1. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 1 Chemistry FIFTH EDITION by Steven S. Zumdahl University of Illinois

2. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 2 Chemistry FIFTH EDITION Chapter 12 Chemical Kinetics Schedule: http://www2.fultonschools.org/teacher/warrene/ http://www2.fultonschools.org/teacher/warrene/

3. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 3 Section 12.3 Determining the Form of the Rate Law CONSIDER THE DECOMPOSITION OF Dinitrogen Pentoxide in CCl 4 SOLUTION 2 N 2 O 5 (SOL’N)  4 NO 2 (SOL’N) + O 2 (g) O 2 bubbles out; therefore, No reverse reaction to worry about. Reverse Rxn. is negligible at all times over the course of this reaction.

4. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 4 Figure 12.3 A Plot of the Concentration of N 2 O 5 as a Function of Time for the Reaction [N 2 O 5 ] 0 = 1.00 M Rate measured at 2 pts. using slope of the tangent line. Two pts.: 0.90 M and 0.45 M

5. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 5 SO WHEN [N 2 O 5 ] IS HALVED, THEN THE RATE IS HALVED. RATE = -  [N 2 O 5 ]/  t = k [N 2 O 5 ] 1 FIRST ORDER REACTION ( n = 1): RATE = k [A] DOUBLE CONC.  DOUBLE RATE HALVE CONC.  HALVE RATE At 0.90 M, Rate = 5.4 x 10 -4 M s -1 and at 0.45 M, Rate = 2.7 x 10 -4 M s -1

6. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 6 RATE ORDER (n = 1) DOES NOT EQUAL COEFFICIENT IN BALANCED EQ’N. 2 N 2 O 5 (SOL’N)  4 NO 2 (SOL’N) + O 2 (g) RATE ORDER MUST BE DETERMINED EXPERIMENTALLY. Measure instantaneous rates at two different reactant concentrations.

7. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 7 Method of Initial Rates Initial Rate: the “instantaneous rate” just after the reaction begins. The initial rate is determined in several experiments using different initial concentrations.

8. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 8 Reaction[A]  [B]  Initial Rate (mol/L s) 1 0.100 0.100 1.53 x 10 -4 2 0.100 0.300 4.59 x 10 -4 3 0.2000.100 6.12 x 10 -4 4 0.1000.200 3.06 x 10 -4 5 0.3000.600 8.26 x 10 -3 A + B  Products Rate = k [A] n [B] m

9. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 9 Homework Let’s do on pages 567 - 568 #25 #27 #29

10. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 10 Section 12.4 INTEGRATED RATE LAW SHOWS HOW THE CONCENTRATION OF A SPECIES IN THE REACTION DEPENDS ON TIME. Use to answer questions such as: How long does it take for the rxn. to go to 50% completion? Or to 80% completion?

11. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 11 FIRST ORDER RATE LAW DIFFERENTIAL RATE LAW: RATE = -  [A] /  t = k [ A] INTEGRATED RATE LAW: ln [A] = -kt + ln [A] ° OR ln ([A] ° /[A]) = kt Concentration of the reactant as a function of time.

12. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 12 FIRST ORDER KINETICS ln [A] = -kt + ln [A] ° OR ln ([A] ° /[A]) = kt Plot ln [A] versus t Plot is a straight line Slope = -k Y-intercept = ln [A] °

13. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 13 Figure 12.4 A Plot of In(N 2 O 5 ) Versus Time Open book to p. 539-540. Read Sample Exercise 12.2.

14. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 14 HALF-LIFE OF A 1 ST ORDER REACTION TIME REQUIRED FOR A REACTANT TO REACH HALF ITS ORIGINAL CONCENTRATION. t = t 1/2 [A] at t 1/2 = [A] ° / 2 t 1/2 = 0.693/ k For a 1 st order reaction, t 1/2 is independent of the initial concentration.

15. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 15 Half-life ln ([A] ° /[A]) = k t At t = t 1/2, [A t ] = ½ [A]  Therefore, ln ([A]  / ½ [A]  ) = k t 1/2 Therefore, t 1/2 = 0.693/ k

16. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 16 First Order Reactions 1 half-life: time to go to 50% completion 2 half-lives: time to go to 75% completion 3 half-lives: time to go to 87.5% completion

17. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 17 100 grams  t 1/2  50 grams 50 grams  t 1/2  25 grams 25 grams  t 1/2  12.5 grams 12.5 grams  t 1/2  6.25 grams Start with 100 grams after 4 half-lives 6.25 grams remaining.

18. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 18 Figure 12.5 A Plot of (N 2 O 5 ) Versus Time for the Decomposition Reaction of N 2 O 5

19. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 19 SECOND ORDER RATE LAWS 2 nd Order DIFFERENTIAL RATE LAW: Rate = -  [ A]/  t = k [A] 2 For a general reaction involving a single reactant, That is, a A  Products

20. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 20 SECOND ORDER RATE LAWS DIFFERENTIAL RATE LAW: Rate = -  [ A]/  t = k [A] 2 INTEGRATED RATE LAW: 1/[A] = kt + 1/[A] ° Plot of 1/[A] versus t gives straight line with Slope = k Conc. doubles Rate quadruples.

21. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 21 HALF-LIFE OF SECOND ORDER REACTION t = t 1/2 = 1/k [A] ° Half-life is dependent on the initial concentration!!

22. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 22 SAMPLE EXERCISE 12.5 PAGE 544 IS THE REACTION 1 ST ORDER or 2 ND ORDER? PLOT ln[A] versus t and 1/[A] versus t Which is a straight line?

23. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 23 Figure 12.6 (a) A Plot of In(C 4 H 6 ) Versus t (b) A Plot of 1/(C 4 H 6 ) Versus t

24. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 24 Half -Life 1 st order: Depends only on k t 1/2 = 0.693/ k Constant time to reduce the conc. of reactant by half, & then half again, etc. 2 nd order: Depends on both k & [A]  t 1/2 = 1/ k [A]  Second half-life is longer that the first. “Each successive half-life is double the preceding one.”

25. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 25 ZERO ORDER RATE LAWS MOST REACTIONS INVOLVING A SINGLE REACTANT SHOW EITHER 1 ST ORDER or 2 ND ORDER KINETICS. BUT THEY CAN BE ZERO ORDER. RATE = k [A] 0 = k(1) = k Therefore Rate is constant It does not change with concentration.

26. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 26 INTEGRATED RATE LAW FOR ZERO ORDER REACTIONS [A] = -kt + [A] ° PLOT OF [A] versus t is a straight line Slope = -k Intercept = [A] °

27. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 27 Figure 12.7 A Plot of (A) Versus t for a Zero-Order Reaction

28. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 28 HALF-LIFE FOR ZERO ORDER REACTIONS t 1/2 = [A] ° / 2k

29. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 29 ZERO ORDER REACTIONS MOSTLY OCCUR WHEN A SUBSTANCE SUCH AS A METAL SURFACE OR ENZYME IS REQUIRED FOR THE REACTION TO OCCUR. EXAMPLE: 2N 2 O (g)  2 N 2 (g) + O 2 (g)

30. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 30 Figure 12.8 Decomposition Reaction of N 2 O When Pt surface is completely covered with N 2 O molecules, an increase in [N 2 O] has no effect on rate. Only N 2 O on the surface can react.

31. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 31 Metal surface can only accommodate so many molecules. Reaction controlled by what happens on the Pt surface rather than by the total concentration of N 2 O.

32. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 32 So far  only simple reactions  with only one reactant!! ☻ What if there is more than one reactant? �� 

33. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 33 More than one Reactant: Consider BrO 3 - (aq) + 5Br - (aq) + 6H +  3 Br 2 (l) + 3H 2 O (l) Experimentally, the rate law has been found to be Rate = -  [BrO 3 -] = k [BrO 3 - ] [Br - ] [H + ] 2  t

34. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 34 MORE THAN ONE REACTANT? KINETICS OF COMPLICATED REACTIONS------- OFTEN STUDIED BY OBSERVING THE BEHAVIOR OF ONE REACTANT AT A TIME.

35. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 35 PSEUDO-1 ST ORDER RATE LAW Technique allows one to determine rate laws for complex reactions. Look at change in concentration with time of a reactant present in a relatively small amount allows one to determine the order of the reaction in that component.

36. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 36 PSEUDO-1 ST ORDER RATE LAW Make initial conc. of one reactant much smaller than the others. Example [BrO 3 - ]  = 1.0 x 10 -3 M [Br - ]  = 1.0 M [H + ]  = 1.0 M

37. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 37 Last two so large compared to [BrO 3 - ] that their conc. will change very little relatively. That is, they will remain constant, So, [Br - ] t ≈ [Br - ]  [H + ] t ≈ [H + ] 

38. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 38 Rate = k [Br - ] [H + ] 2 [BrO 3 - ] Rate = k [Br - ]  [H + ]  2 [BrO 3 - ] Rate = k ’ [BrO 3 - ] This is a Pseudo-first order Rate Law. Plot of ln [BrO 3 - ] vs. t  Straight Line

39. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 39 PSEUDO-1 ST ORDER RATE LAW Technique allows one to determine rate laws for complex reactions. Look at change in concentration with time of a reactant present in a relatively small amount allows one to determine the order of the reaction in that component. Let’s look at #38 on page 570

40. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 40 Time (s)[O] (atoms/cc)ln[O] 0500000000022.3327 0.01190000000021.36512 0.0268000000020.3376 0.0325000000019.33697 Plot of ln[O] vs time #38 page 570

41. Copyright©2000 by Houghton Mifflin Company. All rights reserved. 41 SECTION 12.5 RATE LAWS: SUMMARY GOOD SUMMARY!!! PAGES 548 – 549 TABLE 12.6 p. 548!!! Homework: 31- 47 odd