION I 3 -, BY HYDROGEN PEROXIDE H 2 O 2 (aq) + 3I - (aq) + 2H + (aq) I 3 - (aq) + H 2 O (l) a)OBTAIN THE REACTION ORDER WITH RESPECT TO EACH REACTANT b)FIND THE RATE CONSTANT Initial concentrations (mol/L) Initial Rate [mol/(L.s)] H2O2I-I- H+H+ EXP. 10.010 0.000501.15 x 10 -6 EXP. 20.0200.0100.000502.30 x 10 -6 EXP. 30.0100.0200.000502.30 x 10 -6 EXP. 40.010 0.001001.15 x 10 -6

29. Rate = k[H 2 O 2 ] m [I - ] n [H + ] p (Rate) 2 = k[H 2 O 2 ] m [I - ] n [H + ] p (Rate) 1 = k[H 2 O 2 ] m [I - ] n [H + ] p Rate constant cancels 2.30 x 10 -6 = [0.020] m [0.010] n [0.00050] p 1.15 x 10 -6 = [0.010] m [0.010] n [0.00050] p 2= 2 m m = 1  doubling the H 2 O 2 concentration doubles the rate Rate = k[H 2 O 2 ] [I - ] Reaction order = 1, 1, 0

30. Rate = k[H 2 O 2 ] [I - ] 1.15 x 10 -6 mol = k x 0.010 mol x 0.010 mol L.s L L k = 1.15 x 10 -6 mol L.s 0.010 mol x 0.010 mol L L k = 1.2 x 10 -2 L/(mol.s)

31. EXERCISE 13.4: THE INITIAL-RATE METHODWAS APPLIED TO THE DECOMPOSITION OF NITROGEN DIOXIDE NO 2 (aq) 2NO (g) + O 2 (g) FIND THE RATE LAW AND THE VALUE OF THE RATE CONSTANT WITH RESPECT TO O 2 FORMATION INITIAL NO2 CONCENTRATION (mol/L) INITIAL RATE OF FORMATION OF O2 mol/(L.s) EXP. 10.0107.1 x 10-5 EXP. 20.02028 x 10-5

32. Example 3 The following data was obtained for the gas-phase decomp. of HI Is this reaction zero, first, or second order in HI? Hint: Graph each Conc. Vs. time corresponding to correct [A], ln [A], or 1/[A] Time (h)0246 [HI]1.000.500.330.25

33. [HI] vs. time not linear so it is not zero order

34. ln [HI] vs. time not linear so it is not first order

35. 1/[HI] vs. time is linear so it is second order

36. OrderRate Expression Conc-Time Relation Half-lifeLinear Plot 0 Rate = k[A] 0 – [A] = kt [A] 0 2k [A] vs. t 1 Rate = k[A]ln [A] 0 = kt [A] 0.693 k ln [A] vs. t 2 Rate = k[A] 2 1 – 1 = kt [A] [A] 0 1 k[A] 0 1 vs. t [A]

37. EXAMPLE 13.5: HOW LONG WOULD IT TAKE FOR THE CONC. OF N 2 O 5 TO DECREASE TO 1.00 x 10 -2 mol/L FROM THE INITIAL VALUE? kt = ln [A] 0 [A] t 4.80 x 10 -4 /s (t) = ln [1.65 x 10 -2 mol/L] [1.00 x 10 -2 mol/L] 4.80 x 10 -4 /s (t) = 0.500775 t = 0.500775 4.80 x 10 -4 /s t = 1043.28125s * 1min 60sec t = 17.4min

38. EXERCISE 13.5: A) WHAT WOULD THE CONCENTRATION BE OF DINITROGEN PENTOXIDE IN THE EXPERIMENT DESCRIBED IN EXAMPLE 13.5 AFTER 6.00 x 10 2 s? B) HOW LONG WOULD IT TAKE FOR THE CONCENTRATION OF N 2 O 5 TO DECREASE TO 10% OF ITS INITIAL VALUE?

39. EXAMPLE 13.6: SULFURYL CHLORIDE, SOC 2 Cl 2, IS A COLORLESS CORROSIVE LIQUID WHOSE VAPOR DECOMPOSES IN A FIRST-ORDER REACTION TO SULFUR DIOXIDE AND CHLORINE. SO 2 Cl 2 (g) SO 2 (g) + Cl 2 (g) AT 320 o C, THE RATE CONSTANT IS 2.20 x 10 -5 /s. WHAT IS THE HALF-LIFE OF SO 2 Cl 2 VAPOR AT THIS TEMPERATURE? A) HOW LONG (IN HOURS) WOULD IT TAKE FOR 50.0% OF THE SO 2 Cl 2 TO DECOMPOSE? B) HOW LONG WOULD IT FOR 75.0% OF THE SO 2 Cl 2 ?

40. EXAMPLE 13.6: t½ = 0.693/k t½ = 0.693 2.20 x 10 -5 /s t½ = 3.15 x 10 4 s The half-life is the time required for 50.0% of the SO 2 Cl 2 to decompose. This is 8.75 hours 8.75 hrs x 2 = 17.5 hrs kt = ln [A] 0 [A] t

41. EXERCISE 13.6: THE ISOMERIZATION OF CYCLOPROPANE TO PROPYLENE IS FIRST ORDER IN CYCLOPROPANE AND OVERALL. AT 1000 o C THE RATE CONSTANT IS 9.2/s A) WHAT IS THE HALF-LIFE? B)HOW LONG WOULD IT TAKE FOR THE CONCENTRATION OF CYCLOPROPANE TO DECREASE TO 50% OF ITS INITIAL VALUE? C) TO 25% OF ITS INITIAL VALUE?

42. Activation Energy Activation Energy: E a (kJ) For every reaction there is a certain minimum energy that molecules must possess for collisions to be effective. 1. Positive quantity (E a >0) 2. Depends only upon the nature of reaction 3. Fast reaction = small E a 4. Is independent of temp and concentration

43. For the following reaction: A + B  C If the concentration of A is doubled, and B is constant, the rate doubles. What is the order of the reaction with respect to A? 1. 0 2. 1 3. 2

44. Activation Energy

45. Reaction Rate and Temp 1. As temp increases rate increases, Kinetic Energy increases, and successful collisions increase 2. General rule for every 10°C inc. in temp, rate doubles

46. If I increase the temperature of a reaction from 110 K to 120 K, what happens to the rate of the reaction? 1. Stay the same 2. Doubles 3. Triples

47. The Arrhenius Equation f = e -Ea/RT f = fraction of molecules having an En. equal to or greater than E a R = gas constantA = constantT = temp in K ln k = ln A –E a /RT plot of ln k Vs. 1/T linear slope = -E a /R Two-point equation relating k & T ln k 2 = E a [1/T 1 – 1/T 2 ] k 1 R

48. EXAMPLE 13.7: THE RATE CONSTANT FOR THE FORMATION OF HYDROGEN IODIDE FROM THE ELEMENTS IS 2.7 x 10 -4 L/(mol.s) @600 K AND 3.5 x 10 -3 L/(mol.s) AT 650 K. FIND THE ACTIVATION ENERGY E a ln k 2 = E a [1/T 1 – 1/T 2 ] k 1 R ln 3.5 x 10 -3 L/(mol.s) = E a 1 - 1 2.7 x 10 -4 L/(mol.s) 8.314 J/(mol.K) 600K 650K E a = 2.56 * (8.314 J/mol ) 1.28 x 10 -4 E a = 1.66 x 10 5 J/mol or 166kJ

49. EXERCISE 13.7: ACETALDEHYDE DECOMPOSES WHEN HEATED. THE RATE OF DECOMPOSITION IS 1.05 x 10 -3 L/mol.s AT 759 K AND 2.14 x 10 -2 L/mol.s AT 836 K. CH 3 CHO (g) CH 4 (g) + CO (g) WHAT IS ACTIVATION ENERGY FOR THIS DECOMPOSITION? WHAT IS THE RATE CONSTANT AT 865 K?

50. Example 4 For a certain rxn the rate constant doubles when the temp increases from 15 to 25°C. a) Calc. The activation energy, E a b) Calc. the rate constant at 100°C, taking k at 25°C to be 1.2 x 10 -2 L/mol s

51. Reaction Mechanism Description of a path, or a sequence of steps, by which a reaction occurs at the molecular level. Simplest case- only a single step, collision between two reactant molecules

52. Reaction Mechanism “Mechanism” for the reaction of CO with NO 2 at high temp (above 600 K) CO (g) + NO 2(g)  NO (g) + CO 2(g) “Mechanism” for the reaction of CO with NO 2 at low temp NO 2(g) + NO 2(g)  NO 3(g) + NO (g) slow CO (g) + NO 3(g)  CO 2(g) + NO 2(g) fast CO (g) + NO 2(g)  NO (g) + CO 2(g) overall The overall reaction, obtained by summing the individual steps is identical but the rate expressions are different. High temp: rate = k[CO][NO 2 ] Low temp: rate= k[NO 2 ] 2

53. EXAMPLE 13.8: CCl 4 IS OBTAINED BY CHLORINATING METHANE OR AN INCOMPLETELY CHLORINATED METHANE SUCH AS CHLOROFORM CH 3 Cl. THE MECHANISM FOR THE GAS PHASE CHLORINTATION OF CH 3 Cl IS Cl 2 2Cl Cl + CHCl 3 HCl + CCl 3 Cl + CCl 3 CCl 4 OBTAIN THE NET (OVERALL) CHEMICAL EQUATION Cl 2 2Cl Cl + CHCl 3 HCl + CCl 3 Cl + CCl 3 CCl 4 Cl 2 + CHCl 3 HCl + CCl 4

54. EXERCISE 13.8: THE IODIDE ION CATALYZES THE DECOMPOSITION OF AQUEOUS HYDROGEN PEROXIDE. THIS DECOMPOSITION IS BELIEVED TO OCCUR IN TWO STEPS: H 2 O 2 + I - H 2 O + IO - H 2 O 2 + IO - H 2 O + O 2 + I - WHAT IS THE OVERALL EQUATION REPRESENTING THIS DECOMPOSITION? NOTE THAT THE IO - IS A REACTION INTERMEDIATE. THE IODIDE ION IS NOT AN INTERMEDIATE; IT WAS ADDED TO THE REACTION MIXTURE.

55. Reaction Mechanism Elementary Steps: Individual steps that constitute a reaction mechanism UnimolecularA  B + Crate = k[A] Bimolecular A + A  B + C rate = k[A][A] = [A] 2 TermolecularA + B + C  D + E rate = k[A][B][C] The rate of an elementary step is equal to a rate constant, k, multiplied by the concentration of each reactant molecule. You can treat all reactants as if they were first order. If a reactant is second order it will appear twice in the general equation.

56. EXAMPLE 13.9: WHAT IS THE MOLECULARITY OF EACH STEP IN THE MECHANISM DESCRIBED IN EXAMPLE 13.8 1) Cl 2 2Cl 2)Cl + CHCl 3 HCl + CCl 3 3) Cl + CCl 3 CCl 4 1)UNIMOLECULAR 2)BIMOLECULAR 3)BIMOLECULAR

57. EXAMPLE 13.9: THE FOLLOWING IS AN ELEMENTARY REACTION THAT OCCURS IN THE DECOMPOSITION OF OZONE IN THE STRATOSPHERE BY NITROGEN MONOXIDE NO + O 3 O 2 + NO 2 WHAT IS THE MOLECULARITY OF THIS REACTION?

58. Reaction Mechanism Slow Steps- A step that is much slower than any other in a reaction mechanism. Rate-determining step - The rate of the overall reaction can be taken to be that of the slow step Step 1:A  Bfast Step 2:B  Cslow Step 3:C  Dfast A  D The rate A  D (overall reaction) is approx. equal to the rate of B  C the slow step

59. Deducing a Rate Expression from a Proposed Mechanism 1. Find the slowest step and equate the rate of the overall reaction to the rate of that step. 2. Find the rate expression for the slowest step. NO 2(g) + NO 2(g)  NO 3(g) + NO (g) slow CO (g) + NO 2(g)  CO 2(g) + NO 2(g) fast CO (g) + NO 2(g)  CO 2(g) + NO (g) Rate of overall reaction = rate of 1 st step = k[NO 2 ] [NO 2 ] = k[NO 2 ] 2

60. EXAMPLE 13.10: WRITE RATE EQUATIONS FOR EACH OF THE FOLLOWING ELEMENTARY REACTIONS: OZONE IS CONVERTED TO O 2 BY NO IN A SINGLE STEP 1) O 3 + NO O 2 + NO 2 rate = k[O 3 ][NO] THE RECOMBINATION OF IODINE ATOMS OCCURS AS FOLLOWS 2) I + I + M M* + I 2 rate = k[I] 2 [M] A WATER MOLECULE ABSORBS ENERGY; LATER IN THE PROCESS ENOUGH ENERGY FLOWS INTO O-H BOND TO BREAK IT 3) H 2 O H + O H rate = k[H 2 O]

61. EXAMPLE 13.10: WRITE RATE EQUATIONS FOR THE FOLLOWING ELEMENTARY REACTION: NO 2 + NO 2 O 2 + N 2 O 4

62. EXAMPLE 13.11: OZONE REACTS WITH NITROGEN DIOXIDE TO PRODUCE OXGEN AND DINITROGEN PENTOXIDE. 2NO 2 (g) + O 3 (g) O 2 (g) + N 2 O 5 (g) THE PROPOSED MECHANISM IS O 3 + NO 2 NO 3 + O 2 (slow) NO 3 + NO 2 N 2 O 5 (fast) WHAT IS THE RATE LAW PREDICTED BY THIS MECHANISM? RATE = rate = k[O 3 ][NO 2 ]

63. EXAMPLE 13.11: THE IODIDE-ION-CATALYZED DECOMPOSITION OF HYDROGEN PEROXIDE IS BELIEVED TO HAVE THE FOLLOWING MECHANISM: H 2 O 2 + I - H 2 O + IO - (slow) H 2 O 2 + IO - H 2 O + O 2 + I - (fast) WHAT IS THE RATE LAW PREDICTED BY THIS MECHANISM?

64. EXAMPLE 13.12: NITROGEN MONOXIDE CAN BE REDUCED WITH HYDROGEN GAS TO GIVE NITROGEN AND WATER VAPOR 2NO (g) + 2H 2 (g) N 2 (g) + 2H 2 O (g) (OVERALL RXN) THE PROPOSED MECHANISM IS 2NO N 2 O 2 (fast, eq.) H 2 + N 2 O 2 N 3 O + H 2 O (slow) H 2 + N 2 O N 2 + H 2 O (fast) WHAT IS THE RATE LAW PREDICTED BY THIS MECHANISM? RATE = rate = k[NO] 2 [H 2 ]

65. EXERCISE 13.12: NITROGEN MONOXIDE REACTS WITH OXYGEN TO PRODUCE NITROGEN DIOXIDE 2NO (g) + O 2 (g) 2NO 2 (g) (OVERALL RXN) THE PROPOSED MECHANISM IS O 2 + NO NO 3 (fast, eq.) NO + NO 3 NO 2 + NO 2 (slow) WHAT IS THE RATE LAW PREDICTED BY THIS MECHANISM?

66. Elimination of Intermediates Intermediate 1. A species produced in one step of the mechanism and consumed in a later step. 2. Concentration too small to determine experimentally 3. Must be eliminated from rate expression 4. The final rate expression must include only those species that appear in the balanced equation for the overall reaction

67. Example 5 Find the rate expression for the following reaction mechanism Step1:NO (g) + Cl 2(g)  NOCl 2(g) fast Step2:NOCl 2(g) + NO (g)  2NOCl (g) slow 2 NO (g) + Cl 2(g)  2NOCl (g)

68. Example 6 The decomposition of ozone, O 3, to diatomic oxygen, O 2, is believed to occur by a two-step mechanism: Step 1:O 3(g)  O 2(g) + O (g) fast Step 2:O 3(g) + O (g)  2 O 2(g) slow 2 O 3(g)  3 O 2(g)  Find the rate expression for this reaction

69. Which is an intermediate for the following multi step mechanism? 2 A + 2 B  C + 2 D Step 1 2 B  E Step 2 E + A  D + F Step 3 F + A  C + D 1. E 2. F 3. E & F 4. A

70. Catalysts Catalysis A catalyst increases the rate of a reaction without being consumed by it. Changes the reaction mechanism to one with a lower activation energy. 1. Heterogeneous catalysis a) Catalyst is in a different phase from the reaction mixture. Most common, solid catalyst with gas or liquid mixture. b) Solid catalyst is easily poisoned, foreign material deposited on the surface during reaction reduce or destroy its effectiveness. 2. Homogeneous Catalysis a) Same phase as the reactants

71. Which is the catalyst for the following multi step mechanism? 2 A + 2 B  C + 2 D Step 1 2 B + G  E Step 2 E + A  D + F Step 3 F + A  C + D + G 1. E 2. G 3. E & F 4. A

72. Chapter 12: Kinetics

73. Hydrogen peroxide decomposes to water and oxygen according to the following reaction H 2 O 2(aq)  H 2 O + ½ O 2(g) It’s rate of decomposition is measured by titrating samples of the solution with potassium permanganate (KMnO 4 ) at certain intervals. a) Initial rate determinations at 40  C for the decomposition give the following data: [H 2 O 2 ]Rate (mol/L min) 0.101.93 x 10 -4 0.203.86 x 10 -4 0.305.79 x 10 -4 1.Order of rxn? 2.Rate expression? 3.Calc. k @ 40°C 4.Calc. half-life @ 40°C

74. b) Hydrogen peroxide is sold commercially as a 30.0% solution. If the solution is kept at 40  C, how long will it take for the solution to become 10.0% H 2 O 2 ?

75. c) It has been determined that at 50  C, the rate constant for the reaction is 4.32 x 10 -3 /min. Calculate the activation energy for the decomposition of H 2 O 2

76. d) Manufacturers recommend that solutions of hydrogen peroxide be kept in a refrigerator at 4  C. How long will it take for a 30.0% solution to decompose to 10.0% if the solution is kept at 4  C?

77. e) The rate constant for the uncatalyzed reaction at 25  C is 5.21 x 10 -4 /min. The rate constant for the catalyzed reaction at 25  C is 2.95 x 10 8 /min. 1) What is the half-life of the uncatalyzed reaction at 25  C? 2) What is the half-life of the catalyzed reaction?

78. 1) Express the rate of reaction 2HI (g)  H 2(g) + I 2(g) a) in terms of  [H 2 ] b) in terms of  [HI]

79. 3) Dinitrogen pentaoxide decomposes according to the following equation: 2N 2 O 5(g)  4NO 2(g) + O 2(g) a) write an expression for reaction rate in terms of  [ N 2 O 5 ],  [NO 2 ], and  [O 2 ]

80. 4) What is the order with respect to each reactant and the overall order of the reactions described by the following rate expressions? a) rate = k 1 [A] 3 b) rate = k 2 [A][B] c) rate = k 3 [A][B] 2 d)rate = k 4 [B] e)rate = k

81. 5) Complete the following table for the reaction, which is first order in both reactants A (g) + B (g)  products [A][B]K (L/mol s)Rate (mol/L s).2.31.5.029.78.025.45.520.033

82. 7) The decomposition of ammonia on tungsten at 1100  C is zero-order, with a rate constant of 2.5 x 10 -4 mol/L min a) write the rate expression b) calculate the rate when the concentration of ammonia is 0.080M c) At what concentration of ammonia is the rate equal to the rate constant?

83. 8)In solution at constant H + concentration, I - reacts with H 2 O 2 to produce I 2 H + (aq) + I - (aq) + ½ H 2 O 2(aq)  ½ I 2(aq) + H 2 O The reaction rate can be followed by monitoring iodine production. The following data apply: [I - ][H 2 O 2 ]Rate (mol/L s).02 3.3 x 10 -5.04.026.6 x 10 -5.06.029.9 x 10 -5.04 1.3 x 10 -4 a)Order of I - b)Order of H 2 O 2 c)Calc. k d)Rate? When [I - ] =.01 M [H 2 O 2 ] =.03 M

84. 9) In the first-order decomposition of acetone at 500  C it is found that the concentration is 0.0300 M after 200 min and 0.0200M after 400 min. H 3 C-CO-CH 3(g)  products Calculate a) The rate constant b) The half-life c) The initial concentration

85. 10) The decomposition of hydrogen iodide is second-order. Its half-life is 85 seconds when the initial conc. is 0.15M HI (g)  ½ H 2(g) + ½ I 2(g) a) What is k for the reaction? b) How long will it take to go from 0.300M to 0.100M?

86. 11) Write the rate expression for each of the following elementary steps: a) K + + HCl  KCl + H + b) NO 3 + CO  NO 2 + CO 2 c) 2NO 2  2NO + O 2

87. 12) For the reaction 2H 2(g) + 2NO (g)  N 2(g) + 2H 2 O (g) the experimental rate expression is rate = k[NO] 2 [H 2 ] The following mechanism is proposed: 2NO  N 2 O 2 fast N 2 O 2 + H 2  H 2 O + N 2 Oslow N 2 O + H 2  N 2 + H 2 Ofast Show that the mechanism is consistent with the rate expression.