1. Chemical Kinetics Chapter 14

2. Reminders Assignment 1 due today (end of class) Assignment 2 up on ACME, due Jan. 29 (in class) Assignment 3 will be up Mon., Jan 29 and will be due Mon., Feb. 05 Assignment 4 (Ch. 15) will not be due before Midterm 1, but Ch. 15 will be on the midterm

3. First-Order Reactions 14.3 A product rate = -  [A] tt rate = k [A] k = rate [A] = s -1 M/sM/s M =  [A] tt = k [A] - [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0 Average rate Rate law Differential rate law ln[A] = ln[A] 0 - kt Integrated rate law [A] = [A] 0 exp(-kt) Integrated rate law (linear form) Rate constant

4. First-Order Reactions 14.3 A product rate = -  [A] tt rate = k [A]  [A] tt = k [A] - Average rate Rate law Differential rate law ln[A] = ln[A] 0 - kt Integrated rate law [A] = [A] 0 exp(-kt) Integrated rate law (linear form) [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0

5. The reaction 2A B is first order in A with a rate constant of 2.8 x 10 -2 s -1 at 80 0 C. How long will it take for A to decrease from 0.88 M to 0.14 M ? ln[A] = ln[A] 0 - kt kt = ln[A] 0 – ln[A] ln[A] 0 – ln[A] k = 66 s [A] 0 = 0.88 M [A] = 0.14 M t = ln [A] 0 [A] k = ln 0.88 M 0.14 M 2.8 x 10 -2 s -1 = 14.3 Recall:

6. First-Order Reactions 14.3 The half-life, t ½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. t ½ = t when [A] = [A] 0 /2 ln [A] 0 [A] 0 /2 k = t½t½ ln2 k = 0.693 k = What is the half-life of N 2 O 5 if it decomposes with a rate constant of 5.7 x 10 -4 s -1 ? t½t½ ln2 k = 0.693 5.7 x 10 -4 s -1 = = 1200 s = 20 minutes How do you know decomposition is first order?units of k (s -1 ) t ln [A] 0 [A] k =

7. First-Order Reactions 14.3 The half-life, t ½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. t ½ = t when [A] = [A] 0 /2 What is the half-life of N 2 O 5 if it decomposes with a rate constant of 5.7 x 10 -4 s -1 ? t½t½ ln2 k = 0.693 5.7 x 10 -4 s -1 == 1200 s = 20 minutes How do you know decomposition is first order?units of k (s -1 ) The half-life of a 1 st -order reaction is independent of the initial concentration of the reactant.

8. A product First-order reaction # of half-lives [A] = [A] 0 /n 1 2 3 4 2 4 8 16 14.3 n= 2 for each half-life elapsed

9. 14.3

10. Second-Order Reactions 14.3 A product rate = -  [A] tt rate = k [A] 2 k = rate [A] 2 = 1/M s M/sM/s M2M2 =  [A] tt = k [A] 2 - [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0 1 [A] = 1 [A] 0 + kt t ½ = t when [A] = [A] 0 /2 t ½ = 1 k[A] 0

11. Distinguishing 1 st and 2 nd order reactions Is the rate law 1 st or 2 nd order? –If (a) plot gives a straight line, then 1 st order and rate = k[A] –If (b) plot gives a straight line, then 2 nd order and rate = k[A] 2 ln[A] = ln[A] 0 - kt 1 [A] = 1 [A] 0 + kt NO 2(g) NO (g) + ½ O 2(g) at 300 o C

12. Zero-Order Reactions 14.3 A product rate = -  [A] tt rate = k [A] 0 = k k = rate [A] 0 = M/s  [A] tt = k - [A] is the concentration of A at any time t [A] 0 is the concentration of A at time t=0 t ½ = t when [A] = [A] 0 /2 t ½ = [A] 0 2k2k [A] = [A] 0 - kt

13. Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions OrderRate Law Concentration-Time Equation Half-Life 0 1 2 rate = k rate = k [A] rate = k [A] 2 ln[A] = ln[A] 0 - kt 1 [A] = 1 [A] 0 + kt [A] = [A] 0 - kt t½t½ ln2 k = t ½ = [A] 0 2k2k t ½ = 1 k[A] 0 14.3

14. Temperature and reaction rates Rates of most chemical reactions increase as the temperature rises –Dough rises faster at r.t. than when refrigerated –Plants grow more rapidly in warm weather than in cold

15. Temperature and reaction rates This effect of temperature on reaction rate can be seen directly by observing a chemiluminescent reaction

16. Temperature affects rate of chemiluminescence reaction in Cyalume TM light sticks Left: light stick in hot water Right: light stick in cold water At the higher temperature, the reaction is initially faster and produces a brighter light Although the light stick glows more brightly initially, its luminescence also dies out more rapidly

17. Temperature Dependence of the Rate Constant 14.4 1 st order reaction, ln[CH 3 NC] t = -kt + ln[CH 3 NC] 0 [A] = [A] 0 exp(-kt)ln[A] = ln[A] 0 - kt

18. Temperature Dependence of the Rate Constant 14.4 1 st order reaction, ln[CH 3 NC] t = -kt + ln[CH 3 NC] 0

19. For a successful put, we need: –Enough energy to get over the hill –Directionality For a successful reaction, we need: –Enough energy to get over the activation barrier –Collisions between reacting molecules –Reacting molecules with the right orientation

20. A + B C + D Exothermic Reaction Endothermic Reaction The activation energy (E a ) is the minimum amount of energy required to initiate a chemical reaction. 14.4

21. Cl + NOClNO + Cl 2

24. Temperature Dependence of the Rate Constant k = A exp( -E a /RT ) E a is the activation energy (J/mol) R is the gas constant (8.314 J/Kmol) T is the absolute temperature A is the frequency factor lnk = - EaEa R 1 T + lnA (Arrhenius equation) 14.4

25. lnk = - EaEa R 1 T + lnA

26. 14.5 Reaction Mechanisms The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions. The sequence of elementary steps that leads to product formation is the reaction mechanism. 2NO (g) + O 2 (g) 2NO 2 (g) N 2 O 2 is detected during the reaction! Elementary step:NO + NO N 2 O 2 Elementary step:N 2 O 2 + O 2 2NO 2 Overall reaction:2NO + O 2 2NO 2 +