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Chemical Kinetics

Chemical Kinetics. Chapter 14. Concentration-Time Equation. Order. Rate Law. Half-Life. 1. 1. =. + kt. [A]. [A] o. =. [A] o. t ½ =. t ½ =. t ½. t ½ =. ln 2. 2 k. k. 1. 1. k [A] o. k [A] o. Summary of the Kinetics Reactions. [A] = [A] o - kt. rate = k. 0.

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Chemical Kinetics

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  1. Chemical Kinetics Chapter 14

  2. Concentration-Time Equation Order Rate Law Half-Life 1 1 = + kt [A] [A]o = [A]o t½ = t½ = t½ t½ = ln 2 2k k 1 1 k[A]o k[A]o Summary of the Kinetics Reactions [A] = [A]o - kt rate = k 0 ln[A] = ln[A]o - kt 1 rate = k [A] 2 rate = k [A]2 2 rate = k [A][B]

  3. Temperature and Rate Fig 14.11 Temperature affects the rate of chemiluminescence In light sticks • Generally, as T increases, so does the reaction rate • k is temperature dependent

  4. Fig 14.12 Dependence of rate constant on temperature

  5. The Collision Model (for conc and temp) • 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 Cl + NOCl Cl2 + NO Fig 14.13

  6. Activation Energy Fig 14.14 An energy barrier Activated complex or Transition state Product Reactant

  7. methyl isonitrile acetonitrile Fig 14.15 Activation energy (Ea) ≡ minimum amount of energy required to initiate a chemical reaction Activated complex (transition state) ≡ very short-lived and cannot be removed from reaction mixture

  8. Sample Exercise 14.10 Rank the following series of reactions from slowest to fastest Exothermic Endothermic Exothermic • The lower the Ea the faster the reaction • Slowest to fastest: (2) < (3) < (1)

  9. -Ea RT How does a molecule acquire sufficient energy to overcome the activation barrier? Fig 14.16 Effect of temperature on distribution of kinetic energies Maxwell–Boltzmann Distributions Fraction of molecules with E ≥ Ea f = e R = 8.314 J/(mol ∙ K) T = kelvin temperature

  10. -Ea RT -Ea RT Fraction of molecules with E ≥ Ea f = e R = 8.314 J/(mol ∙ K) T = kelvin temperature e.g., suppose Ea = 100 kJ/mol at T = 300 K: f = e = 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)

  11. ln k = + lnA −Ea 1 T R Temperature Dependence of the Rate Constant Fig 14.12 k = A • exp(− Ea/RT ) (Arrhenius equation) Ea≡ activation energy (J/mol) R ≡ gas constant (8.314 J/K•mol) T ≡ kelvin temperature A ≡ frequency factor y = mx +b

  12. The Arrehenius equation can be used to relate rate constants k1 and k2 at temperatures T1 and T2. k1 = A • exp(−Ea/RT1) combine to give: k2 = A • exp(−Ea/RT2)

  13. ln k =− + lnA Ea 1 T R Arrhenius Equation y = m x + b Fig 14.17 Graphical determination of activation energy Plot of ln k vs 1/T Slope = −Ea/R

  14. Sample Exercise 14.11 The following table shows the rate constants for the rearrangement of methyl isonitrile at various temperatures: • From these data, calculate the activation energy for the reaction. • (b) What is the value of the rate constant at 430.0 K?

  15. Solution Analyze: We are given rate constants, k, measured at several temperatures and asked to determine the activation energy, Ea, and the rate constant, k, at a particular temperature. Plan: We can obtain Ea from the slope of a graph of ln k versus 1/T and the rate constant, k, at a particular temperature. Once we know Ea, we can use Equation 14.21 together with the given rate data to calculate the rate constant at 430.0 K.

  16. 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.

  17. A graph of ln k versus 1/T results in a straight line, as shown in Figure 14.17: slope = − Ea/R = − 1.9 x 104

  18. We use the value for the molar gas constant R in units of J/mol-K (Table 10.2). We thus obtain

  19. (b) To determine the rate constant, k1, at T1 = 430.0 K, we can use Equation 14.21 with Ea = 160 kJ/mol, and one of the rate constants and temperatures from the given data, such as: k2 = 2.52 × 10–5 s–1 and T2 = 462.9 K: Thus, Note that the units of k1 are the same as those of k2.

  20. 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. k2 = 0.349 s-1

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