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Chapter 14

Chapter 14. Chemical Kinetics. Overview:. Reaction Rates Stoichiometry, Conditions, Concentration Rate Equations Order Initial Rate Concentration vs. Time First Order Rxns. Second Order Rxns. Graphical Methods. Cont’d. Molecular Theory Activation Energy Concentration

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Chapter 14

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  1. Chapter 14 Chemical Kinetics

  2. Overview: • Reaction Rates • Stoichiometry, Conditions, Concentration • Rate Equations • Order • Initial Rate • Concentration vs. Time • First Order Rxns. • Second Order Rxns. • Graphical Methods

  3. Cont’d • Molecular Theory • Activation Energy • Concentration • Molecular Orientation • Temperature • Arrhenius Equation • Reaction Mechanisms • Elementary Steps, Reaction Order, Intermediates • Catalysts

  4. Reaction Rates • What Affects Rates of Reactions? • Concentration of the Reactants • Temperature of Reaction • Presence of a Catalyst • Surface Area of Solid or Liquid Reactants

  5. Reaction Rates (graphical): • Average Rate = D[M]Dt [M] for reaction A  B D[M] time Dt

  6. Rates • for A  B- D[A] = D[B] Dt DtRate of the disappearance of A is equal in magnitude but opposite in sign to the rate of the appearance of B

  7. Average Rate--D mol (or concentration) over a period of time, Dt • Instantaneous Rate-- slope of the tangent at a specific time, t • Initial Rate-- instantaneous rate at t = 0 [M] tangent at time, t t time

  8. Average Rate = [A]final time - [A]initial timeDtfinal - Dtinitial for A  B

  9. Instantaneous Ratetime, t = slope of the tangent at time = t [M] tangent at time, t t time

  10. Stoichiometry 4PH3 => P4 + 6H2 - 1D[PH3] = + 1D[P4] = + 1D[H2] 4Dt 1Dt 6Dt - D[PH3] = + 4D[P4] = + 2D[H2] Dt Dt 3Dt

  11. General Relationship Rate = - 1 D[A] = - 1 D[B] = + 1 D[C] = + 1 D[D] aD t b D t cD t dD taA + bB  cC + dD

  12. Conditions which affect rates • Concentration • concentration  rate • Temperature • temperature  rate • Catalyst • substance which increases rate but itself remains unchanged

  13. Rate Equations: • aA + bB  xX rate lawrate = k[A]m[B]n • m, n are orders of the reactants • extent to which rate depends on concentration • m + n = overall rxn order • k is the rate constant for the reaction

  14. Examples: • 2N2O5 => 4NO2 + O2 rate = k[N2O5] 1st order • 2NO + Cl2 => 2NOClrate = k[NO]2[Cl2] 3rd order • 2NH3 => N2 + 3H2rate = k[NH3]0 = k0th order

  15. Determination of Rate Equations: Data for: A + B => C Expt. # [A] [B] initial rate 1 0.10 0.10 4.02 0.10 0.20 4.03 0.20 0.10 16.0 rate = k[A]2[B]0 = k[A]2 k = 4.0 Ms-1 = 400 M-1s-1 (0.10)2 M2

  16. Exponent Values Relative to DRate Exponent Value [conc] rate 0 double same 1 double double 2 double x 4 3 double x 8 4 double x 16

  17. Problem: Data for: 2NO + H2 => N2O + H2O Expt. # [NO] [H2] rate 1 6.4x10-3 2.2x10-3 2.6x10-52 12.8x10-3 2.2x10-3 1.0x10-43 6.4x10-3 4.5x10-3 5.0x10-5 rate = k[NO]2[H2]

  18. Units of Rate Constants • units of rates M/s • units of rate constants will vary depending on order of rxn M = 1 (M)2 (M) for s sM2 rate = k [A]2 [B] • rate constants are independent of the concentration

  19. Concentration vs. Time1st and 2nd order integrated rate equations • First Order: rate = - D[A] = k [A]D t • ln [A]t = - kt A = reactant [A]0 • or ln [A]t - ln [A]0 = - kt

  20. Conversion to base-10 logarithms: ln [A]t = - kt [A]0 tolog [A]t = - kt [A]0 2.303

  21. Problem: The rate equation for the reaction of sucrose in water is, rate = k[C12H22O11]. After 2.57 h at 27°C, 5.00 g/L of sucrose has decreased to 4.50 g/L. Find k. C12H22O11(aq) + H2O(l) => 2C6H12O6 ln 4.50g/L = - k (2.57 h) 5.00g/L k = 0.0410 h-1

  22. Concentration vs. Time • Second Order:rate = - D[A] = k[A]2Dt • 1 - 1 = kt [A]t [A]0 • second order rxn with one reactant: rate = k [A]2

  23. Problem: Ammonium cyanate, NH4NCO, rearranges in water to give urea, (NH2)2CO. If the original concentration of NH4NCO is 0.458 mol/L and k = 0.0113 L/mol min, how much time elapses before the concentration is reduced to 0.300 mol/L?NH4NCO(aq) => (NH2)2CO(aq) rate = k[NH4NCO] 1 - 1 = (0.0113) t (0.300) (0.458) t = 102 min

  24. Graphical Methods • Equation for a Straight Line • y = bx + a • ln[A]t = - kt + ln[A]01st order • 1 = kt + 1 2nd order[A]t [A]0 b = slopea = y interceptx = time

  25. First Order: 2H2O2(aq)® 2H2O(l) + O2(g) [H2O2] time

  26. First Order:2H2O2(aq) ® 2H2O(l) + O2(g) slope, b = -1.06 x 10-3 min-1 = - k ln [H2O2] time

  27. Second Order: 2NO2 ® 2NO + O2 1/[NO2] slope, b = +k time

  28. Half-Life of a 1st order process: 0.020 M t1/2 = 0.693k [M] 0.010 M 0.005 M t1/2 t1/2 time

  29. Problem: The decomposition of SO2Cl2 is first order in SO2Cl2 and has a half-life of 4.1 hr. If you begin with 1.6 x 10-3 mol of SO2Cl2 in a flask, how many hours elapse before the quantity of SO2Cl2 has decreased to 2.00 x 10-4 mol? SO2Cl2(g) => SO2(g) + Cl2(g)

  30. Temperature Effects • Rates typically increase with T increase • Collisions between molecules increase • Energy of collisions increase • Even though only a small fraction ofcollisions lead to reaction • Minimum Energy necessary for reactionis the Activation Energy

  31. Molecular Theory(Collision Theory) Activation Energy, Ea DH reactionEa forward rxn.Ea reverse rxn. Energy Reactant Product Reaction Progress

  32. Activation Energy • Activation Energy varies greatly • almost zero to hundreds of kJ • size of Ea affects reaction rates • Concentration • more molecules, more collisions • Molecular Orientation • collisions must occur “sterically”

  33. The Arrhenius Equation • increase temperature, inc. reaction rates • rxn rates are a to energy, collisions, temp. & orient • k = Ae-Ea/RT k = rxn rate constant A = frequency of collisions -Ea/RT = fraction of molecules with energy necessary for reaction

  34. Graphical Determination of Ea rearrange eqtn to give straight-line eqtn y = bx + a ln k = -Ea1 + ln A RT slope = -Ea/R ln k 1/T

  35. Problem: Data for the following rxn are listed in the table. Calculate Ea graphically, calculate A and find k at 311 K. Mn(CO)5(CH3CN)+ + NC5H5 => Mn(CO)5(NC5H5)+ + CH3CN ln k k, min-1 T (K) 1/T x 10-3 -3.20 0.0409 298 3.35 -2.50 0.0818 308 3.25 -1.85 0.157 318 3.14

  36. slope = -6373 = -Ea/REa = (-6373)(-8.31 x 10-3 kJ/K mol) = 53.0 kJ y intercept = 18.19 = ln A A = 8.0 x 10 7 -3.20 -2.50 -1.85 ln k k = 0.0985 min-1 3.14 3.25 3.35 x 10-3 1/T

  37. Problem: The energy of activation for C4H8(g) => 2C2H4(g) is 260 kJ/mol at 800 K and k = 0.0315 sec Find k at 850 K. ln k2 = - Ea (1/T2 - 1/T1) k1 R k at 850 K = 0.314 sec-1

  38. Reaction Mechanisms • Elementary Step • equation describing a single molecular event • Molecularity • unimolecular • bimolecular • termolecular • 2O3 => 3O2 (1) O3 => O2 + O unimolecular(2) O3 + O => 2 O2 bimolecular

  39. Rate Equations • Molecularity Rate Law unimolecular rate = k[A] bimolecular rate = k[A][B] bimolecular rate = k[A]2 termolecular rate = k[A]2[B] • notice that molecularity for an elementary step is the same as the order

  40. 2O3 => 3O2 O3 => O2 + O rate = k[O3] O3 + O => 2O2 rate = k’[O3][O] 2O3 + O => 3O2 + O O is an intermediate

  41. Problem: • Write the rate equation and give the molecularity of the following elementary steps: NO(g) + NO3(g) => 2NO2(g) rate = k[NO][NO3] bimolecular (CH3)3CBr(aq) => (CH3)3C+(aq)+ Br-(aq) rate = k[(CH3)3CBr] unimolecular

  42. Mechanisms and Rate Equations rate determining step is the slow step --the overall rate is limited by the ratedetermining step step 1 NO2 + F2 => FNO2 + F rate = k1[NO2][F2] k1 slow step 2 NO2 + F => FNO2 rate = k2[NO2][F] k2 fast overall 2NO2 + F2 => 2FNO2 rate = k1[NO2][F2]

  43. Problem: • Given the following reaction and rate law:NO2(g) + CO(g) => CO2(g) + NO(g)rate = k[NO2]2 • Does the reaction occur in a single step? • Given the two mechanisms, which is most likely:NO2 + NO2 =>NO3 + NO NO2 => NO + ONO3 + CO => NO2 + CO2 CO + O => CO2

  44. k1 k2 Reaction Mechanisms & Equilibria 2O3(g) 3O2(g) overall rxn 1: O3(g) O2(g) + O(g) fastequil. rate1 = k1[O3] rate2 = k2[O2][O] 2: O(g) + O3(g) 2O2(g) slow rate3 = k3[O][O3] rate 3 includes the conc. of an intermediate and the exptl. rate law will include only species that are present in measurable quantities k3

  45. Substitution Method at equilibrium k1[O3] = k2[O2][O] rate3 =k3[O][O3] [O] = k1 [O3] k2 [O2] rate3 = k3k1 [O3]2 or k2 [O2] overall rate = k’ [O3]2 [O2] substitute

  46. k1 k2 Problem: Derive the rate law for the following reaction given the mechanism step below: OCl -(aq) + I -(aq) OI -(aq) + Cl -(aq) OCl - + H2O HOCl + OH - fast I - + HOCl HOI + Cl - slow HOI + OH - H2O + OI - fast k3 k4

  47. Cont’d rate1 = k1 [OCl -][H2O] = rate 2 = k2 [HOCl][OH -] [HOCl] = k1[OCl -][H2O] k2[OH -] rate 3 = k3 [HOCl][I -] rate 3 = k3k1[OCl -][H2O][I -] k2 [OH -] overall rate = k’ [OCl -][I -][OH -] solvent

  48. Catalyst • Facilitates the progress of a reaction bylowering the overall activation energy • homogeneous • heterogeneous

  49. Ea Ea DHrxn Energy Reaction Progress catalysts are used in an early rxn step but regenerated in a later rxn step

  50. Uncatalyzed Reaction: O3(g)<=> O2(g) + O(g) O(g) + O3(g) => 2O2(g) Catalyzed Reaction: Step 1: Cl(g) + O3(g) + O(g) => ClO(g) + O2(g) + O(g) Step 2: ClO(g) + O2(g) +O(g) => Cl(g) + 2O2(g) Overall rxn: O3(g) + O(g) => 2O2(g)

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