So, for the reaction A plus B goes to products, the rate law is equal to the concentration of A raised to some power, times the concentration of B raised to some power, times a proportio
Trang 1GENERAL CHEMISTRY TOPICAL:
Kinetics and Equilibrium
Test 1
Time: 23 Minutes*
Number of Questions: 18
* The timing restrictions for the science topical tests are optional If you are using this test for the sole purpose of content reinforcement, you may want to disregard the time limit
Trang 2DIRECTIONS: Most of the questions in the following test
are organized into groups, with a descriptive passage preceding each group of questions Study the passage, then select the single best answer to each question in the group Some of the questions are not based on a descriptive passage; you must also select the best answer
to these questions If you are unsure of the best answer, eliminate the choices that you know are incorrect, then select an answer from the choices that remain Indicate your selection by blackening the corresponding circle on your answer sheet A periodic table is provided below for your use with the questions
PERIODIC TABLE OF THE ELEMENTS
1
H
1.0
2
He
4.0 3
Li
6.9
4
Be
9.0
5
B
10.8
6
C
12.0
7
N
14.0
8
O
16.0
9
F
19.0
10
Ne
20.2 11
Na
23.0
12
Mg
24.3
13
Al
27.0
14
Si
28.1
15
P
31.0
16
S
32.1
17
Cl
35.5
18
Ar
39.9 19
K
39.1
20
Ca
40.1
21
Sc
45.0
22
Ti
47.9
23
V
50.9
24
Cr
52.0
25
Mn
54.9
26
Fe
55.8
27
Co
58.9
28
Ni
58.7
29
Cu
63.5
30
Zn
65.4
31
Ga
69.7
32
Ge
72.6
33
As
74.9
34
Se
79.0
35
Br
79.9
36
Kr
83.8 37
Rb
85.5
38
Sr
87.6
39
Y
88.9
40
Zr
91.2
41
Nb
92.9
42
Mo
95.9
43
Tc
(98)
44
Ru
101.1
45
Rh
102.9
46
Pd
106.4
47
Ag
107.9
48
Cd
112.4
49
In
114.8
50
Sn
118.7
51
Sb
121.8
52
Te
127.6
53
I
126.9
54
Xe
131.3 55
Cs
132.9
56
Ba
137.3
57
La *
138.9
72
Hf
178.5
73
Ta
180.9
74
W
183.9
75
Re
186.2
76
Os
190.2
77
Ir
192.2
78
Pt
195.1
79
Au
197.0
80
Hg
200.6
81
Tl
204.4
82
Pb
207.2
83
Bi
209.0
84
Po
(209)
85
At
(210)
86
Rn
(222) 87
Fr
(223)
88
Ra
226.0
89
Ac †
227.0
104
Unq
(261)
105
Unp
(262)
106
Unh
(263)
107
Uns
(262)
108
Uno
(265)
109
Une
(267)
*
58
Ce
140.1
59
Pr
140.9
60
Nd
144.2
61
Pm
(145)
62
Sm
150.4
63
Eu
152.0
64
Gd
157.3
65
Tb
158.9
66
Dy
162.5
67
Ho
164.9
68
Er
167.3
69
Tm
168.9
70
Yb
173.0
71
Lu
175.0
†
90
Th
232.0
91
Pa
(231)
92
U
238.0
93
Np
(237)
94
Pu
(244)
95
Am
(243)
96
Cm
(247)
97
Bk
(247)
98
Cf
(251)
99
Es
(252)
100
Fm
(257)
101
Md
(258)
102
No
(259)
103
Lr
(260)
GO ON TO THE NEXT PAGE.
Trang 3
Passage I (Questions 1–7)
As Figure 1 illustrates, the reaction of tert-butyl
bromide with aqueous sodium hydroxide results in a
substitution product (Reaction I) and an elimination
product (Reaction II)
CH 3 C CH 3
CH 3
Br
I
II
K=3 × 10 4
K=2 × 10 2
CH 3 C OH
CH 3
CH 3 + Br–
CH 3 C CH 2
CH 3 + H 2 O + Br –
K=1 × 10 2
H + /∆
Figure 1
A student investigated the reaction kinetics for the
conversion of tert-butyl bromide to tert-butyl alcohol
(Reaction I) by varying the concentrations of the reactants
and recording the reaction rate The results are shown in
Table 1
Table 1 Exp
#
Initial
concentration
of tert-Butyl
bromide
(mmol/L)
Initial concentration
of OH–
(mmol/L)
Initial rate of reaction (mmol/[L•sec])
1
2
3
0.02
0.02
0.04
0.05 0.10 0.05
6.0 6.0 12.0
In reaction III, tert-butyl alcohol is converted to
2-methylpropene by the addition of aqueous acid followed by
heating This reaction involves the formation of a
protonated intermediate that loses water to form a
carbonation The alkene is then generated by the loss of a
proton This mechanism is illustrated in Figure 2
CH3 C CH3 O
CH3
H
H +
CH3 C CH3 O H H
CH3
+
CH3 C CH3 O
CH 3
+
CH3 C
CH3
CH3 + H2O
(1)
(2)
CH3 C CH2
CH3
H
+
CH3
(3)
Figure 2
The energy profile for Reaction III is as follows:
Reaction Coordinate
Potential Ener
Figure 3
l Which reaction has proceeded the farthest toward completion once equilibrium has been established?
A Reaction I
B Reaction II
C Reaction III
D It cannot be determined without more information.
GO ON TO THE NEXT PAGE.
Trang 42 Which of the following assumptions can be made
about Reaction III?
A The rate of formation of the carbocation is greater
than the rate of protonation of the alcohol
B The rate of formation of the carbonation is greater
than the deprotonation of the carbonation to form
the alkene
C The rate of deprotonation to form the alkene is
greater than the protonation of the alcohol
D The rate of formation of the carbonation is less
than the rate of protonation of the alcohol
3 Which of the following is the rate equation for
Reaction I?
A Rate = k [tert-butyl bromide] [OH–]
B Rate = k [OH–]
C Rate = k [tert-butyl bromide]
D Rate = k [tert-butyl bromide] [OH–]2
4 What is the rate constant for Reaction I?
A 3000 s–l
B 300 s–l
5 If Reaction I was carried out at a higher temperature,
which of the following would be observed?
A Both the equilibrium constant and the rate
constant would change
B The equilibrium constant would change, but the
rate constant would remain the same
C The rate constant would change, but the
equilibrium constant would remain the same
D Neither the equilibrium constant nor the rate
constant would change
6 For Reaction II, what could be done to shift the
equilibrium to the right, favoring the formation of the alkene?
A Add excess water to the reaction mixture.
B Use tert-butyl chloride as the starting reagent
instead of tert-butyl bromide.
C Use a lower concentration of hydroxide ions.
7 If Reactions I, II, and III were run with an appropriate
catalyst, which of the equilibrium constants would change?
A The equilibrium constant for Reaction I only
B The equilibrium constant for Reaction II only
C The equilibrium constants for Reactions I, II, and
III
D None of the equilibrium constants would change.
GO ON TO THE NEXT PAGE.
Trang 5
Passage II (Questions 8–14)
A chemist interested in the reactivity of iodine
concentrated his study on two reactions: the decomposition
of gaseous hydrogen iodide (Reaction 1) and the reaction
between iodide ions and persulfate ions (Reaction 2)
2HI(g) H2(g) + I2(g)
Reaction 1 3I–(aq) + S2O82–(aq) I3(aq) + 2SO42–(aq)
Reaction 2 The value of the rate constant for Reaction 1 was
studied as a function of temperature The results are shown
below
Table 1
T(K) 1/T (K–1) k (l•mol–1sec–1) log k
555
575
645
700
781
1.80 ×
10–3 1.74 ×
10–3 1.55 ×
10–3 1.43
× 10–3
1.28 ×
10–3
3.52 × 10–7
1.22 × 10–6 8.59 × 10–5
1.16 × 10–3 3.95 × 10–2
–6.453 –5.913 –4.066 –2.936 –1.403
For any reaction, the activation energy (Ea) is related
to the rate constant (k) by the Arrhenius equation
(Equation 1):
k = A × l0(–Ea / 2.303RT)
Equation 1
where R = 8.314 J mol–1K–1, T is the temperature in
Kelvin, and A is a constant, called the frequency factor.
Figure 1 shows a graph of log k vs 1/T for Reaction 1.
slope = -9.71 x 10 3 K
(1/T)
-6 -4 -2
1.6 x 10 -3
1/T
In order to determine the initial rate of Reaction 2, the following data were collected:
Table 2 Experiment [I–]
(M)
[S2O8–2]
(M)
Initial rate
of reaction
(M / sec)
1 2 3
0.21 0.21 0.42
0.15 0.30 0.15
1.14 2.28 2.28
8 What is the rate law for Reaction 2?
A Rate = k[I–]2[S2O82–]
B Rate = k[S2O82–]
C Rate = k[I–][S2O82–]
D Rate = k[I–][S2O82–]2
9 What is the numerical value of the rate constant for
Reaction 2?
A 7.6 mol/L•sec
B 36 mol/L•sec
C 172 mol/L•sec
D 241 mol/L•sec
GO ON TO THE NEXT PAGE.
Trang 61 0 According to Equation 1:
A when the temperature is held constant, the lower
the activation energy, the slower the reaction
B when the activation energy is held constant, the
lower the value of A, the faster the reaction.
C when the activation energy is held constant, the
lower the temperature, the faster the reaction
D when the temperature is held constant, the lower
the activation energy, the faster the reaction
1 1 What is the activation energy for Reaction 1?
A 9.71 kJ/mole
B 22.4 kJ/mole
C 80.7 kJ/mole
D 186 kJ/mole
1 2 In Figure 1, what does the intercept with the y-axis
represent?
A log A
B log k
C –Ea
RT
1 3 If the rate of disappearance of I– in Reaction 2 is 2.5 ×
10–3 mol/(L•s), what is the rate of formation of
SO42–?
A 1.7 × 10–3 mol/(L•s)
B 3.8 × 10–3 mol/(L•s)
C 5.0 × 10–3 mol/(L•s)
D 8.3 × 10–4 mol/(L•s)
1 4 The reaction profile shown below is for an uncatalyzed
reaction
Reaction Coordinate
Potential Ener
120 kJ mole -1
Which of the following is the reaction profile for the same reaction after the addition of a catalyst?
100 kJ mole -1
150 kJ mole -1
95 kJ mole -1
120 kJ mole -1
A.
B.
C.
D.
GO ON TO THE NEXT PAGE.
Trang 7
Questions 15 through 18 are NOT
based on a descriptive passage
1 5 How will the equilibrium of the following reaction be
affected if more chlorine is added?
PCl5(g) PCl3(g) + Cl2(g)
A It will be shifted to the right.
B It will be shifted to the left.
C It will be unaffected.
D The effect on the equilibrium cannot be determined
without more information
1 6 How will the equilibrium of the following reaction be
affected if the temperature is increased?
N2(g) + 3H2(g) 2NH3 (g) ∆H = –30 kJ/mole
A It will be shifted to the right.
B It will be shifted to the left.
C It will be unaffected.
D The effect on the equilibrium cannot be determined
without more information
1 7 What is the value of the equilibrium constant for the
following reaction if the equilibrium concentrations of
nitrogen, hydrogen, and ammonia are 1M, 2M, and
15M, respectively?
N2(g) + 3H2(g) 2NH3 (g)
A 0.035
D 380
1 8 Consider the following gas phase reaction:
H2 (g) + Br2 (g) 2 HBr(g)
The concentrations of H2, Br2, and HBr are 0.05M, 0.03M, and 500.0M, respectively The equilibrium
constant for this reaction at 400°C is 2.5 × 103 Is this system at equilibrium?
A Yes, the system is at equilibrium.
B No, the reaction must shift to the right in order to
reach equilibrium
C No, the reaction must shift to the left in order to
reach equilibrium
D It cannot be determined if this system is at
equilibrium without more information
END OF TEST
Trang 8ANSWER KEY:
Trang 9
KINETICS AND EQUILIBRIUM TEST TRANSCRIPT Passage I (Questions 1–7)
1 The reaction that goes to completion to the greatest degree is the one with the larger equilibrium constant You should recall that for any reaction, the equilibrium constant is equal to the product of the products each raised to their stoichiometric coefficients divided by the product of the reactants each raised to their stoichiometric coefficients So, if the equilibrium
constant is large, it means that a large amount of products have been formed Thus, choice A, Reaction I with a K = 3 x 104, would be the correct choice
2 The answer to question 2 is D For this question you need to be able to understand reaction profiles A reaction profile, such as that shown in Figure 3, is a plot of the potential energy of a reaction versus the reaction coordinate, which represents the progress of the reaction You should know that as reactant molecules with sufficient energy collide, they form what is known as an activated complex This activated complex can either go on to form intermediates, products, or go back to reactants Looking at Figure 3, you can see that there are three peaks representing three activated complexes and that the middle peak is the highest You should know that the higher the peak, the higher the potential energy of the activated complex, and that the higher the potential energy the slower the rate of formation of the activated complex In other words, the slowest step
in a mechanism has the highest peak in the reaction profile Comparing the peaks to the steps of the mechanism (Figure 2), you can see that step two, the formation of the carbocation, has the highest peak All right, let’s look at the answer choices Choice A states that the rate of formation of the carbocation, step 2, is greater than the rate of protonation of the alcohol, step
1 In other words, this choice says that the peak for step 2 is lower than the peak for step 1 This is not true Choice B states that the rate of formation of the carbocation is greater than that of the deprotonation of the carbocation to form the alkene, or the peak for step 2 is smaller than that of step 3 Not true Choice C, the rate of deprotonation to form the alkene is greater than the protonation of the alcohol, or the peak for step 3 is smaller than that of step 1, is also not true That leaves choice D as the correct response: the rate of formation of the carbocation is less than the rate of protonation (The peak for step 2 is taller than that of step 1.)
3 The correct answer to question 3 is choice C For this question you need to be able to construct the rate expression and
be able to determine the order of the reaction from experimental data For a homogeneous reaction, the rate is proportional to the concentrations of the reactants raised to some power So, for the reaction A plus B goes to products, the rate law is equal to the concentration of A raised to some power, times the concentration of B raised to some power, times a proportionality constant The proportionality constant is called the rate constant and the powers that the concentrations of the reactants are raised to can only be determined by experimentation Okay, for the reaction in question, the rate law is equal to the
concentration of tert-butyl bromide raised to some power, times the concentration of hydroxide raised to some power, times a rate constant Now, in order to determine what the exponents in the rate law are, you need to be able to interpret the data presented in Table 1 A couple of things you should remember: if the concentration of a reactant is doubled and the rate doubles, if it is tripled and the rate triples, the exponent is equal to one If the concentration of a reactant is doubled and the rate increases by a factor of 4, if the concentration is tripled and the rate increases by a factor of 9, the exponent is equal to 2 If the concentration of the reactant does not affect the rate, the exponent is equal to zero Armed with this information, let’s look at the data: Looking at experiments number 1 and 2, the concentration of tert-butyl bromide is held constant at 0.02 mmol/liter and the concentration of hydroxide has been doubled We can see that the reaction rate remains unchanged: the exponent for the hydroxide concentration is zero Something raised to the zero power is equal to one, so hydroxide will NOT appear in the rate law Looking at experiments 1 and 3, the concentration of hydroxide is held constant it doesn’t really matter since we just determined it’s not in the rate law and the concentration of tert-butyl bromide has been doubled What happens to the rate? It goes from 6.0 mmol per liter second to 12 mmol per liter second it doubles So the exponent for tert-butyl bromide is equal to one The rate law is, therefore, rate is equal to the rate constant, times the concentration of tert-butyl bromide: Choice C
4 For question number 4 the correct answer is B For this question, what you need to do is solve the rate law for the rate constant Rearranging the rate law that we determined in question 3, you get that the rate constant is equal to the rate divided by the concentration of tert-butyl bromide Taking Experiment 1, the rate constant is equal to 6.0 mmol/liter sec divided by 0.02 mmol /liter 6 divided by 0.02 equals 300: choice B As far as the units are concerned, mmols and liters cancel leaving inverse seconds as the units Whenever you see a rate constant of inverse seconds you can be sure that it is a first-order reaction Second-order rate constants have the units liter per mole second Again the correct answer is B
Trang 105 For question 5 the correct answer is A Rate constants and equilibrium constants are both affected by temperature The temperature dependence of the rate constant is explained by the Arrhenius equation, which is, in logarithmic form, the base-ten log of the rate constant is equal to the log of the frequency constant (A), minus the activation energy (Ea) divided by the product 2.303, times the gas constant, times the temperature If you’re not familiar with this, don’t worry we’ll talk about it in just a while You used this equation to answer some questions in the next passage Basically, what you need to know is that as the temperature increases, so does the rate constant The temperature dependence of the equilibrium constant is explained by the equation DG = –RTlnK, where DG is the standard free energy change of the reaction (remember that standard conditions are observed when each substance is pure and at 1 atmosphere of pressure) This equation is valid for any kind of equilibrium: acid dissociation constants, base dissociation constants, equilibria involving gases, etc You should have this equation memorized Don’t worry about memorizing the Arrhenius equation; just know that the rate constant increases with increasing temperature
If you need to use the Arrhenius equation it will be given to you Again, the correct answer is A
6 The correct answer to question 6 is choice D This question is testing your knowledge of Le Chatelier’s principle, which states if a system is subjected to a stress, it will adjust itself in order to alleviate that stress Let’s talk about of few of these stresses Adding or removing a product or reactant: the reaction will shift in order to consume the reactant that has been added and shifts in order to replace the product or reactant that has been removed In other words, if you add something to the reactant side, the equilibrium will be shifted to the right, in order to consume the excess reactant If a product is added, the equilibrium will shift to the left, in order to consume the excess product If a reactant is removed, the equilibrium will shift to the left, in order to replace the reactant If a product is removed, the equilibrium will shift to the right, in order to replace the product Another type of stress occurs when the volume of a gaseous reaction is reduced When the volume is decreased, the reaction will shift is such a way so as to reduce the total number of gaseous molecules So, if a reaction has 3 moles of gas on the product side and 2 moles of gas on the reactant side, it will shift to the left, favoring the side with fewer molecules
Changes in temperature also stress the system If the temperature of an exothermic reaction is increased, the reaction will shift
to the left In other words, the equilibrium constant becomes smaller If the temperature of an endothermic reaction is
increased, it will be shifted to the right, the equilibrium constant will become larger There are also a couple of additions to a system that do not affect the equilibrium: the addition of an inert gas, and the addition of a catalyst You should know that if a catalyst is added, it affects the forward and reverse reaction equally So it cancels out of the equilibrium expression If an inert gas is added, it also appears on both the product and reactant side, leaving the equilibrium unaffected
All right, for this question you need to look at reaction II and decide what you can do in order to increase the amount of alkene Well, the alkene is a product, and as I just said, if you need to favor the products, you need to remove a product or add a reactant Choice A, adding water to the reaction will not increase the amount of alkene formed Water is a product, and adding a product will shift the equilibrium to the left, in the opposite direction of what we need Choice B, using a different reactant, will not increase the amount of product formed Choice C, decreasing the concentration of hydroxide ions, will also not
increase the amount of product: since this reaction proceeds by an E1 pathway, the concentration of hydroxide won’t affect the reaction at all Choice D, adding a material to complex out the product bromide ion will increase the amount of product formed Bromide ion is a product; if it is removed, the equilibrium will shift to favor the products side Choice D is the correct answer
7 The correct response to question 7 is D As I discussed in question 6, adding an inert gas or a catalyst does not alter the value of the equilibrium constant Again, the correct answer is choice D
Passage II (Questions 8–14)
8 The correct answer to question 8 is C Determining the rate law from experimental data was discussed in question 3
If you had a problem with this question, rewind the tape and listen to the discussion of question 3
All right, let’s recall a couple of helpful hints: if the concentration of a reactant is doubled and the rate law doubles, the reaction is first order with respect to that reactant (In addition, if it triples and the rate law triples, it is first order as well.)
If the concentration of a reactant is doubled or tripled and the rate quadruples or increases by a factor of nine, respectively, the reaction is second order with respect to that reactant
You should know that the concentrations of one or more of the reactants appear in the rate law Are there any choices that can be eliminated? Are there any that contain product concentrations? Unfortunately they are all possible rate laws, so none of the choices can be eliminated Looking at the data for experiments 1 and 2 you can see that the concentration of iodide has been held constant and that the concentration of persulfate has been doubled So, if the concentration of persulfate is
doubled and the rate doubles, it means that the reaction is first order with respect to persulfate Experiments 1 and 3 have the same persulfate concentrations and the concentration of iodide is doubled in experiment 3 What happens to the rate? It doubles
as well The reaction is first order with respect to iodide So, the overall rate law is the rate is equal to the rate constant times the concentration of iodide, times the concentration of persulfate: choice C