1. 1 The Effects of Temperature and Catalyst on Reaction Rate 15.1Activation Energy and Arrhenius Equation 15.2Interpretation of Rates of Gaseous Reactions at Molecular Level Reactions at Molecular Level 15.3Energy Profile 15.4Effect of Catalysts on Rates of Reactions 15

2. 2 Activation Energy and Arrhenius Equation

3. 3 Activation Energy Exothermic reaction Activation energy, E a = energy required to start the reaction related to the rate of reaction

4. 4 Endothermic reaction Activation Energy related to the rate of reaction

5. 5 Most reactions have positive E a since energy is absorbed to break bonds in reactant particles.

6. 6 Arrhenius Equation Since rate = k[A] a [B] b... At fixed concentrations, rate depends on k which in turn depends on A depends on the nature of the reaction and varies with temperature (T) and the nature of the reaction (A and Ea)

7. 7 Q.23 Assume A is a constant = 1.92  2 A 10 K  in T doubles the rate R = 8.31 J K  1 mol  1

8. 8 Arrhenius Equation Rate of reaction  exponentially with temperature

9. 9 Arrhenius Equation T ,  A  (Minor effect)  Rate 

10. 10 Arrhenius Equation T , (Major effect) (less negative)  Rate  (more positive)

11. 11 Arrhenius Equation T ,  A  (Minor effect)  Rate 

12. 12 Arrhenius Equation T , (Major effect) (more negative)  Rate  (less positive)

13. 13 Determination of Activation Energy

14. 14 1/T (K  1 ) log e k log e A Determination of E a by Graphical Method

15. 15 Q.24 log e k1/T (K  1 )  14.91.80  10  3  9.41.55  10  3  6.81.43  10  3  3.21.28  10  3

16. 16 1/T (K  1 ) log e k E a = -slope  R  182 kJ mol  1 log e A  24.8 < E(H – I) before the H – I bond is completely broken (refer to p.32) H – H and I – I bonds are formed

17. 17 If log e rate is plotted against 1/T, since rate = k[A] x [B] y … log e rate = log e k + log e [A] x [B] y … = log e k + constant y-intercept

18. 18 Determination of Activation Energy Using Two Rate Constants

19. 19 Interpretation of Rates of Gaseous Reactions at Molecular Level

20. 20 The Kinetic Theory of Gases - Developed by Maxwell and Boltzman 1.Gas particles are in a state of constant and random motion in all directions, undergoing frequent collisions with one another and with the walls of the container. 2.The pressure exerted on the container is due to the collisions between gas particles and the walls of the containers.

21. 21 The Kinetic Theory of Gases - Developed by Maxwell and Boltzman 3.Gas particles are treated as point masses because their volumes are negligible when compared with the volume of the container. 4.There is no interaction among gas particles except collisions.

22. 22 The Kinetic Theory of Gases - Developed by Maxwell and Boltzman 5.Collisions between gas particles are perfectly elastic, i.e. the total kinetic energy is conserved.

23. 23 Transfer of K.E. among molecules  Distribution of molecular speeds Distribution of Molecular Speeds in a Gas Consider a sample of gas:

24. 24 Distribution of Molecular Speeds in a Gas Area under curve = total no. of gas molecules

25. 25 The Kinetic Theory of Gases - Developed by Maxwell and Boltzman The mean kinetic energy of a sample of gas particles is proportional to its absolute temperature (T).

26. 26 The Kinetic Theory of Gases - Developed by Maxwell and Boltzman n 1 = no. of molecules with velocity c 1 and n 1 + n 2 + n 3 +... = n (total no. of molecules)

27. 27 The distribution of velocity is not symmetrical  the average velocity of a gas sample is best represented by the root mean square velocity

28. 28 The Kinetic Theory of Gases - Developed by Maxwell and Boltzman For a sample of gas containing n molecules, where m is the absolute mass of the gas molecule

29. 29 Variation in the distribution of molecular Speeds with T As T , Molecular speeds  Curve becomes flattened Wider distribution of molecular speeds at a higher temp Area under the curve remains unchanged.

30. 30 M : molar mass of gas in Kg M   r.m.s velocity  If n = 1

31. 31 The areas underneath the curves are the same The lighter molecules are more spread out in molecular speeds.

32. 32 Q.25 H2H2 CO 2 = 1845 ms  1 = 393 ms  1 The lightest gases (H 2, He) can escape from the gravitational pull of small planets  Very rare in the Earth’s atmosphere

33. 33 Simple Collision Theory For a reaction to occur, the reactant particles must collide with (1) kinetic energy  Ea (2) proper orientation.

34. 34 Proper Orientation HCl(g) + NH 3 (g)  NH 4 Cl(s)

35. 35 Improper Orientation

36. 36 Improper Orientation

37. 37 Simple Collision Theory For a reaction to occur, the reactant particles must collide with (1) kinetic energy  Ea (2) proper orientation. No. of effective collisions = Z = collision frequency Effective collision

38. 38 Simple Collision Theory For a reaction to occur, the reactant particles must collide with (1) kinetic energy  Ea (2) proper orientation. Effective collision No. of effective collisions =

39. 39 Simple Collision Theory For a reaction to occur, the reactant particles must collide with (1) kinetic energy  Ea (2) proper orientation. Effective collision No. of effective collisions = p = fraction of collisions with proper orientation

40. 40 Theoretically, (from collision theory and kinetic theory) No. of effective collisions = Experimentally,

41. 41 If molarities of X, Y,… are fixed = no. of collisions with proper orientation

42. 42 Interpretation of the Effect of Temperature Change on Rate of Reaction T   speed of reactant particles   collision frequency (Z)   A   rate  (minor effect)

43. 43 Interpretation of the Effect of Temperature Change on Rate of Reaction T   K.E. of reactant particles   fraction of collisions with K.E.  E a   rate  exponentially (major effect)

44. 44 speed / K.E. No. of molecules concave convex

45. 45 Fraction of particles with K.E. > E = E The shaded area = no. of particles with K.E. > E

46. 46 Fraction of particles with K.E. > E a = EaEa If E = E a

47. 47 As T , the fraction of particles with K.E. > E a increases exponentially.  Rate increases exponentially with T

48. 48 Limitations of Collision Theory In aqueous phase, the interactions between the reactant particles and the solvent molecules have to be considered. The fraction of collisions with proper orientation (the steric factor, p) cannot be predicted. It can only be determined experimentally. Collision theory is based on the calculations from kinetic theory of ideal gases. Thus, it is ONLY applicable to reactions in gas phase.

49. 49 Consider the 2 nd order single-step gas phase rx R(g) + R(g)  products Given : k = 1.00  10  2 mol  1 dm 3 s  1 at 473 K, [R(g)] initial = 1  10  2 mol dm  3, L = 6.02  10 23 mol  1 Gas constant = 8.31 J K  1 mol  1, E a = 100 kJ mol  1 Initial collision frequency(Z) = 7.77  10 32 s  1 (a)Estimate (i)The no. of effective collisions per m 3 per second. (ii)The no. of collisions with K.E.  E a per m 3 per second. (b)Hence, deduce the steric factor, p of the reaction. Q.26

50. 50 = (1.00  10 -2 mol  1 dm 3 s  1 )(1.00  10 -2 mol dm  3 ) 2 = 1.00  10 -6 mol dm  3 s -1 = 1.00  10 -6 mol  6.02  10 23 mol -1 dm -3 s -1 = 6.02  10 17 molecules dm -3 s -1 = 6.02  10 20 molecules m -3 s -1 6.02  10 20 molecules of R are decomposed per cubic meter per second (a)(i)

51. 51 Consider the 2nd order single-step gas phase rx R(g) + R(g)  products Rate = 6.02  10 20 molecules m -3 s -1 One effective collision leads to decomposition of Two molecules of R. Thus, no. of effective collisions per cubic meter per second = 3.01  10 20 (a)(i)

52. 52 No. of effective collisions = = 3.01  10 20 m -3 s -1 No. of collisions with K.E.  E a = 7.77  10 32 m -3 s -1  (8.93  10 -12 ) = 6.94  10 21 m -3 s -1 (a)(ii)

53. 53 No. of effective collisions = = 3.01  10 20 m -3 s -1 No. of collisions with K.E.  E a = 6.94  10 21 m -3 s -1 = 4.34 % (b)

54. 54 Q.27 If E a  0 Rate is independent of T (A-level)

55. 55 K.E. No. of molecules EaEa

56. 56 K.E. No. of molecules EaEa EaEa EaEa No. of effective collisions =  Z  p

57. 57 Energy Profile Transition State Theory

58. 58 Transition State Theory - focuses on what happens after the collisions have started.

59. 59 Energy profile - shows the variation of the potential energy of the reaction mixture as the reaction proceeds. reaction coordinate P.E.

60. 60 Consider the one-step reaction, A–B + X  A + B–X P.E. of the reaction mixture are calculated for any A–B and B–X distances, and the results are plotted on a contour diagram

61. 61 Consider the one-step reaction, A–B + X  A + B–X At R, A-B distance is short B-X distance is long  before reaction

62. 62 Consider the one-step reaction, A–B + X  A + B–X The valley at R represents the potential energy for the initial state of the system, i.e. A–B and X

63. 63 Consider the one-step reaction, A–B + X  A + B–X At P, A-B distance is long B-X distance is short  after reaction

64. 64 Consider the one-step reaction, A–B + X  A + B–X The valley at P represents the potential energy for the final state of the system, i.e. A and B–X

65. 65 Consider the one-step reaction, A–B + X  A + B–X The energy contours rise in all directions from the valleys at R and P, but the ‘easiest’ path is shown by the bold line RTP

66. 66 The transition state is like a col ( 山坳 ) in a mountain region

67. 67 R P T

68. 68 In the transition state, Bond between A and B is partially broken Bond between B and X is partially formed A-B + X  A  B  X  A + B-X Thus, E a is lower than E(A-B)

69. 69 Transition state (Activated complex) is the least stable arrangement of the system in the most probable reaction pathway. R P T

70. 70 Advantages of Transition State Theory 1.Ea and A can be calculated A  Zp  the steric factor p can be predicted 2.It explains why the reaction pathway is specific. 3.It is applicable to gaseous and aqueous reactions.

71. 71 Energy Profile : One-step Mechanism A-B + X  A  B  X  A + B-X

72. 72 Example of One-step Mechanism Rate = k[CH 3 Cl][OH  ] Bimolecular One-step 2 nd Order Nucleophilic Substitution Reaction SN2SN2

73. 73 CH 3 Cl + OH   CH 3 OH + Cl 

74. 74 Energy Profile : Multi-step Mechanism E 1 > E 2  Step 1 is the rate determining step

75. 75 Energy Profile : Multi-step Mechanism

76. 76 Multi-step Mechanism Step 1:A ─ B  A + B (intermediate) Step 2:A + B + X  A + B ─ X Overall reaction: A ─ B + X  A + B ─ X Chemical reactions take place in two or more steps Formation of an intermediate

77. 77 Hydrolysis of 2-chloro-2-methylpropane Example of Multi-step Mechanism (1) (2) carbocation

78. 78 Rate = k[C(CH 3 ) 3 Cl] Unimolecular Two-step 1 st Order Nucleophilic Substitution Reaction SN1SN1

79. 79 + OH  Rate = k[C(CH 3 ) 3 Cl] E1E1 E2E2 E 1 > E 2 Step 1 is r.d.s.

80. 80 Reaction mechanism is the detailed sequence of steps that occur in a reaction. Reaction mechanisms are theoretical proposals used to explain the experimentally determined rate laws. Reaction Mechanism and Rate Law

81. 81 Each of the steps in a mechanism is called an elementary step. Reaction Mechanism and Rate Law The number of reactant particles that takes part in each elementary step is called the molecularity of that step.

82. 82 The number of reactant particles that takes part in each elementary step is called the molecularity of that step. Unimolecular – one particle collides with the wall of the vessel or the excess solvent Reaction Mechanism and Rate Law

83. 83 Pseudo-1st order reaction CH 3 COOCH 3 + H 2 O  CH 3 COOH + CH 3 OH Rate = k[CH 3 COOCH 3 ][H 2 O] If H 2 O is used as solvent (in large excess) [H 2 O]  constant throughout the reaction Rate = k’[CH 3 COOCH 3 ] Unimolecular reaction

84. 84 The number of reactant particles that takes part in each elementary step is called the molecularity of that step. Unimolecular – one particle collides with the wall of the vessel or the excess solvent Bimolecular – two particles collide together Termolecular – three particles collide together simultaneously (very rare) Reaction Mechanism and Rate Law

85. 85 The slowest step in a particular mechanism is called the rate-determining step Requirements for writing reaction mechanisms : 1.The sum of elementary steps must give the overall balanced equation for the reaction. 2.The mechanism must agree with the experimentally determined rate law.

86. 86 Consider the reaction A + B + C D Rate = k[A][B] Only one intermediate R + C D (fast) A + BR(slow) r.d.s Proposed mechanism : -

87. 87 A + B X (slow) r.d.s. X Y (fast) Y + C D (fast) Q.28

88. 88 Reaction coordinate P.E. A + B + C X + C Y + C D E1E1 E2E2 E3E3 E 1 > E 2  E 3 Step one is the r.d.s Q.28 Rate = k[A][B]

89. 89 A + B X (fast) X Y (slow) r.d.s. Y + C D (fast) Q.28 k1k1 k2k2 Rate = k[X] k At equilibrium, k 1 [A][B] = k 2 [X]

90. 90 Effect of Catalysts on Rates of Reactions

91. 91 Working Principle of Catalysts and their Effects on Reaction Rates Catalysts alter the rates of reaction, 1.but remain chemically unchanged at the end of the reaction 2.by providing new, alternative reaction pathways with different activation energies. Catalysis  Catalytic action

92. 92 Positive catalyst: Provides an alternative reaction pathway with a lower activation energy Working Principle of Catalysts and their Effects on Reaction Rates

93. 93 Lower E a Ea’Ea’  Greater fraction of molecules with K.E. greater than or equal to E a  Reaction proceeds faster

94. 94 Negative catalyst: Provides an alternative reaction pathway with a higher activation energy Working Principle of Catalysts and their Effects on Reaction Rates

95. 95 Higher E a Ea”Ea”  Smaller fraction of molecules with K.E. greater than or equal to E a  Reaction proceeds slower

96. 96 Working Principle of Catalysts and their Effects on Reaction Rates With catalysts, the contour diagrams and thus the energy profiles are totally different from those without

97. 97 Catalyst Homogeneous Catalyst Heterogeneous Catalyst Reactants & catalyst are in the same phase Reactants & catalyst are NOT in the same phase

98. 98 Characteristics of Catalysts 1.For a given reversible reaction, Reactants Products k1k1 K -1 catalysts affect the rates of forward reaction and backward reaction to the same extent.

99. 99 Q.29 Reactants Products k1k1 k -1 Without catalyst With catalyst Reactants Products Show that

100. 100 E1E1 E -1 E’1E’1 E ’ -1 = P.E. Reaction coordinate

101. 101 Given : T = 298 K, R = 8.31 J K  1 mol  1 = 5.9  10 8

102. 102 Characteristics of Catalysts 2.Catalysts are chemically unchanged at the end of reactions, but may undergo physical changes. E.g. Lumps of MnO 2 used in the decomposition of H 2 O 2 become powdered at the end of the reaction.

103. 103 Characteristics of Catalysts 3.Only small quantity is sufficient to catalyze a reaction because catalysts can be regenerated. However, if the catalysts are involved in the rate equation, higher concentrations may affect the rate more.

104. 104 Characteristics of Catalysts 4.The effect of heterogeneous catalysts depends on the surface area available for the catalytic action. Surface area of solid catalyst   number of reaction sites   catalytic activity  E.g.Finely divided Fe powder is used as the catalyst in Haber process.

105. 105 Characteristics of Catalysts 5.Catalytic actions are specific especially in biological systems. E.g.Enzymatic actions are highly specific.

106. 106 Characteristics of Catalysts 6.The efficiency of a catalyst is often enhanced by adding promoters. Promoters have no catalytic actions on their own. E.g.Fe 2 O 3, KOH, Al 2 O 3 in Haber process

107. 107 Characteristics of Catalysts 7.The efficiency of a catalyst can be lowered by adding poisons or inhibitors. Catalyst poisons are specific in action. E.g.Arsenic impurities may poison Pt but not V 2 O 5 in Contact process

108. 108 Characteristics of Catalysts 8.Transition metals or compounds/ions containing transition metals show marked catalytic activities. E.g.Pt, Ni, Fe, V 2 O 5, MnO 2, Mn 2+ Fe 3+ The catalytic actions are due to the presence of low-lying partially filled d-orbitals.

109. 109 Heterogeneous Catalysis – Adsorption Occur on the surface of the catalyst. 1.Reactants are adsorbed on the surface, forming new bonds with the catalyst while weakening bonds in reactants 2.Products, once formed, are desorbed from the surface,

110. 110 Examples of heterogeneous catalysis 2H 2 O 2 (aq)  2H 2 O(l) + O 2 (g) MnO 2 (s) 3H 2 (g) + N 2 (g)  2NH 3 (g) Fe(s) C 8 H 18 (g)    C 4 H 10 (g) + C 4 H 8 (g) Al 2 O 3 /SiO 2 (s) CH 2 =CH 2 (g) + H 2 (g)  CH 3 –CH 3 (g) Ni(s)

111. 111 fast CH 2 =CH 2 H–H fast slow H 2 C CH 2 H H CH 2 =CH 2 (g) + H 2 (g)  CH 3 –CH 3 (g) Ni(s)

112. 112 fast CH 2 =CH 2 H–H fast slow H 2 C CH 2 HH CH 2 =CH 2 (g) + H 2 (g)  CH 3 –CH 3 (g) Ni(s)

113. 113 fast CH 2 =CH 2 H–H fast slow H 2 C CH 2 HH CH 2 =CH 2 (g) + H 2 (g)  CH 3 –CH 3 (g) Ni(s)

114. 114 Q.30 CH 2 =CH 2 + H 2 CH 3 -CH 3

115. 115 Homogeneous catalysts participate in certain stages of reactions and are regenerated at the end or later stages of reactions. Homogeneous Catalysis – Intermediate Formation Stage 1: A + catalyst  A ─ catalyst Stage 2: A ─ catalyst + B  A ─ B + catalyst Overall reaction: A + B  A ─ B intermediate

116. 116 Homogeneous Catalysis – Intermediate Formation Acid-catalyzed esterification of ethanoic acid and methanol CH 3 COOH(l) + CH 3 OH(l) CH 3 COOCH 3 (l) + H 2 O(l) H+H+

117. 117 Homogeneous Catalysis – Intermediate Formation

118. 118 r.d.s. Uncatalyzed esterification Q.31 Rate = k[CH 3 COOH][CH 3 OH] ++ nucleophilic attack

119. 119 -H + r.d.s. H+H+ + H 2 O Uncatalyzed esterification Q.31 ++

120. 120 Acid-catalyzed esterification Protonation at carbonyl O rather than hydroxyl O since the former is more electron sufficient due to polarization of pi electron cloud

121. 121 Acid-catalyzed esterification

122. 122 Acid-catalyzed esterification

123. 123 Acid-catalyzed esterification

124. 124 Acid-catalyzed esterification

125. 125 Most probable resonance structure Carbonyl C becomes more electron-deficient  More easily attacked by nucleophile Acid-catalyzed esterification

126. 126 Acid-catalyzed esterification

127. 127 r.d.s. Rate = k[RCOOH][H + ] Acid-catalyzed esterification

128. 128 r.d.s. step 2 Acid-catalyzed esterification

129. 129 r.d.s. step 2 step 3 step 4 step 5 step 6 Acid-catalyzed esterification

130. 130 r.d.s. step 2 step 3 step 4 step 6 Acid-catalyzed esterification

131. 131 r.d.s. step 2 step 3 step 4 step 5 Acid-catalyzed esterification

132. 132 r.d.s. step 2 step 3 step 4 step 5 step 6 H + is regenerated Acid-catalyzed esterification

133. 133 r.d.s. step 2 step 3 step 4 step 5 step 6 For simplicity, steps 3 to 6 are combined Acid-catalyzed esterification

134. 134 Rate = k[CH 3 COCH 3 ][H + ]

135. 135

136. 136 At equilibrium, k 1 [CH 3 COCH 3 ][H + ] = k 2 [ ] Rate = k 3 [ ] = k[CH 3 COCH 3 ][H + ] k1k1 k2k2 k3k3

137. 137 Homogeneous Catalysis Using Transition Metal Ions Principle : - Transition metals exhibit variable oxidation states

138. 138 2I  (aq) + S 2 O 8 2  (aq) I 2 (aq) + 2SO 4 2  (aq) The reaction is slow because colliding particles carry like charges

139. 139 2I  (aq) + S 2 O 8 2  (aq) I 2 (aq) + 2SO 4 2  (aq) Fe 3+ (aq) 2I  (aq) + 2Fe 3+ (aq)  I 2 (aq) + 2Fe 2+ (aq) 2Fe 2+ (aq) + S 2 O 8 2  (aq)  2Fe 3+ (aq) + 2SO 4 2  (aq) Mechanism of catalyzed reaction : - Both steps are fast because colliding particles carry opposite charges.

140. 140 2I  (aq) + S 2 O 8 2  (aq) I 2 (aq) + 2SO 4 2  (aq) Fe 3+ (aq) 2I  (aq) + 2Fe 3+ (aq)  I 2 (aq) + 2Fe 2+ (aq) 2Fe 2+ (aq) + S 2 O 8 2  (aq)  2Fe 3+ (aq) + 2SO 4 2  (aq) Mechanism of catalyzed reaction : - The mechanism is made possible by the variable oxidation states of Fe

141. 141 2I  (aq) + S 2 O 8 2  (aq) I 2 (aq) + 2SO 4 2  (aq) Fe 2+ (aq) 2I  (aq) + 2Fe 3+ (aq)  I 2 (aq) + 2Fe 2+ (aq) 2Fe 2+ (aq) + S 2 O 8 2  (aq)  2Fe 3+ (aq) + 2SO 4 2  (aq) Mechanism of catalyzed reaction : - Q.31

142. 142 2Ce 4+ (aq) + Tl + (aq) 2Ce 3+ (aq) + Tl 3+ (aq) Mn 2+ (aq) Mechanism of catalyzed reaction : - Ce 4+ (aq) + Mn 2+ (aq)  Ce 3+ (aq) + Mn 3+ (aq) Ce 4+ (aq) + Mn 3+ (aq)  Ce 3+ (aq) + Mn 4+ (aq) Mn 4+ (aq) + Tl + (aq)  Mn 2+ (aq) + Tl 3+ (aq) The mechanism is made possible by the variable oxidation states of Fe

143. 143 Applications of Catalysts Industrial Catalysts 1.Iron is used in the Haber process N 2 (g) + 3H 2 (g) 2NH 3 (g) Fe 2.Platinum or vanadium(V) oxide is used in the Contact process 2SO 2 (g) + O 2 (g) 2SO 3 (g) Pt or V 2 O 5

144. 144 3.Nickel, platinium or palladium is used in the hydrogenation of unsaturated oils to make margarine Applications of Catalysts

145. 145 Applications of Catalysts 4. Nickel and nickel(II) oxide are used in the production of town gas in Hong Kong. C 5 H 12 (g) + 5H 2 O(g)  5CO(g) + 11H 2 (g) 2CO(g) + 2H 2 (g)  CO 2 (g) + CH 4 (g) Ni or NiO

146. 146 Catalytic Converters in Car Exhaust Systems Applications of Catalysts

147. 147 Applications of Catalysts 2CO(g) + 2NO(g)  2CO 2 (g) + N 2 (g) C x H y (g) + ( x + y/4) O 2 (g)  xCO 2 (g) + y/2 H 2 O(g) 2CO(g) + O 2 (g)  2CO 2 (g) Pt Rh

148. 148 Enzymes in the Production of Alcoholic Drinks Applications of Catalysts C 6 H 12 O 6 (aq)  2C 2 H 5 OH(aq) + 2CO 2 (g) enzyme Fermentation

149. 149 The END

150. 150 15.1 Activation Energy and Arrhenius Equation (SB p.51) For the following reaction: C 6 H 5 N 2 +Cl – (aq) + H 2 O(l)  C 6 H 5 OH(aq) + N 2 (g) + H + (aq) + Cl – (aq) the rate constants of the reaction at different temperatures were measured and recorded in the following table:

151. 151 15.1 Activation Energy and Arrhenius Equation (SB p.51) Temperature (K)Rate constant (10 -5 s -1 ) 278.00.15 298.14.10 308.220.00 323.0140.00 Determine the activation energy graphically. (Given: R = 8.314 J K –1 mol –1 ) Answer

152. 152 15.1 Activation Energy and Arrhenius Equation (SB p.52) 3.096  10 -3 -6.57 3.245  10 -3 -8.52 3.355  10 -3 -10.10 3.597  10 -3 -13.41 1/T (k -1 )ln k

153. 153 15.1 Activation Energy and Arrhenius Equation (SB p.52) A graph of ln k against gives a straight line with slope.

154. 154 15.1 Activation Energy and Arrhenius Equation (SB p.52) Back  y = -11.8 – (-7) = -4.8  x = (3.48 – 3.13)  10 -3 = 0.35  10 -3 K -1 Slope = = -13.7  10 3 K  = -13.7  10 3 K E a = 13.7  10 3 K  8.314 J K -1 mol -1 = 113.9  10 3 J mol -1 = 113.9 kJ mol -1  The activation energy of the reaction is 113.9 kJ mol -1.

155. 155 The rate constant for a reaction at 110°C is found to be twice the value of that at 100°C. Calculate the activation of the reaction. (Given : R = 8.314 J K -1 mol -1 ) 15.1 Activation Energy and Arrhenius Equation (SB p.53) Answer Since k 110 oC = 2 k 100 oC, E a = 82 327 J mol -1 =82.3 kJ mol -1  The activation energy of the reaction is 82.3 kJ mol -1. Back

156. 156 (a) The reaction 2A(g) + B(g)  C(g) was studied at a number of temperatures, and the following results were obtained: Determine the activation energy of the reaction graphically. (Given: R = 8.314 J K –1 mol –1 ) 15.1 Activation Energy and Arrhenius Equation (SB p.53) Tempera ture ( o C) 1260112203292 Rate constant (dm 6 mol -2 s -1 ) 2.3413.252.53161000 Answer

157. 157 15.1 Activation Energy and Arrhenius Equation (SB p.53) 6.91 1.77  10 -3 565 5.76 2.10  10 -3 476 3.96 2.60  10 -3 385 2.58 3.00  10 -3 333 0.85 3.51  10 -3 285 ln k1 / T (K -1 )T (K)

158. 158 15.1 Activation Energy and Arrhenius Equation (SB p.53) A graph of ln k against gives a straight line with slope.

159. 159 15.1 Activation Energy and Arrhenius Equation (SB p.53) Slope = = -3.48  10 3 E a = 3.48  10 3  8.314 = 28.93 kJ mol -1  The activation energy of the reactions is 28.93 kJ mol -1.

160. 160 15.1 Activation Energy and Arrhenius Equation (SB p.53) (b)Determine the activation energy of the following reaction using the data provided only. A + B  C (Given: R = 8.31 J K –1 mol –1 ) 0.400400 0.096350 Rate constant (mol dm -3 s -1 )Temperature (K) Answer

161. 161 15.1 Activation Energy and Arrhenius Equation (SB p.53) (b) E a = 33 206 J mol-1 = 33.2 kJ mol -1 Back

162. 162 (a) Explain why not all collisions between reactant molecules lead to the formation of products. 15.2 Interpretation of Rates of Gaseous Reactions at Molecular Level (SB p.58) Answer (a)For a reaction to occur, colliding molecules must have kinetic energy equal to or greater than the activation energy to break the bonds in the reactants, so that new bonds can form in the products. Moreover, the collision must be in the right geometrical orientation, and the atoms to be transferred or shared do not come into direct contact with each other, so that the atoms can rearrange to form products. Products cannot be formed if the kinetic energy of the reactant molecules cannot overcome the activation energy, or the collision orientation is not appropriate.

163. 163 (b) Describe the effect of temperature on the distribution of molecular speeds in a gaseous system. 15.2 Interpretation of Rates of Gaseous Reactions at Molecular Level (SB p.58) Answer (b)An increase in temperature will lead to an increase in the most probable speed of the molecules. The peak of the curve of Maxwell-Boltzmann distribution of molecular speeds shifts to the right and the curve becomes flattened. This indicates that the distribution of molecular speed becomes wider and the number of molecules having the most probable speed decreases.

164. 164 (c)Explain why the rates of chemical reactions increase with temperature. 15.2 Interpretation of Rates of Gaseous Reactions at Molecular Level (SB p.58) Answer (c)As temperature rises, the proportion of fast-moving molecules increases. The kinetic energy of the molecules also increases. A greater fraction of molecules can overcome the activation energy required for a reaction to occur. Therefore, the number of effective collisions increases and hence the rates of chemical reactions increase. Back

165. 165 Draw an energy profile of a typical single-stage endothermic reaction. 15.3 Energy Profile (SB p.60) Answer Back

166. 166 The energy profile of a multi-stage reaction is shown below: 15.3 Energy Profile (SB p.61)

167. 167 15.3 Energy Profile (SB p.62) (a)Which stage is the rate determining step? Explain your answer. (b)Is the reaction exothermic or endothermic? Explain your answer. Answer (a)Stage 2 is the rate determining step. It is because stage 2 has the greatest amount of activation energy. (b)The reaction is exothermic. It is because the potential energy of the products is lower than that of the reactants. Back

168. 168 15.3 Energy Profile (SB p.62) Referring to the energy profiles below, answer the questions that follow. A B

169. 169 15.3 Energy Profile (SB p.62) Referring to the energy profiles below, answer the questions that follow. C D

170. 170 15.3 Energy Profile (SB p.62) (a)Which reaction(s) is/are exothermic? (b)Which reaction is the fastest? (c)Which reaction has the greatest amount of activation energy? (a)A, B and C (b)B (c)D Answer Back

171. 171 (a) Explain what a negative homogeneous catalyst is. 15.4 Effect of Catalysts on Rates of Reactions (SB p.69) Answer (a)A negative homogeneous catalyst is a catalyst that slows down a reaction. It exists in the same phase as the reactants and products in the reaction, and involves in the formation of an intermediate in the reaction.

172. 172 (b) Explain what a positive heterogeneous catalyst is. 15.4 Effect of Catalysts on Rates of Reactions (SB p.69) Answer (b)A positive heterogeneous catalyst is a catalyst that speeds up a reaction but it is not in the same phase as the reactant and products. It provides an active surface for the reactant particles to adsorb in a reaction.

173. 173 (c) Give three applications of catalysts. 15.4 Effect of Catalysts on Rates of Reactions (SB p.69) Answer (c)Iron used in the Haber process; Platinum or vanadium(V) oxide used in the Contact process; Nickel, platinum or palladium used in the hydrogenation of unsaturated oils to make margarine; Nickel and nickel(II) oxide used in the production of town gas; Platinum (or palladium) and rhodium used in catalytic converters; Enzymes used in fermentation of glucose to produce ethanol; Enzymes used in the manufacture of biological washing powders. (any 3) Back