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How fast can a reaction go

Chemical Kinetics. How fast can a reaction go. Four things that affect rate a. concentration b. temperature c. catalyst d. surface area. Average rate. Instantaneous rate. Rate Laws. NH 4 + + NO 2 -  N 2 + 2H 2 O R = k [NH 4 + ] m [NO 2 - ] n.

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How fast can a reaction go

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  1. Chemical Kinetics How fast can a reaction go Four things that affect rate a. concentration b. temperature c. catalyst d. surface area

  2. Average rate Instantaneous rate

  3. Rate Laws NH4+ + NO2- N2 + 2H2O R = k [NH4+]m [NO2-]n • k = rate proportionality constant • m = reaction order for the • ammonium ion • n = reaction order for the nitrite • ion

  4. Rate N2 + 3H2→ 2NH3 R or rate = -Δ[N2] = M/s Δt How do the three chemicals compare?

  5. NH4+ + NO2- N2 + 2H2O indep. var control Dep. var

  6. R = k [NH4+]m [NO2]n indep. var control Dep. var Skeleton rate law 2x

  7. R = k [NH4+]1 [NO2]1

  8. 10.8 x 10-7M/s = k [0.02M]1[0.200M]1

  9. 10.8 x 10-7M/s= k[0.02M]1 [0.200M]1 k= 10.8 x 10-7 M/s /[0.02M]1[0.200M]1 k = 10.8 x 10-7 M/s /0.004M2 k = 2.7 x 10-4 1/M • s M = ? M1 M1 s ?

  10. Suppose the concentrations for NH4+ and NO2- are 0.15 M and 0.20 M, respectively Since R = k[NH4+]1 [NO2]1 where k = 2.7 x 10-4 1/M - s at 25 C then R = 2.7 x 10-4 1/M - s [0.15 M] [0.20 M] R = 8.1 x 10-6 M/s

  11. 2NO(g) + Br2(g)  2NOBr Initial [NO] Initial [Br2] Rate of appearance NOBr (M/L) (M/L) (M/s) 1 0.0160 0.0120 3.24 x 10-4 2 0.0160 0.0240 6.38 x 10-4 3 0.0320 0.0060 6.42 x 10-4 R = k[NO]m [Br2]n 3.24 x 10-4 k(0.0160)x (0.0120)1 = k(0.0320)x (0.0060)1 6.42 x 10-4 1 1 (0.0160)x 2 .5x = x = 2 = 4 2 (0.0320)x R = k[NO]2 [Br2]1

  12. H3C N C: H3C C N: CH3NC1 Time (sec) 1mm Hg Using the calculator to find order 0 900 2,500 5,000 10,000 15,000 20,000 30,000 150 140 130 118 90 70 55 35

  13. H3C N C: H3C C N:

  14. H3C N C: H3C C N: CH3NC1 Ln CH3NC2 150 140 130 118 90 70 55 35 5.02 4.94 4.86 4.77 4.49 4.24 4.00 3.55 1mm Hg 2 r = 0.98 ln [A]t = - k t + ln [A]0 y = m x + b

  15. H3C N C: H3C C N: 1 1 = kt + [A]t [A]0 CH3NC1 1/ CH3NC3 150 140 130 118 90 70 55 35 .0067 .0071 .0076 .0084 .0111 .0142 .0182 .0286 y = mx + b 3 r = 0.92 1mm Hg

  16. H3C N C: H3C C N: CH3NC1 ln CH3NC2 1/ CH3NC3 Time (sec) 5.02 4.94 4.86 4.77 4.49 4.24 4.00 3.55 150 140 130 118 90 70 55 35 0 900 2,500 5,000 10,000 15,000 20,000 30,000 .0067 .0071 .0076 .0084 .0111 .0142 .0182 .0286

  17. zero orderRate = k[A]o • plot concentration [A] vs. time (t) – linear, negative slope • integrated rate law - [A]t = -kt + [A]0 • half life – t1/2 = [A]0/2k [A] • 1st order Rate = k[A]1 t • plot of ln conc. vs. time is linear, negative slope • integrated rate law - ln[A]t = -kt + ln[A]0 • half life – t1/2 = .693/k ln[A] • 2nd order Rate = k[A]2 t • plot inverse of conc. vs. time is linear, positive slope • integrated rate law - 1/[A]t = kt + 1/[A]0 • half life – t1/2 = 1/k[A]0 1/[A] • t

  18. Using Rate Law Equations to Determine Change in Concentration Over Time The first order rate constant for the decomposition of a certain insecticide in water at 12 C° is 1.45 yr1-. A quantity of insecticide is washed into a lake in June leading to a concentration of 5.0 x 10-7 g/cm3. Assume that the effective temperature of the lake is 12 C°. (a)What is the concentration of the insecticide in June of the following year? (b) How long will it take for the concentration of the insecticide to drop to 3.0 x 10-7 g/cm3? R = - [A] = k[A] ln [A]t = - k t + ln [A]0 t ln [A]t = - (1.45 yr.-1)(1.00 yr.) + ln( 5.0 x 10-7 g/cm3) ln [A]t = - 15.96, [A]t = e -15.96 [A]t = 1.2 x 10-7 g/cm3

  19. Using Rate Law Equations to Determine the Half-Life of a Reaction [A]t = - kt Let’s begin with ln[A]t - ln[A]0 = ln [A]0 [A]t given ln let [A]t = ½ [A]0 [A]0 ½ [A]0 and ln ½ = -kt ½ = - kt, ln then [A]0 t ½ = -ln ½ = 0.693 = - kt, k k

  20. Using Rate Law Equations to Determine the Half-Life of a Reaction The concentration of an insecticide accidentally spilled into a lake was measured at 1.2 x 10-7 g/cm3. Records from the initial accident show that the concentration of the insecticide was 5.0 x 10-7 g/cm3. Calculate the 1/2 life of the insecticide. 1.2 x 10-7 g/cm3 ln = - k (1.00 yr.) 5.0 x 10-7 g/cm3 k = 1.43 yr-1 t ½ = 0.693 = 0.484 yr. 1.43 yr-1

  21. Examining the Relationship Between the Rate Constant and Temperature

  22. Why does Temperature Have an Effect on Rate? ..it’s because of CollisionTheory

  23. It’s All About Activation Energy and Getting over the HUMP!!

  24. 1010 collisions/sec occur 1 in 1010 collisions succeeds

  25. ...And What Do the Mathematicians Have to Say About Collision Theory? k = Ae-Ea/RT

  26. Rewriting the Arrhenius Equation k = Ae-Ea/RT ln k1 = -Ea/RT1 + ln Ae ln k2 = -Ea/RT2 + ln Ae lnk1 - ln k2 = (-Ea/RT1 + ln Ae) - (Ea/RT2 + ln Ae) k1 Ea 1 1 ln - = k2 R T1 T2

  27. Getting a Line on the Arrhenius Equation ln k1 = -Ea/RT1 + ln Ae R (ideal gas constant) 8.31 J/K-mol y = m x + b Slope = -Ea/R ln k 1/T

  28. Mastering the Arrhenius Equation with your TI ln k1 = -Ea/RT1 + ln Ae Slope = -Ea/R* ln k (L5) y = m x + b 1/T (L3) L1 L2 L3 L4 L5 T(°C) L1 + 273 1/L2 k ln L4

  29. Mastering the Arrhenius Equation with your TI The following table shows the rate constants for the rearrangement of CH3CN (methyl isonitrile) at various temperatures Temperature (°C) k (s-1) 189.7 198.9 230.3 251.2 2.52 x 10-5 5.25 x 10-5 6.30 x 10-4 3.16 x 10-3 From these data calculate the activation energy of this reaction

  30. The following table shows the rate constants for the following reaction at varying temperatures: CO(g) + NO2(g)  CO2(g) + NO(g) Temperature (°C) k (M-1 s-1) 600 650 700 750 800 0.028 0.22 1.3 6 23 From these data calculate the activation energy for this reaction

  31. Reaction Mechanisms A Reaction Mechanism is the process which describes in great detail the order in which bonds are broken and reformed, changes in orientation and the energies involved during those rebondings, and changes in orientations NO(g) + O3(g)  NO2(g) + O2(g) A single reaction event is called an elementary reaction NO2(g) + CO(g)  NO(g) + CO2(g) While this reaction looks like an elementary reaction, it actually takes place in a series of steps NO2(g) +NO2(g) NO3(g) + NO(g) NO3(g) + CO(g)  NO2(g) + CO2(g) NO2(g) + CO(g)  NO(g) + CO2(g)

  32. Reaction Mechanisms Rate Determining step NO2(g) +NO2(g) NO3(g) + NO(g) Slow step NO3(g) + CO(g)  NO2(g) + CO2(g) Fast step NO2(g) + CO(g)  NO(g) + CO2(g) The rate law is constructed from the rate determining step Rate = k1[NO2]2

  33. Reaction Mechanisms: Catalysts NO + O3→ NO2 + O2 O + NO2 → NO + O2 O + O3 → 2O2

  34. Fast step NO(g) +Br2(g) NOBr2(g) Rate Determining step NOBr2(g) + NO(g)  2NOBr(g) Slow step 2NO(g) + Br2(g)  2NOBr(g) If step 2 is the rate determining step, then R = k[NOBr2][Br2]. Keep in mind, however, that one cannot include an intermediate in the rate determining step. We can substitute for NOBr2 using the previous equation NO(g) and Br2(g) which are not intermediates rate = k [NO]2[Br2]

  35. Prove that the following mechanism is consistent with R = k[NO]2[Br2], the rate law which was derived experimentally: NO(g) +NO(g)  N2O2(g) N2O2(g) + Br2(g)  2NOBr(g) 2NO(g) + Br2(g)  2NOBr(g)

  36. Catalyst: a substance that changes the speed of a chemical reaction without it self undergoing a permanent chemical change in the process or it is reconstituted at the end. Br - is a homogeneous catalyst

  37. Heterogeneous Catalysts Ethylene Ethane This metallic substrate is a heterogeneous catalyst

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