Download Answers to Exam I

CHEM 1515.001 - 006
Exam I
John I. Gelder
February 5, 2002
Name ________________________
TA's Name ________________________
Section _______
INSTRUCTIONS:
1. This examination consists of a total of 8 different pages.
The last three pages include a periodic table, a table of
vapor pressures for water, a solubility table and a table of
thermodynamic values. All work should be done in this
booklet.
2. PRINT your name, TA's name and your lab section
number now in the space at the top of this sheet. DO
NOT SEPARATE THESE PAGES.
3. Answer all questions that you can and whenever called
for show your work clearly. Your method of solving
problems should pattern the approach used in lecture.
You do not have to show your work for the multiple
choice or short answer questions.
4. No credit will be awarded if your work is not shown in
4a, 4d and 8.
5. Point values are shown next to the problem number.
6. Budget your time for each of the questions. Some
problems may have a low point value yet be very
challenging. If you do not recognize the solution to a
question quickly, skip it, and return to the question after
completing the easier problems.
7. Look through the exam before beginning; plan your
work; then begin.
8. R e l a x and do well.
SCORES
Page 2
Page 3
Page 4
Page 5
TOTAL
_____
(23)
_____
(35)
_____
(18)
_____
(24)
______
(100)
CHEM 1515 EXAM I
(9)
PAGE 2
1. Write the chemical formula(s) of the product(s) and balance the following reactions. Identify all products
phases as either (g)as, (l)iquid, (s)olid or (aq)ueous. Soluble ionic compounds should be written in the
form of their component ions.
+
2H2O(l) →
a)
2K(s)
b)
Na2CO3(aq)
c)
Ba(OH) 2 · 8H2O (s)
+
2K + (aq)
+
2OH– (aq)
+ H2(g)
Ca(NO3)2(aq) → 2Na+(aq) + 2NO3–(aq) + CaCO3(s)
+ 2NH 4SCN (s) → Ba 2+(aq) + 10H2O (l) + 2NH3(aq) + 2SCN– (aq)
(4) 2a. Write the ionic and net ionic chemical equation for 1a) or 1b).
Ionic equation
2Na + (aq) + CO3 2-(aq) + Ca2+ (aq) + 2NO3 – (aq) → 2Na + (aq) + 2NO3 – (aq) + CaCO3 (s)
Net Ionic equation
CO 3 2- (aq) + Ca2+ (aq) → CaCO 3 (s)
(10) 3. Predict whether the entropy change in the system is positive or negative for each of the following
processes.
a) CH3OH(l) + O2(g) → HCO2H(l) + H2O(l)
The phases are both liquid, while the reactants are a liquid and a gas. There are equal
numbers of moles, and the liquid phase is more ordered compared to the gas phase, so the
products are more ordered compared to the reactants and ∆S is –.
b) 2SO3(g) → 2SO2(g)
+ O2(g)
The reactants and products are all in the same phase, but there are more moles of products
compared to reactants. So there are more possible arrangements in the products compared to
the reactants and ∆ S is +.
CHEM 1515 EXAM I
PAGE 3
4. Methanol, CH3OH, can be synthesized by reacting hydrogen gas with carbon monoxide, according to the
chemical equation,
CO(g) + 2H2(g) → CH3OH(l)
(18) a) Calculate ∆H˚rxn, ∆S˚rxn and ∆G˚rxn.
CO(g)
+
2H2(g)
C(s) +
2H2 (g)
→
CH3OH (l)
°  kJ 
∆Hf mol
-110.5
-0
–239
 J 
198
131
127
S°  .
K mol
°  kJ 
∆Gf mol
-137
-0
–166
°
°
°
∆Hrxn = Σ∆Hf (products) - Σ ∆H f (reactants)
°
°
°
= [∆H f (CH3OH (l)) ] – [∆H f (CO(g)) + 2∆H f (H 2 (g) ) ]
= (-239 kJ) - [(-110.5 kJ) + 2(0 kJ)] = –128 kJ
°
∆S rxn = Σ S°(products) - Σ S°(reactants)
= [S˚(CH3 OH (l) ) ] – [S˚(CO(g) )+ 2S˚(H2 (g) )]
J
J
J
= [(127 K )] - [ (198 K ) + 2(131 K )] = –333 J/K
°
°
°
∆Grxn = Σ∆Gf (products) - Σ ∆G f (reactants)
°
°
°
= [∆G f (CH3OH (l)) ] – [∆G f (CO(g)) + 2∆G f (H 2 (g) ) ]
= (-166 kJ) - [(-137 kJ) + 2(0 kJ)] = –29 kJ
(4) b) Which factor, the change in enthalpy, ∆H˚, or the change in entropy, ∆S˚, provides the principal driving
force for the reaction at 298 K? Explain.
∆ H is the principal driving force, because the sign of ∆ H in the reaction is negative. A –∆
∆H
∆ S does not favor spontaneity.
favors spontaneity. ∆ S is also negative. –∆
∆ S a spontaneous reaction has a negative ∆ G. In this equation spontaneity is
Since ∆ G = ∆ H – T∆
∆
∆ S. In the reaction in part a only the sign of ∆ H favors spontaneity.
favored by a –∆ H and a +∆
(5) c) For the reaction, how is the value of the standard free energy change, ∆G˚, affected by an increase in
temperature? Explain.
∆ S, since ∆ H is negative and ∆ S is negative as the temperature
Using the equation ∆ G = ∆ H – T∆
∆
∆ S is more positive, making ∆ G˚
increases the T∆ S term is becoming more negative, and –T∆
more positive and less spontaneous.
(8) d) Calculate the ∆S˚f for CH3OH(l).
+
1
2
O2(g)
 J 
S°  .
6
131
205
K mol
°
∆S rxn = Σ S°(products) - Σ S°(reactants)
→
CH3OH (l)
127
1
= [S˚(CH3 OH (l) ) ] – [S˚(C(s) )+ 2S˚(H2 (g) ) + 2 S˚(O2 (g) ) ]
J
J
J
J
1
= [(127 K )] - [ (6 K ) + 2(131 K )+ 2 (205 K )] = –243 J/K
CHEM 1515 EXAM I
PAGE 4
(12) 5. A sample of water is in the vacuum flask shown below.
The vacuum flask side arm has vacuum rubber tubing with one end attached to the flask and the other end
attached to a vacuum pump (can not see in this diagram). When the vacuum pump is on any gas/vapor in
the flask is removed.
In an experiment the water placed into the flask initially is at room temperature. The rubber tubing is
attached to the flask and the vacuum pump, and the pump is turned on. After the pump has been on for
several minutes the following two observations are made:
1.
the volume of water in the flask has decreased;
2.
what water remains has turned to ice.
Explain both of these observations.
The volume of liquid water decreases because as the vacuum pump removes vapor from above
the liquid, particles escape the liquid phase to maintain the equilibrium vapor pressure. Since
particles are evaporating and few particles are condensing, the volume of the liquid will
decrease.
The reason the liquid eventually turns to solid is because evaporation is an endothermic
process, and heat is removed from the system (the liquid water). When enough heat is removed
the liquid freezes.
Evaporation is endothermic because the conversion of water in the liquid phase to water in the
vapor phase requires breaking hydrogen-bonding forces. Breaking bonds is always an
endothermic process.
(6)
6. Give the name or draw the complete Lewis structure (showing all C-H(X) bonds) for each of the
following compounds.
4-ethyl-2,3-dimethylhexane
3, 3, 5–trimethylheptane
1,1-dichloro-2,2-difluoroethane
CHEM 1515 EXAM I
PAGE 5
(12) 7. Draw and name six different structural isomers for C8H18. (NOTE: You may use condensed formulas
when representing the different structural isomers.)
n–octane
2-methylheptane
3-methylheptane
4-methylheptane
2,3-dimethylhexane
3,3-dimethylhexane
CHEM 1515 EXAM I
PAGE 6
(12) 8. CH3I has a vapor pressure of 400 mm Hg at 25.3 ˚C. Calculate the temperature that CH3I has a vapor
pressure of 40.0 mm Hg. ∆H˚vap for CH3I is 29.2 kJ mol-1.
P1
∆H
-∆
ln P 2 = R
(T11 - T12)
J
-29200 mol
400
ln 40 =
J
8.314 mol • K
( 298.31 K - T11)
1
– T1 )
1
2.303 = -3512 K (3.35 x 10-3 – T 1 )
1
–6.56 x 10-4 = 3.35 x 10-3 – T 1
1
-3
T1 = 250 K
T 1 = 4.00 x 10
ln (10) = -3512 K (3.35 x 10-3
CHEM 1515 EXAM I
PAGE 7
Periodic Table of the Elements
IA
VIIIA
1
1
H
1.008
3
2
2
He
IIA
IIIA IVA VA VIA VIIA 4.00
4
Li Be
6.94 9.01
11
12
3
Na Mg
22.99 24.30
19
20
4
5
6
7
5
6
7
8
9
10
B
C
N
O
F Ne
10.81 12.01 14.01 16.00 19.00 20.18
13
14
15
16
17
18
VIII
IIIB IVB VB VIB VIIB
21
22
K Ca Sc Ti
23
24
25
26
27
IB
28
29
Al
Si
P
S
Cl Ar
31
32
33
34
35
IIB 26.98 28.09 30.97 32.06 35.45 39.95
30
36
V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
39.10 40.08 44.96 47.88 50.94 52.00 54.94 55.85 58.93 58.69 63.55 65.38 69.72 72.59 74.92 78.96 79.90 83.80
37
38
39
40
41
42
43
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 87.62 88.91 91.22 92.91 95.94 (98) 101.1 102.9 106.4 107.9 112.4 114.8 118.7 121.8 127.6 126.9 131.3
55
56
57
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
Cs Ba La Hf Ta W Re Os Ir
Pt Au Hg Tl Pb Bi Po At Rn
132.9 137.3 138.9 178.5 180.9 183.8 186.2 190.2 192.2 195.1 197.0 200.6 204.4 207.2 209.0 (209) (210) (222)
87
88
89 104 105 106
Fr Ra Ac
(223) 226.0 227.0 (261) (262) (263)
58
Lanthanides
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
140.1 140.9 144.2 (145) 150.4 152.0 157.2 158.9 162.5 164.9 167.3 168.9 173.0 175.0
90
91
92
93
94
95
96
97
98
99 100 101 102 103
Actinides
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) (260)
Useful Information
L·atm
J
PV = nRT
R = 0.0821 mol·K = 8.314 mol·K
 vp2
∆H˚vap  1 1 
g
density of H2O = 1.00 3
lnvp1 = – R T2 – T1
cm
g
density of H2O = 1.00 3
cm
∆H˚rxn = ∑n(∆Hf˚(products)) – ∑ m(∆Hf˚(reactants))
∆S˚rxn = ∑ n(S˚(products)) – ∑ m(S˚(reactants))
∆G˚rxn = ∑ n(∆Gf˚(products)) – ∑ m(∆Gf˚(reactants))
∆G˚ = ∆H˚ - T∆S˚
CHEM 1515 EXAM I
PAGE 8
Temperature (°C)
-5
0
5
10
15
20
25
30
35
40
45
Vapor
Pressure(mmHg)
3.2
4.6
6.52
9.20
12.8
17.5
23.8
31.8
42.1
55.3
71.9
Temperature (°C)
50
55
60
65
70
75
80
85
90
95
100
Vapor
Pressure(mmHg)
92.5
118.0
149.4
187.5
233.7
289.1
355.1
433.6
525.8
633.9
760
Solubility Table
Ion
NO3–
ClO4–
Cl–
I–
SO42–
Solubility
soluble
soluble
soluble
soluble
soluble
Exceptions
none
none
except Ag+, Hg22+, *Pb2+
except Ag+, Hg22+, Pb2+
except Ca2+, Ba2+, Sr2+, Hg2+, Pb2+, Ag+
CO32–
insoluble
except Group IA and NH4
PO43–
-OH
S2–
Na+
NH4+
K+
insoluble
insoluble
insoluble
soluble
soluble
soluble
except Group IA and NH4
except Group IA, *Ca2+, Ba2+, Sr2+
except Group IA, IIA and NH4+
none
none
none
*slightly soluble
+
+
CHEM 1515 EXAM I
PAGE 9
Thermodynamic Values (25 ˚C)
Substance
o
∆H f
and State
 kJ 
 mol
Carb o n
C(s) (graphite)
C(s) (diamond)
CO(g)
CO2(g)
CH4(g)
CH3OH(g)
CH3OH(l)
CH3Cl(g)
CHCl3(g)
CHCl3(l)
H2CO(g)
HCOOH(g)
HCN(g)
C2H2(g)
C2H4(g)
CH3CHO(g)
C2H5OH(l)
C2H6(g)
C3H6(g)
C3H8(g)
0
2
-110.5
-393.5
?
-201
-239
-80.8
-100.8
-131.8
-116
-363
135.1
227
52
-166
-278
-84.7
20.9
-104
Bromine
Br2(l)
BrCl(g)
0
14.64
o
∆G f
 kJ 
 mol
So
Substance
 J 
 K.mol
0
3
-137
-394
-51
-163
-166
-57.4
6
2
198
214
186
240
127
234
-110
-351
125
209
68
-129
-175
-32.9
62.7
-24
219
249
202
201
219
250
161
229.5
266.9
270
0
-0.96
152.
240
Chlorine
Cl 2 (g)
Cl2(aq)
Cl-(aq)
HCl(g)
0
-23
0
7
223
121
-167
-92
-131
-95
57
187
Fluorine
F2(g)
F-(aq)
HF(g)
0
-333
-271
0
-279
-273
203
-14
174
Hydrogen
H2(g)
H(g) 217
H+(aq)
OH-(aq)
H2O(l)
H2O(g)
Magnesium
Mg(s)
Mg(aq)
MgO(s)
0
203
0
-230
0
115
0
-157
131
-242
-229
189
0
-492
-601
0
-456
-569
33
-118
26.9
0
-11
o
o
So
∆H f
and State
 kJ 
 mol
∆G f
 kJ 
 mol
Oxygen
O2(g)
O(g) 249
O3(g)
0
232
143
0
161
163
239
Nitrogen
N2(g)
NCl3(g)
NF3(g)
NH3(g)
NH3(aq)
NH2CONH2(aq)
NO(g)
NO2(g)
N2O(g)
N2O4(g)
N2O5(g)
HNO3(aq)
HNO3(l)
NH4Cl(s)
NH4ClO4(s)
0
230
-125
?
?
?
90
32
82
10
-42
-207
-174
-314
-295
0
271
-83.6
-17
-27
?
87
52
104
98
134
-111
-81
-201
-89
192
-137
-139
193
111
174
211
240
220
304
178
146
156
95
186
Silver
Ag(s)
Ag+(aq)
Ag(S2O3)3-(aq)
AgBr(s)
AgCl(s)
0
105.6
-1285.7
-100.4
-127.1
0
77.1
--96.9
-109.8
42.6
72.7
-107.1
96.2
 J 
 K.mol
205
Sulfur
S(rhombic)
SO2(g)
SO3(g)
H2S(g)
0
-296.8
-395.7
-20.17
0
-300.2
-371.1
-33.0
31.8
248.8
256.3
205.6
Titanium
TiCl4(g)
TiO2(s)
-763
-945
-727
-890
355
50
Aluminum
AlCl3(s)
-526
-505
184
Barium
BaCl2(aq)
-872
Ba(OH)2·8H2O(s) -3342
-823
-2793
123
427
Iodine
I2(s)
HI(g)
0
25.94
0
1.30
116.7
206.3
CHEM 1515 EXAM I
H 2O
→
PAGE 10