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AP Chapter 14

AP Chapter 14. Chemical Kinetics HW : 1 4 10 15 23 25 29 31 33 37 45 51 63 69 73. Kinetics = Study of the rate at which a chemical process occurs. 14.1 – Factors Affecting Rate. Physical State of the Reactants In order to react, molecules must come in contact with each other.

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AP Chapter 14

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  1. AP Chapter 14 Chemical Kinetics HW: 1 4 10 15 23 25 29 31 33 37 45 51 63 69 73

  2. Kinetics = Study of the rate at which a chemical process occurs.

  3. 14.1 – Factors Affecting Rate • Physical State of the Reactants • In order to react, molecules must come in contact with each other. • The more homogeneous the mixture of reactants, the faster the molecules can react.

  4. 14.1 – Factors Affecting Rate • 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.

  5. 14.1 – Factors Affecting Rate • Presence of a Catalyst • Catalysts speed up reactions by changing the mechanism of the reaction. • Catalysts are not consumed during the course of the reaction.

  6. Average rate = [B] t 14.2 - Reaction Rates A->B Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time.

  7. Average Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) In this reaction, the concentration of butyl chloride, C4H9Cl, was measured at various times.

  8. Average rate = [C4H9Cl] t Reaction Rates 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:

  9. -[C4H9Cl] t Rate = = [C4H9OH] t Average 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 • Since rate cannot be negative, the change in concentration of the reactants (final – initial) needs a negative sign to make the answer +.

  10. 1 2 [HI] t = Rate = − [I2] t Reaction Rates and Stoichiometry • What if the ratio is not 1:1? 2 HI(g) H2(g) + I2(g) • Therefore,

  11. aA + bB cC + dD = − = = Rate = − 1 a 1 b 1 c 1 d [C] t [D] t [A] t [B] t Reaction Rates and Stoichiometry • To generalize, then, for the reaction

  12. Instantaneous Reaction Rates C4H9Cl(aq) + H2O(l) C4H9OH(aq) + HCl(aq) • Instantaneous Rate = The rate at a specific instant in time. • 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 (wow, Calculus!).

  13. 14.3 – The Rate Law – The Effect of [ ] on Rate Relates rate changes with changes in concentration of reactants. aA + bB -> cC + dD Rate = k[A]x[B]y x and y must be EXPERIMENTALLY determined

  14. Rate Order • The exponents tell the order of the reaction with respect to each reactant. • Can be with respect to one – the exponent of the individual reactant • Can be an overall rate order – the total of all of the exponents • Individual rate orders can be: zero order (0), first order (1), second order (2), third order (3), or fractions

  15. N2(g) + 2 H2O(l) NH4+(aq) + NO2−(aq) Example Experimental data must be provided: -”Isolate” one substance by finding two experiments that hold the other substance constant…

  16. N2(g) + 2 H2O(l) NH4+(aq) + NO2−(aq) Example -Compare Trial 1 and 2: -[Nitrite] is held constant -[Ammonium] doubles -Rate doubles =FIRST ORDER

  17. N2(g) + 2 H2O(l) NH4+(aq) + NO2−(aq) Example -Compare Trial 5 and 6: -[Ammonium] is held constant -[Nitrate] doubles -Rate doubles =FIRST ORDER

  18. Concentration and Rate • This means Rate  [NH4+] Rate  [NO2−] Rate  [NH+] [NO2−] or Rate = k [NH4+] [NO2−] Overall Rate Order = 2

  19. Example • Page 584 – Sample Exercise 14.6

  20. Example • Page 585 – Practice Exercise

  21. ln = −kt [A]t [A]0 14.4-Relationship between [Reactant] and Time - Integrated Rate Laws Using calculus to integrate the rate law for a first-order process gives us Where [A]0 is the initial concentration of A. [A]t is the concentration of A at some time, t, during the course of the reaction.

  22. CH3NC CH3CN First-Order Processes This data was collected for this reaction at 198.9°C. Equivalent to Concentration

  23. ln = −kt [A]t [A]0 First Order - Graphing Manipulating the equation produces… ln [A]t− ln [A]0 = −kt ln [A]t = −kt + ln [A]0 …which is in the form y = mx + b

  24. First-Order Processes Therefore, if a reaction is first-order, a plot of ln [A] vs. twill yield a straight line, and the slope of the line will be -k. ln [A]t = -kt + ln [A]0

  25. CH3NC CH3CN First-Order Processes Consider the process in which methyl isonitrile is converted to acetonitrile.

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

  27. Half-Life • 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.

  28. 0.5 [A]0 [A]0 ln = −kt1/2 ln = −kt [A]t [A]0 0.693 k = t1/2 Half-Life – First Order For a first-order process, this becomes ln 0.5 = −kt1/2 −0.693 = −kt1/2 NOTE: For a first-order process, the half-life does not depend on [A]0.

  29. 1 [A]0 1 [A]t = kt+ Second-Order Processes Similarly, integrating the rate law for a process that is second-order in reactant A, we get also in the form y = mx + b

  30. 1 [A]0 1 [A]t = kt+ Second-Order Processes So if a process is second-order in A, a plot of 1/[A] vs. t will yield a straight line, and the slope of that line is k.

  31. NO2(g) NO (g) + 1/2 O2(g) Second-Order Processes The decomposition of NO2 at 300°C is described by the equation and yields data comparable to this:

  32. Second-Order Processes • Graphing ln 1/[NO2] vs. t, however, gives this plot. • Because this is a straight line, the process is second-order in [A]. • m = k • b = 1/[A]0

  33. 1 k[A]0 1 [A]0 1 [A]0 1 [A]0 2 [A]0 1 0.5 [A]0 2 − 1 [A]0 = kt1/2 + 1 [A]0 1 [A]t = kt1/2 + = kt+ = = kt1/2 = t1/2 Half-Life For a second-order process,

  34. Zero Order Rate = k Concentration has NO effect Straight line plot: [A] vs. t • Slope = -k

  35. 14.5 – Activation Energy, Temperature and Rate • Generally, as temperature increases, so does the reaction rate. • This is because k is temperature dependent.

  36. 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.

  37. The Collision Model Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation.

  38. The Orientation Factor • Molecules must be oriented properly three-dimensionally to bond during collisions

  39. Activation Energy • 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.

  40. CH3NC CH3CN Example: Consider the process in which methyl isonitrile is converted to acetonitrile.

  41. Don’t worry about the other pictures yet – focus on shape of graph and DE and Ea. This is an EXOTHERMIC reaction because the products are at a LOWER energy than the reactants. DE= Eproduct – Ereactant = -

  42. Reaction Coordinate Diagrams • 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.

  43. 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.

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

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

  46. Arrhenius Equation • Svante Arrhenius developed a mathematical relationship between k and Ea: • k = A e where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction.

  47. Ea R 1 T Arrhenius Equation lnk = - () + lnA y = m x + b Therefore, if k is determined experimentally at several temperatures, Eacan be calculated from the slope of a plot of ln k vs. .

  48. Reaction Mechanisms The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism.

  49. Reaction Mechanisms • Reactions may occur all at once or through several discrete steps. • Each of these processes is known as an elementary reaction or elementary process. • The “sum” of all of the elementary reactions must give the overall balanced equation.

  50. Reaction Mechanisms MOST Common The molecularity of a process tells how many molecules are involved in the process.

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