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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
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–3
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.
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–4
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]
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–5
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.
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–6
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
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–7
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
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–8
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.
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–9
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?
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–10
QUESTION (continued)
1) 0 2) 1 3) 2 4) 3 5) 4
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 reactant as long as some is present does not alter the rate of the reaction.
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–12
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
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–13
QUESTION (continued)
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
–
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–14
QUESTION (continued)
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.
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 Mechanism cannot be a rate determining step due to one of the reactants being an intermediate species.
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–16
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
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–17
QUESTION (continued)
The half-life of this reaction is approximately: 1) 15 minutes. 2) 18 minutes. 3) 23 minutes. 4) 36 minutes. 5) 45 minutes.
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.
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–19
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.
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.
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–21
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.
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.
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–23
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 )
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–24
QUESTION (continued)
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)
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–25
QUESTION (continued)
Which of the steps would be called the rate-determining step? 1) 1 2) 2 3) 3 4) 4 5) 5
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–26
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.
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–27
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 )
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–28
QUESTION (continued)
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)
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–29
QUESTION (continued)
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
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–30
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.
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–31
QUESTION
Refer to the following diagram:
Why is this reaction considered to be exothermic?
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–32
QUESTION (continued)
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
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–33
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.
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–34
QUESTION
Refer to the following diagram:
At what point on the graph is the activated complex present?
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–35
QUESTION (continued)
1) Point W 2) Point X 3) Point Y 4) Point Z 5) None of these
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.
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–37
QUESTION
Refer to the following diagram:
If the reaction were reversible, would the forward or the reverse reaction have a higher activation energy?
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–38
QUESTION (continued)
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
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–39
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.
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–40
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
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–41
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.
Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 12–42
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
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.