2. Kinetics Ch 15

3. Kinetics Thermodynamics and kinetics are not directly related Investigate the rest of the reaction coordinate Rate is important!

4. Chemical Kinetics Kinetics – the study of the rates of chemical reactions Rate of reaction – change in concentration per unit time rate = Δ conc / Δ time Rate is generally not constant. It changes over the course of a reaction A  B

5. 101824283133 What is happening to the rate of the reaction as time progresses? Why?

6. Rate = Δ[B]/Δt = -Δ[A]/Δt Rate = Δ[product]/Δt = -Δ[reactant]/Δt A  B Rate of Reaction

7. 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) Rate = Δ[O 2 ]/ Δt Stoichiometry important! Rate = Δ[NO 2 ]/ 4Δt = - Δ[N 2 O 5 ]/ 2Δt 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) Defining Rate Rate is defined arbitrarily by one pdt or rxt To be self consistent, Example:

8. Another Example Data Calculated Rates

9. Collect concentration data for reactants and products, then graph Effect of stoichiometry Average rate Instantaneous rate

10. Rate Law Study rates to understand mechanism of reaction True rate depends on forward and reverse reactions (remember equilibrium?) But we can write rate law based on reactants –Many reactions functionally irreversible –Use initial rates (reverse rate is negligible)

11. Rate Laws Two forms of rate law Differential Rate Law (Rate Law) –How rate depends on concentration of reactants –Experiment: Initial Rates of multiple trials Integrated Rate Law –How concentrations of species depend on time –Experiment: One trial sampled at multiple times

12. Relationship Between Rate and Concentration 2 NO 2 (g) + F 2 (g)  2 NO 2 F (g) Rate = Δ[NO 2 F]/ 2Δt = -Δ[F 2 ]/ Δt = -Δ[NO 2 ]/ 2Δt Rate α [NO 2 ] and [F 2 ] Rate = k [NO 2 ] x [F 2 ] y k = rate constant x and y are the orders of reaction, these are determined experimentally – not from stoichiometry!

13. 2 NO 2 (g) + F 2 (g)  2 NO 2 F (g) Rate = k [NO 2 ] x [F 2 ] y From experiment, x = 1, y = 1 =Rate = k [NO 2 ] [F 2 ] = Rate Law 1 st order in NO 2, 1 st order in F 2, 2 nd order overall One way to determine the rate law is from initial rates.

14. H 2 O 2 (aq) + 3 I - (aq) + 2 H + (aq)  I 3 - (aq) + 2 H 2 O (l) Expt # [H 2 O 2 ][I - ][H + ]Initial rate M/s 10.010 0.00051.15 x 10 -6 20.0200.0100.00052.30 x 10 -6 30.0100.0200.00052.30 x 10 -6 40.010 0.0011.15 x 10 -6 Rate = k [H 2 O 2 ] x [I - ] y [H + ] z

15. H 2 O 2 (aq) + 3 I - (aq) + 2 H + (aq)  I 3 - (aq) + 2 H 2 O (l) Expt # [H 2 O 2 ] Rel [I - ] Rel [H + ] Rel Initial rate M/s Rel 10.0101 10.000511.15 x 10 -6 1 20.02020.01010.000512.30 x 10 -6 2 30.01010.02020.000512.30 x 10 -6 2 40.0101 10.00121.15 x 10 -6 1 Rate = k [H 2 O 2 ] x [I - ] y [H + ] z Rate = k [H 2 O 2 ] [I - ]

16. Example Problem 2 NO 2 (g)  2 NO (g) + O 2 (g) Expt #[NO 2 ]Rate, M/s 10.0107.1 x 10 -5 20.0202.8 x 10 -4

17. Relationship Between Concentration and Time We want to use a single experiment to determine the rate law. We will do this by plotting concentration versus time. We will deal with simplest cases initially— only one reactant, generally “A”

18. Zero Order Reaction How can a reaction rate be concentration independent?

19. Zero Order Reactions Zero order reaction A  B – Rate = k[A] 0 = k – Rate = -d[A]/dt – k = -d[A]/dt –Rearrange and integrate from time = 0 to time = t – [A] t – [A] o = -kt – [A] t = -kt + [A] o

20. Graphing Zero Order – [A] t = -kt + [A] o – y = mx + b –Plot of conc. vs. time gives straight line with slope of -k –Units of k are M/s

21. First Order Reactions Plot of conc vs time does not give straight line (not 0 order) Rate changes over time: Doubling concentration of A doubles the rate A  products Rate = k[A] Rate = -d[A]/ dt k[A] = -d[A]/ dt UTMOC ln[A]/[A] 0 = - kt ln[A] = -kt + ln[A] 0

22. Example Problem The decomposition of N 2 O 5 to NO 2 and O 2 is first order with k = 4.80 x 10 -4 s -1 at 45 o C. –If the initial concentration is 1.65 x 10 -2 M, what is the concentration after 825 sec? –How long would it take for the concentration of N 2 O 5 to decrease to 1.00 x 10 -2 M?

23. Graphing First Order Plot of ln[A] vs t gives straight line with a slope of -k and a y-intercept of ln[A] 0 Units of k = s -1 2 N 2 O 5  4 NO 2 + O 2

24. First Order Reactions Determine Order of Reaction by plotting data!

25. T (sec)[N 2 O 5 ] (M) 02.15 x 10 -3 40001.88 x 10 -3 80001.64 x 10 -3 120001.43 x 10 -3 160001.25 x 10 -3 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) 0.27 0.24 0.18

26. NOT Zero order!

27. T (sec)[N 2 O 5 ] (M)ln [N2O 5 ] 00.00215-6.14 40000.00188-6.28 80000.00164-6.41 120000.00143-6.55 160000.00125-6.68 0.14 0.13 0.14 0.13

28. This is First Order

29. Second Order Reactions Plot of [conc] vs t and plot of ln[conc] vs t do not give straight lines. (not 0 or 1 st ) A  products Rate = k[A] 2 Rate = -Δ[A]/ Δt k[A] 2 = -Δ[A]/ Δt UTMOC 1/[A] = kt + 1/[A] 0

30. Graphing Second Order Plot of 1/[A] vs t gives straight line, with a slope of k and a y-intercept of 1/[A] 0

31. Time, sec[HI], M 01.000 10000.112 20000.061 30000.041 40000.031 2 HI (g)  H 2 (g) + I 2 (g) @ 580K

32. Not Zero Order

33. Time, sec[HI], Mln [HI] 01.0000 10000.112-2.19 20000.061-2.80 30000.041-3.19 40000.031-3.47

34. Not First Order

35. Time, sec[HI], Mln [HI]1/[HI] 01.00001 10000.112-2.198.93 20000.061-2.8016.4 30000.041-3.1924.4 40000.031-3.4732.3

36. Second Order Rate = k[HI] 2

37. Reactions Involving Gases A (g)  products PV = n A RT [A] = n A / V = P/RT ln[A]/[A] 0 = -kt ln(P/RT) /(P 0 /RT) = -kt ln P/P 0 = -kt Can use the pressures of gases for the concentrations.

38. Half Life k = describes speed of the reaction –Large k = fast reaction Another way to describe speed is to use t ½, the half life. This is the time needed to decrease to ½ [A] 0. For a first order reaction, t = (1/k)ln[A] 0 /[A] t ½ = (1/k) ln[A] 0 /([A] 0 /2) t ½ = (1/k) ln2 = 0.693/k

40. Comparison of Half-Lives Use same procedure to derive each half-life For zero order, each half life is half as long as previous one For first order, each half-life is the same For second order, each half life is twice as long as the previous one OrderZeroFirstSecond t 1/2 [A] o / 2k0.693 / k 1/ k[A] o

42. Application Question Kinetic data were plotted for A  2B + C What can these data tell you about this reaction?

43. More Complicated System So far we have assumed one reactant How do we study A + 2B  C + 3D Run experiment with B in huge excess –Rate = k [A] x [B] y but [B] remains constant –Called “psuedo” kinetics—can be used to determine order of A Repeat with excess A to get pseudo-order of B Combine experiments to get real rate law

44. Reaction Mechanisms A reaction may be more complex that 1 simple collision – may form intermediates. It is unlikely that 3 or more molecules will collide simultaneously. Elementary stepsElementary steps – describe a molecular event.

45. NO 2 + CO  NO + CO 2 2 elementary steps NO 2 + NO 2 + NO 3 + CO  NO 3 + NO + NO 2 + CO 2 NO 2 + CO  NO + CO 2 NO 3 is an intermediate

46. Cl 2  2 Cl Cl + CHCl 3  HCl + CCl 3 Cl + CCl 3  CCl 4 Cl 2 + CHCl 3  HCl + CCl 4 (overall) Rate laws for elementary steps can be written from stoichiometry. (unlike overall) 1.Rate = k 1 [Cl 2 ]unimolecular 2.Rate = k 2 [Cl][CHCl 3 ]bimolecular 3.Rate = k 3 [Cl][CCl 3 ]bimolecular

47. Rate Determining Step Rate Determining Step (rate limiting step) – is the slowest step leading to the formation of the products (slow step). The rates of any steps after the slow step are not important.

48. O 3 + 2 NO 2  O 2 + N 2 O 5 1. O 3 + NO 2  NO 3 + O 2 slow 2.NO 3 + NO 2  N 2 O 5 fast Rate = k[O 3 ][NO 2 ] This explains why stoichiometry and rate law are independent

49. 2 NO 2 + O 3  N 2 O 5 + O 2 Rate = k[NO 2 ][O 3 ] 2 NO + Cl 2  2 NOCl Rate = k[NO] 2 [Cl 2 ]

50. Temperature Changes Rate 2NO (g) + Cl 2 (g)  2NOCl (g) Rate = k[NO] 2 [Cl 2 ] k @ 25 o C = 4.9 x 10 -6 M -1 s -1 k @ 35 o C = 1.5 x 10 -5 M -1 s -1 This is more than 3x increase! Why is there a temperature dependence on k? Can k and temperature be related theoretically and quantitatively?

51. Dependence of Rate Constant on Temperature Exponential of rate constant on absolute temperature Every curve different This one represents double rate every 10 K

52. Collision Theory Collision Theory – molecules must collide in order to react. Rate α # collisions / sec α [reactants] From KMT Increase temp, increase speed Accounts for higher rate Kinetic energy made into potential energy to break bonds

53. Activation Energy Arrhenius – expanded collision theory (1888) Molecules must collide with enough energy to rearrange bonds. If not, they just bounce off. Activation Energy = E a = minimum amount of energy required to initiate a chemical reaction. Activated Complex (transition state) – temporary species in reaction sequence, least stable, highest energy, often undetectable.

54. Activation Energy and Transition States

55. Molecules with Enough KE to Overcome Activation Energy

56. Steric Factor A is pre-exponential factor, or frequency factor, for the Arrhenius equation

57. Orientation of Molecules

58. Arrhenius Equation k = Ae -Ea/RT ln k = ln Ae -Ea/RT y = m x + b If you experimentally determine ___ and ____, then you can graphically determine ____ and ____. Can also be used to determine k at any ____.

59. m = -E a /R b = ln A ln k vs. 1/T

60. Useful Form of Arrhenius Equation In principle, best to do many experiments, graph line, and determine E a In practice, can get decent value from only two experiments Rearrange Arrhenius equation to get

61. Example Problem H 2 + I 2  2 HI k = 2.7 x 10 -4 M -1 s -1 @ 600 K k = 3.5 x 10 -3 M -1 s -1 @ 650 K a. Find E a b. Calculate k @ 700 K

62. Catalysis To increase the rate of a reaction –1. Increase temperature –2. Add a catalyst Catalyst – a substance which increases the rate of a reaction but is not consumed. Catalysts are involved in the course of a reaction. Usually by lowering E a Homogeneous – catalyst the same phase as the reactants Heterogeneous – catalyst in different phase (usually a solid)

64. Ammonia Formation with a Catalyst N 2 + 3 H 2  2 NH 3

65. Catalytic Hydrogenation Food processing Trans fatty acids