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GC Chemical Kinetics

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Chemical Kinetics PDF PowerPoint Presentation

Chemical Kinetics PDF PowerPoint Presentation


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  • 1. Chemical Kinetics• Study of speed with which a chemical reaction occurs and the factors affecting that speed• Provides information about the feasibility of a chemical reaction• Provides information about the time it takes for a chemical reaction to occur• Provides information about the series of elementary steps which lead to the formation of product
  • 2. Rate Data for A + B → Ctime (seconds) Concentration A Concentration B, Concentration C, mol/L mol/L mol/L 0 0.76 0.38 0 1 0.31 0.16 0.20 2 0.13 6.5 x 10-2 0.40 3 5.2 x 10-2 2.6 x 10-2 0.58 4 2.1 x 10-2 1.1 x 10-2 0.73 5 8.8 x 10-3 4.4 x 10-3 0.86 6 3.6 x 10-3 1.8 x 10-3 0.95 7 1.4 x 10-3 7.0 x 10-4 1.02 8 6.1 x 10-4 3.1 x 10-4 1.07 9 2.5 x 10-4 1.3 x 10-4 1.07 10 1.0 x 10-4 5.0 x 10-5 1.07
  • 3. A + B→C 1.2 C 1 0.8Concentration (mol/L) 0.6 A 0.4 B 0.2 0 0 2 4 6 8 10 12 time (seconds)
  • 4. The Rate of a Chemical Reaction• The speed of a reaction can be examined by the decrease in reactants or the increase in products.• aA +bB →cC + d D Rate = k  A  B m n Where m and n are determined experimentally, and not necessarily Equal to the stiochiometry of the reaction
  • 5. Reaction A → 2 B A A A B B B B B B B B A A A B B B B A A A B B B B A B B B BA = 6.022 x 1022 molecules B = 6.022 x 1022 moleculesin a 1.00 L container in a 1.00 Liter container 1 mol/L 2 mol/L Δ  A 1 Δ  B - = t 2 t
  • 6. Average Rate Δ A• Rate of A disappearing is - t• Let’s suppose that after 20 seconds ½ half of A disappears.• Then Δ  A A f  Ai 0.50 mol/L - 1.00 mol/L mol- =- =- = - 2.5 x 10-2 M/s or - 2.5 x 10-2 t tf  ti 20 s - 0 s L-s 1 Δ  B• And Rate of B appearing is 2 t• Then 1 Δ  B Bf -Bi 1 1.00 mol/L - 0.00 mol/L mol = = x = 2.5 x 10-2 M/s or 2.5 x 10-2 2 t tf - ti 2 20 s - 0 s L-s
  • 7. Average Rate Law for the General Equation aA+bB→cC + dD 1 Δ  A 1 Δ  B 1 Δ C 1 Δ  D- x = - x = x = x a Δt b Δt c Δt d Δt For Example: N2O5 (g) → 2 NO2 (g) + ½ O2 (g) Δ  N 2O5 (g)    1 Δ  NO2 (g)    = 2 x Δ O2 (g)   - = x Δt 2 Δt Δt
  • 8. Determination of the Rate Equation• Determined Experimentally• Can be obtained by examining the initial rate after about 1% or 2% of the limiting reagent has been consumed.
  • 9. Consider the Reaction:CH3CH2CH2CH2Cl (aq) + H2O (l) → CH3CH2CH2CH2OH (aq) + HCl (aq) time (seconds) Concentration n-butyl chloride mol/L 0 0.10 50 9.05 x 10-2 100 8.2 x 10-2 150 7.41 x 10-2 200 6.71 x 10-2 300 5.49 x 10-2 400 4.48 x 10-2 500 3.68 x 10-2 800 2.00 x 10-2
  • 10. mol Average Rates, L-stime, seconds [CH3CH2CH2CH2Cl] Average Rate, mol L-s [C4 H9Cl]  t 0.0 0.1000 1.90 x 10-4 50.0 0.0905time, seconds [CH3CH2CH2CH2Cl] Average Rate, mol L-s [C4 H9Cl]  t 50.0 0.0905 1.70 x 10-4 100.0 0.0820
  • 11. mol Average Rates, L-stime, seconds [CH3CH2CH2CH2Cl] Average Rate, mol L-s [C4 H9Cl]  t 100.0 0.0820 1.58 x 10-4 150.0 0.0741time, seconds [CH3CH2CH2CH2Cl] Average Rate, mol L-s [C4 H9Cl]  t 150.0 0.0741 1.74 x 10-4 200.0 0.0671
  • 12. mol Average Rates, L-stime, seconds [CH3CH2CH2CH2Cl] Average Rate, mol L-s [C4 H9Cl]  t 200.0 0.0671 1.22 x 10-4 300.0 0.0549time, seconds [CH3CH2CH2CH2Cl] Average Rate, mol L-s [C4 H9Cl]  t 300.0 0.0549 1.01 x 10-4 400.0 0.0448
  • 13. mol Average Rates, L-stime, seconds [CH3CH2CH2CH2Cl] Average Rate, mol L-s [C4 H9Cl]  t 400.0 0.0448 8.00 x 10-5 500.0 0.0368time, seconds [CH3CH2CH2CH2Cl] Average Rate, mol L-s [C4 H9Cl]  t 500.0 0.0368 5.60 x 10-5 800.0 0.0200
  • 14. 0.12 0.1 0.08Concentration (mol/L) Instantaneous Rate or initial rate at t=0 s 0.06 Instantaneous Rate at t = 500 s 0.04 0.02 0 0 100 200 300 400 500 600 700 800 900 time (seconds)
  • 15. 0.10 M - 0.060 M Instaneous Rateat 0 s = 190 s - 0 s 0.040 M Instaneous Rateat 0 s = = 2.1 x 10-4 M s 190 s 0.042 M - 0.020 MInstaneous Rateat 500 s = 800 s - 400 s 0.022 MInstaneous Rateat 500 s = = 5.5 x 10-5 M s 400 s
  • 16. Order of Reaction• Zero order –independent of the concentration of the reactants, e.g, depends on light• First order - depends on a step in the mechanism that is unimolecular• Pseudo first order reaction – one of the reactants in the rate determining step is the solvent• Second order – depends on a step in the mechanism that is bimolecular• Rarely third order – depends on the step in the mechanism that is termolecular
  • 17. Data from the hydrolysis of n-butyl chloride time (seconds) Concentration n-butyl chloride mol/L 0 0.10 50 9.05 x 10-2 100 8.2 x 10-2 150 7.41 x 10-2 200 6.71 x 10-2 300 5.49 x 10-2 400 4.48 x 10-2 500 3.68 x 10-2 800 2.00 x 10-2
  • 18. IF Zero Ordertime (seconds) [C4H9Cl] 0 0.10 50 9.05 x 10-2 100 8.2 x 10-2 150 7.41 x 10-2 200 6.71 x 10-2 300 5.49 x 10-2 400 4.48 x 10-2 500 3.68 x 10-2 800 2.00 x 10-2
  • 19. 60 50 40[n-butyl chloride] 30 20 10 0 0 100 200 300 400 500 600 700 800 900 time (seconds) Therefore, the reaction is not zero order
  • 20. If Second Ordertime (seconds) 1/[C4H9Cl] 0 10 50 11.0 100 12.2 150 13.5 200 14.9 300 18.2 400 22.3 500 27.2 800 50
  • 21. 60 50 401/[n-butyl chloride] 30 20 10 0 0 100 200 300 400 500 600 700 800 900 time (seconds) Therefore, the reaction is not second order
  • 22. IF First Order Reaction time log [C4H9Cl] ln[C4H9Cl](seconds) 0 -1 -2.3 50 -1.04 -2.4 100 -1.09 -2.51 150 -1.13 -2.60 200 -1.17 -2.69 300 -1.26 -2.90 400 -1.35 -3.11 500 -1.43 -3.29 800 -1.7 -3.92
  • 23. First Order Plot 0 0 100 200 300 400 500 600 700 800 900 -0.2 -0.4 -0.6log [n-butyl chloride] -0.8 -1 -1.2 y = -0.0009x - 0.9985 -1.4 -1.6 -1.8 time (seconds)
  • 24. First Order Plot 0 0 100 200 300 400 500 600 700 800 900 -0.5 -1 -1.5ln[C4H9Cl] -2 -2.5 -3 y = -0.002x - 2.2987 -3.5 -4 -4.5 time (seconds)
  • 25. Slope k slope = - 2.303 -2.303 x slope = k-2.303 x - 9.0 x 10-4 = k 2.1 x 10-3 = k
  • 26. Slope slope = - k 3 1- (- 2 x 10 s ) = k2 x 103 s 1 = k
  • 27. Rate of the ReactionRate = k [n-butylchloride]
  • 28. For the ReactionN2O5 (g) → 2 NO2 (g) + ½ O2 (g) Rate = k [N2O5 ]The rate can be used to explain the mechanism
  • 29. (1) Slow Step .. .. .. .. .. :O O: :O .. + O .. + slow .+ .O .. + O: N N N + N :O : :O : :O : .. .. .. :O : - - .. - - NO2 NO3 N2O5 (2) Fast Step .. .. .. :O .O + O: .. fast .+ 1/2 O2 N N + :O : :O : .. .. - - NO3 NO2
  • 30. Sum of the two steps: .. .. .. .. :O :O:O + O: .+ + O .. .+ N N N N + + 1/2 O2 :O : :O : :O : :O : .. .. .. .. - - - - NO2 NO2 N2O5 N2O5 → 2 NO2 + ½ O2 or 2 N2O5 → 4 NO2 + O2
  • 31. Application Mechanism of a Chemical Reaction(a) Suggest a possible mechanism for NO2 (g) + CO (g) → NO (g) + CO2 (g) Given that Rate = k [NO2(g) ]2(b) Suggest a possible mechanism for 2 NO2 (g) + F2 (g) → 2 NO2F (g) Given that Rate = k [NO2 (g) ] [F2 (g) ]
  • 32. Factors Affecting the Rate of a Chemical Reaction• The Physical State of Matter• The Concentration of the Reactants• Temperature• Catalyst
  • 33. For A Reaction to Occur• Molecules Must Collide• Molecules must have the Appropriate Orientation• Molecules must have sufficient energy to overcome the energy barrier to the reaction-• Bonds must break and bonds must form
  • 34. A Second Order ReactionH2O2 (aq) + I (aq)  H2O (l) + O2 (g) - - Rate = k [H2O2(aq) ] [I(aq) ]
  • 35. Rate Constant “k”• Must be determined experimentally• Its value allows one to find the reaction rate for a new set of concentrations
  • 36. The following data were collected for the rate of the reactionBetween A and B, A + B → C , at 25oC. Determine the rate lawfor the reaction and calculate k. Experiment [A], moles/L [B], moles/L Initial Rate, M/s 1 0.1000 0.1000 5.500 x 10-6 2 0.2000 0.1000 2.200 x 10-5 3 0.4000 0.1000 8.800 x 10-5 4 0.1000 0.3000 1.650 x 10-5 5 0.1000 0.6000 3.300 x 10-5
  • 37. From Experiments 1 and 2 Rate = k  A  B m n Solution A:(1) 5.5 x 10-6 M/s = k 0.1000 M 0.1000 M m n k 0.2000 M 0.1000 M m n(2) 2.2 x 10-5 M/s = Divide equation (1) into equation (2) k  0.2000 M  0.1000 M  -5 m n 2.2 x 10 M/s = k  0.1000 M   0.1000 M  -6 m n 5.5 x 10 M/s 4   2 m 2=m
  • 38. Solution B: (1) log (5.5 x 10-6 ) = log k + m log 0.1000 + n log 0.1000 (2) log (2.2 x 10-5 ) = log k + m log 0.2000 + n log 0.1000Subtract equation (2) from equation (1)log (5.5 x 10-6 ) -log (2.2 x 10-5 ) = m [log  0.1000 -log  0.2000] -5.3 - (-4.7) = m [-1 - (-0.7)] -0.6 = m [-0.3] -0.6 =m -0.3 2=m
  • 39. From Experiments 4 and 5 Solution A:(1) 1.65 x 10 M/s = k 0.1000 M 0.3000 M -5 m n(2) 3.3 x 10-5 M/s = k 0.1000 M 0.6000 M m n Divide equation (1) into equation (2) k  0.1000 M  0.6000 M  -5 m n 3.3 x 10 M/s = k  0.1000 M   0.3000 M  -6 m n1.65 x 10 M/s2   2 n1=n
  • 40. Solution B: (1) log (1.65 x 10-5 ) = log k + m log 0.1000 + n log 0.3000 (2) log (3.3 x 10-5 ) = log k + m log 0.1000 + n log 0.6000Subtract equation (2) from equation (1)log (1.65 x 10-5 ) -log (3.3 x 10-5 ) = n [log  0.3000 -log  0.6000]-4.78 - (-4.5) = n [-0.5227 - (-0.2218)]-0.3 = n [-0.3]-0.3 =n-0.31=n
  • 41. Rate Constant kRate = k  A  B m nRate = k  A  B 2 1 Rate =k  A   B 2
  • 42. Rate Constant k From Experiment 3 Rate =k  A   B 2 M 8.800 x 10-5 s =k0.4000 M  0.1000 M  2 1-35.500 x 10 2 =k M s L25.500 x 10-3 2 =k mol s
  • 43. Rate Constant k From Experiment 1 Rate =k  A   B 2 M 5.500 x 10-5 s =k0.1000 M  0.1000 M  2 -3 15.500 x 10 2 =k M s L25.500 x 10-3 2 =k mol s
  • 44. Your Understanding of this Process Consider the Data for the Following Reaction: O O _CH3 C + OH CH3 C + CH3OH _ OCH3 O Experiment [CH3CO2CH3] [-OH] Initial Rate, M M M/s 1 0.050 0.050 0.00034 2 0.050 0.100 0.00069 3 0.100 0.100 0.00137 Determine the Rate Law Expression and the value of k consistent With these data.
  • 45. From Experiments 1 and 2 nRate = k [CH3CO2CH3 ] m  - OH   Solution :(1) 3.4 x 10 M/s = k 0.050 M 0.050 M -4 m n(2) 6.9 x 10-4 M/s = k 0.50 M 0.100 M m n Divide equation (1) into equation (2) k  0.050 M  0.100 M  -4 m n 6.9 x 10 M/s = k  0.050 M   0.050 M  -5 m n 3.4 x 10 M/s 2   2 n1=n
  • 46. From Experiments 2 and 3 Solution :(1) 6.9 x 10 M/s = k 0.050 M 0.050 M -4 m n(2) 1.37 x 10-3 M/s = k 0.100 M 0.100 M m n Divide equation (1) into equation (2) k  0.100 M  0.100 M  -3 m n 1.37 x 10 M/s = k  0.050 M   0.100 M  -5 m n 6.9 x 10 M/s 2   2 m 1=m
  • 47. Rate ExpressionRate = k [CH3CO2CH3 ] [ - OH] Rate - =k [CH3CO2CH3 ] [ OH] M 0.00137 s =k [0.100 M] [0.100 M] 1 0.137 =k Ms L 0.137 =k mol s
  • 48. AssignmentDetermine the Rate Law for the following reaction from the given data: 2 NO (g) + O2 (g) → 2 NO2 (g) Experiment [NO (g)] [O2 (g)] Initial Rate, M M M/s 1 0.020 0.010 0.028 2 0.020 0.020 0.057 3 0.020 0.040 0.114 4 0.040 0.020 0.227 5 0.010 0.020 0.014
  • 49. Relationship Between Concentration and Time A  productFirst Order Reaction ao  x xdx = k (a o -x)dt -ln (a o -x) = kt + C dx at x = o and t = o, C = -ln a o = k dta o -x -ln (a o -x) = kt - ln a o dx a o -x = k  dt ln a o - ln (a o -x) = kt aolet s = a o -x ln = kt a o -xds = -dx or ds = k  dt log ao = kt s a o -x 2.303-ln s = kt + C
  • 50. A plot of ao ln versus t a o -x Gives a Straight line The Following Reaction is a First Order Reaction: H H HH C H H C H C C C  H CH H H HPlot the linear graph for concentration versus timeand obtain the rate constant for the reaction.
  • 51. Data for the Transformation of cylcpropane to propene ao X ao –x Ln[ao]/[ ao –x] t M M M seconds0.050 0 0.050 0 00.050 0.0004 0.0496 9.0 x 10-3 6000.050 0.0009 0.0491 0.0180 12000.050 0.0015 0.0485 0.0300 20000.050 0.0022 0.0478 0.045 30000.050 0.0036 0.0464 0.075 50000.050 0.0057 0.0443 0.120 80000.050 0.0070 0.0430 0.150 100000.050 0.0082 0.0418 0.180 12000
  • 52. 0.2 0.18 y = 2E-05x - 4E-17 0.16ln (ao /(ao – x)) 0.14 0.12 0.1 0.08 0.06 0.04 0.02 0 0 2000 4000 6000 8000 10000 12000 14000 -0.02 time (seconds) slope = 2 x 10-5 s-1
  • 53. Data for the Transformation of cylcpropane to propene ao X ao –x Ln [ ao –x] t M M M seconds0.050 0 0.050 -2.996 00.050 0.0004 0.0496 -3.0038 6000.050 0.0009 0.0491 -3.014 12000.050 0.0015 0.0485 -3.026 20000.050 0.0022 0.0478 -3.0407 30000.050 0.0036 0.0464 -3.070 50000.050 0.0057 0.0443 -3.1167 80000.050 0.0070 0.0430 -3.1466 100000.050 0.0082 0.0418 -3.1748 12000
  • 54. -2.98 0 2000 4000 6000 8000 10000 12000 14000 -3 -3.02 -3.04 -3.06 y = -2E-05x - 2.9957Ln (ao – x) -3.08 -3.1 -3.12 -3.14 -3.16 -3.18 -3.2 time (seconds) -slope = -2 x 10-5s-1 Slope = 2 x 10-5 s-1
  • 55. Relationship Between Concentration and Time A  productSecond Order Reaction ao  x xdx = k (a o -x) 2dt 1 dx = kt + C 2 = k dt a o -x(a o -x) dx 1 at x = o and t = o, C = (a o -x)2 = k  dt aolet s = a o -x 1 1 = kt +ds = -dx a o -x ao ds = k  dt 1 1 s 2 - = kt1 a o -x a o = kt + Cs
  • 56. A plot of 1 versus t a o -x Gives a Straight lineThe Following Reaction is a Second Order Reaction: 2 HI (g) → H2 (g) + I2 (g)Plot the linear graph for concentration versus timeand obtain the rate constant for the reaction.
  • 57. Data for the Transformation of hydrogen iodide gas to hydrogen and iodine ao X ao –x 1/[ ao –x] t M M M M-1 Minutes 0.0100 0 0.0100 100 0.00 0.0100 0.0060 0.00400 250 5.00 0.0100 0.0075 0.00250 400 10.0 0.0100 0.0086 0.00143 700 20.0 0.0100 0.0090 0.0010 1000 30.0 0.0100 0.0099 0.00077 1300 40.0 0.0100 0.0094 0.00063 1600 50.0 0.0100 0.0095 0.00053 1900 60.0
  • 58. 2000 1800 1600 1400 y = 30x + 1001/(ao – x) 1200 1000 800 600 400 200 0 0 10 20 30 40 50 60 70 time (minutes) slope = 30. L mol-1 min-1
  • 59. Graphical Method for Determining the Order of a Reaction• First Order Reaction: y = ln (ao – x); x = t; slope = -k; and the intercept is ln ao or y = ln (ao /(ao – x)); x = t; slope = k; and the intercept =0• Second Order Reaction: y = 1/ (ao – x); x = t; slope = k; and the intercept = 1/ ao• Zero Order Reaction: y = x; x = t; slope = k and the intercept = 0 or y = ao – x ; x = t; slope = -k and the intercept = ao
  • 60. Zero Order Reactiondx a o -x t  dx =  dt =kdtdx = k dt ao 0 dx = k  dt a o - (a o -x) = ktx = kt + C - (a o -x) = kt - a oat t = 0 and x =0; C = 0 (a o -x) = - kt + a ox = kt
  • 61. Application of the Graphical Method for Determining the Order of a Reaction N2O5 (g) → 2 NO2 (g) + ½ O2 (g) [ N 2O 5 ] t M minutes 2.08 3.07 1.67 8.77 1.36 14.45 0.72 31.28 Tabulate the data so that each order may be tested
  • 62. Data tabulation to determine which order will give a linear graph [ N2O5 ] t ln[ N2O5 ] 1/[ N2O5 ] M minutes (first order) M-1(zero order) (second order) 2.08 3.07 0.732 0.481 1.67 8.77 0.513 0.599 1.36 14.45 0.307 0.735 0.72 31.28 -0.329 1.390
  • 63. Test for zero order reaction 2.5 2 1.5 [N2O5] 1 0.5 0 0 5 10 15 20 25 30 35 time (minutes) Not linear; therefore, the reaction is not zero order
  • 64. Test for first order reaction 0.8 0.6 y = -0.0376x + 0.8462 0.4 ln[N2O5] 0.2 0 0 5 10 15 20 25 30 35 -0.2 -0.4 time (minutes) Linear; therefore, the reaction is first order
  • 65. Test for second order reaction 1.6 1.4 1.2 1 1/[N2O5] 0.8 0.6 0.4 0.2 0 0 5 10 15 20 25 30 35 time (minutes) Non-linear; therefore, the reaction is not second order
  • 66. Half Life for a First Order Reaction ao ln = kt 1 1 2 x ao 2 ln 2= k t 1 2 0.693 = k t 1 2 0.693 = t1 k 2
  • 67. Half Life for a Second Order Reaction 1 1 - = kt 1 1 ao ao 2 2 2 1 - = kt 1 ao ao 2 1 = kt 1 ao 2 1 = t1 k ao 2
  • 68. Half Life for a Zero Order Reaction 1 a o = - kt 1 + a o 2 2 1 a o - a o = - kt 1 2 2 1  a o = - kt 1 2 2 ao = t1 2k 2
  • 69. Application of Half Life• The rate constant for transforming cyclopropane into propene is 0.054 h-1• Calculate the half-life of cyclopropane.• Calculate the fraction of cyclopropane remaining after 18.0 hours.• Calculate the fraction of cyclopropane remaining after 51.5 hours.
  • 70. Half-Life Fraction of cyclopropane Fraction of cyclopropane Remaining after 18.0 hours Remaining after 51.5 hours ao ao 0.693 ln = kt ln = kt = t1 a o -x a o -x0.054/h 2 a o -x a o -x12.8 h = t 1 ln = - kt ln = - kt 2 ao ao a o -x a o -x = e- kt = e- kt ao ao a o -x a o -x = e- (0.054/h) x 18.0 h = e- (0.054/h) x 51.5 h ao ao a o -x a o -x = e- 0.972 = e- 2.8 ao ao a o -x a o -x = 0.38 = 0.061 ao ao
  • 71. Effect of Temperature on the Reaction RateArrhenius Equation Eact - RT k=Ae Eact ln k = - + ln A RTof the form y = mx + b
  • 72. Use the Following Data to Determine the Eact for 2 N2O (g)  2 N2 (g) + O2 (g) T k Ln k 104(1/T) K M-1/s 1125 11.5900 2.450 8.890 1053 1.6700 0.510 9.50 1001 0.3800 -0.968 9.99 838 0.0010 -6.810 11.9
  • 73. 5 0 0 2 4 6 8 10 12 14 16 18 20 -5 ln k -10 -15 -20 y = -3.0712x + 29.721 -25 -30 104(1/T) E actslope = -3.07 x 104 K = - R-3.07 x 104 K x - R = E act J-3.07 x 104 K x - 8.314 = E act K mol2.55 x 105 J/mol = E act 255 kJ/mol = Eact
  • 74. Effect of a Catalyst on the Rate of a Reaction• Lowers the energy barrier to the reaction via lowering the energy of activation• Homogeneous catalyst- in the same phase as the reacting molecules• Herterogeneous catalyst – in a different phase from the reacting molecules
  • 75. Example of a Homogeneous Catalyst 1 11. H 2O2 (aq) + Br-  (aq) Br2 (aq) + H 2O (l) + O 2 (g) 2 2 1 12. H 2O2 (aq) + Br2 (aq)  Br(aq) + H 2O (l) + O 2 (g) - 2 2
  • 76. PE intermediate reactants products reaction coordinates
  • 77. Example of Heterogeneous Catalyst H H H H C C H H Finely divided metal
  • 78. An interesting problem:The reaction between propionaldehyde and hydrocyanic acidhave been observed by Svirbely and Roth and reported in theJournal of the American Chemical Society. Use this data toascertain the order of the reaction and the value of the rateconstant for this reaction. .. H H O: C C CH3CH2 C + H C N: CH3CH2 C N : + :N C CH2CH3 H :OH .. HO : ..
  • 79. time, minutes [HCN] [CH3CH2CHO] 2.78 0.0990 0.0566 5.33 0.0906 0.0482 8.17 0.0830 0.0406 15.23 0.0706 0.0282 19.80 0.0653 0.0229 0.0424 0.0000 ∞
  • 80. Check to determine first order in HCN -2.35 0 2 4 6 8 10 12 14 16 18 -2.4 -2.45 -2.5 ln([HCN]-x) -2.55 -2.6 -2.65 -2.7 -2.75 time (minutes)
  • 81. Check to determine first order in propionaldehyde 0 0 2 4 6 8 10 12 14 16 18 -0.5 time (minutes) -1 ln ([propionaldehyde] -x) -1.5 -2 -2.5 -3 -3.5 -4
  • 82. So close; therefore, let’s take another approach. Let [HCN] = ao and[propionaldehyde] = boThen, dx = k (a o -x) (b o -x) dt dx = k dt (a o -x) (b o -x) dx  (a o -x) (bo -x) = k  dt Solution: 1 (a o -x) 1 bo ln = kt - ln (a o -bo ) b o -x (a o -b o ) a o
  • 83. 1 (a o -x) 1 0.0566 ln = kt - ln0.0424 M b o -x 0.0424 M 0.0990 -1 (a o -x)23.6 M ln = kt - 23.6 M -1 ln 0.572 bo -x -1 (a o -x)23.6 M ln = kt - 23.6 M -1 (-0.559) bo -x -1 (a o -x)23.6 M ln = kt + 13.2 M -1 bo -x
  • 84. 1 (a o -x) 1 0.0566 ln = kt - ln0.0424 M b o -x 0.0424 M 0.0990 -1 (a o -x)23.6 M ln = kt - 23.6 M -1 ln 0.572 bo -x (a o -x)23.6 M -1 ln = kt - 23.6 M -1 (-0.559) bo -x (a o -x)23.6 M -1 ln = kt + 13.2 M -1 bo -x
  • 85. Let’s construct the data in a different format time, minutes [HCN] - x [CH3CH2CHO]-x ([HCN]-x) (23.6) ln ([CH3CH2 CHO]-x) 2.55 0.0906 0.0482 14.9 5.39 0.0830 0.0406 16.9 12.76 0.0706 0.0282 21.7 17.02 0.0653 0.0229 24.8
  • 86. 30 25 20 ([HCN] - x) y = 0.6778x + 13.18423.6 ln ([CH3CH 2CHO]-x) 15 10 5 0 0 2 4 6 8 10 12 14 16 18 time, minutes Slope = 0.678; therefore, k = 0.678 M-1min-1
  • 87. Rate = k [HCN] [CH3CH2CHO] Mechanism: k1 + -1. HCN + H2O H3O + CN k -1 + O OH k22. + + H2O CH3CH2 C + H3O CH3CH2 C k-2 H H + H3. OH k3 - C CH3CH2 C + CN CH3CH2 CN H HO
  • 88. Steps 1 and 2 are fast equilibrium steps; and step 3 is the rate determining step + OH Rate = k3 CH3CH2 [ CN- ] C Hk 1 [HCN] [H2O] = k -1 [H3O+] [CN-]k 1 [HCN] [H2O] = [CN-]k -1 [H3O+]
  • 89. O CH3CH2 C + [H3O ]k +2 H OH = CH3CH2 C k [H2O] H -2 O CH3CH2 C + k3 k1 k2 [HCN] [H2O] H [H3O ]Rate = + k -1 k [H3O ] [H2O] -2
  • 90. ORate = k [HCN] CH3CH2 C H
  • 91. Revisit the kinetics for2 NO (g) + O2 (g) → 2 NO2 (g) Rate = k [NO]2 [O2 ]