Chapter13 chemical kinetics

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Chapter13 chemical kinetics

  1. 1. Chemical Kinetics Tro, Chemistry: A Molecular Approach
  2. 2. <ul><li>kinetics is the study of the factors that affect the speed of a reaction and the mechanism by which a reaction proceeds. </li></ul><ul><li>experimentally it is shown that there are 4 factors that influence the speed of a reaction: </li></ul><ul><ul><li>nature of the reactants, </li></ul></ul><ul><ul><li>temperature, </li></ul></ul><ul><ul><li>catalysts, </li></ul></ul><ul><ul><li>concentration </li></ul></ul>Kinetics Tro, Chemistry: A Molecular Approach
  3. 3. Defining Rate <ul><li>rate is how much a quantity changes in a given period of time </li></ul><ul><li>the speed you drive your car is a rate – the distance your car travels (miles) in a given period of time (1 hour) </li></ul><ul><ul><li>so the rate of your car has units of mi/hr </li></ul></ul>Tro, Chemistry: A Molecular Approach
  4. 4. Defining Reaction Rate <ul><li>the rate of a chemical reaction is generally measured in terms of how much the concentration of a reactant decreases in a given period of time </li></ul><ul><ul><li>or product concentration increases </li></ul></ul><ul><li>for reactants, a negative sign is placed in front of the definition </li></ul>Tro, Chemistry: A Molecular Approach
  5. 5. Reaction Rate Changes Over Time <ul><li>as time goes on, the rate of a reaction generally slows down </li></ul><ul><ul><li>because the concentration of the reactants decreases. </li></ul></ul><ul><li>at some time the reaction stops, either because the reactants run out or because the system has reached equilibrium. </li></ul>Tro, Chemistry: A Molecular Approach
  6. 6. Tro, Chemistry: A Molecular Approach at t = 0 [A] = 8 [B] = 8 [C] = 0 at t = 0 [X] = 8 [Y] = 8 [Z] = 0 at t = 16 [A] = 4 [B] = 4 [C] = 4 at t = 16 [X] = 7 [Y] = 7 [Z] = 1
  7. 7. Tro, Chemistry: A Molecular Approach at t = 16 [A] = 4 [B] = 4 [C] = 4 at t = 16 [X] = 7 [Y] = 7 [Z] = 1 at t = 32 [A] = 2 [B] = 2 [C] = 6 at t = 32 [X] = 6 [Y] = 6 [Z] = 2
  8. 8. Tro, Chemistry: A Molecular Approach at t = 32 [A] = 2 [B] = 2 [C] = 6 at t = 32 [X] = 6 [Y] = 6 [Z] = 2 at t = 48 [A] = 0 [B] = 0 [C] = 8 at t = 48 [X] = 5 [Y] = 5 [Z] = 3
  9. 9. Hypothetical Reaction Red  Blue in this reaction, one molecule of Red turns into one molecule of Blue the number of molecules will always total 100 the rate of the reaction can be measured as the speed of loss of Red molecules over time, or the speed of gain of Blue molecules over time Time (sec) Number Red Number Blue 0 100 0 5 84 16 10 71 29 15 59 41 20 50 50 25 42 58 30 35 65 35 30 70 40 25 75 45 21 79 50 18 82
  10. 10. Hypothetical Reaction Red  Blue Tro, Chemistry: A Molecular Approach
  11. 11. Hypothetical Reaction Red  Blue Tro, Chemistry: A Molecular Approach
  12. 12. Reaction Rate and Stoichiometry <ul><li>in most reactions, the coefficients of the balanced equation are not all the same </li></ul><ul><li>H 2 ( g ) + I 2 ( g )  2 HI ( g ) </li></ul><ul><li>for these reactions, the change in the number of molecules of one substance is a multiple of the change in the number of molecules of another </li></ul><ul><ul><li>for the above reaction, for every 1 mole of H 2 used, 1 mole of I 2 will also be used and 2 moles of HI made </li></ul></ul><ul><ul><li>therefore the rate of change will be different </li></ul></ul><ul><li>in order to be consistent, the change in the concentration of each substance is multiplied by 1/coefficient </li></ul>Tro, Chemistry: A Molecular Approach
  13. 13. Average Rate <ul><li>the average rate is the change in measured concentrations in any particular time period </li></ul><ul><ul><li>linear approximation of a curve </li></ul></ul><ul><li>the larger the time interval, the more the average rate deviates from the instantaneous rate </li></ul>Tro, Chemistry: A Molecular Approach
  14. 14. Hypothetical Reaction Red  Blue Avg. Rate Avg. Rate Avg. Rate Time (sec) Number Red Number Blue (5 sec intervals) (10 sec intervals) (25 sec intervals) 0 100 0       5 84 16 3.2     10 71 29 2.6 2.9   15 59 41 2.4     20 50 50 1.8 2.1   25 42 58 1.6   2.3 30 35 65 1.4 1.5   35 30 70 1     40 25 75 1 1   45 21 79 0.8     50 18 82 0.6 0.7 1
  15. 15. H 2 I 2 HI Stoichiometry tells us that for every 1 mole/L of H 2 used, 2 moles/L of HI are made. Assuming a 1 L container, at 10 s, we used 0.181 moles of H 2 . Therefore the amount of HI made is 2(0.181 moles) = 0.362 moles At 60 s, we used 0.699 moles of H 2 . Therefore the amount of HI made is 2(0.699 moles) = 1.398 moles The average rate is the change in the concentration in a given time period. In the first 10 s, the  [H 2 ] is -0.181 M, so the rate is Avg. Rate, M/s Avg. Rate, M/s Time (s) [H 2 ], M [HI], M -  [H 2 ]/  t 1/2  [HI]/  t 0.000 1.000 10.000 0.819 20.000 0.670 30.000 0.549 40.000 0.449 50.000 0.368 60.000 0.301 70.000 0.247 80.000 0.202 90.000 0.165 100.000 0.135 Avg. Rate, M/s Avg. Rate, M/s Time (s) [H 2 ], M [HI], M -  [H 2 ]/  t 1/2  [HI]/  t 0.000 1.000 0.000 10.000 0.819 0.362 20.000 0.670 0.660 30.000 0.549 0.902 40.000 0.449 1.102 50.000 0.368 1.264 60.000 0.301 1.398 70.000 0.247 1.506 80.000 0.202 1.596 90.000 0.165 1.670 100.000 0.135 1.730 Avg. Rate, M/s Time (s) [H 2 ], M [HI], M -  [H 2 ]/  t 0.000 1.000 0.000 10.000 0.819 0.362 0.0181 20.000 0.670 0.660 0.0149 30.000 0.549 0.902 0.0121 40.000 0.449 1.102 0.0100 50.000 0.368 1.264 0.0081 60.000 0.301 1.398 0.0067 70.000 0.247 1.506 0.0054 80.000 0.202 1.596 0.0045 90.000 0.165 1.670 0.0037 100.000 0.135 1.730 0.0030 Avg. Rate, M/s Avg. Rate, M/s Time (s) [H 2 ], M [HI], M -  [H 2 ]/  t 1/2  [HI]/  t 0.000 1.000 0.000 10.000 0.819 0.362 0.0181 0.0181 20.000 0.670 0.660 0.0149 0.0149 30.000 0.549 0.902 0.0121 0.0121 40.000 0.449 1.102 0.0100 0.0100 50.000 0.368 1.264 0.0081 0.0081 60.000 0.301 1.398 0.0067 0.0067 70.000 0.247 1.506 0.0054 0.0054 80.000 0.202 1.596 0.0045 0.0045 90.000 0.165 1.670 0.0037 0.0037 100.000 0.135 1.730 0.0030 0.0030
  16. 16. Tro, Chemistry: A Molecular Approach average rate in a given time period =  slope of the line connecting the [H 2 ] points; and ½ +slope of the line for [HI] the average rate for the first 10 s is 0.0181 M/s the average rate for the first 40 s is 0.0150 M/s the average rate for the first 80 s is 0.0108 M/s
  17. 17. Instantaneous Rate <ul><li>the instantaneous rate is the change in concentration at any one particular time </li></ul><ul><ul><li>slope at one point of a curve </li></ul></ul><ul><li>determined by taking the slope of a line tangent to the curve at that particular point </li></ul><ul><ul><li>first derivative of the function </li></ul></ul><ul><ul><ul><li>for you calculus fans </li></ul></ul></ul>Tro, Chemistry: A Molecular Approach
  18. 18. H 2 ( g ) + I 2 ( g )  2 HI ( g ) Using [H 2 ], the instantaneous rate at 50 s is: Using [HI], the instantaneous rate at 50 s is:
  19. 19. Ex 13.1 - For the reaction given, the [I  ] changes from 1.000 M to 0.868 M in the first 10 s. Calculate the average rate in the first 10 s and the  [H + ]. H 2 O 2 ( aq ) + 3 I  ( aq ) + 2 H + ( aq )  I 3  ( aq ) + 2 H 2 O ( l ) Solve the equation for the Rate (in terms of the change in concentration of the Given quantity) Solve the equation of the Rate (in terms of the change in the concentration for the quantity to Find) for the unknown value
  20. 20. Measuring Reaction Rate <ul><li>in order to measure the reaction rate you need to be able to measure the concentration of at least one component in the mixture at many points in time </li></ul><ul><li>there are two ways of approaching this problem (1) for reactions that are complete in less than 1 hour, it is best to use continuous monitoring of the concentration, or (2) for reactions that happen over a very long time, sampling of the mixture at various times can be used </li></ul><ul><ul><li>when sampling is used, often the reaction in the sample is stopped by a quenching technique </li></ul></ul>Tro, Chemistry: A Molecular Approach
  21. 21. Continuous Monitoring <ul><li>polarimetry – measuring the change in the degree of rotation of plane-polarized light caused by one of the components over time </li></ul><ul><li>spectrophotometry – measuring the amount of light of a particular wavelength absorbed by one component over time </li></ul><ul><ul><li>the component absorbs its complimentary color </li></ul></ul><ul><li>total pressure – the total pressure of a gas mixture is stoichiometrically related to partial pressures of the gases in the reaction </li></ul>Tro, Chemistry: A Molecular Approach
  22. 22. Sampling <ul><li>gas chromatography can measure the concentrations of various components in a mixture </li></ul><ul><ul><li>for samples that have volatile components </li></ul></ul><ul><ul><li>separates mixture by adherence to a surface </li></ul></ul><ul><li>drawing off periodic aliquots from the mixture and doing quantitative analysis </li></ul><ul><ul><li>titration for one of the components </li></ul></ul><ul><ul><li>gravimetric analysis </li></ul></ul>Tro, Chemistry: A Molecular Approach
  23. 23. Factors Affecting Reaction Rate Nature of the Reactants <ul><li>nature of the reactants means what kind of reactant molecules and what physical condition they are in. </li></ul><ul><ul><li>small molecules tend to react faster than large molecules; </li></ul></ul><ul><ul><li>gases tend to react faster than liquids which react faster than solids; </li></ul></ul><ul><ul><li>powdered solids are more reactive than “blocks” </li></ul></ul><ul><ul><ul><li>more surface area for contact with other reactants </li></ul></ul></ul><ul><ul><li>certain types of chemicals are more reactive than others </li></ul></ul><ul><ul><ul><li>e.g., the activity series of metals </li></ul></ul></ul><ul><ul><li>ions react faster than molecules </li></ul></ul><ul><ul><ul><li>no bonds need to be broken </li></ul></ul></ul>Tro, Chemistry: A Molecular Approach
  24. 24. <ul><li>increasing temperature increases reaction rate </li></ul><ul><ul><li>chemist’s rule of thumb - for each 10°C rise in temperature, the speed of the reaction doubles </li></ul></ul><ul><ul><ul><li>for many reactions </li></ul></ul></ul><ul><li>there is a mathematical relationship between the absolute temperature and the speed of a reaction discovered by Svante Arrhenius which will be examined later </li></ul>Factors Affecting Reaction Rate Temperature Tro, Chemistry: A Molecular Approach
  25. 25. <ul><li>catalysts are substances which affect the speed of a reaction without being consumed. </li></ul><ul><li>most catalysts are used to speed up a reaction, these are called positive catalysts </li></ul><ul><ul><li>catalysts used to slow a reaction are called negative catalysts. </li></ul></ul><ul><li>homogeneous = present in same phase </li></ul><ul><li>heterogeneous = present in different phase </li></ul><ul><li>how catalysts work will be examined later </li></ul>Factors Affecting Reaction Rate Catalysts Tro, Chemistry: A Molecular Approach
  26. 26. <ul><li>generally, the larger the concentration of reactant molecules, the faster the reaction </li></ul><ul><ul><li>i ncreases the frequency of reactant molecule contact </li></ul></ul><ul><ul><li>concentration of gases depends on the partial pressure of the gas </li></ul></ul><ul><ul><ul><li>higher pressure = higher concentration </li></ul></ul></ul><ul><li>concentration of solutions depends on the solute to solution ratio (molarity) </li></ul>Factors Affecting Reaction Rate Reactant Concentration Tro, Chemistry: A Molecular Approach
  27. 27. The Rate Law <ul><li>the Rate Law of a reaction is the mathematical relationship between the rate of the reaction and the concentrations of the reactants </li></ul><ul><ul><li>and homogeneous catalysts as well </li></ul></ul><ul><li>the rate of a reaction is directly proportional to the concentration of each reactant raised to a power </li></ul><ul><li>for the reaction aA + bB  products the rate law would have the form given below </li></ul><ul><ul><li>n and m are called the orders for each reactant </li></ul></ul><ul><ul><li>k is called the rate constant </li></ul></ul>Tro, Chemistry: A Molecular Approach
  28. 28. Reaction Order <ul><li>the exponent on each reactant in the rate law is called the order with respect to that reactant </li></ul><ul><li>the sum of the exponents on the reactants is called the order of the reaction </li></ul><ul><li>The rate law for the reaction: </li></ul>Tro, Chemistry: A Molecular Approach 2 NO( g ) + O 2 ( g )  2 NO 2 ( g ) is Rate = k[NO] 2 [O 2 ] The reaction is second order with respect to [NO], first order with respect to [O 2 ], and third order overall
  29. 29. Sample Rate Laws Tro, Chemistry: A Molecular Approach The reaction is autocatalytic, because a product affects the rate. Hg 2+ is a negative catalyst, increasing its concentration slows the reaction.
  30. 30. Reactant Concentration vs. Time A  Products Tro, Chemistry: A Molecular Approach
  31. 31. Half-Life <ul><li>the half-life , t 1/2 , of a reaction is the length of time it takes for the concentration of the reactants to fall to ½ its initial value </li></ul><ul><li>the half-life of the reaction depends on the order of the reaction </li></ul>Tro, Chemistry: A Molecular Approach
  32. 32. Zero Order Reactions <ul><li>Rate = k [A] 0 = k </li></ul><ul><ul><li>constant rate reactions </li></ul></ul><ul><li>[A] = - k t + [A] 0 </li></ul><ul><li>graph of [A] vs. time is straight line with slope = - k and y-intercept = [A] 0 </li></ul><ul><li>t ½ = [A 0 ]/2k </li></ul><ul><li>when Rate = M/sec, k = M/sec </li></ul>Tro, Chemistry: A Molecular Approach [A] 0 [A] time slope = - k
  33. 33. First Order Reactions <ul><li>Rate = k [A] </li></ul><ul><li>ln[A] = - k t + ln[A] 0 </li></ul><ul><li>graph ln[A] vs. time gives straight line with slope = - k and y-intercept = ln[A] 0 </li></ul><ul><ul><li>used to determine the rate constant </li></ul></ul><ul><li>t ½ = 0.693/ k </li></ul><ul><li>the half-life of a first order reaction is constant </li></ul><ul><li>the when Rate = M/sec, k = sec -1 </li></ul>Tro, Chemistry: A Molecular Approach
  34. 34. Tro, Chemistry: A Molecular Approach ln[A] 0 ln[A] time slope = − k
  35. 35. Half-Life of a First-Order Reaction Is Constant Tro, Chemistry: A Molecular Approach
  36. 36. Rate Data for C 4 H 9 Cl + H 2 O  C 4 H 9 OH + HCl Tro, Chemistry: A Molecular Approach Time (sec) [C 4 H 9 Cl], M 0.0 0.1000 50.0 0.0905 100.0 0.0820 150.0 0.0741 200.0 0.0671 300.0 0.0549 400.0 0.0448 500.0 0.0368 800.0 0.0200 10000.0 0.0000
  37. 37. C 4 H 9 Cl + H 2 O  C 4 H 9 OH + 2 HCl Tro, Chemistry: A Molecular Approach
  38. 38. C 4 H 9 Cl + H 2 O  C 4 H 9 OH + 2 HCl Tro, Chemistry: A Molecular Approach
  39. 39. C 4 H 9 Cl + H 2 O  C 4 H 9 OH + 2 HCl Tro, Chemistry: A Molecular Approach slope = -2.01 x 10 -3 k = 2.01 x 10 -3 s -1
  40. 40. Second Order Reactions <ul><li>Rate = k [A] 2 </li></ul><ul><li>1/[A] = k t + 1/[A] 0 </li></ul><ul><li>graph 1/[A] vs. time gives straight line with slope = k and y-intercept = 1/[A] 0 </li></ul><ul><ul><li>used to determine the rate constant </li></ul></ul><ul><li>t ½ = 1/(k[A 0 ]) </li></ul><ul><li>when Rate = M/sec, k = M -1 ∙ sec -1 </li></ul>Tro, Chemistry: A Molecular Approach
  41. 41. Tro, Chemistry: A Molecular Approach l/[A] 0 1/[A] time slope = k
  42. 42. Rate Data For 2 NO 2  2 NO + O 2 Time (hrs.) Partial Pressure NO 2 , mmHg ln(P NO2 ) 1/(P NO2 ) 0 100.0 4.605 0.01000 30 62.5 4.135 0.01600 60 45.5 3.817 0.02200 90 35.7 3.576 0.02800 120 29.4 3.381 0.03400 150 25.0 3.219 0.04000 180 21.7 3.079 0.04600 210 19.2 2.957 0.05200 240 17.2 2.847 0.05800
  43. 43. Rate Data Graphs For 2 NO 2  2 NO + O 2 Tro, Chemistry: A Molecular Approach
  44. 44. Rate Data Graphs For 2 NO 2  2 NO + O 2 Tro, Chemistry: A Molecular Approach
  45. 45. Rate Data Graphs For 2 NO 2  2 NO + O 2 Tro, Chemistry: A Molecular Approach
  46. 46. Determining the Rate Law <ul><li>can only be determined experimentally </li></ul><ul><li>graphically </li></ul><ul><ul><li>rate = slope of curve [A] vs. time </li></ul></ul><ul><ul><li>if graph [A] vs time is straight line, then exponent on A in rate law is 0, rate constant = -slope </li></ul></ul><ul><ul><li>if graph ln[A] vs time is straight line, then exponent on A in rate law is 1, rate constant = -slope </li></ul></ul><ul><ul><li>if graph 1/[A] vs time is straight line, exponent on A in rate law is 2, rate constant = slope </li></ul></ul><ul><li>initial rates </li></ul><ul><ul><li>by comparing effect on the rate of changing the initial concentration of reactants one at a time </li></ul></ul>Tro, Chemistry: A Molecular Approach
  47. 47. Tro, Chemistry: A Molecular Approach
  48. 48. Practice - Complete the Table and Determine the Rate Equation for the Reaction A  2 Prod Tro, Chemistry: A Molecular Approach
  49. 49. Practice - Complete the Table and Determine the Rate Equation for the Reaction A  2 Prod Tro, Chemistry: A Molecular Approach
  50. 50. Tro, Chemistry: A Molecular Approach
  51. 51. Tro, Chemistry: A Molecular Approach
  52. 52. Tro, Chemistry: A Molecular Approach
  53. 53. Practice - Complete the Table and Determine the Rate Equation for the Reaction A  2 Prod Tro, Chemistry: A Molecular Approach the reaction is second order, Rate = -  [A]  t = 0.1 [A] 2
  54. 54. Ex. 13.4 – The reaction SO 2 Cl 2( g )  SO 2( g ) + Cl 2( g ) is first order with a rate constant of 2.90 x 10 -4 s -1 at a given set of conditions. Find the [SO 2 Cl 2 ] at 865 s when [SO 2 Cl 2 ] 0 = 0.0225 M the new concentration is less than the original, as expected [SO 2 Cl 2 ] 0 = 0.0225 M, t = 865, k = 2.90 x 10 -4 s -1 [SO 2 Cl 2 ] Check: Solution: Concept Plan: Relationships: Given: Find: [SO 2 Cl 2 ] [SO 2 Cl 2 ] 0 , t, k
  55. 55. Initial Rate Method <ul><li>another method for determining the order of a reactant is to see the effect on the initial rate of the reaction when the initial concentration of that reactant is changed </li></ul><ul><ul><li>for multiple reactants, keep initial concentration of all reactants constant except one </li></ul></ul><ul><ul><li>zero order = changing the concentration has no effect on the rate </li></ul></ul><ul><ul><li>first order = the rate changes by the same factor as the concentration </li></ul></ul><ul><ul><ul><li>doubling the initial concentration will double the rate </li></ul></ul></ul><ul><ul><li>second order = the rate changes by the square of the factor the concentration changes </li></ul></ul><ul><ul><ul><li>doubling the initial concentration will quadruple the rate </li></ul></ul></ul>Tro, Chemistry: A Molecular Approach
  56. 56. Ex 13.2 – Determine the rate law and rate constant for the reaction NO 2( g ) + CO ( g )  NO ( g ) + CO 2( g ) given the data below. Comparing Expt #1 and Expt #2, the [NO 2 ] changes but the [CO] does not Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033 Write a general rate law including all reactants Examine the data and find two experiments in which the concentration of one reactant changes, but the other concentrations are the same Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033
  57. 57. Ex 13.2 – Determine the rate law and rate constant for the reaction NO 2( g ) + CO ( g )  NO ( g ) + CO 2( g ) given the data below. Tro, Chemistry: A Molecular Approach Determine by what factor the concentrations and rates change in these two experiments. Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033
  58. 58. Ex 13.2 – Determine the rate law and rate constant for the reaction NO 2( g ) + CO ( g )  NO ( g ) + CO 2( g ) given the data below. Tro, Chemistry: A Molecular Approach Determine to what power the concentration factor must be raised to equal the rate factor. Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033
  59. 59. Ex 13.2 – Determine the rate law and rate constant for the reaction NO 2( g ) + CO ( g )  NO ( g ) + CO 2( g ) given the data below. Tro, Chemistry: A Molecular Approach Repeat for the other reactants Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033 Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033
  60. 60. Ex 13.2 – Determine the rate law and rate constant for the reaction NO 2( g ) + CO ( g )  NO ( g ) + CO 2( g ) given the data below. Tro, Chemistry: A Molecular Approach n = 2, m = 0 Substitute the exponents into the general rate law to get the rate law for the reaction Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033
  61. 61. Ex 13.2 – Determine the rate law and rate constant for the reaction NO 2( g ) + CO ( g )  NO ( g ) + CO 2( g ) given the data below. Tro, Chemistry: A Molecular Approach Substitute the concentrations and rate for any experiment into the rate law and solve for k Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1. 0.10 0.10 0.0021 2. 0.20 0.10 0.0082 3. 0.20 0.20 0.0083 4. 0.40 0.10 0.033
  62. 62. Practice - Determine the rate law and rate constant for the reaction NH 4 +1 + NO 2 -1      given the data below. Tro, Chemistry: A Molecular Approach Expt. No. Initial [NH 4 + ], M Initial [NO 2 - ], M Initial Rate, (x 10 -7 ), M/s 1 0.0200 0.200 10.8 2 0.0600 0.200 32.3 3 0.200 0.0202 10.8 4 0.200 0.0404 21.6
  63. 63. Practice - Determine the rate law and rate constant for the reaction NH 4 +1 + NO 2 -1      given the data below. Rate = k [NH 4 + ] n [NO 2  ] m Expt. No. Initial [NH 4 + ], M Initial [NO 2 - ], M Initial Rate, (x 10 -7 ), M/s 1 0.0200 0.200 10.8 2 0.0600 0.200 32.3 3 0.200 0.0202 10.8 4 0.200 0.0404 21.6
  64. 64. The Effect of Temperature on Rate <ul><li>changing the temperature changes the rate constant of the rate law </li></ul><ul><li>Svante Arrhenius investigated this relationship and showed that: </li></ul>Tro, Chemistry: A Molecular Approach R is the gas constant in energy units, 8.314 J/(mol∙K) where T is the temperature in kelvins A is a factor called the frequency factor E a is the activation energy , the extra energy needed to start the molecules reacting
  65. 65. Tro, Chemistry: A Molecular Approach
  66. 66. Activation Energy and the Activated Complex <ul><li>energy barrier to the reaction </li></ul><ul><li>amount of energy needed to convert reactants into the activated complex </li></ul><ul><ul><li>aka transition state </li></ul></ul><ul><li>the activated complex is a chemical species with partially broken and partially formed bonds </li></ul><ul><ul><li>always very high in energy because partial bonds </li></ul></ul>Tro, Chemistry: A Molecular Approach
  67. 67. Isomerization of Methyl Isonitrile Tro, Chemistry: A Molecular Approach methyl isonitrile rearranges to acetonitrile in order for the reaction to occur, the H 3 C-N bond must break; and a new H 3 C-C bond form
  68. 68. Energy Profile for the Isomerization of Methyl Isonitrile the collision frequency is the number of molecules that approach the peak in a given period of time the activation energy is the difference in energy between the reactants and the activated complex the activated complex is a chemical species with partial bonds As the reaction begins, the C-N bond weakens enough for the C  N group to start to rotate
  69. 69. The Arrhenius Equation: The Exponential Factor <ul><li>the exponential factor in the Arrhenius equation is a number between 0 and 1 </li></ul><ul><li>it represents the fraction of reactant molecules with sufficient energy so they can make it over the energy barrier </li></ul><ul><ul><li>the higher the energy barrier (larger activation energy), the fewer molecules that have sufficient energy to overcome it </li></ul></ul><ul><li>that extra energy comes from converting the kinetic energy of motion to potential energy in the molecule when the molecules collide </li></ul><ul><ul><li>increasing the temperature increases the average kinetic energy of the molecules </li></ul></ul><ul><ul><li>therefore, increasing the temperature will increase the number of molecules with sufficient energy to overcome the energy barrier </li></ul></ul><ul><ul><li>therefore increasing the temperature will increase the reaction rate </li></ul></ul>Tro, Chemistry: A Molecular Approach
  70. 70. Tro, Chemistry: A Molecular Approach
  71. 71. Arrhenius Plots <ul><li>the Arrhenius Equation can be algebraically solved to give the following form: </li></ul>Tro, Chemistry: A Molecular Approach this equation is in the form y = m x + b where y = ln( k ) and x = (1/T) a graph of ln( k ) vs. (1/T) is a straight line (-8.314 J/mol∙K)(slope of the line) = E a , (in Joules) e y- intercept = A , (unit is the same as k )
  72. 72. Ex. 13.7 Determine the activation energy and frequency factor for the reaction O 3( g )  O 2( g ) + O ( g ) given the following data: Tro, Chemistry: A Molecular Approach Temp, K k, M -1 ∙s -1 Temp, K k, M -1 ∙s -1 600 3.37 x 10 3 1300 7.83 x 10 7 700 4.83 x 10 4 1400 1.45 x 10 8 800 3.58 x 10 5 1500 2.46 x 10 8 900 1.70 x 10 6 1600 3.93 x 10 8 1000 5.90 x 10 6 1700 5.93 x 10 8 1100 1.63 x 10 7 1800 8.55 x 10 8 1200 3.81 x 10 7 1900 1.19 x 10 9
  73. 73. Ex. 13.7 Determine the activation energy and frequency factor for the reaction O 3( g )  O 2( g ) + O ( g ) given the following data: Tro, Chemistry: A Molecular Approach use a spreadsheet to graph ln( k ) vs. (1/T)
  74. 74. Ex. 13.7 Determine the activation energy and frequency factor for the reaction O 3( g )  O 2( g ) + O ( g ) given the following data: Tro, Chemistry: A Molecular Approach E a = m∙(- R ) solve for E a A = e y -intercept solve for A
  75. 75. Arrhenius Equation: Two-Point Form <ul><li>if you only have two (T, k ) data points, the following forms of the Arrhenius Equation can be used: </li></ul>Tro, Chemistry: A Molecular Approach
  76. 76. Ex. 13.8 – The reaction NO 2( g ) + CO ( g )  CO 2( g ) + NO ( g ) has a rate constant of 2.57 M -1 ∙ s -1 at 701 K and 567 M -1 ∙ s -1 at 895 K. Find the activation energy in kJ/mol most activation energies are tens to hundreds of kJ/mol – so the answer is reasonable T 1 = 701 K, k 1 = 2.57 M -1 ∙s -1 , T 2 = 895 K, k 2 = 567 M -1 ∙s -1 E a , kJ/mol Check: Solution: Concept Plan: Relationships: Given: Find: E a T 1 , k 1 , T 2 , k 2
  77. 77. Collision Theory of Kinetics <ul><li>for most reactions, in order for a reaction to take place, the reacting molecules must collide into each other. </li></ul><ul><li>once molecules collide they may react together or they may not, depending on two factors - </li></ul><ul><li>whether the collision has enough energy to &quot;break the bonds holding reactant molecules together&quot;; </li></ul><ul><li>whether the reacting molecules collide in the proper orientation for new bonds to form. </li></ul>Tro, Chemistry: A Molecular Approach
  78. 78. Effective Collisions <ul><li>collisions in which these two conditions are met (and therefore lead to reaction) are called effective collisions </li></ul><ul><li>the higher the frequency of effective collisions, the faster the reaction rate </li></ul><ul><li>when two molecules have an effective collision, a temporary, high energy (unstable) chemical species is formed - called an activated complex or transition state </li></ul>Tro, Chemistry: A Molecular Approach
  79. 79. Effective Collisions Kinetic Energy Factor Tro, Chemistry: A Molecular Approach for a collision to lead to overcoming the energy barrier, the reacting molecules must have sufficient kinetic energy so that when they collide it can form the activated complex
  80. 80. Effective Collisions Orientation Effect Tro, Chemistry: A Molecular Approach
  81. 81. Collision Theory and the Arrhenius Equation <ul><li>A is the factor called the frequency factor and is the number of molecules that can approach overcoming the energy barrier </li></ul><ul><li>there are two factors that make up the frequency factor – the orientation factor ( p ) and the collision frequency factor ( z ) </li></ul>Tro, Chemistry: A Molecular Approach
  82. 82. Orientation Factor <ul><li>the proper orientation results when the atoms are aligned in such a way that the old bonds can break and the new bonds can form </li></ul><ul><li>the more complex the reactant molecules, the less frequently they will collide with the proper orientation </li></ul><ul><ul><li>reactions between atoms generally have p = 1 </li></ul></ul><ul><ul><li>reactions where symmetry results in multiple orientations leading to reaction have p slightly less than 1 </li></ul></ul><ul><li>for most reactions, the orientation factor is less than 1 </li></ul><ul><ul><li>for many, p << 1 </li></ul></ul><ul><ul><li>there are some reactions that have p > 1 in which an electron is transferred without direct collision </li></ul></ul>Tro, Chemistry: A Molecular Approach
  83. 83. Reaction Mechanisms <ul><li>we generally describe chemical reactions with an equation listing all the reactant molecules and product molecules </li></ul><ul><li>but the probability of more than 3 molecules colliding at the same instant with the proper orientation and sufficient energy to overcome the energy barrier is negligible </li></ul><ul><li>most reactions occur in a series of small reactions involving 1, 2, or at most 3 molecules </li></ul><ul><li>describing the series of steps that occur to produce the overall observed reaction is called a reaction mechanism </li></ul><ul><li>knowing the rate law of the reaction helps us understand the sequence of steps in the mechanism </li></ul>Tro, Chemistry: A Molecular Approach
  84. 84. An Example of a Reaction Mechanism <ul><li>Overall reaction: </li></ul><ul><li>H 2( g ) + 2 ICl ( g )  2 HCl ( g ) + I 2( g ) </li></ul><ul><li>Mechanism: </li></ul><ul><ul><li>H 2( g ) + ICl ( g )  HCl ( g ) + HI ( g ) </li></ul></ul><ul><ul><li>HI ( g ) + ICl ( g )  HCl ( g ) + I 2( g ) </li></ul></ul><ul><li>the steps in this mechanism are elementary steps , meaning that they cannot be broken down into simpler steps and that the molecules actually interact directly in this manner without any other steps </li></ul>Tro, Chemistry: A Molecular Approach
  85. 85. Elements of a Mechanism Intermediates <ul><li>notice that the HI is a product in Step 1, but then a reactant in Step 2 </li></ul><ul><li>since HI is made but then consumed, HI does not show up in the overall reaction </li></ul><ul><li>materials that are products in an early step, but then a reactant in a later step are called intermediates </li></ul>Tro, Chemistry: A Molecular Approach <ul><ul><li>H 2 ( g ) + 2 ICl( g )  2 HCl( g ) + I 2 ( g ) </li></ul></ul><ul><ul><li>H 2 ( g ) + ICl( g )  HCl( g ) + HI( g ) </li></ul></ul><ul><ul><li>HI( g ) + ICl( g )  HCl( g ) + I 2 ( g ) </li></ul></ul>
  86. 86. Molecularity <ul><li>the number of reactant particles in an elementary step is called its molecularity </li></ul><ul><li>a unimolecular step involves 1 reactant particle </li></ul><ul><li>a bimolecular step involves 2 reactant particles </li></ul><ul><ul><li>though they may be the same kind of particle </li></ul></ul><ul><li>a termolecular step involves 3 reactant particles </li></ul><ul><ul><li>though these are exceedingly rare in elementary steps </li></ul></ul>Tro, Chemistry: A Molecular Approach
  87. 87. Rate Laws for Elementary Steps <ul><li>each step in the mechanism is like its own little reaction – with its own activation energy and own rate law </li></ul><ul><li>the rate law for an overall reaction must be determined experimentally </li></ul><ul><li>but the rate law of an elementary step can be deduced from the equation of the step </li></ul>Tro, Chemistry: A Molecular Approach <ul><ul><li>H 2 ( g ) + 2 ICl( g )  2 HCl( g ) + I 2 ( g ) </li></ul></ul><ul><ul><li>H 2 ( g ) + ICl( g )  HCl( g ) + HI( g ) Rate = k 1 [H 2 ][ICl] </li></ul></ul><ul><ul><li>HI( g ) + ICl( g )  HCl( g ) + I 2 ( g ) Rate = k 2 [HI][ICl] </li></ul></ul>
  88. 88. Rate Laws of Elementary Steps Tro, Chemistry: A Molecular Approach
  89. 89. Rate Determining Step <ul><li>in most mechanisms, one step occurs slower than the other steps </li></ul><ul><li>the result is that product production cannot occur any faster than the slowest step – the step determines the rate of the overall reaction </li></ul><ul><li>we call the slowest step in the mechanism the rate determining step </li></ul><ul><ul><li>the slowest step has the largest activation energy </li></ul></ul><ul><li>the rate law of the rate determining step determines the rate law of the overall reaction </li></ul>Tro, Chemistry: A Molecular Approach
  90. 90. Another Reaction Mechanism Tro, Chemistry: A Molecular Approach <ul><ul><li>NO 2( g ) + CO ( g )  NO ( g ) + CO 2( g ) Rate obs = k [NO 2 ] 2 </li></ul></ul><ul><ul><li>NO 2( g ) + NO 2( g )  NO 3( g ) + NO ( g ) Rate = k 1 [NO 2 ] 2 slow </li></ul></ul><ul><ul><li>NO 3( g ) + CO ( g )  NO 2( g ) + CO 2( g ) Rate = k 2 [NO 3 ][CO] fast </li></ul></ul>The first step in this mechanism is the rate determining step. The first step is slower than the second step because its activation energy is larger. The rate law of the first step is the same as the rate law of the overall reaction.
  91. 91. Validating a Mechanism <ul><li>in order to validate (not prove) a mechanism, two conditions must be met: </li></ul><ul><li>the elementary steps must sum to the overall reaction </li></ul><ul><li>the rate law predicted by the mechanism must be consistent with the experimentally observed rate law </li></ul>Tro, Chemistry: A Molecular Approach
  92. 92. Mechanisms with a Fast Initial Step <ul><li>when a mechanism contains a fast initial step, the rate limiting step may contain intermediates </li></ul><ul><li>when a previous step is rapid and reaches equilibrium, the forward and reverse reaction rates are equal – so the concentrations of reactants and products of the step are related </li></ul><ul><ul><li>and the product is an intermediate </li></ul></ul><ul><li>substituting into the rate law of the RDS will produce a rate law in terms of just reactants </li></ul>Tro, Chemistry: A Molecular Approach
  93. 93. An Example Tro, Chemistry: A Molecular Approach 2 H 2 (g ) + 2 NO ( g )  2 H 2 O ( g ) + N 2( g ) Rate obs = k [H 2 ][NO] 2 2 NO ( g )  N 2 O 2( g ) Fast H 2 (g ) + N 2 O 2( g )  H 2 O ( g ) + N 2 O ( g ) Slow Rate = k 2 [H 2 ][N 2 O 2 ] H 2 (g ) + N 2 O ( g )  H 2 O ( g ) + N 2( g ) Fast k 1 k - 1 for Step 1 Rate forward = Rate reverse
  94. 94. Ex 13.9 Show that the proposed mechanism for the reaction 2 O 3( g )  3 O 2( g ) matches the observed rate law Rate = k [O 3 ] 2 [O 2 ] -1 Tro, Chemistry: A Molecular Approach O 3( g )  O 2( g ) + O ( g ) Fast O 3 (g ) + O ( g )  2 O 2( g ) Slow Rate = k 2 [O 3 ][O] k 1 k - 1 for Step 1 Rate forward = Rate reverse
  95. 95. Catalysts <ul><li>catalysts are substances that affect the rate of a reaction without being consumed </li></ul><ul><li>catalysts work by providing an alternative mechanism for the reaction </li></ul><ul><ul><li>with a lower activation energy </li></ul></ul><ul><li>catalysts are consumed in an early mechanism step, then made in a later step </li></ul>Tro, Chemistry: A Molecular Approach mechanism without catalyst O 3 (g ) + O ( g )  2 O 2( g ) V. Slow mechanism with catalyst Cl ( g ) + O 3( g )  O 2( g ) + ClO ( g ) Fast ClO (g ) + O ( g )  O 2( g ) + Cl ( g ) Slow
  96. 96. Ozone Depletion over the Antarctic Tro, Chemistry: A Molecular Approach
  97. 97. Energy Profile of Catalyzed Reaction Tro, Chemistry: A Molecular Approach polar stratospheric clouds contain ice crystals that catalyze reactions that release Cl from atmospheric chemicals
  98. 98. Catalysts <ul><li>homogeneous catalysts are in the same phase as the reactant particles </li></ul><ul><ul><li>Cl ( g ) in the destruction of O 3( g ) </li></ul></ul><ul><li>heterogeneous catalysts are in a different phase than the reactant particles </li></ul><ul><ul><li>solid catalytic converter in a car’s exhaust system </li></ul></ul>Tro, Chemistry: A Molecular Approach
  99. 99. Types of Catalysts Tro, Chemistry: A Molecular Approach
  100. 100. Catalytic Hydrogenation H 2 C=CH 2 + H 2 -> CH 3 CH 3 Tro, Chemistry: A Molecular Approach
  101. 101. Enzymes <ul><li>because many of the molecules are large and complex, most biological reactions require a catalyst to proceed at a reasonable rate </li></ul><ul><li>protein molecules that catalyze biological reactions are called enzymes </li></ul><ul><li>enzymes work by adsorbing the substrate reactant onto an active site that orients it for reaction </li></ul>Tro, Chemistry: A Molecular Approach
  102. 102. Enzyme-Substrate Binding Lock and Key Mechanism Tro, Chemistry: A Molecular Approach
  103. 103. Enzymatic Hydrolysis of Sucrose Tro, Chemistry: A Molecular Approach
  104. 104. Chapter 13 Chemical Kinetics 2008, Prentice Hall Chemistry: A Molecular Approach , 1 st Ed. Nivaldo Tro Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

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