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Chemical Kinetics Chapter 14

Chemical Kinetics Chapter 14. Chemistry, The Central Science , 7 th & 8 th editions Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten. 1. What do we mean by kinetics?. Kinetics refers to the rate at which chemical reactions occur.

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Chemical Kinetics Chapter 14

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  1. Chemical Kinetics Chapter 14 Chemistry, The Central Science, 7th& 8theditions Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten 1.

  2. What do we mean by kinetics? Kinetics refers to • the rate at which chemical reactions occur. • The reaction mechanism or pathway through which a reaction proceeds. 2.

  3. Topics for study in kinetics 3.

  4. Factors That Affect Reaction Rates • The Nature of the Reactants • Chemical compounds vary considerably in their chemical reactivities. • Concentration of Reactants • As the concentration of reactants increases, so does the likelihood that reactant molecules will collide. • Temperature • At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy. • Catalysts • Change the rate of a reaction by changing the mechanism. 4.

  5. Reaction Rates The rate of a chemical reaction can be determined by monitoring the change in concentration of either reactants or the products as a function of time. [A] vs t 5.

  6. Example 1: Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) [C4H9Cl] M In this reaction, the concentration of butyl chloride, C4H9Cl, was measured at various times, t. [C4H9Cl] t Rate = 6.

  7. Reaction Rates Calculation C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl (aq) The average rate of the reaction over each interval is the change in concentration divided by the change in time: Average Rate, M/s 7.

  8. Reaction Rate Determination C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • Note that the average rate decreases as the reaction proceeds. • This is because as the reaction goes forward, there are fewer collisions between the reacting molecules. 8.

  9. Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • A plot of concentration vs. time for this reaction yields a curve like this. • The slope of a line tangent to the curve at any point is the instantaneous rate at that time. 9.

  10. Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • The reaction slows down with time because the concentration of the reactants decreases. 10.

  11. -[C4H9Cl] t Rate = = [C4H9OH] t Reaction Rates and Stoichiometry C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • In this reaction, the ratio of C4H9Cl to C4H9OH is 1:1. • Thus, the rate of disappearance of C4H9Cl is the same as the rate of appearance of C4H9OH. 11.

  12. Reaction Rates & Stoichiometry Suppose that the mole ratio is not 1:1? Example H2(g) + I2(g)  2 HI(g) 2 moles of HI are produced for each mole of H2 used. The rate at which H2 disappears is only half of the rate at which HI is generated 12.

  13. Concentration and Rate Each reaction has its own equation that gives its rate as a function of reactant concentrations. This is called its Rate Law The general form of the rate law is Rate = k[A]x[B]y Where k is the rate constant, [A]and [B]are the concentrations of the reactants. X and y are exponents known as rate orders that must be determined experimentally To determine the rate law we measure the rate at different starting concentrations. 13.

  14. Concentration and Rate NH4+ (aq) + NO2- (aq)  N2 (g) + 2H2O (l) Compare Experiments 1 and 2:when [NH4+] doubles, the initial rate doubles. 14.

  15. Concentration and Rate NH4+ (aq) + NO2- (aq)  N2 (g) + 2H2O (l) Likewise, compare Experiments 5 and 6: when [NO2-] doubles, the initial rate doubles. 15.

  16. Concentration and Rate NH4+ (aq) + NO2- (aq)  N2 (g) + 2H2O (l) This equation is called the rate law, andk is the rate constant. 16.

  17. The Rate Law • A rate law shows the relationship between the reaction rate and the concentrations of reactants. • For gas-phase reactants use PA instead of [A]. • k is a constant that has a specific value for each reaction. • The value of k is determined experimentally. Rate = K [NH4+][NO2- ] “Constant” is relative here- the rate constant k is unique for each reaction and the value of k changes with temperature 17.

  18. The Rate Law • Exponents tell the order of the reaction with respect to each reactant. • This reaction is First-order in [NH4+] First-order in [NO2−] • The overall reaction order can be found by adding the exponents on the reactants in the rate law. • This reaction is second-order overall. Rate = K [NH4+]1[NO2- ]1 18.

  19. [NO] mol dm-3 [O2] mol dm-3 Rate mol dm-3 s-1 0.40 0.50 1.6 x 10-3 0.40 0.25 8.0 x 10-4 0.20 0.25 2.0 x 10-4 Determining the Rate constant and Order The following data was collected for the reaction of substances A and B to produce products C and D. Deduce the order of this reaction with respect to A and to B. Write an expression for the rate law in this reaction and calculate the value of the rate constant. 19.

  20. The First Order Rate Equation Consider a simple 1st order reaction: A  B Rate = k[A] How much reactant A is left after time t? The rate equation as a function of time can be written as Where [A]t = the concentration of reactant A at time t [A]o = the concentration of reactant A at time t = 0 K = the rate constant [A]t =[A]o e-kt Ln [A]t - Ln[A]o = - kt 20.

  21. CH3NC CH3CN First-Order Processes Consider the process in which methyl isonitrile is converted to acetonitrile. How do we know this is a first order reaction? 21.

  22. CH3NC CH3CN First-Order Processes This data was collected for this reaction at 198.9°C. Does rate=k[CH3NC] for all time intervals? 22.

  23. First-Order Processes • When Ln P is plotted as a function of time, a straight line results. • The process is first-order. • k is the negative slope: 5.1  10-5 s-1. 23.

  24. Half-Life of a Reaction • Half-life is defined as the time required for one-half of a reactant to react. • Because [A] at t1/2 is one-half of the original [A], [A]t = 0.5 [A]0. 24.

  25. Half-Life of a First Order Reaction For a first-order process, set [A]t=0.5 [A]0 in integrated rate equation: NOTE: For a first-order process, the half-life does not depend on the initial concentration, [A]0. 25.

  26. First Order Rate Calculation Example 1: The decomposition of compound A is first order. If the initial [A]0 = 0.80 mol dm-3. and the rate constant is 0.010 s-1, what is the concentration of [A] after 90 seconds? 26.

  27. First Order Rate Calculation Example 1: The decomposition of compound A is first order. If the initial [A]0 = 0.80 mol dm-3. and the rate constant is 0.010 s-1, what is the concentration of [A] after 90 seconds? Ln[A]t – Ln[A]o = -kt Ln[A]t – Ln[0.80] = - (0.010 s-1 )(90 s) Ln[A]t = - (0.010 s-1 )(90 s) + Ln[0.80] Ln[A]t = -0.90 - 0.2231 Ln[A]t = -1.1231 [A]t = 0.325 mol dm-3 27.

  28. First Order Rate Calculations Example 2:A certain first order chemical reaction required 120 seconds for the concentration of the reactant to drop from 2.00 M to 1.00 M. Find the rate constant and the concentration of reactant [A] after 80 seconds. 28.

  29. First Order Rate Calculations Example 2:A certain first order chemical reaction required 120 seconds for the concentration of the reactant to drop from 2.00 M to 1.00 M. Find the rate constant and the concentration of reactant [A] after 80 seconds. Solution k =0.693/t1/2 =0.693/120s =0.005775 s-1 Ln[A] – Ln(2.00) = -0.005775 s-1 (80 s)= -0.462 Ln A = - 0.462 + 0.693 = 0.231 A = 1.26 mol dm-3 29.

  30. First Order Rate Calculations Example 3: Radioactive decay is also a first order process. Strontium 90 is a radioactive isotope with a half-life of 28.8 years. If some strontium 90 were accidentally released, how long would it take for its concentration to fall to 1% of its original concentration? 30.

  31. First Order Rate Calculations Example 3: Radioactive decay is also a first order process. Strontium 90 is a radioactive isotope with a half-life of 28.8 years. If some strontium 90 were accidentally released, how long would it take for its concentration to fall to 1% of its original concentration? Solution k =0.693/t1/2 =0.693/28.8 yr =0.02406 yr-1 Ln[1] – Ln(100) = - (0.02406 yr-1)t = - 4.065 t = - 4.062 . - 0.02406 yr-1 t = 168.8 years 31.

  32. Second-Order Processes Similarly, integrating the rate law for a process that is second-order in reactant A: Rate = k[A]2 1 [A]o 1 [A]t = kt + Where [A]t = the concentration of reactant A at time t [A]o = the concentration of reactant A at time t = 0 K = the rate constant 32.

  33. Second-Order Rate Equation So if a process is second-order in A, a graph of 1/[A] vs. twill yield a straight line with a slope of k. 33.

  34. NO2(g) NO (g) + 1/2 O2(g) Determining Reaction OrderDistinguishing Between 1st and 2nd Order The decomposition of NO2 at 300°C is described by the equation: A experiment with this reaction yields this data: 34.

  35. Determining Reaction OrderDistinguishing Between 1st and 2nd Order Graphing ln [NO2] vs.t yields: • The graph is not a straight line, so this process cannot be first-order in [A]. Does not fit the first order equation: 35.

  36. Second-Order Reaction Kinetics A graph of 1/[NO2] vs. t gives this plot. • This is a straight line. Therefore, the process is second-order in [NO2]. • The slope of the line is the rate constant, k. 36.

  37. Half-Life for 2nd Order Reactions For a second-order process, set [A]t=0.5 [A]0 in 2nd order equation. In this case the half-life depends on the initial concentration of the reactant A. 37.

  38. Sample Problem 1: Second Order Acetaldehyde, CH3CHO, decomposes by second-order kinetics with a rate constant of 0.334 mol-1dm3s-1 at 500oC. Calculate the amount of time it would take for 80% of the acetaldehyde to decompose in a sample that has an initial concentration of 0.00750 M. The final concentration will be 20% of the original 0.00750 M or = 0.00150 1 . .00150 1 . .00750 = 0.334 mol-1dm3s-1t + 666.7 = 0.334 t + 133.33 0.334 t = 533.4 t = 1600 seconds 38.

  39. Sample Problem 2: Second Order Acetaldehyde, CH3CHO, decomposes by second-order kinetics with a rate constant of 0.334 mol-1dm3s-1 at 500oC. If the initial concentration of acetaldehyde is 0.00200 M. Find the concentration after 20 minutes (1200 seconds) Solution 1 . [A]t 1 . 0.00200 mol dm-3 = 0.334 mol-1dm3s-1(1200s) + 1 . [A]t = 0.334 mol-1dm3 s-1(1200s) + 500 mol-1dm3 = 900.8 mol-1dm3 1 _____. 900.8 mol-1dm3 [A]t = = 0.00111 moldm-3 39.

  40. Summary of Kinetics Equations 40.

  41. Temperature and Rate • Generally speaking, the reaction rate increases as the temperature increases. • This is because k is temperature dependent. • As a rule of thumb a reaction rate increases about 10 fold for each 10oC rise in temperature 41.

  42. The Collision Model • In a chemical reaction, bonds are broken and new bonds are formed. • Molecules can only react if they collide with each other. • These collisions must occur with sufficient energy and at the appropriate orientation. 42.

  43. The Collision Model Furthermore, molecules must collide with the correct orientation and with enough energy to cause bonds to break and new bonds to form 43.

  44. Activation Energy • In other words, there is a minimum amount of energy required for reaction: the activation energy, Ea. • Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier. 44.

  45. Reaction Coordinate Diagrams It is helpful to visualize energy changes throughout a process on a reaction coordinate diagram like this one for the rearrangement of methyl isonitrile. 45.

  46. Reaction Coordinate Diagrams • It shows the energy of the reactants and products (and, therefore, E). • The high point on the diagram is the transition state. • The species present at the transition state is called the activated complex. • The energy gap between the reactants and the activated complex is the activation energy barrier. 46.

  47. Maxwell–Boltzmann Distributions • Temperature is defined as a measure of the average kinetic energy of the molecules in a sample. • At any temperature there is a wide distribution of kinetic energies. 47.

  48. Maxwell–Boltzmann Distributions • As the temperature increases, the curve flattens and broadens. • Thus at higher temperatures, a larger population of molecules has higher energy. 48.

  49. Maxwell–Boltzmann Distributions • If the dotted line represents the activation energy, as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier. • As a result, the reaction rate increases. 49.

  50. Maxwell–Boltzmann Distributions This fraction of molecules can be found through the expression: where R is the gas constant and T is the temperature in Kelvin . 50.

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