1. Supplemental Instruction
CHEM 115b
Professors Krylov and Parr
Perry Kumagai
pkumagai@usc.edu
www.usc.edu/si
Final Exam Review
Chemical Kinetics, Light and Absorbance
1. In a bimolecular second-order decomposition reaction, 53.8% of the reactant is converted to products in 6.39 hours.
a. Calculate the rate constant.
b. Before beginning this experiment, you shined 240 nm light at the solution and 46% of the light was absorbed.
Shining this light through a reference cell resulted in transmittance of 90% of the light. If your reactant has a
molar extinction coefficient of 1.5 L mol-1
cm-1
and you’ve been using a cuvette with a path length of 1.0 cm, find
the half-life of this reaction.
c. How long would it take to have only 7.2% reactant left?
2. Take a look at the following elementary reaction: CH3Br (aq) + OH-
(aq) ! CH3OH (aq) + Br-
(aq). An experiment
is run using 0.100 M of OH-
and 0.00100 M of CH3Br. A graph of ln[CH3Br] versus time yields a straight line, with a
slope of -403 s-1
. Find the k-value for this pseudo-first-order reaction.
3. Consider the following two reaction mechanisms and propose a rate expression for both:
A: NO2 "! O + NO Fast equilibrium, k = k1, k-1
O3 + O ! 2 O2 Fast, k = k2

2. Supplemental Instruction
CHEM 115b
Professors Krylov and Parr
Perry Kumagai
pkumagai@usc.edu
www.usc.edu/si
B: A2 "! A + A Fast equilibrium, k = k1, k-1
A + B2 "! AB + B Fast equilibrium, k = k2, k-2
B + A2 ! AB + A Slow, k = k3
Chemical Equilibrium
4. At a brisk temperature of 1200 K, it was found that the reaction 3 Fe (s) + 4 H2O (g) "! Fe3O4 (s) + 4 H2 (g) has an
equilibrium water vapor pressure of 0.0197 atm and a total pressure of 0.0478 atm. Calculate Kp at 1200 K. What is
Kp for the reverse reaction?
5. Given the following reaction at equilibrium, how would the following reaction shift and how would K change if:
2 NO2 (g) + 7 H2 (g) "! 2 NH3 (g) + 4 H2O (g) ∆H = -87kJ/mol
a. Total pressure is doubled at constant V, T
b. Decrease pressure at constant T
c. You increase the temperature
d. A catalyst is added
e. [NH3] is decreased
f. Add helium gas at constant P, T
Acid-Base Reactions
6. What is the pH of a 1.0 L buffer solution made with 0.400 moles of sodium formate (NaCOOH) and 0.200 moles of
formic acid (HCOOH, Ka = 1.77 x 10-4
)?

3. Supplemental Instruction
CHEM 115b
Professors Krylov and Parr
Perry Kumagai
pkumagai@usc.edu
www.usc.edu/si
a. Your good friend Perry prepares this buffer but then accidentally drops a bunch of solid HCl into the solution
without changing the volume. The new pH is equal to 3.65. How much acid did Perry spill?
7. Suppose you have 1.0 L of 2.0 M HC2H3O2 (Ka = 1.8 x 10-5
). How many moles of NaOH would you have to add,
assuming no volume change, to have a solution buffered at pH = 4.00?
Solubility Equilibrium
8. A solution contains 3.5 x 10-4
M Ag+
and 6.3 x 10-2
M Pb2+
.
a. If Cl-
is gradually added to the solution, will AgCl (Ksp = 1.6 x 10-10
) or PbCl2 (Ksp = 1.6 x 10-5
) precipitate first?
What concentration of Cl-
is necessary to precipitate each salt?

4. Supplemental Instruction
CHEM 115b
Professors Krylov and Parr
Perry Kumagai
pkumagai@usc.edu
www.usc.edu/si
9. Calculate how many moles of AgCl will dissolve in 1.0 L of 1.0 M NH3 solution, if the diamminesilver (I) ion forms
simultaneously. Ksp = 1.6 x 10-10
; Kf = 1.7 x 107
.
Thermodynamics
10. For the reaction A (g) + 2 B (g) "! C (g), the initial pressures are all 0.100 atm. If equilibrium is established, one
finds that PC = 0.040 atm. What is ∆Gº for the reaction at 25 ºC?
a. If the initial pressures of A, B, and C were 0.100 atm, 0.500 atm, and 0.0100 atm, respectively, what would ∆G be
for this reaction?
11. For the autoionization of water, H2O (l) "! H+
(aq) + OH-
(aq), Kw = 1.139 x 10-15
at 0 ºC and 9.614 x 10-14
at 60 ºC.
Assuming ∆Hº and ∆Sº don’t change much over this temperature range, calculate ∆Hº and ∆Sº for this reaction.

5. Supplemental Instruction
CHEM 115b
Professors Krylov and Parr
Perry Kumagai
pkumagai@usc.edu
www.usc.edu/si
Electrochemistry
12. A galvanic cell consists of a silver cathode suspended in 0.500 M AgNO3 and a nickel anode suspended in 0.100 M
Ni(NO3)2 at 0 ºC.
a. Calculate the standard cell voltage, if the standard reduction potentials of Ag+
and Ni2+
are 0.80 V and -0.25 V,
respectively.
b. Draw a diagram of this cell and label the anode, cathode, and the direction of electron flow.
c. Calculate the cell potential at the given conditions.
d. Calculate ∆Gº and K for the reaction at 25ºC.
13. A buffer with pH = 4.05 is connected to a solution with pH = 0 to form an electrochemical cell. What is the voltage
of this cell at 25 ºC?

6. Supplemental Instruction
CHEM 115b
Professors Krylov and Parr
Perry Kumagai
pkumagai@usc.edu
www.usc.edu/si
Nuclear Chemistry
14. Calculate the amount of energy released, in kJ/nucleon of uranium, in the reaction used in the first generation atomic
bombs: 235
92 U + 1
0 n ! 94
36 Kr + 139
56 Ba + 3 1
0 n. Atomic masses of U, Kr, and Ba are 235.044 amu, 93.919 amu, and
138.909 amu, respectively.
mp = 1.00728 amu; mn = 1.00866 amu; me = 5.48580 x 10-4
amu; 1 eV = 1.602 x 10-19
J; 1 amu = 1.6605 x 10-27
kg
Coordination Compounds, Isomerism, Crystal Field Theory
15. Draw all isomers for [Zn(NO2)2Br4]-4
. Write its name out. Is it paramagnetic or diamagnetic?
Organic Chemistry
16. Draw structures for the organic compounds below. If you have an alcohol or amine, classify it.
a. 2-ethylbutanoic acid
b. 4-fluoro-2-propylpentanal

7. Supplemental Instruction
CHEM 115b
Professors Krylov and Parr
Perry Kumagai
pkumagai@usc.edu
www.usc.edu/si
c. 2-amino-4-sec-butyl-3,5-dichlorooctane
17. Name the organic compounds below. If you have an alcohol or amine, classify it.
a.
b.
18. Identify all the functional groups in the following molecule:
OH
O Br
Cl
Cl
O
NH
O
O
N
OH
F
OH
O
O
O O

8. Supplemental Instruction
CHEM 115b
Professors Krylov and Parr
Perry Kumagai
pkumagai@usc.edu
www.usc.edu/si
19. Identify all the chiral centers in the following molecule:
20. Complete the following reactions:
a. CH3CH3 + Cl2 !
b.
c.
21. Polyvinyl chloride (PVC) is formed by addition polymerization of vinyl chloride . Draw a portion of
PVC that contains three monomer subunits.
Cl
Cl
OH
F
O
+
HO
HO
O
Dehydration
OH
Cl

9. Supplemental Instruction
CHEM 115b
Professors Krylov and Parr
Perry Kumagai
pkumagai@usc.edu
www.usc.edu/si
22. Given the two monomers shown below, draw the structure of one dimer unit formed if the two undergo condensation
copolymerization.
H2N
NH2
HO
OH
O
O
Initials: __________
I VIII
1
H
1.01
II III IV V VI VII
2
He
4.003
3
Li
6.94
4
Be
9.01
5
B
10.81
6
C
12.01
7
N
14.01
8
O
16.00
9
F
19.00
10
Ne
20.18
11
Na
22.99
12
M g
24.31
13
Al
26.98
14
Si
28.09
15
P
30.97
16
S
32.07
17
Cl
35.45
18
Ar
39.95
19
K
39.10
20
Ca
40.08
21
Sc
44.97
22
Ti
47.88
23
V
50.94
24
Cr
52.00
25
M n
54.94
26
Fe
55.85
27
Co
58.93
28
Ni
58.69
29
C u
63.55
30
Zn
65.38
31
Ga
69.72
32
Ge
72.59
33
As
74.92
34
Se
78.96
35
Br
79.90
36
Kr
83.80
37
Rb
85.47
38
Sr
87.62
39
Y
88.91
40
Zr
91.22
41
Nb
92.91
42
M o
95.94
43
Tc
(99)
44
R u
101.1
45
Rh
102.9
46
Pd
106.4
47
Ag
107.9
48
Cd
112.4
49
I n
114.8
50
S n
118.7
51
S b
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
A u
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
R n
(222)
87
Fr
(223)
88
Ra
226.0
89
Ac
227.0
104
Rf
(261)
105
Db
(262)
106
Sg
(263)
107
Bh
(262)
108
Hs
(265)
109
Mt
(268)
Lanthanides
58
Ce
140.1
59
Pr
140.9
60
Nd
144.2
61
Pm
(145)
62
S m
150.4
63
E u
152.0
64
Gd
157.3
65
Tb
158.9
66
Dy
162
67
Ho
164.9
68
Er
167.3
69
Tm
168.9
70
Yb
173.0
71
L u
175
Actinides
90
Th
232.0
91
Pa
231.0
92
U
238.0
93
Np
237.0
94
P u
(244)
95
Am
(243)
96
Cm
(247)
97
Bk
(247)
98
Cf
(251)
99
Es
(252)
100
Fm
(257)
101
M d
(258)
102
No
(259)
103
Lr
(26)
Useful constants:
c = 3.00 x 108
m/s R = 8.314 J/K*mol = 0.08206 L*atm/K*mol
F = 96485 C/mol e-
1 A = 1 C/s
-14 o

10. Supplemental Instruction
CHEM 115b
Professors Krylov and Parr
Perry Kumagai
pkumagai@usc.edu
www.usc.edu/si
1
Lanthanides
58
Ce
140.1
59
Pr
140.9
60
Nd
144.2
61
Pm
(145)
62
S m
150.4
63
E u
152.0
64
Gd
157.3
65
Tb
158.9
66
Dy
162
67
Ho
164.9
68
Er
167.3
69
Tm
168.9
70
Yb
173.0
71
L u
175
Actinides
90
Th
232.0
91
Pa
231.0
92
U
238.0
93
Np
237.0
94
P u
(244)
95
Am
(243)
96
Cm
(247)
97
Bk
(247)
98
Cf
(251)
99
Es
(252)
100
Fm
(257)
101
M d
(258)
102
No
(259)
103
Lr
(26)
Useful constants:
c = 3.00 x 108
m/s R = 8.314 J/K*mol = 0.08206 L*atm/K*mol
F = 96485 C/mol e-
1 A = 1 C/s
Kw = 1 x 10-14
@ 25 o
C 1 atm = 760 torr
1 eV = 1.6022 x 10-19
J
Useful equations:
ΔGo
= -RT ln(K) ΔG = -nFε
ε = εo
– (RT/nF) ln Q ε = εo
– (0.0591/n) log Q
ΔG = ΔH – TΔS ΔSuniv = ΔSsys + ΔSsurr
ΔSsurr = -ΔH/T ΔG = ΔGo
+ RT ln(Q)
pH = pKa + log([A-
]/[HA]) ln(K) = -(ΔHo
/RT) + (ΔSo
/R)
ΔHo
rxn = Σnp[ΔHf
o
products] – Σnr[ΔHf
o
reactants] ΔGo
rxn = Σnp[ΔGf
o
products] – Σnr[ΔGf
o
reactants]
ΔSo
rxn = Σnp[ΔSo
products] – Σnr[ΔSo
reactants] RT
Ea
Ae
k /
a
ac
b
b
x
2
4
2
2
1
1
2 1
1
ln
T
T
R
E
k
k a
ln[A] = -kt + ln[A]o (1/[A]) = kt + (1/[A]o) [A] = -kt + [A]o
t1/2 = 0.693/k t1/2 = 1/(k[A]o) t1/2 = [A]o/2k
1
2
1
2 1
1
ln
T
T
R
H
K
K o
rxn
15
Lanthanides
58
Ce
140.1
59
Pr
140.9
60
Nd
144.2
61
Pm
(145)
62
Sm
150.4
63
E u
152.0
64
Gd
157.3
65
Tb
158.9
66
Dy
162
67
Ho
164.9
68
Er
167.3
69
Tm
168.9
70
Yb
173.0
71
L u
175
Actinides
90
Th
232.0
91
Pa
231.0
92
U
238.0
93
Np
237.0
94
P u
(244)
95
Am
(243)
96
Cm
(247)
97
Bk
(247)
98
Cf
(251)
99
Es
(252)
100
Fm
(257)
101
M d
(258)
102
No
(259)
103
Lr
(26)
Useful constants:
c = 3.00 x 108
m/s R = 8.314 J/K*mol = 0.08206 L*atm/K*mol
F = 96485 C/mol e-
1 A = 1 C/s
Kw = 1 x 10-14
@ 25 o
C 1 atm = 760 torr
1 eV = 1.6022 x 10-19
J
Useful equations:
ΔGo
= -RT ln(K) ΔG = -nFε
ε = εo
– (RT/nF) ln Q ε = εo
– (0.0591/n) log Q
ΔG = ΔH – TΔS ΔSuniv = ΔSsys + ΔSsurr
ΔSsurr = -ΔH/T ΔG = ΔGo
+ RT ln(Q)
pH = pKa + log([A-
]/[HA]) ln(K) = -(ΔHo
/RT) + (ΔSo
/R)
ΔHo
rxn = Σnp[ΔHf
o
products] – Σnr[ΔHf
o
reactants] ΔGo
rxn = Σnp[ΔGf
o
products] – Σnr[ΔGf
o
reactants]
ΔSo
rxn = Σnp[ΔSo
products] – Σnr[ΔSo
reactants] RT
Ea
Ae
k /
a
ac
b
b
x
2
4
2
2
1
1
2 1
1
ln
T
T
R
E
k
k a
ln[A] = -kt + ln[A]o (1/[A]) = kt + (1/[A]o) [A] = -kt + [A]o
t1/2 = 0.693/k t1/2 = 1/(k[A]o) t1/2 = [A]o/2k
1
2
1
2 1
1
ln
T
T
R
H
K
K o
rxn
Supplemental Materials for Exam 1 (Chem115b)
Summary of reaction kinetics:
Zero order:
0
1/ 2
2
C
t
First order:
Second order: for 2A → products
for A → products
Arrhenius equaiton:
R= 0.0082 L atm mol-1K-1 = 8.314 J mol-1K-1
( )
a
E
RT
k T Ae
−
=
0
C C kt
= −
k
=
e 1/ 2
ln 2
t =
0
kt
C C −
= k
1/ 2
0
1
0
1 1
2kt
2
t =
C C
= +
kC
1/ 2
0
1
0
1 1
kt t
kC
=
C C
= +