Chapter 12 Chemical Kinetics

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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.

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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.

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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.

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

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.

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

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.

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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.

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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.

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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.

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

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.

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

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.

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

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.

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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.

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

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.

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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.

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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.