Chemistry 105 B
Practice Exam 1
PLEASE PRINT YOUR NAME IN BLOCK LETTERS
First Letter of
last Name
Name: __________________________________
Last 4 Digits of USC ID:____ ____ ____ ____
Dr. Jessica Parr
Lab TA’s Name: _________________________________
Question
1
2
3
4
5
6
7
8
9
Points
18
14
9
10
9
7
12
10
11
Total
100
Score
Grader
Please Sign Below:
I certify that I have observed all the rules of Academic Integrity while taking this examination.
Signature: _______________________________________________________________
Instructions:
1. You must show work to receive credit.
2. If necessary, please continue your solutions on the back of the preceding page (facing you).
3. YOU MUST use black or blue ink. (No pencil, no whiteout, no erasable ink.)
4. There are 9 problems on 11 pages. Please count them before you begin. A periodic table and
some useful equations can be found on the last page.
5. Good luck!! =)
1
1. (18 pt) Circle the correct answers for the following questions.
i. The catalyzed pathway in a reaction mechanism has a __________ activation energy that
causes a _____________ reaction rate.
a) higher, lower
b) higher, higher
d) lower, steady
e) higher, steady
c) lower, higher
ii. Which of the following statements best describes the condition(s) needed for a successful
formation of a product according to the collision model.
a) The collision must involve a sufficient amount of energy, provided by the motion of
the particles, to overcome the activation energy.
b) The relative orientation of the particles has little or no effect on the formation of the
product.
c) The relative orientation of the particles has an effect only if the kinetic energy of the
particles is below some minimum value.
d) The relative orientation of the particles must allow for formation of the new bonds in
the product.
e) The energy of the incoming particles must be above a certain minimum value and the
relative orientation of the particles must allow for formation of new bonds in the product.
iii. For a particular system at a particular temperature there _________ equilibrium constant(s)
and there __________ equilibrium position(s).
a) are infinite, is one
b) is one, are infinite
d) are infinite, is one
e) none of these
c) is one, is one
iv. Which of the following is true for a system with a relatively small equilibrium constant?
a) It will take a short time to reach equilibrium
b) The equilibrium lies to the left
c) It will take a long time to reach equilibrium
d) The equilibrium lies to the right
e) Two of these
2
v. Equilibrium is reached in chemical reactions when:
a) the rates of the forward and reverse reactions become equal.
b) the concentrations of reactants and products become equal.
c) the temperature shows a sharp rise.
d) all chemical reaction stop.
e) the forward reaction stops.
vi. Given the reaction A (g) + B (g) ↔ C (g) + D (g). You have the gases A, B, C, and D at
equilibrium. Upon adding more of gas A, the value of K:
a) increase because by adding A, more products are made, increasing the product to
reactant ratio.
b) decreases because A is a reactant so the product to reactant ratio decreases.
c) does not change because A does not figure into the product to reactant ratio.
d) does not change as long as the temperature remains constant.
e) depends on whether the reaction is endothermic or exothermic.
3
2. (14 pt) Answer the following questions using the following data for the reaction:
4 HBr (g) + O2 (g) → 2 H2O (g) + 2 Br2 (g)
Time (s)
0
10
20
30
40
50
60
70
80
90
100
[HBr] (mol/L)
1.00
0.85
0.73
0.64
0.56
0.50
0.45
0.41
0.38
0.35
0.33
(a)
What is the half-life for this reaction?
____________________________________
(b)
What is the concentration of H2O after 30 s? ____________________________________
(c)
What is the concentration of Br2 after 80 s? ____________________________________
(d)
If the initial concentration of O2 is 1.0 M, what is the concentration of O2 after 40 s?
______________________________________________________________________________
(e)
The rate of disappearance of HBr is ____________ times the rate of disappearance of O2.
(f)
The rate of appearance of H2O is ____________ times the rate of disappearance of O2.
(g)
The rate of appearance of Br2 is ____________ times the rate of disappearance of O2.
4
3. (9 pt) Consider the following data concerning the equation: H2O2 + 3 I- + 2 H+ ↔ I3- + 2 H2O
[I-]
5.00 x 10-4 M
1.00 x 10-3 M
1.00 x 10-3 M
1.00 x 10-3 M
[H2O2]
0.100 M
0.100 M
0.200 M
0.400 M
Experiment
1
2
3
4
[H+]
1.00 x 10-2 M
1.00 x 10-2 M
1.00 x 10-2 M
2.00 x 10-2 M
rate
0.137 M/s
0.268 M/s
0.542 M/s
1.084 M/s
i. The rate law for the reaction is:
a) rate = k[H2O2][I-][H+]
b) rate = k[H2O2]2[I-]2[H+]2
c) rate = k[I-][H+]
d) rate = k[H2O2][H+]
e) rate = k[H2O2][I-]
ii. The average value of the rate constant k (without units) is
a) 2710
iii.
b) 2.74 x 104
c) 137
d) 108
e) none of these
Two mechanisms are proposed:
I.
H2O2 + I- Æ H2O + OI-
II.
H2O2 + I- +H+ Æ H2O + HOI
OI- + H+ Æ HOI
HOI + I- + H+ Æ I2 + H2O
HOI + I- + H+ Æ I2 + H2O
I2 + I- Æ I3-
I2 + I- Æ I3Which of the following describes a potentially correct mechanism?
a) Mechanism I with the first step the rate determining step.
b) Mechanism I with the second step the rate determining step.
c) Mechanism II with the first step the rate determining step.
d) Mechanism II with the second step the rate determining step.
e) None of the above is correct.
5
4. (10 pt) For a reaction aA Æ products, [A]o = 6.0 M, and the first two half-lives are 56 and 28
minutes, respectively.
a. Determine k, with appropriate units.
b. Calculate [A] at t = 99 minutes.
6
5. (9 pt) A first order reaction with an initial concentration of 1.0 M is found to have a half-life of
1.26 s at 57.0 oC, and a half-life of 0.0277 s at 99 oC. What is the activation energy for this
reaction?
7
6. (7 pt) Consider the reaction: 4 PH3 (g) Æ P4 (g) + 6 H2 (g). If, in a certain experiment, over a
specific time period, 0.0048 mol PH3 is consumed in a 2.0 L container each second of reaction,
what are the rates of production of P4 and H2 in this experiment?
7. (12 pt) For the reaction: H2 (g) + Br2 (g) ↔ 2 HBr (g) Kp = 3.5 x 104 at 1495 K. What is the value
of Kp for the following reactions at 1495 K?
a. HBr (g) ↔ ½ H2 (g) + ½ Br2 (g)
b. 2 HBr (g) ↔ H2 (g) + Br2 (g)
c. 2 H2 (g) + 2 Br2 (g) ↔ 4 HBr (g)
8
8. (10 pt) The following equilibrium pressures were observed at a certain temperature for the
reaction: N2 (g) + 3 H2 (g) ↔ 2 NH3 (g): PNH3 = 0.031 atm, PN2 = 0.85 atm, PH2 = 0.0031 atm.
a. What is the value of Kp for this reaction?
b. If PN2 = 0.525 atm, PNH3 = 0.0167 atm, and PH2 = 0.00761 atm, does this represent a system at
equilibrium?
9
9. (11 pt) At 327 oC, the equilibrium concentrations are [CH3OH] = 0.15 M, [CO] = 0.24 M, and
[H2] = 1.1 M for the reaction: CH3OH (g) ↔ CO (g) + 2 H2 (g). Calculate Kp at this temperature.
10
1
2
H
He
1.008
4.003
3
4
5
6
7
8
9
10
Li
Be
B
C
N
O
F
Ne
6.941
11
9.012
12
10.81
13
12.01
14
14.01
15
16.0
16
19.00
17
20.18
18
Na
Mg
Al
Si
P
S
Cl
Ar
22.99
19
24.31
20
21
22
23
24
25
26
27
28
29
30
26.98
31
28.09
32
30.97
33
32.07
34
35.45
35
39.95
36
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
39.10
40.08
44.969
47.88
50.94
51.996
54.9380
58.9332
58.69
63.546
65.377
69.72
72.59
74.9216
78.96
79.90
83.80
37
38
39
40
41
42
43
55.84
7
44
45
46
47
48
49
50
51
52
53
54
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
85.47
55
87.62
56
88.91
57
91.22
72
92.91
73
95.94
74
(99)
75
101.1
76
102.9
77
106.4
78
107.9
79
112.4
80
114.8
81
118.7
82
121.8
83
127.6
84
126.9
85
131.3
86
Cs
Ba
La
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
132.9
87
137.3
88
138.9
89
178.5
104
180.9
105
183.85
106
186.2
107
190.2
108
192.2
109
195.09
197.0
200.6
204.4
207.2
209.0
(209)
(210)
(222)
Fr
Ra
Ac
Rf
Db
Sg
Bh
Hs
Mt
(223)
226.0
227.0
(261)
(262)
(263)
(262)
(265)
(268)
Lanthanides
Actinides
58
59
60
61
62
63
64
65
66
67
68
69
70
71
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
140.1
90
140.9
91
144.2
92
(145)
93
150.4
94
151.96
95
157.3
96
158.9
97
162
98
164.9
99
167.3
100
168.9
101
173.0
102
175
103
Th
Pa
U
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
232.0
231.0
238.0
237.0
(244)
(243)
(247)
(247)
(251)
(252)
(257)
(258)
(259)
(26)
R = 0.08206 L*atm/K*mol = 8.314 J/K*mol
PV = nRT
k = A exp(-Ea/RT)
⎛k ⎞ E ⎛ 1 1 ⎞
ln⎜⎜ 2 ⎟⎟ = a ⎜⎜ − ⎟⎟
⎝ k1 ⎠ R ⎝ T1 T2 ⎠
[ A] = −kt + [ A]0
ln[ A] = −kt + ln[ A]0
1
1
= kt +
[ A]
[ A]0
11