Chapter 15 Chemical Kinetics: The Rates of Chemical Reactions - PowerPoint PPT Presentation

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Chapter 15 Chemical Kinetics: The Rates of Chemical Reactions

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  1. Chapter 15Chemical Kinetics: The Rates of Chemical Reactions

  2. Chemical Kinetics: The Rates of Chemical Reactions Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed? Chemical Kinetics will now provide information about the arrow! This gives us information on HOWa reaction occurs! Reactants Products

  3. Chemical Kinetics • Kinetics is the study of how fast (Rates) chemical reactions occur. • Important factors that affect the rates of chemical reactions: • reactant concentration or surface area in solids • temperature • action of catalysts • Our goal:Use kinetics to understand chemical reactions at the particle or molecular level.

  4. Rate of Reactions • Reactants go away with time. • Products appear with time. • The rate of a reaction can be measured by either. In this example by the loss of color with time.

  5. A B Determining a Reaction Rate Blue dye is oxidized with bleach. Its concentration decreases with time. The rate — the change in dye conc. with time — can be determined from the plot. Dye Conc Time

  6. Reaction Rate & Stoichiometry aA + bBcC + dD In general for the reaction: reactants go away with time therefore the negative sign… Reaction rate is the change in the concentration of a reactant or a product with time (M/s).

  7. Determining a Reaction Rate

  8. Determining a Reaction Rate The rate of appearance or disappearance is measured in units of concentration vs. time. Rate = = Ms1or Mmin1 etc... time • There are three “types” of rates • initial rate • average rate • instantaneous rate

  9. Reaction Rates The concentration of a reactant decreases with time. Reactant concentration (M) Time

  10. Reaction Rates Initial rate Reactant concentration (M) Time

  11. Reaction Rates Reactant concentration (M) Time

  12. Instantaneous rate (tangent line) Reaction Rates Reactant concentration (M) Time

  13. Instantaneous rate (tangent line) Reaction Rates Initial rate During the beginning stages of the reaction, the initial rate is very close to the instantaneous rate of reaction. Reactant concentration (M) Time

  14. Problem: Consider the reaction: Over a period of 50.0 to 100.0 s, the concentration of NO(g) drops from 0.0250M to 0.0100M. a) What is the average rate of disappearance of NO(g) during this time?

  15. D[NO] RATE= - 1 Dt 2 Problem: Consider the reaction: Over a period of 50.0 to 100.0 s, the concentration of NO(g) drops from 0.0250M to 0.0100M. What is the rate of rxn? What is the average rate of disappearance of NO(g) during this time? a)  0.0250M) (0.0100M = 1.50104 Ms1 100.0 s  50.0 s b) D[NO]/ Dt = -0.0150/50.0 M/s = -3.00 X 10-4M/s

  16. D[CO2] = Dt D[CH4] rate = - Dt D[H2O] = Dt D[O2] = - 1 1 Dt 2 2 Practice example Write the rate expression for the following reaction: CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (g)

  17. Reaction Conditions & Rate • There are several important factors that will directly affect the rate of a reaction: • Temperature • The physical state of the reactants • Addition of a catalyst All of the above can have a dramatic impact on the rate of a chemical process.

  18. Reaction Conditions & Rate: Temperature Bleach at 54 ˚C Bleach at 22 ˚C

  19. Reaction Conditions & Rate: Physical State of Reactants

  20. Reaction Conditions & Rate: Catalysts Catalyzed decomposition of H2O2 2 H2O2 2 H2O + O2

  21. Effect of Concentration on Reaction Rate: Concentration Mg(s) + 2 HCl(aq)  MgCl2(aq) + H2(g) 0.3 M HCl 6 M HCl

  22. The Rate Law Expression • The rate of reaction must be a function of concentration: • As concentration increases, so do the number ofcollisions… • As the number of collisions increase, so does the probability of a reaction. • This in turn increases the rate of conversion of reactants to products. • The relationship between reaction rate and concentration is given by the reaction Rate Law Expression. • The reaction rate law expression relates the rate of a reaction to the concentrations of the reactants.

  23. Reaction Order aA + bB cC + dD For the general reaction: Each concentration is expressed with an order (exponent). The rate constant converts the concentration expression into the correct units of rate (Ms1). x and y are the reactant orders determined from experiment. x and y are NOT the stoichiometric coefficients.

  24. Reaction Orders A reaction order can be zero, or positive integer and fractional number. Order Name Rate Law 0 zeroth rate = k[A]0 = k 1 first rate = k[A] 2 second rate = k[A]2 0.5 one-half rate = k[A]1/2 1.5 three-half rate = k[A]3/2 0.667 two-thirds rate = k[A]2/3

  25. Reaction Order Overall order: 1 + ½ + 2 = 3.5 = 7/2 or sevenhalves order note: when the order of a reaction is 1 (first order) no exponent is written.

  26. Reaction Constant To find the units of the rate constant, divide the rate units by the Molarity raised to the power of the overall reaction order. k = if (x+y) = 1 k has units of s-1 if (x+y) = 2 k has units of M-1s-1

  27. Reaction Constant • The rate constant, k, is a proportionality constant that relates rate and concentration. • It is found through experiment that the rate constant is a function temperature. • Rate constants must therefore be reported at the temperature with which they are measured. • The rate constant also contains information about the energetics and collision efficiency of the reaction.

  28. EXAMPLE: The reaction, 2 NO (g) + 2 H2 (g)  N2 (g) + 2 H2O (g) is experimentally found to be first order in H2 and third order in NO a) Write the rate law.

  29. EXAMPLE: The reaction, 2 NO (g) + 2 H2 (g)  N2 (g) + 2 H2O (g) is experimentally found to be first order in H2 and third order in NO a) Write the rate law. 3 Rate(Ms-1) = k [H2] [NO] b) What is the overall order of the reaction?

  30. EXAMPLE: The reaction, 2 NO (g) + 2 H2 (g)  N2 (g) + 2 H2O (g) is experimentally found to be first order in H2 and third order in NO a) Write the rate law. 3 Rate(Ms-1) = k [H2] [NO] b) What is the overall order of the reaction? “4th order” Overall order = 1 + 3 = 4 c) What are the units of the rate constant?

  31. EXAMPLE: The reaction, 2 NO (g) + 2 H2 (g)  N2 (g) + 2 H2O (g) is experimentally found to be first order in H2 and third order in NO a) Write the rate law. 3 Rate(Ms-1) = k [H2] [NO] b) What is the overall order of the reaction? “4th order” Overall order = 1 + 3 = 4 c) What are the units of the rate constant?

  32. Rate Law Expression If the rate doubles when [A] doubles and [B] stays constant, the order for [A] is?

  33. Rate Law Expression If the rate doubles when [A] doubles and [B] stays constant, the order for [A] is? one… 1

  34. Rate Law Expression If the rate doubles when [A] doubles and [B] stays constant, the order for [A] is? one… 1

  35. Instantaneous rate (tangent line) Determining a Rate Equation Initial rate During the beginning stages of the reaction, the initial rate is very close to the instantaneous rate of reaction. Reactant concentration (M)

  36. Determining Reaction Order: The Method of Initial Rates The reaction of nitric oxide with hydrogen at 1280 °C is as follows: 2NO (g) + 2H2 (g)  N2 (g) + 2H2O (g) From the following experimental data, determine the rate law and rate constant.

  37. Determining Reaction Order: The Method of Initial Rates The reaction of nitric oxide with hydrogen at 1280 °C is as follows: 2NO (g) + 2H2 (g)  N2 (g) + 2H2O (g) Notice that in Trial 1 and 3, the initial concentration of NO is held constant while H2 is changed.

  38. Determining Reaction Order: The Method of Initial Rates The reaction of nitric oxide with hydrogen at 1280 °C is as follows: 2NO (g) + 2H2 (g)  N2 (g) + 2H2O (g) This means that any changes to the rate must be due to the changes in H2 which is related to the concentration of H2 & its order!

  39. Determining Reaction Order: The Method of Initial Rates 2NO(g) + 2H2(g)  N2(g) + 2H2O(g) The rate law for the reaction is given by: Rate(M/min) = k [NO]x [H2]y Taking the ratio of the rates of Trials 3 and 1 one finds: Rate (Trial 3) = Rate (Trial 1) Plugging in the values from the data:

  40. Determining Reaction Order: The Method of Initial Rates 2.00 Take the log of both sides of the equation: log 2.00 log(2.00) log(2.00) y = 1 Rate(M/min) = k [NO]x[H2]

  41. Determining Reaction Order: The Method of Initial Rates Similarly for x: Rate(M/min) = k [NO]x[H2]y k k x = 3

  42. Determining Reaction Order: The Method of Initial Rates The Rate Law expression is: 2NO(g) + 2H2(g)  N2(g) + 2H2O(g) The order for NO is 3 The order for H2 is 1 The over all order is 3 + 1 =4

  43. Determining Reaction Order: The Method of Initial Rates The Rate constant Rate(M/min) = k [NO]3[H2] To find the rate constant, choose one set of data and solve:

  44. Concentration–Time Relationships:Integrated Rate Laws • It is important know how long a reaction must proceed to reach a predetermined concentration of some reactant or product. • We need a mathematical equation that relates time and concentration: • This equation would yield concentration of reactants or products at a given time. • It will all yield the time required for a given amount of reactant to react.

  45. Integrated Rate Laws For a zeroorder process where “A” goes onto products, the rate law can be written: A  products = k k has units of Ms1 For a zero order process, the rate is the rate constant!

  46. Integrated Rate Laws Zero order kinetics A  products = k This is the “average rate” If one considers the infinitesimal changes in concentration and time the rate law equation becomes: This is the “instantaneous rate”

  47. Integrated Rate Laws Zero order kinetics where [A] = [A]o at time t = 0 and [A] = [A] at time t = t [A]t [A]o k(t  0) = kt = [A]t [A]o = kt

  48. Integrated Rate Laws Zero order kinetics what’s this look like? [A]o [A]t = kt [A]t = kt + [A]o rearranging… y = mx + b a plot of [A]tvstlooks like…  the y-intercept is [A]o k has units of M×(time)1 slope = k [A]t (mols/L) Conclusion: If a plot of reactant concentration vs. time yields a straight line, then the reactant order is ZERO! t (time)

  49. Integrated Rate Laws For a first orderprocess, the rate law can be written: A  products This is the “average rate” If one considers the infinitesimal changes in concentration and time the rate law equation becomes: This is the “instantaneous rate”

  50. Integrated Rate Laws First order kinetics Taking the exponent to each side of the equation: or Conclusion:The concentration of a reactant governed by first order kinetics falls off from an initial concentration exponentially with time.