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Chemistry 114: Fundamental Chemistry
Third sample test, 2014.
Name:
ID:
R = 8.31447 J/(mol. K)
ln(K2/K1) = –(ΔH°/R)(1/T2 – 1/T1)
ΔG = ΔG° + RT ln Q
ΔG° = –RT ln K
G = H –TS
S = kBlnW
Sm = S° – R ln p (for an ideal gas)
Group**
Period
1
1
5
6
7
8
9
10 11 12 13
17
18
4.003
5
6
7
8
9
10
Li Be
B
C
N
O
F
Ne
6.941 9.012
10.81
12.01
14.01
16.00
19.00
20.18
13
14
15
16
17
18
4
12
Na Mg
Al Si
P
S
22.99 24.31
26.98
28.09
30.97
32.07
35.45
39.95
31
32
33
34
35
36
20
21
22
23
24
25
26
27
28
29
30
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As
38
39
40
41
42
43
44
45
46
47
48
49
Cl Ar
Se
Br Kr
72.59
74.92
78.96
79.90
83.80
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 87.62 88.91 91.22 92.91 95.94
127.6
126.9
131.3
85
86
55
56
72
73
74
(98)
75
101.1 102.9 106.4 107.9 112.4 114.8
76
77
78
79
80
81
118.7
121.8
82
83
84
Cs Ba * Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi
Po
At Rn
132.9 137.3
87
7
16
1.008
37
6
15
2
39.10 40.08 44.96 47.88 50.94 52.00 54.94 55.85 58.47 58.69 63.55 65.39 69.72
5
14
He
19
4
4
1
11
3
3
H
3
2
2
178.5 180.9 183.9 186.2 190.2 190.2 195.1 197.0 200.5 204.4
88
104
105
106
107
108
109
110
111
112
113
207.2
209.0
(210)
(210)
(222)
114
115
116
117
118
Fr Ra ** Rf Db Sg Bh Hs Mt Ds Rg Cn Uut Uuq Uup Uuh Uus Uuo
(223) (226)
Lanthanide
Series*
Actinide
Series**
(257) (260) (263)
57
58
59
60
(262) (265) (266) (271) (272) (277)
61
62
63
64
65
66
(?)
(285)
67
68
(?)
69
(289)
70
71
La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu
138.9 140.1 140.9 144.2 (147) 150.4 152.0 157.3 158.9 162.5 164.9
89
90
91
92
93
94
95
96
97
98
99
167.3
100
168.9 173.0 175.0
101 102 103
Ac Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr
(227) 232.0 (231) (238)
(237) (242) (243) (247) (247) (249) (254)
(253)
(256) (254) (257)
(?)
(?)
Short answer section (40 points). Underline the correct alternative. There is
only one in each case. Answer all questions.
(1) When we burn lithium in air, the major oxide formed is (a)Li2O (b)Li2O2 (c)LiO2
(d)Li2O3
(2) Which of these compounds is a Lewis acid (a) NH3 (b) BCl3 (b) CCl4 (d) SF6
(3) Aluminum oxide is (a) basic (b) acidic (c) amphoteric (d) none of the above.
(3) At 0 K, the entropy of a pure crystalline substance becomes equal to /zero / the
Gibbs free energy / the enthalpy/ infinity
(4) Plotted against temperature, the entropy of a pure substance /always slopes
upward / always slopes downward / usually but not always slopes upward / usually
but not always slopes downward.
(5) If you double the pressure of a mole of an ideal gas, keeping the temperature
constant, the free energy relative to the standard free energy /doubles /decreases by
a factor of 2/ increases by a factor RT ln 2 / decreases by a factor RT ln 2
(6) A solution of AlCl3 is expected to be /basic/neutral/acid/blue
(7) When water boils at 1 bar pressure and approximately 100°C, which does not
increase? /the entropy of the system/ the enthalpy of the system/ the Gibbs free
energy of the system / the internal energy of the system.
(8) In the solid state, lead monoxide PbO is (a) ionic (b) network covalent (c) molecular
covalent (d) metallic.
(9) For the reaction 2Na(s) + 2H2O (l) → 2 NaOH (aq) + H2 (g) at 1 bar, the entropy
change is / positive /zero /negative/ I need a table of S° values to decide.
(10) Which is not an expression of the second law of thermodynamics? /heat always
flows spontaneously from hot to cold/ the entropy of the universe is always
increasing/ one cannot extract work from a system by cooling it below the
temperature of its surroundings/ the Gibbs free energy is always positive.
(11) (extra credit) Which topic has Chancellor Harvey Perlman not addressed in his
Perls of Knowledge video series (A clear sign the topic has jumped the shark) / the
zombie apocalypse / ice cream from the Dairy Store / YOLO / Lady Gaga (thank
heavens)
Long answer section (60 points): complete 5 of 8 questions. Mark the ones you
want graded. If you do not, we will grade the first three you in any way attempt.
1) Bromine data at 25°C.
ΔHf° kJ/mol
ΔGf° kJ/mol
S° J/(mol.K)
Br2 (g.)
30.91
3.11
245.5
Br2 (l.)
0
0
152.2
Predict the vapor pressure of bromine liquid at 25°C and the boiling point at 1 bar
pressure.
RT ln K = RT ln pv = –3110 J mol so p = 0.285 bar.
At the boiling point ΔG = 0, and ( G/ T)p = –S, so ΔT ~ –ΔG/ΔS =
3110/(245.5 – 152.2) = 33.3, so Tb = 33.3 + 25 = 58.5°C.
2) Chlorate ions (ClO3–) disproportionate in strong acid solution to give chlorine dioxide
(ClO2) and perchlorate (ClO4–). Write a balanced chemical equation for the reaction.
3 ClO3– + 2 H+ → 2 ClO2 + ClO4– + H2O
3) (The rich get heavy metal poisoning too!) You are feeding your baby with orange
juice via a silver spoon (you know how that is!) The orange juice is at pH 3.
Assuming each spoonful of the orange juice comes to equilibrium with the spoon and
with atmospheric oxygen at 0.2 bar, what will the concentration of silver ions be in
the orange juice? (hint; write down a balanced equation for the oxidation of silver by
oxygen under acid conditions first)
Ag+ + e− ⇌ Ag(s)
:
E° = +0.7996 V
O2(g) + 4 H+ + 4 e− ⇌ 2 H2O
:
E° = +1.229
4Ag(s) ⇌ 4Ag+ + 4e−
:
E° = –0.7996 V
O2(g) + 4 H+ + 4Ag(s) ⇌ 2H2O + 4Ag+: E° = 0.4294 V
Q = [Ag+]4/(pO2[H+]4) = [Ag+]4/2.0 × 10-13
At equilibrium (RT/NF)ln Q = E° = 0.4294 V, so ln Q = 66.9, so Q = 1.08 × 1029
so [Ag+]4 = 2.16 × 1016, so [Ag+] =1.2 × 104. The spoon will dissolve completely!
(given enough time)
4) Maltose binds to maltose binding protein with the following equilibrium
Maltose + MBP ⇌ Maltose-MBP.
It has an equilibrium constant of 180,000 at 300 K, changing to 44,000 at 277 K.
Determine ΔH° and ΔS° for the reaction, and ΔG° at 300 K.
ΔG° = –RT ln K = –3.0 × 104 J/mol
ΔH° = –R(ln 180000/44000)/(1/300-1/277) = 4.2 × 104 J/mol
ΔS° = (ΔH° – ΔG°)/T = 241.7 J mol–1 K–1
5) In the synthesis of gaseous methanol from CO and H2 according to the reaction
CO(g) + 2 H2(g) ⇌ 2 CH3OH
at 483 K, the equilibrium pressures measured were pCO = 0.911 bar; pH2 = .822 bar;
pCH3OH = 0.0892 bar. Calculate the equilibrium constant Kp and the standard Gibbs
free energy change for the reaction.
K = 0.08922/(.911 × .8222) = .0129
ΔG° = –RT ln K = 17.4 kJ/mol
6) What will be the peak voltage of a battery composed of solid lithium and solid MnO2
plates, and an electrolyte containing an aqueous solution of 0.01 M H+, 1 M Li+ and
0.1 M Mn2+? If one plate is composed of 100 g of pure MnO2, what should the mass
of Li on the other plate be, and how much total energy in J will the battery provide?
Li+ (aq) + e– → Li (s) E° = –3.040 V
MnO2 (s) + 4 H+ (aq) + 2 e– → Mn2+ (aq) + 2 H2O (l)
E° = 1.23V
+
2+
MnO2 (s) + 4 H (aq) + 2 Li (s) → Mn (aq) + 2 H2O (l) + 2 Li+ (aq) E° = 4.27V
E = E° – (RT/NF) ln Q = 4.06 V
100 g = 100/(54.94+2× 16)mol = 1.15 mol, reacts with 2.30 mol Li = 16.0 g
gives 2.30 mol electrons = 2.22 × 105 C × 4.06 V = 9.02 × 105 J
7) Tl3I4 is the empirical formula of a compound of unknown structure. Write formulas of
it in terms of the oxidation state of thallium [e.g. Fe3O4 is actually Fe(II) Fe(III)2O4 ]
The total oxidation number of I is –4; there is no way 3 Tl atoms can have an
oxidation number of +4. So let’s try doubling it to Tl6I8; now we can make it
Tl(I)5Tl(III)1I8