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Chapter 16

Chapter 16. Chemical Kinetics 化學 動力學. Outline. The Rate of a Reaction Nature of the Reactants Concentrations of the Reactants: The Rate-Law Expression Concentration Versus Time: The Integrated Rate Equation Collision Theory of Reaction Rates Transition State Theory

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Chapter 16

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  1. Chapter 16 Chemical Kinetics 化學動力學

  2. Outline • The Rate of a Reaction • Nature of the Reactants • Concentrations of the Reactants: The Rate-Law Expression • Concentration Versus Time: The Integrated Rate Equation • Collision Theory of Reaction Rates • Transition State Theory • Reaction Mechanisms and the Rate-Law Expression • Temperature: The Arrhenius Equation • Catalysts

  3. The Rate of a Reaction • Kinetics is the study of rates of chemical reactions and the mechanisms by which they occur. • Thereaction 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.

  4. 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.

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

  6. [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. The Rate of Reaction

  7. 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.

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

  9. 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. The Rate of Reaction

  10. 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: • The same as the initial time. • Twice the initial time. • Half the initial time. • If the initial concentration of A is halved the initial rate of reaction is cut in half.

  11. 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.

  12. 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: • rate laws • rate equations or rate expressions

  13. The Rate of Reaction • Important terminology for kinetics. • The order of a reaction can beexpressed in terms of either: • each reactant in the reaction or • the overall reaction. • Order for the overall reaction is the sum of the orders for each reactant in the reaction. • For example:

  14. The Rate of Reaction • A second example is:

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

  16. 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. 22 = 4 If the [A] is halved the rate of the reaction will decrease by a factor of 4. (1/2)2 = 1/4 The Rate of Reaction

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

  18. 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.

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

  20. 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.

  21. However, Mg reacts with steam rapidly to liberate H2 and form magnesium oxide. The differences in the rate of these three reactions can be attributed to the changing “nature of the reactants”. Nature of Reactants

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

  23. A B A B A B B A B A B A B A B 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 4 different possible A-B collisions 6 different possible A-B collisions 9 different possible A-B collisions

  24. Concentrations of Reactants: The Rate-Law Expression Example 16-1: The following rate data were obtained at 25oC 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)

  25. Concentrations of Reactants: The Rate-Law Expression

  26. Concentrations of Reactants: The Rate-Law Expression Example 16-2: The following data were obtained for the following reaction at 25oC. 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)

  27. Concentrations of Reactants: The Rate-Law Expression

  28. 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!

  29. Concentrations of Reactants: The Rate-Law Expression

  30. 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 1st order in the reactant and 1st order overall.

  31. Concentration vs. Time: The Integrated Rate Equation • An example of a reaction that is 1st order in the reactant and 1st 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 N2O5(g) 2 N2O4(g) + O2(g) 238U  234Th + 4He

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

  33. Solve the first order integrated rate equation for t. Define the half-life, t1/2, 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. Concentration vs. Time: The Integrated Rate Equation

  34. Concentration vs. Time: The Integrated Rate Equation • At time t = t1/2, the expression becomes:

  35. 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 10000C. Calculate the half life of cyclopropane at 10000C. Concentration vs. Time: The Integrated Rate Equation

  36. 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.

  37. Concentration vs. Time: The Integrated Rate Equation Example 16-6: The half-life for the following first order reaction is 688 hours at 10000C. Calculate the specific rate constant, k, at 10000C and the amount of a 3.0 g sample of CS2 that remains after 48 hours. CS2(g) ® CS(g) + S(g) You do it!

  38. Concentration vs. Time: The Integrated Rate Equation

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

  40. 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, t1/2 [A] = 1/2[A]0.

  41. If we solve for t1/2: Note that the half-life of a second order reaction depends on the initial concentration of A. Concentration vs. Time: The Integrated Rate Equation

  42. Example 16-7: Acetaldehyde, CH3CHO, undergoes gas phase thermal decomposition to methane and carbon monoxide. The rate-law expression is Rate = k[CH3CHO]2, and k = 2.0 x 10-2 L/(mol.hr) at 527oC. (a) What is the half-life of CH3CHO if 0.10 mole is injected into a 1.0 L vessel at 527oC? Concentration vs. Time: The Integrated Rate Equation

  43. Example 16-7: The rate-law expression is Rate = k[CH3CHO]2, and k = 2.0 x 10-2 L/(mol.hr) at 527oC. (a) What is the half-life of CH3CHO if 0.10 mole is injected into a 1.0 L vessel at 527oC? Concentration vs. Time: The Integrated Rate Equation

  44. Concentration vs. Time: The Integrated Rate Equation (b) How many moles of CH3CHO remain after 200 hours?

  45. Concentration vs. Time: The Integrated Rate Equation (c) What percent of the CH3CHO remains after 200 hours?

  46. Concentration vs. Time: The Integrated Rate Equation Example 16-8: Refer to Example 16-7. (a) What is the half-life of CH3CHO if 0.10 mole is injected into a 10.0 L vessel at 527oC? • Note that the vessel size is increased by a factor of 10 which decreases the concentration by a factor of 10! You do it!

  47. Concentration vs. Time: The Integrated Rate Equation • (b) How many moles of CH3CHO remain after 200 hours? You do it!

  48. Concentration vs. Time: The Integrated Rate Equation • (c) What percent of the CH3CHO remains after 200 hours? You do it!

  49. Concentration vs. Time: The Integrated Rate Equation • Let us now summarize the results from the last two examples.

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

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