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Chemical Kinetics Chapter 14
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Summary of the Kinetics Reactions OrderRate Law Concentration-Time Equation Half-Life 0 1 2 rate = k rate = k [A] rate = k [A] 2 ln[A] = ln[A] o - kt 1 [A] = 1 [A] o + kt [A] = [A] o - kt t½t½ ln 2 k = t ½ = [A] o 2k2k t ½ = 1 k[A] o rate = k [A][B] t ½ = 1 k[A] o 2
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Generally, as T increases, so does the reaction rate k is temperature dependent Fig 14.11 Temperature affects the rate of chemiluminescence In light sticks
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Fig 14.12 Dependence of rate constant on temperature
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In a chemical rxn, bonds are broken and new bonds are formed Molecules can only react if they collide with each other Furthermore, molecules must collide with the correct orientation and with enough energy Fig 14.13 Cl + NOCl Cl 2 + NO
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Activation Energy Fig 14.14 An energy barrier Reactant Activated complex or Transition state Product
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Activation energy (E a ) ≡ minimum amount of energy required to initiate a chemical reaction Fig 14.15 Activated complex (transition state) ≡ very short-lived and cannot be removed from reaction mixture methyl isonitrile acetonitrile
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Sample Exercise 14.10 Rank the following series of reactions from slowest to fastest The lower the E a the faster the reaction Slowest to fastest: (2) < (3) < (1) Exothermic Endothermic
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How does a molecule acquire sufficient energy to overcome the activation barrier? Fig 14.16 Effect of temperature on distribution of kinetic energies f = e - E a RT Fraction of molecules with E ≥ E a R = 8.314 J/(mol ∙ K) T = kelvin temperature Maxwell–Boltzmann Distributions
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f = e - E a RT Fraction of molecules with E ≥ E a R = 8.314 J/(mol ∙ K) T = kelvin temperature e.g., suppose E a = 100 kJ/mol at T = 300 K: f = e - E a RT = 3.9 x 10 -18 (implies very few energetic molecules) at T = 300 K:f = 1.4 x 10 -17 (about 3.6 times more molecules)
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Temperature Dependence of the Rate Constant k = A exp(− E a /RT ) E a ≡ activation energy (J/mol) R ≡ gas constant (8.314 J/Kmol) T ≡ kelvin temperature A ≡ frequency factor ln k = −E a R 1 T + ln A (Arrhenius equation) y = mx +b Fig 14.12
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k 1 = A exp(−E a /RT 1 ) The Arrehenius equation can be used to relate rate constants k 1 and k 2 at temperatures T 1 and T 2. k 2 = A exp(−E a /RT 2 ) combine to give:
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Plot of ln k vs 1/T Slope = −E a /R ln k =− EaEa R 1 T + ln A Fig 14.17 Graphical determination of activation energy y = m x + b Arrhenius Equation
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Sample Exercise 14.11 The following table shows the rate constants for the rearrangement of methyl isonitrile at various temperatures: (a) From these data, calculate the activation energy for the reaction. (b) What is the value of the rate constant at 430.0 K?
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Solution Analyze: We are given rate constants, k, measured at several temperatures and asked to determine the activation energy, E a, and the rate constant, k, at a particular temperature. Plan: We can obtain E a from the slope of a graph of ln k versus 1/T and the rate constant, k, at a particular temperature. Once we know E a, we can use Equation 14.21 together with the given rate data to calculate the rate constant at 430.0 K.
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Solve: (a) We must first convert the temperatures from degrees Celsius to kelvins. We then take the inverse of each temperature, 1/T, and the natural log of each rate constant, ln k.
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A graph of ln k versus 1/T results in a straight line, as shown in Figure 14.17: slope = − E a /R = − 1.9 x 10 4
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We use the value for the molar gas constant R in units of J/mol-K (Table 10.2). We thus obtain
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(b) To determine the rate constant, k 1, at T 1 = 430.0 K, we can use Equation 14.21 with E a = 160 kJ/mol, and one of the rate constants and temperatures from the given data, such as: k 2 = 2.52 × 10 –5 s –1 and T 2 = 462.9 K: Thus, Note that the units of k 1 are the same as those of k 2.
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Practice Exercise The first-order rate constant for the reaction of methyl chloride with water to produce methanol and hydrochloric acid is 3.32 x 10 -10 s -1 at 25 °C. Calculate the rate constant at 40 °C if the activation energy is 116 kJ/mol. k 2 = 0.349 s -1
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