1. Chapter 12 Chemical Kinetics

2. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–2 QUESTION

3. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–3 ANSWER ) 2)2 : 1 Section12.1Reaction Rates(p. 527 Forevery mole of oxygen reacted, two moles of water are produced. Thecoefficients are the key to this type of problem.

4. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–4 QUESTION

5. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–5 ANSWER k ) 1)Rate =[X] Section12.2Rate Laws: An Introduction(p. 532 The rate law is dependent on the concentration of reactants, not products. Rate lawscontaining concentrations of products are not dealt with at this level.

6. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–6 QUESTION

7. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–7 QUESTION (continued)

8. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–8 ANSWER k 2 2 ] ) 3)Rate =[NO][O Section12.3Determining the Form of the Rate Law(p. 534 Remember that the concentration of only one reactant can be allowed to change at a time. The other reactant(s) must bekept constant.

9. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–9 QUESTION

10. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–10 QUESTION (continued)

11. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–11 ANSWER — — 2)1 Section 12.3 Determining the Form of the Rate Law (p. 534) An order of zero is possible. This occurs when the concentration of a reactantas long as some is presentdoes not alter the rate of the reaction.

12. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–12 QUESTION

13. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–13 QUESTION (continued)

14. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–14 QUESTION (continued)

15. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–15 ANSWER a - 1)Mechanism a with the first step the rate determining step. Section 12.6 Reaction Mechanisms (p. 549) The second step in Mechanismcannot be a ratedetermining step due to one of the reactants being an intermediate species.

16. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–16 QUESTION

17. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–17 QUESTION (continued)

18. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–18 ANSWER. 3)23 minutes Section 12.4 The Integrated Rate Law (p. 538) A good way to approach this problem would be to plot the data on a graphing calculator.

19. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–19 QUESTION

20. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–20 ANSWER U 1)nimolecular Section 12.6 Reaction Mechanisms (p. 549) This is an elementary reaction, so the molecularity can be found directly from the reactants and their coefficients.

21. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–21 QUESTION

22. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–22 ANSWER. 4)reaction intermediate Section 12.6 Reaction Mechanisms (p. 549) A reaction intermediate plays a role in the complete reaction, but is not shown in the overall reaction.

23. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–23 QUESTION

24. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–24 QUESTION (continued)

25. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–25 QUESTION (continued)

26. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–26 ANSWER ) 2)2 Section12.6Reaction Mechanisms(p. 549 The fast steps of a reaction mechanism very often take so little time to occur that they cannot be noticed. It is the slow step that allows the reaction rate to be measured.

27. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–27 QUESTION

28. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–28 QUESTION (continued)

29. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–29 QUESTION (continued)

30. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–30 ANSWER kk 2 ) 3) = K Section12.6Reaction Mechanisms(p. 549 The slow step in this mechanism contains an intermediate. This intermediate must be removed and the equilibrium step of the mechanism is used for this purpose.

31. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–31 QUESTION

32. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–32 QUESTION (continued)

33. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–33 ANSWER ) 2)Because energy difference B is greater than energy difference A Section12.7A Model for Chemical Kinetics (p. 552 Z is more stable than W, allowing the energy needed to stabilize W to be released as heat.

34. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–34 QUESTION

35. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–35 QUESTION (continued)

36. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–36 ANSWER 3) Point Y Section 12.7 A Model for Chemical Kinetics (p. 552) The activated complex is a halfway point between reactants and products, where bonds are breaking as new bonds are forming. It is not a species that can be isolated.

37. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–37 QUESTION

38. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–38 QUESTION (continued)

39. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–39 ANSWER ) 4)The reverse activation energy Section12.7A Model for Chemical Kinetics (p. 552 Looking at the graph we can see that Z must absorb more energy than W to reach the activation complex, Y.

40. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–40 QUESTION

41. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–41 ANSWER 4) Two of these Section12.8Catalysis(p. 557) The catalyst provides a new pathway in the reaction mechanism and the catalyst speeds up the reaction.

42. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–42 QUESTION

43. Copyright © Houghton Mifflin Company. All rights reserved.CRS Question, 12–43 ANSWER ; 3)lower higher Section 12.8 Catalysis (p. 557) Catalysts often bind to reactants altering their structure and bringing them into close contact.