chapter 12 chemical kinetics. copyright © houghton mifflin company. all rights reserved.crs...

Chapter 12

Chemical KineticsChemical Kinetics

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

QUESTION

Consider the reaction 2H2 + O2 2H2O. What is

the ratio of the initial rate of the appearance of water to the initial rate of disappearance of oxygen? 1) 1 : 1 2) 2 : 1 3) 1 : 2 4) 2 : 2 5) 3 : 2


ANSWER

)

2) 2 : 1

Section 12.1 Reaction Rates (p. 527

For every mole of oxygen reacted, two moles of water are produced. The coefficients are the key to this type of problem.


QUESTION

Consider the reaction X Y + Z. Which of the following is a possible rate law? 1) Rate = k[X] 2) Rate = k[Y] 3) Rate = k[Y][Z] 4) Rate = k[X][Y] 5) Rate = k[Z]


ANSWER

k

)

1) Rate = [X]

Section 12.2 Rate Laws: An Introduction (p. 532

The rate law is dependent on the concentration of reactants, not products. Rate laws containing concentrations of products are not dealt with at this level.


QUESTION

The following data were obtained for the reaction of NO with O2. Concentrations are in molecules/cm3 and rates are in molecules/cm3

s.

[NO]0 [O2]0 Initial Rate 1 10

18 1 10

18 2.0 10

16

2 1018

1 1018

8.0 1016

3 10

18 1 10

18 18.0 10

16

1 1018

2 1018

4.0 1016

1 10

18 3 10

18 6.0 10

16


QUESTION (continued)

Which of the following is the correct rate law? 1) Rate = k[NO][O2] 2) Rate = k[NO][O2]

2

3) Rate = k[NO]2[O2] 4) Rate = k[NO]

2

5) Rate = k[NO]2[O2]

2


ANSWER

k2

2]

)

3) Rate = [NO] [O

Section 12.3 Determining 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 be kept constant.


QUESTION

A general reaction written as 1A + 2B C + 2D is studied and yields the following data:

[A]0 [B]0 Initial [C]/t 0.150 M 0.150 M 8.00 10

–3 mol/L s

0.150 M 0.300 M 1.60 10–2

mol/L s 0.300 M 0.150 M 3.20 10

–2 mol/L s

What is the order of the reaction with respect to B?



1) 0 2) 1 3) 2 4) 3 5) 4


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 reactant as long as some is present does not alter the rate of the reaction.


QUESTION

Consider the following data concerning the equation: H2O2 + 3I

– + 2H

+ I3

– + 2H2O

[H2O2] [I

–] [H

+] rate

a. 0.100 M 5.00 10–4 M 1.00 10

–2 M 0.137 M/sec b. 0.100 M 1.00 10

–3 M 1.00 10–2 M 0.268 M/sec

c. 0.200 M 1.00 10–3 M 1.00 10

–2 M 0.542 M/sec d. 0.400 M 1.00 10

–3 M 2.00 10–2 M 1.084 M/sec



Two mechanisms are proposed:

a. H2O2 + I– H2O + OI

–

OI– + H

+ HOI

HOI + I– + H

+ I2 + H2O

I2 + I– I3

–

b. H2O2 + I– + H

+ H2O + HOI

HOI + I– + H+ I2 + H2O

I2 + I– I3

–



Which of the following describes a potentially correct mechanism? 1) Mechanism a with the first step the rate

determining step. 2) Mechanism a with the second step the rate

determining step. 3) Mechanism b with the first step rate determining. 4) Mechanism b with the second step rate

determining. 5) None of these could be correct.


ANSWER

a

-

1) Mechanism a with the first step the rate determining step.

Section 12.6 Reaction Mechanisms (p. 549)

The second step in Mechanism cannot be a rate determining step due to one of the reactants being an intermediate species.


QUESTION

For the reaction 2N2O5(g) 4NO2(g) + O2(g), the following data were collected: t (minutes) [N2O5] (mol/L) 0 1.24 10

–2

10. 0.92 10–2

20. 0.68 10

–2

30. 0.50 10–2

40. 0.37 10

–2

50. 0.28 10–2

70. 0.15 10

–2



The half-life of this reaction is approximately: 1) 15 minutes. 2) 18 minutes. 3) 23 minutes. 4) 36 minutes. 5) 45 minutes.


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.


QUESTION

Determine the molecularity of the following elementary reaction: O3

O2 + O. 1) Unimolecular 2) Bimolecular 3) Termolecular 4) Quadmolecular 5) The molecularity cannot be determined.


ANSWER

U

1) nimolecular


This is an elementary reaction, so the molecularity can be found directly from the reactants and their coefficients.


QUESTION

The decomposition of ozone may occur through the two-step mechanism shown:

step 1 O3 O2 + O

step 2 O3 + O 2O2

The oxygen atom is considered to be a(n): 1) reactant. 2) product. 3) catalyst. 4) reaction intermediate. 5) activated complex.


ANSWER

.

4) reaction intermediate


A reaction intermediate plays a role in the complete reaction, but is not shown in the overall reaction.


QUESTION

U n d e r c e r t a i n c o n d i t i o n s t h e r e a c t i o n H 2 O 2 + 3 I–

+ 2 H

+ I 3

– + 2 H 2 O o c c u r s b y t h e f o l l o w i n g

s e r i e s o f s t e p s : S t e p 1 . H 2 O 2 + H

+ H 3 O 2+

r a p i d e q u i l i b r i u m c o n s t a n t K = k 1

k – 1

S t e p 2 . H 3 O 2+

+ I–

H 2 O + H O I ( s l o w , r a t e c o n s t a n t k 2 )



Step 3. HOI + I– OH

– + I2 (fast, rate constant k3)

Step 4. OH– + H

+ H2O (fast, rate constant k4)

Step 5. I2 + I– I3

– (fast, rate constant k5)



Which of the steps would be called the rate-determining step? 1) 1 2) 2 3) 3 4) 4 5) 5


ANSWER

)

2) 2

Section 12.6 Reaction 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.


QUESTION

U n d e r c e r t a i n c o n d i t i o n s t h e r e a c t i o n H 2 O 2 + 3 I–

+ 2 H

+ I 3

– + 2 H 2 O o c c u r s b y t h e f o l l o w i n g

s e r i e s o f s t e p s : S t e p 1 . H 2 O 2 + H

+ H 3 O 2+

r a p i d e q u i l i b r i u m c o n s t a n t K = k 1

k – 1

S t e p 2 . H 3 O 2+

+ I–

H 2 O + H O I ( s l o w , r a t e c o n s t a n t k 2 )



Step 3. HOI + I– OH

– + I2 (fast, rate constant k3)

Step 4. OH– + H

+ H2O (fast, rate constant k4)

Step 5. I2 + I– I3

– (fast, rate constant k5)



The rate constant k for the reaction would be given by: 1) k = k2 2) k = k2k3 3) k = k2K 4) k = k5 5) k = Kk2k3k4k5


ANSWER

k k2

)

3) = K

Section 12.6 Reaction 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.


QUESTION

Refer to the following diagram:

Why is this reaction considered to be exothermic?



1) Because energy difference B is greater than energy difference C

2) Because energy difference B is greater than energy difference A

3) Because energy difference A is greater than energy difference C

4) Because energy difference B is greater than energy difference C plus energy difference A

5) Because energy difference A and energy difference C are about equal


ANSWER

)

2) Because energy difference B is greater than energy difference A

Section 12.7 A Model for Chemical Kinetics(p. 552

Z is more stable than W, allowing the energy needed to stabilize W to be released as heat.


QUESTION


At what point on the graph is the activated complex present?



1) Point W 2) Point X 3) Point Y 4) Point Z 5) None of these


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.


QUESTION


If the reaction were reversible, would the forward or the reverse reaction have a higher activation energy?



1) The diagram shows no indication of any activation energy.

2) The forward and reverse activation energies are equal.

3) The forward activation energy 4) The reverse activation energy 5) None of these


ANSWER

)

4) The reverse activation energy

Section 12.7 A 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.


QUESTION

Which of the following statements is typically true for a catalyst? 1) The concentration of the catalyst will go

down as a reaction proceeds. 2) The catalyst provides a new pathway in the

reaction mechanism. 3) The catalyst speeds up the reaction. 4) Two of these 5) None of these


ANSWER

4) Two of these Section 12.8 Catalysis (p. 557 ) The catalyst provides a new pathway in the reaction mechanism and the catalyst speeds up the reaction.


QUESTION

The catalyzed pathway in a reaction mechanism has a(n) __________ activation energy and thus causes a(n) __________ reaction rate. 1) higher; lower 2) higher; higher 3) lower; higher 4) lower; steady 5) higher; steady


ANSWER

;

3) lower higher

Section 12.8 Catalysis (p. 557)

Catalysts often bind to reactants altering their structure and bringing them into close contact.

chapter 12 chemical kinetics. copyright © houghton mifflin company. all rights reserved.crs...

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