1. CHAPTER 14 Chemical Kinetics

2. 2 Chapter Goals 1. The Rate of a Reaction Factors That Affect Reaction Rates 2. Nature of the Reactants 3. Concentrations of the Reactants: The Rate- Law Expression 4. Concentration Versus Time: The Integrated Rate Equation 5. Collision Theory of Reaction Rates

3. 3 Chapter Goals 6. Transition State Theory 7. Reaction Mechanisms and the Rate-Law Expression 8. Temperature: The Arrhenius Equation 9. Catalysts

4. 4 The Rate of a Reaction Kinetics is the study of rates of chemical reactions and the mechanisms by which they occur. The reaction rate is the increase in concentration of a product per unit time or decrease in concentration of a reactant per unit time. A reaction mechanism is the series of molecular steps by which a reaction occurs.

5. 5 The Rate of a Reaction Thermodynamics (Chapter 15) determines if a reaction can occur. Kinetics (Chapter 16) determines how quickly a reaction occurs. Some reactions that are thermodynamically feasible are kinetically so slow as to be imperceptible.

6. 6 The Rate of Reaction Consider the hypothetical reaction, A (g) + B (g)  C (g) + D (g) equimolar amounts of reactants, A and B, will be consumed while products, C and D, will be formed as indicated in this graph:

7. 7 [A] is the symbol for the concentration of A in M ( mol/L). Note that the reaction does not go entirely to completion. The [A] and [B] > 0 plus the [C] and [D] < 1.

8. 8 The Rate of Reaction Reaction rates are the rates at which reactants disappear or products appear. This movie is an illustration of a reaction rate.

9. 9 The Rate of Reaction Mathematically, the rate of a reaction can be written as:

10. 10 The Rate of Reaction The rate of a simple one-step reaction is directly proportional to the concentration of the reacting substance. [A] is the concentration of A in molarity or moles/L. k is the specific rate constant. k is an important quantity in this chapter.

11. 11 The Rate of Reaction For a simple expression like Rate = k[A] If the initial concentration of A is doubled, the initial rate of reaction is doubled. If the reaction is proceeding twice as fast, the amount of time required for the reaction to reach equilibrium would be: A. The same as the initial time. B. Twice the initial time. C. Half the initial time. If the initial concentration of A is halved the initial rate of reaction is cut in half.

12. 12 The Rate of Reaction If more than one reactant molecule appears in the equation for a one-step reaction, we can experimentally determine that the reaction rate is proportional to the molar concentration of the reactant raised to the power of the number of molecules involved in the reaction.

13. 13 The Rate of Reaction Rate Law Expressions must be determined experimentally. The rate law cannot be determined from the balanced chemical equation. This is a trap for new students of kinetics. The balanced reactions will not work because most chemical reactions are not one-step reactions. Other names for rate law expressions are: 1. rate laws 2. rate equations 3. rate expressions

14. 14 The Rate of Reaction Important terminology for kinetics. The order of a reaction can be expressed in terms of either: 1 each reactant in the reaction or 2 the overall reaction.  Order for the overall reaction is the sum of the orders for each reactant in the reaction. For example:

15. 15 The Rate of Reaction A second example is:

16. 16 The Rate of Reaction A final example of the order of a reaction is:

17. 17 The Rate of Reaction Given the following one step reaction and its rate-law expression. Remember, the rate expression would have to be experimentally determined. Because it is a second order rate-law expression: If the [A] is doubled the rate of the reaction will increase by a factor of 4. 2 2 = 4 If the [A] is halved the rate of the reaction will decrease by a factor of 4. (1/2) 2 = 1/4

18. 18 Factors That Affect Reaction Rates There are several factors that can influence the rate of a reaction: 1. The nature of the reactants. 2. The concentration of the reactants. 3. The temperature of the reaction. 4. The presence of a catalyst. We will look at each factor individually.

19. 19 Nature of Reactants This is a very broad category that encompasses the different reacting properties of substances. For example sodium reacts with water explosively at room temperature to liberate hydrogen and form sodium hydroxide.

20. 20 Nature of Reactants Calcium reacts with water only slowly at room temperature to liberate hydrogen and form calcium hydroxide.

21. 21 Nature of Reactants The reaction of magnesium with water at room temperature is so slow that that the evolution of hydrogen is not perceptible to the human eye.

22. 22 Nature of Reactants However, Mg reacts with steam rapidly to liberate H 2 and form magnesium oxide. The differences in the rate of these three reactions can be attributed to the changing “nature of the reactants”.

23. 23 Concentrations of Reactants: The Rate-Law Expression This movie illustrates how changing the concentration of reactants affects the rate.

24. 24 Concentrations of Reactants: The Rate-Law Expression This is a simplified representation of the effect of different numbers of molecules in the same volume. The increase in the molecule numbers is indicative of an increase in concentration. A (g) + B (g)  Products A B B A B 4 different possible A-B collisions 6 different possible A-B collisions 9 different possible A-B collisions

25. 25 Concentrations of Reactants: The Rate-Law Expression Example 16-1: The following rate data were obtained at 25 o C for the following reaction. What are the rate-law expression and the specific rate-constant for this reaction? 2 A (g) + B (g)  3 C (g) Experiment Number Initial [A] (M) Initial [B] (M) Initial rate of formation of C (M/s) 10.10 2.0 x 10 -4 20.200.304.0 x 10 -4 30.100.202.0 x 10 -4

26. 26 Concentrations of Reactants: The Rate-Law Expression

27. 27 Concentrations of Reactants: The Rate-Law Expression

28. 28 Concentrations of Reactants: The Rate-Law Expression

29. 29 Concentrations of Reactants: The Rate-Law Expression Example 16-2: The following data were obtained for the following reaction at 25 o C. What are the rate-law expression and the specific rate constant for the reaction? 2 A (g) + B (g) + 2 C (g)  3 D (g) + 2 E (g) Experiment Initial [A] (M) Initial [B] (M) Initial [C] (M) Initial rate of formation of D (M/s) 10.200.10 2.0 x 10 -4 20.200.300.206.0 x 10 -4 30.200.100.302.0 x 10 -4 40.600.300.401.8 x 10 -3

30. 30 Concentrations of Reactants: The Rate-Law Expression

31. 31 Concentrations of Reactants: The Rate-Law Expression

32. 32 Concentrations of Reactants: The Rate-Law Expression

33. 33 Concentrations of Reactants: The Rate-Law Expression

34. 34 Concentrations of Reactants: The Rate-Law Expression Example 16-3: consider a chemical reaction between compounds A and B that is first order with respect to A, first order with respect to B, and second order overall. From the information given below, fill in the blanks. You do it! Experiment Initial Rate (M/s) Initial [A] (M) Initial [B] (M) 14.0 x 10 -3 0.200.050 21.6 x 10 -2 ?0.050 33.2 x 10 -2 0.40?

35. 35 Concentrations of Reactants: The Rate-Law Expression

36. 36 Concentrations of Reactants: The Rate-Law Expression

37. 37 Concentrations of Reactants: The Rate-Law Expression

38. 38 Concentration vs. Time: The Integrated Rate Equation The integrated rate equation relates time and concentration for chemical and nuclear reactions. From the integrated rate equation we can predict the amount of product that is produced in a given amount of time. Initially we will look at the integrated rate equation for first order reactions. These reactions are 1 st order in the reactant and 1 st order overall.

39. 39 Concentration vs. Time: The Integrated Rate Equation An example of a reaction that is 1 st order in the reactant and 1 st order overall is: a A  products This is a common reaction type for many chemical reactions and all simple radioactive decays. Two examples of this type are: 2 N 2 O 5(g)  2 N 2 O 4(g) + O 2(g) 238 U  234 Th + 4 He

40. 40 Concentration vs. Time: The Integrated Rate Equation where: [A] 0 = mol/L of A at time t=0.[A] = mol/L of A at time t. k = specific rate constant. t = time elapsed since beginning of reaction. a = stoichiometric coefficient of A in balanced overall equation. The integrated rate equation for first order reactions is:

41. 41 Concentration vs. Time: The Integrated Rate Equation Solve the first order integrated rate equation for t. Define the half-life, t 1/2, of a reactant as the time required for half of the reactant to be consumed, or the time at which [A]=1/2[A] 0.

42. 42 Concentration vs. Time: The Integrated Rate Equation At time t = t 1/2, the expression becomes:

43. 43 Concentration vs. Time: The Integrated Rate Equation Example 16-4: Cyclopropane, an anesthetic, decomposes to propene according to the following equation. The reaction is first order in cyclopropane with k = 9.2 s -1 at 1000 0 C. Calculate the half life of cyclopropane at 1000 0 C.

44. 44 Concentration vs. Time: The Integrated Rate Equation Example 16-5: Refer to Example 16-4. How much of a 3.0 g sample of cyclopropane remains after 0.50 seconds? The integrated rate laws can be used for any unit that represents moles or concentration. In this example we will use grams rather than mol/L.

45. 45 Concentration vs. Time: The Integrated Rate Equation

46. 46 Concentration vs. Time: The Integrated Rate Equation Example 16-6: The half-life for the following first order reaction is 688 hours at 1000 0 C. Calculate the specific rate constant, k, at 1000 0 C and the amount of a 3.0 g sample of CS 2 that remains after 48 hours. CS 2(g)  CS (g) + S (g) You do it!

47. 47 Concentration vs. Time: The Integrated Rate Equation

48. 48 Concentration vs. Time: The Integrated Rate Equation

49. 49 Concentration vs. Time: The Integrated Rate Equation For reactions that are second order with respect to a particular reactant and second order overall, the rate equation is: Where: [A] 0 = mol/L of A at time t=0.[A] = mol/L of A at time t. k = specific rate constant. t = time elapsed since beginning of reaction. a = stoichiometric coefficient of A in balanced overall equation.

50. 50 Concentration vs. Time: The Integrated Rate Equation Second order reactions also have a half-life. Using the second order integrated rate-law as a starting point. At the half-life, t 1/2 [A] = 1/2[A] 0.

51. 51 Concentration vs. Time: The Integrated Rate Equation If we solve for t 1/2 : Note that the half-life of a second order reaction depends on the initial concentration of A.

52. 52 Concentration vs. Time: The Integrated Rate Equation Example 16-7: Acetaldehyde, CH 3 CHO, undergoes gas phase thermal decomposition to methane and carbon monoxide. The rate-law expression is Rate = k[CH 3 CHO] 2, and k = 2.0 x 10 -2 L/(mol. hr) at 527 o C. (a) What is the half-life of CH 3 CHO if 0.10 mole is injected into a 1.0 L vessel at 527 o C?

53. 53 Concentration vs. Time: The Integrated Rate Equation

54. 54 Concentration vs. Time: The Integrated Rate Equation (b) How many moles of CH 3 CHO remain after 200 hours?

55. 55 Concentration vs. Time: The Integrated Rate Equation (c) What percent of the CH 3 CHO remains after 200 hours?

56. 56 Concentration vs. Time: The Integrated Rate Equation Example 16-8: Refer to Example 16-7. (a) What is the half-life of CH 3 CHO if 0.10 mole is injected into a 10.0 L vessel at 527 o C? Note that the vessel size is increased by a factor of 10 which decreases the concentration by a factor of 10! You do it!

57. 57 Concentration vs. Time: The Integrated Rate Equation

58. 58 Concentration vs. Time: The Integrated Rate Equation (b) How many moles of CH 3 CHO remain after 200 hours? You do it!

59. 59 Concentration vs. Time: The Integrated Rate Equation

60. 60 Concentration vs. Time: The Integrated Rate Equation (c) What percent of the CH 3 CHO remains after 200 hours? You do it!

61. 61 Concentration vs. Time: The Integrated Rate Equation

62. 62 Concentration vs. Time: The Integrated Rate Equation Let us now summarize the results from the last two examples. Initial Moles CH 3 CHO [CH 3 CHO] 0 (M) [CH 3 CHO] (M) Moles of CH 3 CHO at 200 hr. % CH 3 CHO remaining Ex. 16-70.10 0.071 71% Ex. 16-80.010 0.00960.09696%

63. 63 Enrichment - Derivation of Integrated Rate Equations For the first order reaction a A  products the rate can be written as:

64. 64 Enrichment - Derivation of Integrated Rate Equations For a first-order reaction, the rate is proportional to the first power of [A].

65. 65 Enrichment - Derivation of Integrated Rate Equations In calculus, the rate is defined as the infinitesimal change of concentration d[A] in an infinitesimally short time dt as the derivative of [A] with respect to time.

66. 66 Enrichment - Derivation of Integrated Rate Equations Rearrange the equation so that all of the [A] terms are on the left and all of the t terms are on the right.

67. 67 Enrichment - Derivation of Integrated Rate Equations Express the equation in integral form.

68. 68 Enrichment - Derivation of Integrated Rate Equations This equation can be evaluated as:

69. 69 Enrichment - Derivation of Integrated Rate Equations Which rearranges to the integrated first order rate equation.

70. 70 Enrichment - Derivation of Integrated Rate Equations Derive the rate equation for a reaction that is second order in reactant A and second order overall. The rate equation is:

71. 71 Enrichment - Derivation of Integrated Rate Equations Separate the variables so that the A terms are on the left and the t terms on the right.

72. 72 Enrichment - Derivation of Integrated Rate Equations Then integrate the equation over the limits as for the first order reaction.

73. 73 Enrichment - Derivation of Integrated Rate Equations Which integrates the second order integrated rate equation.

74. 74 Enrichment - Derivation of Integrated Rate Equations For a zero order reaction the rate expression is:

75. 75 Enrichment - Derivation of Integrated Rate Equations Which rearranges to:

76. 76 Enrichment - Derivation of Integrated Rate Equations Then we integrate as for the other two cases:

77. 77 Enrichment - Derivation of Integrated Rate Equations Which gives the zeroeth order integrated rate equation.

78. 78 Enrichment - Rate Equations to Determine Reaction Order Plots of the integrated rate equations can help us determine the order of a reaction. If the first-order integrated rate equation is rearranged. This law of logarithms, ln (x/y) = ln x - ln y, was applied to the first-order integrated rate-equation.

79. 79 Enrichment - Rate Equations to Determine Reaction Order The equation for a straight line is: Compare this equation to the rearranged first order rate-law.

80. 80 Enrichment - Rate Equations to Determine Reaction Order Now we can interpret the parts of the equation as follows: y can be identified with ln[A] and plotted on the y-axis. m can be identified with –ak and is the slope of the line. x can be identified with t and plotted on the x-axis. b can be identified with ln[A] 0 and is the y-intercept.

81. 81 Enrichment - Rate Equations to Determine Reaction Order Example 16-9: Concentration-versus-time data for the thermal decomposition of ethyl bromide are given in the table below. Use the following graphs of the data to determine the rate of the reaction and the value of the rate constant.

82. 82 Enrichment - Rate Equations to Determine Reaction Order Time (min)012345 [C 2 H 5 Br]1.000.820.670.550.450.37 ln [C 2 H 5 Br]0.00-0.20-0.40-0.60-0.80-0.99 1/[C 2 H 5 Br]1.01.21.51.82.22.7

83. 83 Enrichment - Rate Equations to Determine Reaction Order We will make three different graphs of the data. 1 Plot the [C 2 H 5 Br] (y-axis) vs. time (x-axis) If the plot is linear then the reaction is zero order with respect to [C 2 H 5 Br]. 2 Plot the ln [C 2 H 5 Br] (y-axis) vs. time (x-axis) If the plot is linear then the reaction is first order with respect to [C 2 H 5 Br]. 3 Plot 1/ [C 2 H 5 Br] (y-axis) vs. time (x-axis) If the plot is linear then the reaction is second order with respect to [C 2 H 5 Br].

84. 84 Enrichment - Rate Equations to Determine Reaction Order Plot of [C 2 H 5 Br] versus time. Is it linear or not?

85. 85 Enrichment - Rate Equations to Determine Reaction Order Plot of ln [C 2 H 5 Br] versus time. Is it linear or not?

86. 86 Enrichment - Rate Equations to Determine Reaction Order Plot of 1/[C 2 H 5 Br] versus time. Is it linear or not?

87. 87 Enrichment - Rate Equations to Determine Reaction Order Note that the only graph which is linear is the plot of ln[C 2 H 5 Br] vs. time. Thus this is a first order reaction with respect to [C 2 H 5 Br]. Next, we will determine the value of the rate constant from the slope of the line on the graph of ln[C 2 H 5 Br] vs. time. Remember slope = y 2 -y 1 /x 2 -x 1.

88. 88 Enrichment - Rate Equations to Determine Reaction Order From the equation for a first order reaction we know that the slope = -a k. In this reaction a = 1.

89. 89 Enrichment - Rate Equations to Determine Reaction Order The integrated rate equation for a reaction that is second order in reactant A and second order overall. This equation can be rearranged to:

90. 90 Enrichment - Rate Equations to Determine Reaction Order Compare the equation for a straight line and the second order rate-law expression. Now we can interpret the parts of the equation as follows: y can be identified with 1/[A] and plotted on the y-axis. m can be identified with a k and is the slope of the line. x can be identified with t and plotted on the x-axis b can be identified with 1/[A] 0 and is the y-intercept.

91. 91 Enrichment - Rate Equations to Determine Reaction Order Example 16-10: Concentration-versus-time data for the decomposition of nitrogen dioxide are given in the table below. Use the graphs to determine the rate of the reaction and the value of the rate constant

92. 92 Enrichment - Rate Equations to Determine Reaction Order Time (min)012345 [NO 2 ]1.00.530.360.270.220.18 ln [NO 2 ]0.0-0.63-1.3-1.5-1.7 1/[NO 2 ]1.01.92.83.74.65.5

93. 93 Enrichment - Rate Equations to Determine Reaction Order Once again, we will make three different graphs of the data. 1. Plot [NO 2 ] (y-axis) vs. time (x-axis). If the plot is linear then the reaction is zero order with respect to NO 2. 2. Plot ln [NO 2 ] (y-axis) vs. time (x-axis). If the plot is linear then the reaction is first order with respect to NO 2. 3. Plot 1/ [NO 2 ] (y-axis) vs. time (x-axis). If the plot is linear then the reaction is second order with respect to NO 2.

94. 94 Enrichment - Rate Equations to Determine Reaction Order Plot of [NO 2 ] versus time. Is it linear or not?

95. 95 Enrichment -Rate Equations to Determine Reaction Order Plot of ln [NO 2 ] versus time. Is it linear or not?

96. 96 Enrichment - Rate Equations to Determine Reaction Order Plot of 1/[NO 2 ] versus time. Is it linear or not?

97. 97 Enrichment - Rate Equations to Determine Reaction Order Note that the only graph which is linear is the plot of 1/[NO 2 ] vs. time. Thus this is a second order reaction with respect to [NO 2 ]. Next, we will determine the value of the rate constant from the slope of the line on the graph of 1/[NO 2 ] vs. time.

98. 98 Enrichment - Rate Equations to Determine Reaction Order From the equation for a first order reaction we know that the slope = a k In this reaction a = 2.

99. 99 Collision Theory of Reaction Rates Three basic events must occur for a reaction to occur the atoms, molecules or ions must: 1. Collide. 2. Collide with enough energy to break and form bonds. 3. Collide with the proper orientation for a reaction to occur.

100. 100 Collision Theory of Reaction Rates One method to increase the number of collisions and the energy necessary to break and reform bonds is to heat the molecules. As an example, look at the reaction of methane and oxygen: We must start the reaction with a match. This provides the initial energy necessary to break the first few bonds. Afterwards the reaction is self-sustaining.

101. 101 Collision Theory of Reaction Rates Illustrate the proper orientation of molecules that is necessary for this reaction. X 2(g) + Y 2(g)  2 XY (g) Some possible ineffective collisions are : XXXX Y YYYY X Y

102. 102 Collision Theory of Reaction Rates An example of an effective collision is: X Y + X Y

103. 103 Collision Theory of Reaction Rates This picture illustrates effective and ineffective molecular collisions.

104. 104 Transition State Theory Transition state theory postulates that reactants form a high energy intermediate, the transition state, which then falls apart into the products. For a reaction to occur, the reactants must acquire sufficient energy to form the transition state. This energy is called the activation energy or E a. Look at a mechanical analog for activation energy

105. 105 Transition State Theory  E pot = mg  h Cross section of mountain Boulder E activation hh h2h2 h1h1 E pot =mgh 2 E pot =mgh 1 Height

106. 106 Transition State Theory Potential Energy Reaction Coordinate X 2 + Y 2 2 XY E activation - a kinetic quantity  E  H a thermodynamic quantity Representation of a chemical reaction.

107. 107 Transition State Theory

108. 108 Transition State Theory The relationship between the activation energy for forward and reverse reactions is Forward reaction = E a Reverse reaction = E a +  E difference =  E

109. 109 Transition State Theory The distribution of molecules possessing different energies at a given temperature is represented in this figure.

110. 110 Reaction Mechanisms and the Rate-Law Expression Use the experimental rate-law to postulate a molecular mechanism. The slowest step in a reaction mechanism is the rate determining step.

111. 111 Reaction Mechanisms and the Rate-Law Expression Use the experimental rate-law to postulate a mechanism. The slowest step in a reaction mechanism is the rate determining step. Consider the iodide ion catalyzed decomposition of hydrogen peroxide to water and oxygen.

112. 112 Reaction Mechanisms and the Rate-Law Expression This reaction is known to be first order in H 2 O 2, first order in I -, and second order overall. The mechanism for this reaction is thought to be:

113. 113 Reaction Mechanisms and the Rate-Law Expression Important notes about this reaction: 1. One hydrogen peroxide molecule and one iodide ion are involved in the rate determining step. 2. The iodide ion catalyst is consumed in step 1 and produced in step 2 in equal amounts. 3. Hypoiodite ion has been detected in the reaction mixture as a short-lived reaction intermediate.

114. 114 Reaction Mechanisms and the Rate-Law Expression Ozone, O 3, reacts very rapidly with nitrogen oxide, NO, in a reaction that is first order in each reactant and second order overall.

115. 115 Reaction Mechanisms and the Rate-Law Expression One possible mechanism is:

116. 116 Reaction Mechanisms and the Rate-Law Expression A mechanism that is inconsistent with the rate-law expression is:

117. 117 Reaction Mechanisms and the Rate-Law Expression Experimentally determined reaction orders indicate the number of molecules involved in: 1. the slow step only or 2. the slow step and the equilibrium steps preceding the slow step.

118. 118 Temperature: The Arrhenius Equation Svante Arrhenius developed this relationship among (1) the temperature (T), (2) the activation energy (E a ), and (3) the specific rate constant (k).

119. 119 Temperature: The Arrhenius Equation This movie illustrates the effect of temperature on a reaction.

120. 120 Temperature: The Arrhenius Equation If the Arrhenius equation is written for two temperatures, T 2 and T 1 with T 2 >T 1.

121. 121 Temperature: The Arrhenius Equation 1. Subtract one equation from the other.

122. 122 Temperature: The Arrhenius Equation 2. Rearrange and solve for ln k 2 /k 1.

123. 123 Temperature: The Arrhenius Equation Consider the rate of a reaction for which E a =50 kJ/mol, at 20 o C (293 K) and at 30 o C (303 K). How much do the two rates differ?

124. 124 Temperature: The Arrhenius Equation For reactions that have an E a  50 kJ/mol, the rate approximately doubles for a 10 0 C rise in temperature, near room temperature. Consider: 2 ICl (g) + H 2(g)  I 2(g) + 2 HCl (g) The rate-law expression is known to be R=k[ICl][H 2 ].

125. 125 Catalysts Catalysts change reaction rates by providing an alternative reaction pathway with a different activation energy.

126. 126 Catalysts Homogeneous catalysts exist in same phase as the reactants. Heterogeneous catalysts exist in different phases than the reactants. Catalysts are often solids.

127. 127 Catalysts Examples of commercial catalyst systems include:

128. 128 Catalysts This movie shows catalytic converter chemistry on the Molecular Scale

129. 129 Catalysts A second example of a catalytic system is:

130. 130 Catalysts A third examples of a catalytic system is:

131. 131 Catalysts Look at the catalytic oxidation of CO to CO 2 Overall reaction 2 CO (g) + O 2(g)  2CO 2(g) Absorption CO (g)  CO (surface) + O 2(g) O 2(g)  O 2(surface) Activation O 2(surface)  O (surface) Reaction CO (surface) +O (surface)  CO 2(surface) Desorption CO 2(surface)  CO 2(g)

132. 132 Synthesis Question The Chernobyl nuclear reactor accident occurred in 1986. At the time that the reactor exploded some 2.4 MCi of radioactive 137 Cs was released into the atmosphere. The half- life of 137 Cs is 30.1 years. In what year will the amount of 137 Cs released from Chernobyl finally decrease to 100 Ci? A Ci is a unit of radioactivity called the Curie.

133. 133 Synthesis Question

134. 134 Group Question 99m Tc has a half-life of 6.02 hours and is often used in nuclear medical diagnostic tests. Patients are injected with approximately 10 mCi of 99m Tc that is then directed to specific sites in the patient’s body to detect gallstones, brain tumors and function, and other medical conditions. How long will the patient have a higher than normal radioactivity level after they have been injected with 10 mCi of 99m Tc?

135. 135 End of Chapter 16 A great deal of chemistry is a competition between Thermodynamics (Chapter 15) and Kinetics (Chapter 16).