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

3. Average rate Instantaneous rate

4. NH 4 + + NO 2 - N 2 + 2H 2 O R = k [NH 4 + ] m [NO 2 - ] n k = rate proportionality constant m = reaction order for the ammonium ion n = reaction order for the nitrite ion Rate Laws

5. Rate N 2 + 3H 2 2NH 3 R or rate = -Δ[N 2 ] = M/s Δt How do the three chemicals compare?

6. NH 4 + + NO 2 - N 2 + 2H 2 O indep. var control Dep. var

7. indep. var controlDep. var R = k [NH 4 + ] m [NO 2 ] n 2x Skeleton rate law

8. R = k [NH 4 + ] 1 [NO 2 ] 1

9. 10.8 x 10 -7 M/s = k [0.02M] 1 [0.200M] 1

10. k= 10.8 x 10 -7 M/s / [0.02M] 1 [0.200M] 1 k = 10.8 x 10 -7 M/s / 0.004M 2 k = 2.7 x 10 -4 1/M s M = ? M 1 M 1 s ?

11. Suppose the concentrations for NH 4 + and NO 2 - are 0.15 M and 0.20 M, respectively Since R = k[NH 4 + ] 1 [NO 2 ] 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

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

13. CH 3 NC 1 Time (sec) 1 mm Hg 150 140 130 118 90 70 55 35 0 900 2,500 5,000 10,000 15,000 20,000 30,000 H 3 C N C: H 3 C C N: Using the calculator to find order

14. H 3 C N C: H 3 C C N:

15. CH 3 NC 1 1 mm Hg 150 140 130 118 90 70 55 35 Ln CH 3 NC 2 5.02 4.94 4.86 4.77 4.49 4.24 4.00 3.55 2 r = 0.98 ln [A] t = - k t + ln [A] 0 y = m x + b H 3 C N C: H 3 C C N:

16. CH 3 NC 1 1 mm Hg 150 140 130 118 90 70 55 35 1/ CH 3 NC 3.0067.0071.0076.0084.0111.0142.0182.0286 3 r = 0.92 y = mx + b 1 = kt + [A] t 1 [A] 0 H 3 C N C: H 3 C C N:

17. CH 3 NC 1 Time (sec) 150 140 130 118 90 70 55 35 0 900 2,500 5,000 10,000 15,000 20,000 30,000 H 3 C N C: H 3 C C N: 1/ CH 3 NC 3. 0067.0071.0076.0084.0111.0142.0182.0286 ln CH 3 NC 2 5.02 4.94 4.86 4.77 4.49 4.24 4.00 3.55

18. §zero order Rate = k[A] o §plot concentration [A] vs. time (t) – linear, negative slope §integrated rate law - [A] t = -kt + [A] 0 §half life – t 1/2 = [A] 0 /2k [A] §1 st 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 – t 1/2 =.693/k ln[A] §2 nd 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 – t 1/2 = 1/k[A] 0 1/[A] § t

19. 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 yr 1-. A quantity of insecticide is washed into a lake in June leading to a concentration of 5.0 x 10 -7 g/cm 3. 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/cm 3 ? R = - [A] = k[A] t ln [A] t = - k t + ln [A] 0 ln [A] t = - (1.45 yr. -1 )(1.00 yr.) + ln( 5.0 x 10 -7 g/cm 3 ) ln [A] t = - 15.96, [A] t = e -15.96 [A] t = 1.2 x 10 -7 g/cm 3

20. Using Rate Law Equations to Determine the Half-Life of a Reaction Lets begin with ln[A] t - ln[A] 0 = ln [A] t [ A] 0 = - kt ln [A] t [A] 0 = - kt, let [A] t = ½ [A] 0 given ln ½ [A] 0 [A] 0 = - kt, then and ln ½ = -kt ½ t ½ = -ln ½ = 0.693 k k

21. 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/cm 3. Records from the initial accident show that the concentration of the insecticide was 5.0 x 10 -7 g/cm 3. Calculate the 1/2 life of the insecticide. ln 1.2 x 10 -7 g/cm 3 5.0 x 10 -7 g/cm 3 = - k (1.00 yr.) k = 1.43 yr -1 t ½ = 0.693 = 0.484 yr. 1.43 yr -1

22. Examining the Relationship Between the Rate Constant and Temperature

23. Why does Temperature Have an Effect on Rate?..its because of Collision Theory

24. Its All About Activation Energy and Getting over the HUMP !!

25. 10 10 collisions/sec occur 1 in 10 10 collisions succeeds

26. ...And What Do the Mathematicians Have to Say About Collision Theory? k = A e -E a /RT

27. ln k 1 = -E a /RT 1 + ln A e ln k 2 = -E a /RT 2 + ln A e lnk 1 - ln k 2 = (-E a /RT 1 + ln A e ) - (E a /RT 2 + ln A e ) ln k1k1 k2k2 = EaEa R 1 T2T2 1 T1T1 - Rewriting the Arrhenius Equation

28. ln k 1 = -E a /RT 1 + ln A e y = m x + b ln k 1/T Slope = -E a /R R (ideal gas constant) 8.31 J/K-mol Getting a Line on the Arrhenius Equation

29. ln k 1 = -E a /RT 1 + ln A e y = m x + b L1 L2 L3 L4 L5 T(°C) L1 + 273 1/L2 k ln L 4 ln k (L 5 ) 1/T (L 3 ) Slope = -E a /R * Mastering the Arrhenius Equation with your TI

30. The following table shows the rate constants for the rearrangement of CH 3 CN (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 Mastering the Arrhenius Equation with your TI

31. The following table shows the rate constants for the following reaction at varying temperatures: CO(g) + NO 2 (g) CO 2 (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

32. 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) + O 3 (g) NO 2 (g) + O 2 (g) elementary reaction A single reaction event is called an elementary reaction NO 2 (g) + CO(g) NO(g) + CO 2 (g) elementary reaction, While this reaction looks like an elementary reaction, it actually takes place in a series of steps NO 2 (g) +NO 2 (g) NO 3 (g) + NO(g) NO 3 (g) + CO(g) NO 2 (g) + CO 2 (g) NO 2 (g) + CO(g) NO(g) + CO 2 (g)

33. Reaction Mechanisms NO 2 (g) +NO 2 (g) NO 3 (g) + NO(g) NO 3 (g) + CO(g) NO 2 (g) + CO 2 (g) NO 2 (g) + CO(g) NO(g) + CO 2 (g) Slow step Fast step Rate Determining step The rate law is constructed from the rate determining step Rate = k 1 [NO 2 ] 2

34. Reaction Mechanisms: Catalysts NO + O 3 NO 2 + O 2 O + NO 2 NO + O 2 O + O 3 2O 2

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

36. Prove that the following mechanism is consistent with R = k[NO] 2 [Br 2 ], the rate law which was derived experimentally: NO(g) +NO(g) N 2 O 2 (g) N 2 O 2 (g) + Br 2 (g) 2NOBr(g) 2NO(g) + Br 2 (g) 2NOBr(g)

37. 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

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