1. 1 Rate Equations and Order of Reactions 14.1Rate Equations and Order of Reactions 14.2Zeroth, First and Second Order Reactions 14.3Determination of Simple Rate Equations from Initial Rate Method 14.4Determination of Simple Rate Equations from Differential Rate Equations 14.5Determination of Simple Rate Equations from Integrated Rate Equations 14

2. 2 Rate Equations and Order of Reactions

3. 3 For the reaction aA + bB  cC + dD Rate  k[A] x [B] y rate law or rate equation

4. 4 For the reaction aA + bB  cC + dD Rate  k[A] x [B] y where x and y are the orders of reaction with respect to A and B x and y can be  integers or fractional x  y is the overall order of reaction.

5. 5 For the reaction aA + bB  cC + dD Rate  k[A] x [B] y For multi-step reactions, x, y have no direct relation to the stoichiometric coefficients and can ONLY be determined experimentally. For single-step reactions (elementary reactions), x = a and y = b (refer to p.35)

6. 6 For the reaction aA + bB  cC + dD Rate  k[A] x [B] y x = 0  zero order w.r.t. A x = 1  first order w.r.t. A x = 2  second order w.r.t. A y = 0  zero order w.r.t. B y = 1  first order w.r.t. B y = 2  second order w.r.t. B

7. 7 For the reaction aA + bB  cC + dD Rate  k[B] 2 Describe the reaction with the following rate law. The reaction is zero order w.r.t. A and second order w.r.t. B.

8. 8 Rate  k[A] x [B] y k is the rate constant For the reaction aA + bB  cC + dD Temperature-dependent Can only be determined from experiments

9. 9 Rate  k[A] x [B] y units of k : - mol dm  3 s  1 /(mol dm  3 ) x+y or, mol dm  3 min  1 /(mol dm  3 ) x+y For the reaction aA + bB  cC + dD

10. 10 Rate  k[A] 0 [B] 0 units of k = mol dm  3 s  1 /(mol dm  3 ) 0+0 = mol dm  3 s  1 = units of rate For the reaction aA + bB  cC + dD

11. 11 Rate  k[A][B] 0 units of k = mol dm  3 s  1 /(mol dm  3 ) 1+0 = s  1 For the reaction aA + bB  cC + dD

12. 12 Rate  k[A][B] units of k = mol dm  3 s  1 /(mol dm  3 ) 1+1 = mol  1 dm 3 s  1 For the reaction aA + bB  cC + dD The overall order of reaction can be deduced from the units of k

13. 13 Rate  k[A] x [B] y [C] z … For the reaction aA + bB + cC + …  products units of k : - mol dm  3 s  1 /(mol dm  3 ) x+y+z+…

14. 14 Determination of rate equations To determine a rate equation is to find k, x, y, z,… Rate  k[A] x [B] y [C] z … Two approaches : - 1.Initial rate method (pp.17-18) 2.Graphical method (pp.19-26)

15. 15 Determination of Rate Equations by Initial Rate Methods

16. 16 5Cl  (aq) + ClO 3  (aq) + 6H + (aq)  3Cl 2 (aq) + 3H 2 O(l) Expt [Cl  (aq)] / mol dm  3 [ClO 3  (aq)] / mol dm  3 [H + (aq)] / mol dm  3 Initial rate / mol dm  3 s  1 10.150.080.201.0  10  5 20.150.080.404.0  10  5 30.150.160.408.0  10  5 40.300.080.202.0  10  5 rate  k[Cl  (aq)] x [ClO 3  (aq)] y [H + (aq)] z

17. 17 Expt [Cl  (aq)] / mol dm  3 [ClO 3  (aq)] / mol dm  3 [H + (aq)] / mol dm  3 Initial rate / mol dm  3 s  1 10.150.080.201.0  10  5 20.150.080.404.0  10  5 30.150.160.408.0  10  5 40.300.080.202.0  10  5 From experiments 1 and 2, 4 = 2 z  z = 2 = 2 z

18. 18 Expt [Cl  (aq)] / mol dm  3 [ClO 3  (aq)] / mol dm  3 [H + (aq)] / mol dm  3 Initial rate / mol dm  3 s  1 10.150.080.201.0  10  5 20.150.080.404.0  10  5 30.150.160.408.0  10  5 40.300.080.202.0  10  5 From experiments 2 and 3, 2 = 2 y  y = 1 = 2 y

19. 19 Expt [Cl  (aq)] / mol dm  3 [ClO 3  (aq)] / mol dm  3 [H + (aq)] / mol dm  3 Initial rate / mol dm  3 s  1 10.150.080.201.0  10  5 20.150.080.404.0  10  5 30.150.160.408.0  10  5 40.300.080.202.0  10  5 From experiments 1 and 4, 2 = 2 x  x = 1 = 2 x

20. 20 rate  k[Cl  (aq)][ClO 3  (aq)][H + (aq)] 2 Expt [Cl  (aq)] / mol dm  3 [ClO 3  (aq)] / mol dm  3 [H + (aq)] / mol dm  3 Initial rate / mol dm  3 s  1 10.150.080.201.0  10  5 20.150.080.404.0  10  5 30.150.160.408.0  10  5 40.300.080.202.0  10  5 From experiment 1, 1.0  10  5  k(0.15)(0.08)(0.20) 2 k = 0.02 mol  3 dm 9 s  1

21. 21 rate  k[Cl  (aq)][ClO 3  (aq)][H + (aq)] 2 Expt [Cl  (aq)] / mol dm  3 [ClO 3  (aq)] / mol dm  3 [H + (aq)] / mol dm  3 Initial rate / mol dm  3 s  1 10.150.080.201.0  10  5 20.150.080.404.0  10  5 30.150.160.408.0  10  5 40.300.080.202.0  10  5 From experiment 2, 4.0  10  5  k(0.15)(0.08)(0.40) 2 k = 0.02 mol  3 dm 9 s  1

22. 22 Q.152C + 3D + E  P + 2Q Expt [C] / mol dm  3 [D] / mol dm  3 [E] / mol dm  3 Initial rate / mol dm  3 s  1 10.10 3.0  10  3 20.200.10 2.4  10  2 30.100.200.103.0  10  3 40.10 0.302.7  10  2 (a)rate  k[C] x [D] y [E] z

23. 23 Expt [C] / mol dm  3 [D] / mol dm  3 [E] / mol dm  3 Initial rate / mol dm  3 s  1 10.10 3.0  10  3 20.200.10 2.4  10  2 30.100.200.103.0  10  3 40.10 0.302.7  10  2 (a)rate  k[C] x [D] y [E] z From experiments 1 and 2, 8 = 2 x  x = 3 = 2 x

24. 24 Expt [C] / mol dm  3 [D] / mol dm  3 [E] / mol dm  3 Initial rate / mol dm  3 s  1 10.10 3.0  10  3 20.200.10 2.4  10  2 30.100.200.103.0  10  3 40.10 0.302.7  10  2 (a)rate  k[C] x [D] y [E] z From experiments 1 and 3, 1 = 2 y  y = 0 = 2 y

25. 25 Expt [C] / mol dm  3 [D] / mol dm  3 [E] / mol dm  3 Initial rate / mol dm  3 s  1 10.10 3.0  10  3 20.200.10 2.4  10  2 30.100.200.103.0  10  3 40.10 0.302.7  10  2 (a)rate  k[C] x [D] y [E] z From experiments 1 and 4, 9 = 3 z  z = 2 = 3 z

26. 26 Expt [C] / mol dm  3 [D] / mol dm  3 [E] / mol dm  3 Initial rate / mol dm  3 s  1 10.10 3.0  10  3 20.200.10 2.4  10  2 30.100.200.103.0  10  3 40.10 0.302.7  10  2 (a)rate  k[C] 3 [D] 0 [E] 2 = k[C] 3 [E] 2

27. 27 Expt [C] / mol dm  3 [D] / mol dm  3 [E] / mol dm  3 Initial rate / mol dm  3 s  1 10.10 3.0  10  3 20.200.10 2.4  10  2 30.100.200.103.0  10  3 40.10 0.302.7  10  2 (b)rate  k[C] 3 [E] 2 From experiment 1, 3.0  10  3  k(0.10) 3 (0.10) 2 k = 300 mol  4 dm 12 s  1

28. 28 Q.16 H+H+ CH 3 COCH 3 (aq) + I 2 (aq) CH 3 COCH 2 I(aq) + H + (aq) + I  (aq) Initial rate / mol dm  3 s  1 Initial concentration / mol dm  3 [I 2 (aq)][CH 3 COCH 3 (aq)][H + (aq)] 3.5  10  5 2.5  10  4 2.0  10  1 5.0  10  3 3.5  10  5 1.5  10  4 2.0  10  1 5.0  10  3 1.4  10  4 2.5  10  4 4.0  10  1 1.0  10  2 7.0  10  5 2.5  10  4 4.0  10  1 5.0  10  3 (a)rate  k[I 2 (aq)] x [CH 3 COCH 3 (aq)] y [H + (aq)] z

29. 29 Initial rate / mol dm  3 s  1 Initial concentration / mol dm  3 [I 2 (aq)][CH 3 COCH 3 (aq)][H + (aq)] 3.5  10  5 2.5  10  4 2.0  10  1 5.0  10  3 3.5  10  5 1.5  10  4 2.0  10  1 5.0  10  3 1.4  10  4 2.5  10  4 4.0  10  1 1.0  10  2 7.0  10  5 2.5  10  4 4.0  10  1 5.0  10  3 (a) rate  k[I 2 (aq)] x [CH 3 COCH 3 (aq)] y [H + (aq)] z From experiments 1 and 2, 1 = 1.67 x  x = 0 = 1.67 x

30. 30 Initial rate / mol dm  3 s  1 Initial concentration / mol dm  3 [I 2 (aq)][CH 3 COCH 3 (aq)][H + (aq)] 3.5  10  5 2.5  10  4 2.0  10  1 5.0  10  3 3.5  10  5 1.5  10  4 2.0  10  1 5.0  10  3 1.4  10  4 2.5  10  4 4.0  10  1 1.0  10  2 7.0  10  5 2.5  10  4 4.0  10  1 5.0  10  3 (a) rate  k[I 2 (aq)] x [CH 3 COCH 3 (aq)] y [H + (aq)] z From experiments 1 and 4, 2 = 2 y  y = 1 = 2 y

31. 31 Initial rate / mol dm  3 s  1 Initial concentration / mol dm  3 [I 2 (aq)][CH 3 COCH 3 (aq)][H + (aq)] 3.5  10  5 2.5  10  4 2.0  10  1 5.0  10  3 3.5  10  5 1.5  10  4 2.0  10  1 5.0  10  3 1.4  10  4 2.5  10  4 4.0  10  1 1.0  10  2 7.0  10  5 2.5  10  4 4.0  10  1 5.0  10  3 (a) rate  k[I 2 (aq)] x [CH 3 COCH 3 (aq)] y [H + (aq)] z From experiments 3 and 4, 2 = 2 z  z = 1 = 2 z

32. 32 Initial rate / mol dm  3 s  1 Initial concentration / mol dm  3 [I 2 (aq)][CH 3 COCH 3 (aq)][H + (aq)] 3.5  10  5 2.5  10  4 2.0  10  1 5.0  10  3 3.5  10  5 1.5  10  4 2.0  10  1 5.0  10  3 1.4  10  4 2.5  10  4 4.0  10  1 1.0  10  2 7.0  10  5 2.5  10  4 4.0  10  1 5.0  10  3 (a)Rate= k[I 2 (aq)] 0 [CH 3 COCH 3 (aq)][H + (aq)] = k[CH 3 COCH 3 (aq)][H + (aq)]

33. 33 Initial rate / mol dm  3 s  1 Initial concentration / mol dm  3 [I 2 (aq)][CH 3 COCH 3 (aq)][H + (aq)] 3.5  10  5 2.5  10  4 2.0  10  1 5.0  10  3 3.5  10  5 1.5  10  4 2.0  10  1 5.0  10  3 1.4  10  4 2.5  10  4 4.0  10  1 1.0  10  2 7.0  10  5 2.5  10  4 4.0  10  1 5.0  10  3 (b)Rate = k[CH 3 COCH 3 (aq)][H + (aq)] From experiment 1, 3.5  10  5  k(2.0  10  1 )(5.0  10  3 ) k = 0.035 mol  1 dm 3 s  1

34. 34 Determination of Rate Equations by Graphical Methods

35. 35 Two types of rate equations : - (1)Differential rate equation (2)Integrated rate equation

36. 36 A  products (Differential rate equation) shows the variation of rate with [A] Two types of plots to determine k and n

37. 37 [A] rate n = 0 k rate = k

38. 38 Examples of zero-order reactions : - 2NH 3 (g) N 2 (g) + 3H 2 (g) Fe or W as catalyst Decomposition of NH 3 /HI can take place only on the surface of the catalyst. Once the surface is covered completely (saturated) with NH 3 /HI molecules at a given concentration of NH 3 /HI, further increase in [NH 3 ]/[HI] has no effect on the rate of reaction. 2HI(g) H 2 (g) + I 2 (g) Au as catalyst

39. 39 [A] rate n = 0 k rate = k

40. 40 [A] rate n = 1 slope = k linear

41. 41 [A] rate n = 2 k cannot be determined directly from the graph parabola

42. 42 [A] rate n = 2 n = 1 n = 0

43. 43 log 10 [A] log 10 rate slope y-intercept n = 0 n = 1 log 10 k n = 2 slope = 1 slope = 2 slope = 0

44. 44 (Differential rate equation) [A] t = [A] 0 – kt (Integrated rate equation) If n = 0 Derivation not required

45. 45 [A] t = [A] 0 – kt (Integrated rate equation) shows variation of [A] with time time [A] t [A] 0 constant rate

46. 46 (Differential rate equation) (Integrated rate equation) If n = 1, log e [A] t – log e [A] 0 =  kt Or [A] t  [A] 0 e  kt log e [A] t = log e [A] 0  kt ln

47. 47 Two types of plots to determine k and n Or [A] t  [A] 0 e  kt log e [A] t = log e [A] 0  kt time log e [A] t log e [A] 0 slope =  k linear  n = 1

48. 48 Two types of plots to determine k and n Or [A] t  [A] 0 e  kt log e [A] t = log e [A] 0  kt time [A] t [A] t varies exponentially with time constant half life  n = 1

49. 49

50. 50 log e [A] t = log e [A] 0  kt

51. 51 = 6.9  10  3 s  1

52. 52 Q.17 sucrose  fructose + glucose Rate = k[sucrose] k = 0.208 h  1 at 298 K (a)

53. 53 Q.17 sucrose  fructose + glucose Rate = k[sucrose] k = 0.208 h  1 at 298 K (b)[A] t  [A] 0 e  kt 87.5% decomposed  [A] t = 0.125[A] 0 0.125 = e  kt ln0.125 =  0.208  t t = 9.99 h = e  0.208  t

54. 54 (Differential rate equation) (Integrated rate equation) If n = 2, Or

55. 55 Or Linear  n = 2 Slope = k time

56. 56 Or time [A] t n = 2 Variable half life n = 1 [A] t  more rapidly with time in the early stage

57. 57 time [A] t n = 1 n = 2 n = 0 Plotting based on integrated rate equations More common because [A] t and time can be obtained directly from expereiments.

58. 58 [A] rate n = 2 n = 1 n = 0 Plotting based on differential rate equations Less common because rate cannot be obtained directly from expereiments.

59. 59 log 10 [A] log 10 rate n = 0 n = 1 log 10 k n = 2 slope = 1 slope = 2 slope = 0 Plotting based on differential rate equations Less common because rate cannot be obtained directly from expereiments.

60. 60 mol  1 dm 3 s  1 k against t2 s1s1 kkln[A] t against t1 mol dm  3 s  1 kk[A] t against t [A] t  [A] 0 – kt 0 Units of kSlope Straight line plot Integrated rate equation Order Summary : - For reactions of the type A  Products

61. 61 2H 2 O 2 (aq)  2H 2 O(l) + O 2 (g) Rate = k[H 2 O 2 (aq)] Examples of First Order Reactions

62. 62 Examples of First Order Reactions ReactionRate equation 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g)Rate = k[N 2 O 5 (g)] SO 2 Cl 2 (l)  SO 2 (g) + Cl 2 (g)Rate = k[SO 2 Cl 2 (l)] (CH 3 ) 3 CCl(l) + OH - (aq)  (CH 3 ) 3 COH(l) + Cl - (aq) Rate = k[(CH 3 ) 3 CCl(l)] (S N 1) All radioactive decayse.g. Rate = k[Ra] S N 1 : 1 st order Nucleophilic Substitution Reaction

63. 63 1.For a reaction involving one reactant only: 2NOCl(g)  2NO(g) + Cl 2 (g) Rate = k[NOCl(g)] 2 2NO 2 (g)  2NO(g) + O 2 (g) Rate = k[NO 2 (g)] 2 Examples of Second Order Reactions

64. 64 Examples of Second Order Reactions ReactionRate equation H 2 (g) + I 2 (g)  2HI(g)Rate = k[H 2 (g)][I 2 (g)] CH 3 Br(l) + OH  (aq)  CH 3 OH(l) + Br  (aq) Rate = k[CH 3 Br(l)][OH  (aq)] (S N 2) CH 3 COOC 2 H 5 (l) + OH  (aq)  CH 3 COO  (aq) + C 2 H 5 OH(l) Rate = k[CH 3 COOC 2 H 5 (l)][OH  (aq)] S N 2 : 2 nd order Nucleophilic Substitution Reaction 2.For a reaction involving one reactant only:

65. 65 2.For a reaction involving two reactants: A + B  products Rate = k[A][B] To determine the rate equation, the concentration of one of the reactants must be kept constant (in large excess) such that the order of reaction w.r.t. the other reactant can be determined.

66. 66 2.For a reaction involving two reactants: A + B  products Rate = k[A][B] When [B] is kept constant, excess rate = k’[A] (where k’ = k[B] excess )

67. 67 Rate = k[A][B] excess = k’[A] k can be determined from k’ if [B] excess is known Linear  first order

68. 68 2.For a reaction involving two reactants: A + B  products Rate = k[B][A] When [A] is kept constant, rate = k”[B] (where k” = k[A] excess ) excess

69. 69 Rate = k[A] excess [B] = k’’[B] k can be determined from k’’ if [A] excess is known Linear  first order

70. 70 Q.18(a)2NO 2 (g)  2NO(g) + O 2 (g) first-order reaction, k  3.6  10  3 s  1 at 573 K =  (3.6  10  3 s  1 )(150s) = 0.58

71. 71 Q.18(b)2NO 2 (g)  2NO(g) + O 2 (g) first-order reaction, k  3.6  10  3 s  1 at 573 K = 0.01 ln0.01 =  (3.6  10  3 s  1 )t t = 1279 s

72. 72 Q.18(c)2NO 2 (g)  2NO(g) + O 2 (g) first-order reaction, k  3.6  10  3 s  1 at 573 K = 0.021 mol dm  3 Calculate the rate of consumption of NO 2 when the partial pressure of NO 2 is 1.0 atm. (Gas constant, R = 0.082 atm dm 3 K  1 mol  1 )

73. 73 Q.18(c)2NO 2 (g)  2NO(g) + O 2 (g) first-order reaction, k  3.6  10  3 s  1 at 573 K Calculate the rate of consumption of NO 2 when the partial pressure of NO 2 is 1.0 atm. (Gas constant, R = 0.082 atm dm 3 K  1 mol  1 ) [NO 2 (g)] = 0.021 mol dm  3 Rate of reaction = k[NO 2 (g)] = (3.6  10  3 s  1 )(0.021 mol dm  3 ) = 7.6  10  5 mol dm  3 s  1

74. 74 Q.18(c)2NO 2 (g)  2NO(g) + O 2 (g) first-order reaction, k  3.6  10  3 s  1 at 573 K Calculate the rate of consumption of NO 2 when the partial pressure of NO 2 is 1.0 atm. (Gas constant, R = 0.082 atm dm 3 K  1 mol  1 ) =  2  (7.6  10  5 mol dm  3 s  1 ) =  1.5  10  4 mol dm  3 s  1

75. 75 Q.19 =  (0.104 y  1 )(5 y) = 0.595 Half life = 6.67 years Mass of Ra remaining after 5 years = 0.297 g

76. 76 Q.20 Half life = 4.51  10 9 years Let 1.000x be the mass of U left behind at time t  Mass of Pb produced at time t = 0.231x  Mass of U consumed at time t = 0.231x  Mass of U at t 0 = 1.231x =  0.208 kt = (1.54  10  10 y  1 )  t = 0.208 t = 1.35  10 9 years

77. 77 Q.21(a)

78. 78 Q.21(b) No. of moles of U decayed = No. of moles of He formed= (3.20  10  3 mol) = 4.00  10  4 mol No. of moles of Pb produced = 4.00  10  4 mol No. of moles of U at t 0 = (4.00  10  4 + 4.40  10  4 ) mol = 8.40  10  4 mol

79. 79 Q.21(b) =  0.647 kt = 0.647 (1.54  10  10 y  1 )  t = 0.647 t = 4.20  10 9 y

80. 80 Q.22(a) 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 1.12.74.45.97.28.910.913.3 / kPa 12508005504003002001000 t / minute 52 ON P

81. 81 Q.22(a) 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) 1.12.74.45.97.28.910.913.3 / kPa 12508005504003002001000 t / minute 52 ON P Constant half life  350 minutes  1 st order

82. 82 Q.22(a)

83. 83 time k =  slope = 1.99  10  3 min  1 linear  first order Q.22(a)/(b) 0.100.991.481.771.972.192.392.59 12508005504003002001000 t / minute

84. 84 14.1 Rate Equations and Order of Reactions (SB p.27) (a)The reaction between tyrosine (an amino acid) and iodine obeys the rate law: rate = k [Tyr] [I 2 ]. Write the orders of the reaction with respect to tyrosine and iodine respectively, and hence the overall order. Answer (a) The order of the reaction with respect to tyrosine is 1, and the order of the reaction with respect to iodine is also 1. Therefore, the overall order of the reaction is 2.

85. 85 14.1 Rate Equations and Order of Reactions (SB p.27) (b)Determine the unit of the rate constant ( k ) of the following rate equation: Rate = k [A] [B] 3 [C] 2 (Assume that all concentrations are measured in mol dm –3 and time is measured in minutes.) Answer (b)  k =  Unit of k = = mol -5 dm 15 min -1 Back

86. 86 The initial rate of a second order reaction is 8.0 × 10 –3 mol dm –3 s –1. The initial concentrations of the two reactants, A and B, are 0.20 mol dm –3. Calculate the rate constant of the reaction and state its unit. 14.2 Zeroth, First and Second Order Reactions (SB p.29) Answer 8.0  10 -3 = k  (0.20) 2  k = 0.2 mol -1 dm 3 s -1 Back

87. 87 14.3 Determination of Simple Rate Equations from Initial Rate Method (SB p.30) For a reaction between two substances A and B, experiments with different initial concentrations of A and B were carried out. The results were shown as follows: ExptInitial conc. of A (mol dm -3 ) Initial conc. of B (mol dm -3 ) Initial rate (mol dm -3 s -1 ) 10.010.020.0005 20.02 0.001 0 30.010.040.002 0

88. 88 14.3 Determination of Simple Rate Equations from Initial Rate Method (SB p.30) (a)Calculate the order of reaction with respect to A and that with respect to B. Answer

89. 89 14.3 Determination of Simple Rate Equations from Initial Rate Method (SB p.30) (a)Let x be the order of reaction with respect to A, and y be the order of reaction with respect to B. Then, the rate equation for the reaction can be expressed as: Rate = k [A] x [B] y Therefore, 0.0005 = k (0.01) x (0.02) y.......................... (1) 0.0010 = k (0.02) x (0.02) y.......................... (2) 0.002 0 = k (0.01) x (0.04) y.......................... (3) Dividing (1) by (2),  x = 1

90. 90 14.3 Determination of Simple Rate Equations from Initial Rate Method (SB p.30) (a)Dividing (1) by (3),  y = 2

91. 91 14.3 Determination of Simple Rate Equations from Initial Rate Method (SB p.30) (b)Using the result of experiment (1), Rate = k [A] [B] 2 0.000 5 = k  0.01  0.02 2 k = 125 mol -2 dm 6 s -1 (c)Rate = 125 [A] [B] 2 Back (b)Calculate the rate constant using the result of experiment 1. (c)Write the rate equation for the reaction. Answer

92. 92 In the kinetic study of the reaction, CO(g) + NO 2 (g)  CO 2 (g) + NO(g) four experiments were carried out to determine the initial reaction rates using different initial concentrations of reactants. The results were as follows: 14.3 Determination of Simple Rate Equations from Initial Rate Method (SB p.31) ExptInitial conc. of CO(g) (mol dm -3 ) Initial conc. of NO 2 (g) (mol dm -3 ) Initial rate (mol dm -3 s -1 ) 10.1 0.015 20.20.10.030 30.10.20.030 40.40.10.060

93. 93 (a)Calculate the rate constant of the reaction, and hence write the rate equation for the reaction. 14.3 Determination of Simple Rate Equations from Initial Rate Method (SB p.31) Answer

94. 94 14.3 Determination of Simple Rate Equations from Initial Rate Method (SB p.31) (a)Let m be the order of reaction with respect to CO, and n be the order of reaction with respect to NO 2. Then, the rate equation for the reaction can be expressed as: Rate = k [CO] m [NO 2 ] n Therefore, 0.015 = k (0.1) m (0.1) n.......................... (1) 0.030 = k (0.2) m (0.1) n.......................... (2) 0.030 = k (0.1) m (0.2) n.......................... (3) Dividing (1) by (2),  m = 1

95. 95 14.3 Determination of Simple Rate Equations from Initial Rate Method (SB p.31) (a)Dividing (1) by (3),  n = 1  Rate = k [CO] [NO 2 ] Using the result of experiment (1), 0.015 = k (0.1) 2 k = 1.5 mol -1 dm 3 s -1  Rate = 1.5 [CO] [NO 2 ]

96. 96 14.3 Determination of Simple Rate Equations from Initial Rate Method (SB p.31) (b) Determine the initial rate of the reaction when the initial concentrations of both CO( g) and NO 2 ( g) are 0.3 mol dm –3. Answer (b)Initial rate = 1.5  0.3  0.3 = 0.135 mol dm -3 s -1 Back

97. 97 (a) Write a chemical equation for the decomposition of hydrogen peroxide solution. 14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.34) Answer (a)2H 2 O 2 (aq)  2H 2 O(l) + O 2 (g)

98. 98 (b) Explain how you could find the rate of decomposition of hydrogen peroxide solution in the presence of a solid catalyst using suitable apparatus. 14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.34) Answer

99. 99 14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.34) (b)In the presence of a suitable catalyst such as manganese(IV) oxide, hydrogen peroxide decomposes readily to give oxygen gas which is hardly soluble in water. A gas syringe can be used to collect the gas evolved. To minimize any gas leakage, all apparatus should be sealed properly. A stopwatch is used to measure the time. The volume of gas evolved per unit time (i.e. the rate of evolution of the gas) can then be determined.

100. 100 (c) The table below shows the initial rates of decomposition of hydrogen peroxide solution of different concentrations. Plot a graph of the initial rate against [H 2 O 2 (aq)]. 14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.34) Answer [H 2 O 2 (aq)] (mol dm -3 ) 0.1000.1750.2500.300 Initial rate (10 -4 mol dm -3 s -1 ) 0.591.041.501.80

101. 101 14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.34) (c)

102. 102 (d)From the graph in (c), determine the order and rate constant of the reaction. 14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.34) Answer

103. 103 14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.34) (d) There are two methods to determine the order and rate constant of the reaction. Method 1: When the concentration of hydrogen peroxide solution increases from 0.1 mol dm –3 to 0.2 mol dm –3, the reaction rate increases from 0.59 × 10 –4 mol dm –3 s –1 to about 1.20 × 10 –4 mol dm –3 s –1. ∴ Rate  [H 2 O 2 (aq)] Therefore, the reaction is of first order. The rate constant (k) is equal to the slope of the graph. k = = 6.0  10 -4 s -1

104. 104 14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.34) (d) Method 2: The rate equation can be expressed as: Rate = k [H 2 O 2 (aq)] x where k is the rate constant and x is the order of reaction. Taking logarithms on both sides of the rate equation, log (rate) = log k + x log [H 2 O 2 (aq)]................. (1) -3.74-3.82-3.98-4.23log (rate) -0.523-0.602-0.757log [H 2 O 2 (aq)]

105. 105 14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.34) (d)A graph of log (rate) against log [H 2 O 2 (aq)] gives a straight line of slope x and y-intercept log k.

106. 106 14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.34) (d)Slope of the graph = =1.0  The reaction is of first order. Substitute the slope and one set of value into equation (1): -4.23 = log k + (1.0) (-1.000) log k = -3.23 k = 5.89  10 -4 s -1 Back

107. 107 14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.36) (a)Decide which curve in the following graph corresponds to (i)a zeroth order reaction; (ii)a first order reaction. (a)(i)(3) (ii)(2) Answer

108. 108 14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.36) (b) The following results were obtained for the decomposition of nitrogen(V) oxide. 2N 2 O 5 (g)  4NO 2 (g) + O 2 (g) Concentration of N 2 O 5 (mol dm -3 ) Initial rate (mol dm -3 s -1 ) 1.6  10 -3 0.12 2.4  10 -3 0.18 3.2  10 -3 0.24

109. 109 14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.36) (i) Write the rate equation for the reaction. Answer (i)The rate equation for the reaction can be expressed as: Rate = k [N 2 O 5 (g)] m where k is the rate constant and m is the order of reaction.

110. 110 14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.36) (ii) Determine the order of the reaction. Answer

111. 111 14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.36) (ii)Method 1: A graph of the initial rates against [N 2 O 5 (g)] is shown as follows:

112. 112 14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.36) As shown in the graph, when the concentration of N 2 O 5 increases from 1.0  10 –3 mol dm –3 to 2.0  10 –3 mol dm –3, the rate of the reaction increases from 0.075 mol dm –3 s –1 to 0.15 mol dm –3 s –1.  Rate  [N 2 O 5 (g)]  The reaction is of first order. Then, the rate constant k is equal to the slope of the graph. k = = 75 s-1  The rate equation for the reaction is: Rate = 75 [N 2 O 5 (g)]

113. 113 14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.36) Method 2: Taking logarithms on both sides of the rate equation, we obtain: log (rate) = log k + m log [N 2 O 5 (g)].......... (1) A graph of log (rate) against log [N 2 O 5 (g)] gives a straight line of slope m and y-intercept log k. -0.62-0.74-0.92log (rate) -2.50-2.62-2.80log [N 2 O 5 (g)]

114. 114 14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.36) Slope of the graph = =1.0  The reaction is of first order. Substitute the slope and one set of value into equation (1): -0.92 = log k + (1.0) (-2.80) log k = 1.88 k = 75.86 s -1  The rate equation for the reaction is: Rate = 75.86 [N 2 O 5 (g)]

115. 115 14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.36) (iii)Determine the initial rate of reaction when the concentration of nitrogen(V) oxide is: (1)2.0 × 10 –3 mol dm –3. (2)2.4 × 10 –2 mol dm –3. Answer

116. 116 14.4 Determination of Simple Rate Equations from Differential Rate Equations (SB p.36) (iii)The rate equation, rate = 75 [N 2 O 5 (g)], is used for the following calculation. (1)Rate = 75 [N 2 O 5 (g)] = 75 s –1  2.0  10 –3 mol dm –3 = 0.15 mol dm –3 s –1 (2)Rate = 75 [N 2 O 5 (g)] = 75 s –1  2.4  10 –2 mol dm –3 = 1.8 mol dm –3 s –1 Back

117. 117 The half-life of a radioactive isotope A is 1 997 years. How long does it take for the radioactivity of a sample of A to drop to 20% of its original level? 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.39) Answer

118. 118 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.39) As radioactive decay is a first order reaction, = 3.47  10 -4 year -1   t = 4638 years  It takes 4638 years for the radioactivity of a sample of A to dropt to 20 % of its original level. Back

119. 119 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.40) (a)At 298 K, the rate constant for the first order decomposition of nitrogen(V) oxide is 0.47 × 10 –4 s –1. Determine the half-life of nitrogen(V) oxide at 298 K. N 2 O 5  2NO 2 + O 2 Answer

120. 120 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.40) (a)Let the half-life of nitrogen(V) oxide be.  The half-life of nitrogen(V) oxide is 14 745 s.

121. 121 (b)The decomposition of CH 3 N = NCH 3 to form N 2 and C 2 H 6 follows first order kinetics and has a half-life of 0.017 minute at 573 K. Determine the amount of CH 3 N = NCH 3 left if 1.5 g of CH 3 N = NCH 3 was decomposed for 0.068 minute at 573 K. 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.40) Answer

122. 122 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.40) (b) Let m be the amount of CH 3 N=NCH 3 left after 0.068 minute. m = 0.094 g Back

123. 123 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.42) In the decomposition of gaseous hydrogen iodide, the following experimental data were obtained. Determine the order of decomposition of gaseous hydrogen iodide graphically. You may try to plot graphs of [HI(g)] against time, ln[HI(g)] against time and against time. 0.1000.1250.1670.2500.500[HI(g)] (mol dm -3 ) 4803602401200Time (min) Answer

124. 124 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.42) 10.000-2.3030.100480 8.000-2.0790.125360 5.988-1.7900.167240 4.000-1.3860.250120 2.000-0.6930.5000 1/[HI(g)] (mol -1 dm 3 ) ln [HI(g)][HI(g)] (mol dm -3 ) Time (min)

125. 125 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.42) The order of decomposition can be determined by plotting: (a)[HI(g)] against time,

126. 126 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.42) (b)ln [HI(g)] against time,

127. 127 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.43) (c) against time,

128. 128 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.43) In graph (a), the plot of [HI(g)] against time is not a straight line, thus the decomposition reaction is not of zeroth order. Similarly, in graph (b), the plot of ln [HI(g)] against time is not a straight line, thus the decomposition reaction is not of first order. However, in graph (c), the plot of against time gives a straight line, thus the decomposition of gaseous hydrogen iodide is of second order. Back

129. 129 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.43) The change in concentration of substance X as it decomposed at 698 K was recorded in the following table: Determine the order of the reaction graphically. 0.0560.0630.0720.0830.100[X] (mol dm -3 ) 200150100500Time (s) Answer

130. 130 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.43) 17.86-2.880.056200 15.87-2.760.063150 13.89-2.630.072100 12.05-2.490.08350 10.00-2.300.1000 1 / [X] (mol -1 dm 3 ) ln [X][X] (mol dm -3 ) Time (s)

131. 131 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.43) As the graph of [X] against time is not a straight line, the reaction is not of zeroth order.

132. 132 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.43) Similarly, the plot of ln [X} against time is not a straight line, thus the reaction is not of first order.

133. 133 14.5 Determination of Simple Rate Equations from Integrated Rate Equations (SB p.43) The plot of Against time gives a straight line, therefore the reaction is of second order. Back