Ap chem unit 12 presentation

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Chemical Kinetics, AP Chem Unit 12

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Ap chem unit 12 presentation

  1. 1.  Reaction Rates Rate Laws: An Introduction Determining the Form of the Rate Law The Integrated Rate Law Reaction Mechanisms A Model for Chemical Kinetics Catalysis
  2. 2.  Identities of products and reactants Stoichiometric quantities Spontaneity › Refers to the inherent tendency for the process to occur. Does not imply anything about speed. Spontaneous does not mean fast. A reaction can be considered spontaneous but take years to occur.
  3. 3.  The area of chemistry that concerns rates is called chemical kinetics. › One of the main goals of chemical kinetics is to understand the steps by which a reaction takes place. › This series of steps is called the reaction mechanism. › Understanding the mechanism allows us to find ways to change or improve the rate of a reaction.
  4. 4. We start with a flask of NO2 gas at 300°C.But NO2 dioxide decomposes to nitricoxide (a source of air pollution) and O2. 2NO2(g)  2NO(g) + O2(g) If we were to measure the concentrations of the three gases over time we would see a change in the amount of reactants and products over time.
  5. 5.  The reactant NO2 decreases with time and the concentrations of the products (NO and O2) increase with time. 2NO2(g)  2NO(g) + O2(g)
  6. 6. The speed, or rate, of a process is definedas the change in a given quantity(concentration in Molarity) over a specificperiod of time. Reaction rate = [concentration of A at time (t2) – concentration of A at time (t1)]/ (t2 – t1) (Final – initial) D[A] Rate = - Dt
  7. 7. D[A] Rate = - Dt The square brackets indicate concentration in mol/l In Kinetics, rate is always defined as positive. Since the concentration of reactants decreases over time, a negative sign is added to the equation.
  8. 8.  Looking at the NO2 table, calculate the average rate at which NO2 changes over the first 50 seconds of the reaction. 4.2 x 10-5 mol/ls
  9. 9. Reaction rates areoften not constantthrough the course of areaction. For example,average rates for NO2are not constant butdecreases with time.
  10. 10. The value of the rate ata particular time canbe obtained bycomputing the slope ofa line tangent to thecurve at that point intime. Dy slope = Dx
  11. 11.  When considering rates of a reaction you must also take into account the coefficients in the balanced equation for the reaction. The balanced reaction determines the relative rates of consumption of reactants and generation of products.
  12. 12. 2NO2(g)  2NO(g) + O2(g) In this example, both the reactant NO2 and the product NO have a coefficient of 2, so NO is produced at the same rate NO2 is consumed.
  13. 13. 2NO2(g)  2NO(g) + O2(g) The product O2 has a coefficient of 1, which means it is produced half as fast as NO.
  14. 14. Chemical reactions are reversible. Often times as products are formed, they accumulate and react to form what was the reactant(s). The previous example: › 2NO2(g) <-> 2NO(g) + O2(g) › As NO and O2 accumulate, they can react to re-form NO2.
  15. 15. Chemical reactions are reversible. Often times as products are formed, they accumulate and react to form what was the reactant(s). Now the Δ[NO2] depends on the difference in the rates of the forward and reverse reactions. In this unit we will not take into account the reverse reactions.
  16. 16. If the reverse reaction can be neglected, the reaction rate will depend only on the concentrations of the reactants. A rate law shows how concentrations relate to the rate of a reaction. D[A] Rate = - = k[A]n Dt › A is a reactant. k is a proportionality constant and n is called the order of the reactant. Both are usually determined by experiment.
  17. 17. Most simple reactions, the rate orders are often positive integers, but they can be 0 or a fraction. The concentrations of the products do not appear in the rate law because the reaction rate is being studied under conditions where the reverse reaction does not contribute to the overall rate.
  18. 18. Most simple reactions, the orders are often positive integers, but they can be 0 or a fraction. The value of the exponent n must be determined by experiment; it cannot be written from the balanced equation. The rate law constant is dependent upon species in a reaction.
  19. 19. There are two types of rate laws:1. The differential rate law (often called simply the rate law) shows how the rate of a reaction depends on concentration.2. The integrated rate law shows how the concentrations of species in the reaction depend on time.
  20. 20.  The differential and integrated rate laws for a given reaction are related and knowing the rate law for a reaction is important because we can usually infer the individual steps involved in a reaction from the specific form of the rate law.
  21. 21. The first step in understanding how a givenchemical reaction occurs is to determinethe form of the rate law. Reaction rate form is described as orders. › Example: first order, second order, zero order  A first order reaction: concentration of the reactants are reduced by half, the overall rate of the reaction will also be half.  First Order: A direct relationship exists between concentration and rate.
  22. 22.  One common method for experimentally determining the form of the rate law for a reaction is the method of initial rates. › The initial rate of a reaction is the instantaneous rate determined just after the reaction begins. Before the initial concentrations of reactants have changed significantly.
  23. 23. NH4+(aq)+ NO2-(aq) → N2(g)+ 2H2O(l) + D[NH 4 ] Rate = = k[NH + ]n [NO2 ]m - Dt 4 Rate 1 = 1.35 x10-7 mol/ls = k(.100M)n(.0050M)m
  24. 24. NH4+(aq)+ NO2-(aq) → N2(g)+ 2H2O(l) n and m can be determined by dividing known rates.
  25. 25. NH4+(aq)+ NO2-(aq) → N2(g)+ 2H2O(l) Example: Find the form of the rate law for each reactant and the overall reaction order: n and m are both 1 (unrelated), overall reaction order is 2 (n+m=2)
  26. 26. NH4+(aq)+ NO2-(aq) → N2(g)+ 2H2O(l) Example: Find the rate constant for this reaction k = 2.7 x 10-4 L/mols
  27. 27. The reaction between bromate ions andbromide ions in acidic aqueous solution isgiven by the equation: BrO3-(aq) + 5Br-(aq) + 6H+(aq)  3Br2(l) + 3H2O(l) Using the experimental data, determine the orders for all three reactants, the overall reaction order, and the value of the rate constant.
  28. 28.  BrO3-(aq) + 5Br-(aq) + 6H+(aq)  3Br2(l) + 3H2O(l) n=1, m=1, p=2, overall = 4, k=8.0 L3/mol3s
  29. 29. The rate laws we have considered so farexpress the rate as a function of thereactant concentrations. The integratedrate law expresses the reactantconcentrations as a function of time.
  30. 30. D[A]First order rate law: Rate = - = k[A]1 Dt The above rate law can be put into a different form using calculus (integration) Integrated rate law for first order: ln[A] = -kt + ln[Ao ] › this equation is of the form y = mx + b › first order slope = -k, ln[A] vs t is a straight line.
  31. 31. D[A]First order rate law: Rate = - = k[A]1 Dt Integrated rate law for first order can also be written: æ [Ao ] ö ln ç è [A] ø ÷ = kt
  32. 32. The decomposition of N2O5 in the gas phasewas studied at constant temperature:2N2O5(g)  4NO2(g) + O2(g)The following results were collected: [N2O5] (mol/l) Time (s)Verify that this is first order 0.1000 0for N2O5. Calculate k. 0.0707 50 0.0500 100 0.0250 200 0.0125 300 0.00625 400
  33. 33. 2N2O5(g)  4NO2(g) + O2(g) [N2O5] ln[N2O5] Time (s) (mol/l) 0.1000 0 0.0707 50 0.0500 100 0.0250 200 0.0125 300 0.00625 400 k= -slope= 6.93 x 10-3s-1
  34. 34. [N2O5] Time (s) (mol/l) 0.1000 0 0.0707 502N2O5(g)  4NO2(g) + O2(g) 0.0500 100 0.0250 200Using the data from practice 0.0125 300problem 2, calculate [N2O5] at 0.00625 400150 s after the start of the reaction.k= 6.93 x 10-3s-1 .0353 mol/L
  35. 35. The time required for a reactant to reachhalf its original concentration is called thehalf life of a reactant. t1/2
  36. 36. Using the data from the [N2O5] (mol/l) Time (s)previous example we 0.1000 0can observe half-life. 0.0707 50 0.0500 100 0.0250 200 0.0125 300 0.00625 400
  37. 37. The general equation for the half-life of afirst-order reaction is 0.693 t1/2 = k The half-life does not depend on concentration
  38. 38. A certain first-order reaction has a half-life of20.0 minutes.Calculate the rate constant for this reactionand determine how much time is required forthis reaction to be 75% complete? k=3.47 x 10-2min-1 or 5.78 x 10-4s-1 t= 40 min.
  39. 39. D[A] second order rate law: Rate = - = k[A]2 Dt integrated second order rate law: 1 1 = kt + [A] [Ao ] › A plot of 1/[A] vs t is a straight line, slope =k half-life of a second order reaction: 1 t1/2 = k[Ao ]
  40. 40. Butadiene reacts to form its dimeraccording to the equation: [C H ] 4 6 Time (+/- (mol/l) 1s) 2C4H6(g)C8H12(g) 0.01000 0 0.00625 1000The following data was 0.00476 1800 collected: 0.00370 2800 0.00313 3600Is this reaction first or second 0.00270 4400order? What is the rate 0.00241 5200 0.00208 6200constant? What is the half-life?
  41. 41. [C4H6] 1/[C4 ln Time (mol/l) H6] [C4H6] (+/-1s)2C4H6(g)C8H12(g) 0.01000 0 0.00625 1000 0.00476 1800 0.00370 2800 0.00313 3600 0.00270 4400 0.00241 5200 0.00208 6200
  42. 42. [C4H6] 1/[C4 ln Time (mol/l) H6] [C4H6] (+/-1s) 2C4H6(g)C8H12(g) 0.01000 0 0.00625 1000 0.00476 1800 0.00370 2800 0.00313 3600 0.00270 4400 0.00241 5200 0.00208 6200 second order, k =6.14x10-2L/mols half life= 1630s
  43. 43. It is important to recognize the differencebetween the half-life for a first-orderreaction and the half-life for a second-order reaction. First order half life depends only on k › half life remains constant throughout the rxn. Second order half life depends on k and the initial concentration › each successive half-life doubles the preceding one.
  44. 44.  First order  Second order [N2O5] Time (s) [C4H6] Time (+/- (mol/l) (mol/l) 1s) 0.1000 0 0.01000 0 0.00625 1000 0.0707 50 0.00476 1800 0.0500 100 0.00370 2800 0.0250 200 0.00313 3600 0.0125 300 0.00270 4400 0.00625 400 0.00241 5200 0.00208 6200
  45. 45. D[A]zero order rate law: Rate = - = k[A]0 = k Dt integrated zero order rate law: [A] = -kt +[A0 ] › A plot of [A] vs t is a straight line, slope = -k half-life of a zero order reaction: [A0 ] t1/2 = 2k
  46. 46.  Zero-order reactions are most often encountered when a substance such as a metal surface or enzyme (catalyst) is required for the reaction to occur.
  47. 47. Most reactions with more than onereactant are simplified by varying theconcentrations of the reactants. the reactant in which the form of the rate law is being determined will have a low concentration. the other reactants are given a much higher concentration so that the use of the other reactants are not limiting the first.
  48. 48. The reaction from the first example problem:BrO3-(aq) + 5Br-(aq) + 6H+(aq)  3Br2(l) + 3H2O(l) If the rate form for BrO3- is being determined, the concentration of BrO3- is set at 0.001M The other two reactants have concentrations of 1.0M so that their initial and final concentrations are very similar and can be disregarded.
  49. 49.  Review p561
  50. 50. Most chemical reactions occur by a seriesof steps called the reaction mechanism. to really understand a reaction, the mechanism is studied. Kinetics includes the study of possible steps in a reaction
  51. 51. NO2(g) + CO(g)  NO(g) + CO2(g)The rate law for this reaction= k[NO2]2 The reaction is actually more complicated than it appears › The balanced equation only gives us products, reactants and the stoichiometry, but does not give direct information about the mechanism.
  52. 52. NO2(g) + CO(g)  NO(g) + CO2(g) For this reaction, the mechanism involves the following steps: › NO2(g) + NO2(g)  NO3(g) + NO(g) , k1 › NO3(g) + CO(g)  NO2(g) + CO2(g) , k2 k1 and k2 are the rate constants for the individual reactions
  53. 53. NO2(g) + CO(g)  NO(g) + CO2(g) For this reaction, the mechanism involves the following steps: › NO2(g) + NO2(g)  NO3(g) + NO(g) , k1 › NO3(g) + CO(g)  NO2(g) + CO2(g) , k2 NO3(g) is an intermediate, a species that is neither a reactant nor a product but is formed and consumed during the reaction sequence.
  54. 54. NO2(g) + CO(g)  NO(g) + CO2(g) For this reaction, the mechanism involves the following steps: › NO2(g) + NO2(g)  NO3(g) + NO(g) , k1 › NO3(g) + CO(g)  NO2(g) + CO2(g) , k2 Each of these two reactions is called an elementary step. Elementary steps have a rate law that is written from its molecularity.
  55. 55. Molecularity is defined as the number ofspecies that must collide to produce thereaction. A unimolecular step involves one molecule A bimolecular reaction involves the collision of two species A termolecular reaction involves the collision of three species › These steps are rare due to the probability of three molecules colliding simultaneously.
  56. 56. A reaction mechanism is a series ofelementary steps that must satisfy tworequirements:1. The sum of the elementary steps must give the overall balanced equation for the reaction.2. The mechanism must agree with the experimentally determined rate law
  57. 57. Example: NO2(g) + CO(g)  NO(g) + CO2(g)For this reaction, the mechanism involvesthe following steps: › NO2(g) + NO2(g)  NO3(g) + NO(g) , k1 › NO3(g) + CO(g)  NO2(g) + CO2(g) , k2
  58. 58. In order for a mechanism to meet rule 2,the rate-determining step is considered. Multistep reactions often have one step that is much slower than all the others. The overall reaction cannot be faster than the slowest, or rate-determining step.
  59. 59. Example: NO2(g) + CO(g)  NO(g) + CO2(g)If we assume the first step is rate-determining: › NO2(g) + NO2(g)  NO3(g) + NO(g) , k1 › NO3(g) + CO(g)  NO2(g) + CO2(g) , k2 The overall rate is equal to the rate of production of NO3.
  60. 60. Example: NO2(g) + CO(g)  NO(g) + CO2(g)If we assume the first step is rate-determining: › NO2(g) + NO2(g)  NO3(g) + NO(g) , k1 › NO3(g) + CO(g)  NO2(g) + CO2(g) , k2 The rate law for the first step is written from its molecularity: › Overall Rate = k1[NO2]2
  61. 61. The balanced equation for the reaction ofthe gases nitrogen dioxide and fluorine is: 2NO2(g) + F2(g)  2NO2F(g)The experimentally determined rate law is: Rate = k[NO2][F2]A suggested mechanism for this reaction is: NO2 + F2 NO2F + F (slow, k1) F + NO2 NO2F (fast, k2)Is this an acceptable mechanism?
  62. 62. 2NO2(g) + F2(g)  2NO2F(g)The experimentally determined rate law is: Rate = k[NO2][F2]A suggested mechanism for this reaction is: NO2 + F2 NO2F + F (slow, k1) F + NO2 NO2F (fast, k2) The mechanisms satisfy both requirements.
  63. 63. Kinetics depend on concentration, timeand reaction mechanisms. There are otherfactors that affect reaction rates: Temperature: chemical reactions speed up when the temperature is increased
  64. 64.  Rate constants show an exponential increase with absolute temperature.
  65. 65. This observed increase in reaction rate withtemperature is explained with the Collisionmodel. Molecules must collide to react. › higher concentration means more molecules to collide with each other. › KMT states that an increase in temperature raises molecular velocities and increases collision frequency.
  66. 66. Only a small amount of collisions producea reaction Activation energy is the minimum amount of energy required for a collision to produce a reaction. The kinetic energy of a molecule is changed into potential energy as the molecules are distorted during a collision  The arrangement of this distortion is called the activated complex or transition state.
  67. 67. In addition to the activation energyrequired for a reaction to take place,another variable must be considered. Molecular orientations must be correct for collisions to lead to reactions. A correction factor is included in our activation energy formula to account for nonproductive molecular collisions
  68. 68. Two requirements must be satisfied forreactants to collide successfully andrearrange to form products:1. The collision must involve enough energy to produce the reaction; the collision energy must equal or exceed activation energy.2. The relative orientation of the reactants must allow formation of new bonds.
  69. 69. - Ea /RT k = Ae Ea æ R ö ln(k) = ç ÷ + ln(A) R èTø The first equation can be rewritten in a linear equation (y=mx+b). Slope = -Ea/R, x=1/T, y=ln(k) R = 8.3145 J/Kmol
  70. 70. The most common procedure for finding Eais to plot ln(k) vs 1/T and find the slope It can also be found using only two temperatures: æ k2 ö E a æ 1 1 ö ln ç ÷ = ç - ÷ è k1 ø R è T1 T2 ø
  71. 71. The reaction 2N2O5(g)  4NO2(g) + O2(g) wasstudied at several temperatures, and thefollowing values of k were obtained:Calculate the value of Ea for this reaction. k(s-1) T (°C) 2.0x10-5 20 7.3x10-5 30 2.7x10-4 40 1.0x105 J/mol 9.1x10-4 50 2.9x10-3 60
  72. 72. A catalyst is a substance that speeds up areaction without being consumed in thereaction.There are two different types of catalysts:1. homogeneous – a catalyst that is present in the same phase as the reacting molecules2. heterogeneous – exists in a different phase as the reacting molecules (usually a solid).
  73. 73. Heterogeneous catalysis most ofteninvolves gaseous reactants beingadsorbed on the surface of a solidcatalyst. This process is calledadsorption. Adsorption is the collection of one substance on the surface of another substance.
  74. 74. Biologically important reactions in our bodyour usually assisted by a catalyst/enzyme. specific proteins are needed by the human body, the proteins in food must be broken into their constituent amino acids that are then used to construct new proteins in the body’s cells. Without enzymes in human cells these reactions would be much to slow to be useful.

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