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FORM A
Principles of Chemistry II
3150:153-431
EXAM I
Wednesday, 11:30AM
July 22, 2009
120
Instructions:
1.
Each student is responsible for following instructions. Read this page carefully.
2.
Write your name on this page and on your computer answer sheet.
3.
Enter your information on your computer answer sheet using an ordinary (No. 2) pencil. It is very
important to code the information correctly!! SEE BELOW for details.
4.
Put all calculations on the examination pages. Do not make any extra marks on the computer answer
sheet!!
5.
This exam consists of 28 multiple-choice questions worth 4.5 points each. For each multiple-choice
question, choose the ONE best or correct answer and write it both on your exam paper and on the
answer sheet.
6.
This exam booklet consists of 6 pages (including this one), a Periodic Table & Formula/Data Sheet,
and a sheet of blank paper. Please check to be sure that you have them all!
KEEP YOUR EXAM BOOKLET AND ANSWER SHEET COVERED TO PROTECT THE
INTEGRITY OF YOUR WORK!
PROHIBITED DURING EXAM:
cell phones, pagers, laptops, PDAs, headphones, hats worn low over the eyes
Please put these items away until you have left the room.
_____ 1.
A student determined the value of the rate constant, k, for a chemical reaction at several
different temperatures. Which graph of the student's data would give a straight line?
(a)
(b)
(c)
(d)
(e)
_____ 2.
k versus T
k versus (1/T)
ln k versus (1/T)
ln k versus T
none of these
Consider this reaction and the following initial rate data.
2 ClO2 (aq) + 2 OH– (aq) → ClO3– (aq) + ClO2– (aq) + H2O (l)
(ClO2)o, mol/L
(OH–)o, mol/L
initial rate, mol/L•s
0.0500
0.100
5.77 × 10-2
0.100
0.100
2.32 × 10-1
0.100
0.050
Determine the rate law for the reaction.
(a) rate = k (ClO2) (OH–)
(b) rate = k (ClO2)2 (OH–)
(c) rate = k (ClO2) (OH–)2
(d) rate = k (ClO2)2 (OH–)2
(e) rate = k
1.15 × 10-1
The next two questions deal with this reaction: 2 N2O5 (g) → 4 NO2 (g) + O2 (g)
ΔH = + 111 kJ
EA = + 122 kJ
_____ 3.
What is EA of the reverse reaction (4 NO2 (g) + O2 (g) → 2 N2O5 (g))?
(a)
(b)
(c)
(d)
_____ 4.
– 122 kJ
+ 122 kJ
+ 233 kJ
+ 11 kJ
If the reaction is first-order in N2O5, which plot should give a straight line?
(a) (N2O5) vs time
(b) ln(N2O5) vs time
(c) 1/(N2O5) vs time
(d) (N2O5)2 vs time
(e) none of these
_____ 5.
Which factor(s) affect the effectiveness of collisions between particles in a chemical reaction?
(a) orientation of molecules
(b) concentration
(c) energy of colliding molecules
(d) all of these
_____ 6.
Under which set of conditions would a gas be most soluble in a liquid?
(a)
(b)
(c)
(d)
high pressure, low temperature
low pressure, low temperature
high pressure, high temperature
low pressure, high temperature
CHEM 153 Exam I
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2/21/08
_____ 7.
Consider this reaction: 2 N2O5 (g) → 4 NO2 (g) + O2 (g)
Which correctly relates the rate of appearance of O2 to the rate of appearance of NO2?
(a)
(b)
(c)
(d)
(e)
Δ(O 2 )
Δt
Δ(O 2 )
Δt
Δ(O 2 )
Δt
Δ(O 2 )
Δt
Δ(O 2 )
Δt
1 ⎛ Δ(NO 2 ) ⎞
⎜
⎟
4 ⎝ Δt ⎠
1 ⎛ Δ(NO 2 ) ⎞
= ⎜−
⎟
4⎝
Δt ⎠
Δ(NO 2 )
=–
Δt
⎛ Δ(NO 2 ) ⎞
= 4⎜
⎟
⎝ Δt ⎠
⎛ Δ(NO 2 ) ⎞
= 4⎜ −
⎟
Δt ⎠
⎝
=
_____ 8.
Consider this reaction: (CH3)3CBr (aq) + OH– (aq) → (CH3)3COH (aq) + Br– (aq)
This mechanism has been proposed for the reaction:
(CH3)3CBr (aq) → (CH3)3C+ (aq) + Br– (aq)
slow step
+
+
(CH3)3C (aq) + H2O (l) → (CH3)3COH2 (aq)
fast step
+
–
(CH3)3COH2 (aq) + OH (aq) → (CH3)3COH (aq) + H2O (l)
fast step
What rate law should be observed if this mechanism were correct?
(a) rate = k ((CH3)3CBr) (OH–)
(b) rate = k ((CH3)3CBr) ((CH3)3C+) ((CH3)3COH2+) (H2O) (OH–)
(c) rate = k ((CH3)3CBr)
(d) rate = k ((CH3)3CBr) (H2O) (OH–)
(e) rate = k ((CH3)3CBr) (H2O)
_____ 9.
Which aqueous solution will freeze at the highest temperature? (Recall that common
compounds of the Cs+ ion are soluble.)
(a)
(b)
(c)
(d)
(e)
_____ 10.
What is the mole fraction of water in a solution that contains 9.0 g of water (18 g/mol) and 92 g
of ethyl alcohol (46 g/mol)?
(a)
(b)
(c)
(d)
(e)
_____ 11.
A 0.10 m solution of CsNO3
A 0.10 m solution of Cs2SO4
A 0.10 m solution of Cs3PO4
A 0.10 m solution of Cs4[CoBr6] (contains the [CoBr6]4– ion.)
All would freeze at the same temperature since they are all 0.10 m solutions.
0.16
0.20
0.33
0.50
0.87
What is the molarity of a solution of hydrogen peroxide (H2O2, 34 g/mol) containing 3.0%
H2O2 by mass? The density of the aqueous solution is 1.0 g/mL.
(a) 0.088 M
(b) 0.30 M
(c) 0.88 M
(d) 1.0 M
(e) 1.3 M
CHEM 153 Exam I
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7/22/09
_____ 12.
Which step(s) in the solution process is/are endothermic?
I. Separation of solute molecules
II. Separation of solvent molecules
III. Formation of solute-solvent interactions
(a) I only.
(b) II only.
(c) III only.
(d) I and II.
(e) I, II, and III.
_____ 13.
Below is the structure of Vitamin A1. Vitamin A1 should be
(a) more soluble in water than in fats.
(b) less soluble in water than in fats.
(c) equally soluble in water and in fats.
_____ 14.
Which represents a substitutional alloy? (Filled circles represent one kind of atom, and open
circles represent a different type of atom.)
(a)
(b)
(c) Both represent a substitutional alloy equally well.
_____ 15.
Which type(s) of intermolecular force(s) would a sample of pure Compound “A” exhibit that a
sample of pure Compound “B” would NOT exhibit? H
F
H
F
(a)
(b)
(c)
(d)
(e)
_____ 16.
dispersion forces
dipole-dipole forces
hydrogen bonding
(b) and (c)
(a), (b), and (c)
C
H
C
C
C
H
F
Compound A
F
Compound B
Consider acetic acid (structure at right). How many of the hydrogen atoms in an acetic acid
molecule can form hydrogen bonds with atoms on other molecules?
(a) 0
(b) 1
(c) 2
(d) 3
(e) 4
H
H
O
C
C
O
H
H
_____ 17.
Compounds that have relatively weak intermolecular forces are likely to have:
(a) relatively low melting points, low boiling points, and low vapor pressures.
(b) relatively high melting points, high boiling points, and high vapor pressures.
(c) relatively low melting points, low boiling points, and high vapor pressures.
(d) relatively high melting points, high boiling points, and low vapor pressures.
CHEM 153 Exam I
4
7/22/09
_____ 18.
What is the overall order of a reaction that has a rate constant with units L/mol•s?
(a) –1
(b) 0
(c) 1
(d) 2
(e) none of these
_____ 19.
Quartz and diamond are examples of what class of solid crystalline substances?
(a) atomic
(b) molecular
(c) ionic
(d) metallic
(e) network covalent
_____ 20.
What is the molecular formula of this compound?
(a)
(b)
(c)
(d)
(e)
_____ 21.
_____ 23.
O
2.
3.
6.
12.
18.
Which correctly lists the three alcohols in order of INCREASING solubility in water?
(a) CH3(CH2)2CH2OH < CH3(CH2)7CH2OH < CH3(CH2)5CH2OH
(b) CH3(CH2)7CH2OH < CH3(CH2)5CH2OH < CH3(CH2)2CH2OH
(c) CH3(CH2)5CH2OH < CH3(CH2)7CH2OH < CH3(CH2)2CH2OH
(d) CH3(CH2)2CH2OH < CH3(CH2)5CH2OH < CH3(CH2)7CH2OH
(e) None of the above responses is correct.
What is the activation energy of a reaction if k at 500 K is 2.9 × 10–1 L/mol•s and k at 400 K is
6.6 × 10–4 L/mol•s?
(a)
(b)
(c)
(d)
(e)
_____ 24.
NH2
A reaction is second-order with respect to the concentration of reactant A and first-order with
respect to the concentration of reactant B. Doubling the concentration of A and tripling the
concentration of B will cause the rate of the reaction rate to increase by a factor of
(a)
(b)
(c)
(d)
(e)
_____ 22.
C9H17O2N
C9H19O2N
C9H21O2N
C7H15O2N
none of these
O
1.0 × 102 J/mol
1.2 × 103 J/mol
1.0 × 105 J/mol
1.3 × 106 J/mol
1.2 × 108 J/mol
The presence of a non-volatile solute in water will
(a) lower the vapor pressure and freezing point and raise the boiling point.
(b) lower the freezing point, vapor pressure, and boiling point.
(c) lower the freezing point and raise the vapor pressure and boiling point.
(d) raise the boiling point, freezing point, and vapor pressure.
(e) lower the vapor pressure and boiling point and raise the freezing point.
CHEM 153 Exam I
5
7/22/09
_____ 25.
Consider converting 1.5 mol of ice at –25 °C to liquid water at 35 °C. What calculation would
give the quantity of heat required by this process in J?
The molar heat capacity of ice is 37.6 J/mol °C. The molar heat capacity of liquid water is
75.4 J/mol °C. The heat of fusion of water is 6.02 kJ/mol.
(a) q = (1.50 mol)(
37.6 J
)(35 °C − ( − 25 °C))
mol °C
(b) q = (1.50 mol)(
75.4 J
)(35 °C − ( − 25 °C))
mol °C
37.6 J
6.02 kJ ⎞
⎛
⎞ ⎛
)(0 °C − ( − 25 °C)) ⎟ + ⎜ (1.50 mol)(
)⎟ +
(c) q = ⎜ (1.50 mol)(
mol °C
mol ⎠
⎝
⎠ ⎝
75.4 J
⎛
⎞
)(35 °C − 0 °C) ⎟
⎜ (1.50 mol)(
mol °C
⎝
⎠
37.6 J
6.02 kJ 1000 J ⎞
⎛
⎞ ⎛
)(0 °C − ( − 25 °C)) ⎟ + ⎜ (1.50 mol)(
)(
)⎟ +
(d) q = ⎜ (1.50 mol)(
mol °C
mol
kJ ⎠
⎝
⎠ ⎝
75.4 J
⎛
⎞
)(35 °C − 0 °C) ⎟
⎜ (1.50 mol)(
mol °C
⎝
⎠
6.02 kJ 1000 J ⎞
⎛
)(
)⎟
(e) q = ⎜ (1.50 mol)(
mol
kJ ⎠
⎝
Use the figure below for questions 26 – 28.
_____ 26. Which point on the phase diagram lies entirely in the
gas phase?
(a) A
(b) B
(c) C
(d) D
(e) E
_____ 27. Which point on the phase diagram is the critical point?
(a) D
(b) E
(c) B
(d) G
(e) none of these
_____ 28. Which state of the substance in the phase diagram is more dense?
(a) liquid
(b) solid
(c) It is impossible to tell from this phase diagram.
END OF EXAM
Check to see that you have entered answers to questions 1-28 on your scan sheet. Proofread your work.
You may take this exam booklet with you but you must return your scan sheet before leaving the room!
CHEM 153 Exam I
6
7/22/09
CHEM 153 Exam I
7
7/22/09
Equations/Constants/Conversion Factors
TK = TC + 273.15
TF = (9/5)TC + 32
TC = (5/9) [TF – 32]
Avogadro: 6.022 × 1023
R = 8.314 J/mol·K
R = 0.08206 L·atm/mol·K
PA = XA PEA
rate = k (A) , n=0, 1 or 2
k = Ae
KP = KC (RT)Δn
ΔTB = i kB m
(A) = !kt + (A)o
pH = – log[H3O+]
ΔTF = – i kF m
ln(A) = !kt + ln(A)o
EA ⎛ 1 ⎞
ln k = –
⎜ ⎟ + ln A
R ⎝T⎠
TF = TF° + ΔTF
1
1
= kt +
( A)
( A)o
n
TB = TB° + ΔTB
kF for water = 1.86 °C/m
o
[H 3O + ] = K W + K A Cacid
[OH _ ] = K BCobase
[OH _ ] = K W + K BCobase
ln((A)o / (A)) = kt
ΔSSURR =
ΔG = –nFE
−ΔHSYS
T
KW = 1.0 × 10–14 at 25°C
(at constant P)
pH + pOH = 14.00 at 25°C
ΔHE = ΣnΔH°F(products) – ΣnΔH°F(reactants)
ΔG = ΔH ! TΔS
o
[H 3O + ] = K A Cacid
⎛k ⎞ E ⎛ 1
1 ⎞
ln ⎜ 2 ⎟ = A ⎜ −
⎟
R ⎝ T1 T2 ⎠
⎝ k1 ⎠
ln((A) / (A)o) = !kt
kB for water = 0.512 °C/m
ΔG° = –RT ln K
–EA/RT
KW = KAKB
ΔG = ΔG° + RT ln Q ΔSE = ΣnS°(products) – ΣnS° (reactants)
[H+] = [H3O+]
ΔG = wMAX
pH = pK A + log
ΔG° = ΣnΔG°F(products) – ΣnΔG°F(reactants)
ECELL = E°CELL !
1A
(1)
1
H
RT
ln Q
nF
ln K =
nFE D
RT
F = 96485 coul/mol e–
[A _ ]
[HA]
cubic cells 52%, 68%, 74%
8A
(18)
2
He
3
Li
2A
(2)
4
Be
3A 4A 5A 6A 7A
(13) (14) (15) (16) (17)
5
6
7
8
9
B
C
N
O
F
6.941
9.012
10.81
12.01
14.01
16.00
19.00
20.18
11
Na
12
Mg
13
Al
14
Si
15
P
16
S
17
Cl
18
Ar
22.99
24.31
26.98
28.09
30.97
32.07
35.45
39.95
19
K
31
Ga
32
Ge
33
As
34
Se
35
Br
36
Kr
1.008
20
Ca
3B
(3)
21
Sc
4B
(4)
22
Ti
5B
(5)
23
V
6B
(6)
24
Cr
7B
(7)
25
Mn
39.10
40.08
44.96
47.88
50.94
52.00
37
Rb
38
Sr
39
Y
40
Zr
41
Nb
42
Mo
85.47
87.62
88.91
91.22
92.91
55
Cs
56
Ba
71
Lu
72
Hf
73
Ta
132.91
137.33
174.97
178.49
180.95
87
Fr
88
Ra
103 104 105 106 107 108 109 110 111 112
Lr Rf Db Sg Bh Hs Mt
114
116
(223)
[226]
(260)
(261)
(262)
(263)
(262)
(265)
(266)
(269)
(272)
(266)
(285)
( )
57
La
58
Ce
59
Pr
60
Nd
61
Pm
62
Sm
63
Eu
64
Gd
65
Tb
66
Dy
67
Ho
68
Er
69
Tm
70
Yb
138.91
140.12
140.91
144.24
(145)
150.36
151.96
157.25
158.93
162.50
164.93
167.26
168.93
173.04
89
Ac
90
Th
91
Pa
92
U
93
Np
94
Pu
95 96
Am Cm
97
Bk
98
Cf
99
Es
100 101 102
Fm Md No
[227]
232.04
231.04
238.03
[237]
(244)
(243)
(247)
(251)
(252)
(257)
4.003
10
Ne
(8)
26
Fe
8B
(9)
27
Co
1B 2B
(10) (11) (12)
28 29 30
Ni Cu Zn
54.94
55.85
58.93
58.69
63.55
65.39
69.72
72.61
74.92
78.96
79.90
83.80
43
Tc
44
Ru
45
Rh
46
Pd
47
Ag
48
Cd
49
In
50
Sn
51
Sb
52
Te
53
I
54
Xe
95.94
(98)
101.07
102.91
106.42
107.87
112.41
114.82
118.71
121.76
127.60
126.90
131.29
74
W
75
Re
76
Os
77
Ir
78
Pt
79
Au
80
Hg
81
Tl
82
Pb
83
Bi
84
Po
85
At
86
Rn
183.84
186.21
190.23
192.22
195.08
197.00
200.59
204.38
207.2
208.98
(209)
(210)
(222)
(247)
(258)
(259)