Download Answers Exam V Spring 2002

CHEM 1515.001 - 006
Exam V
John V. Gelder
May 6, 2002
Name ________________________
TA's Name ________________________
Section _______
INSTRUCTIONS:
1. This examination consists of a total of 11 different pages.
The last four pages include a periodic table; many useful
equations and constants; a table of vapor pressures for
water; a solubility table; a table of equilibrium constants for
acids and bases; 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 6.
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
Page 6
Page 7
TOTAL
_____
(18)
_____
(16)
_____
(16)
_____
(18)
_____
(20)
_____
(12)
______
(100)
CHEM 1515 EXAM V
(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.
a)
Zn(NO3)2(aq)
b)
Na2CO 3(s)
c)
(CH3)2NH(aq)
Na2S(aq) → ZnS(s) + 2NO3-(aq) + 2+(aq)
+
+
HCl(aq) → CO 2(g) + H2O (l)
+
+ 2Cl-(aq) + 2Na+(aq)
HCN(aq) → (CH3)2NH2+(aq)
+
CN-(aq)
(4) 2a. Write the ionic and net ionic chemical equation for 1a), 1b) or 1c).
Ionic equation
2Na + (aq) + CO3 2-(g) + 2H+ (aq)
+ 2Cl-(aq) → CO 2 (g) + H2 O (l)
+ 2Cl-(aq) + 2Na+ (aq)
Net Ionic equation
CO 3 2- (g) + 2H+ (aq) → CO 2 (g) + H2 O (l)
(5)
3. Account for the following observations about NH3 and NF3. In your answer, use appropriate principles
of intermolecular forces. Your answer must include reference to both substances.
a)
NH3 has a normal boiling point of –33.4 ˚C where as NF3 has a boiling point of –128.8 ˚C.
NH3 is polar and can hydrogen bond. Hydrogen-bonding is a strong intermolecular
attractive force and causes the boiling point of NH3 to be high. NF3 is polar but can not
hydrogen bond. The dipole-dipole forces are not as strong an intermolecular attractive force
so the boiling point of NF3 is much lower compared to NH3.
CHEM 1515 EXAM V
PAGE 3
(16) 4.
0.10 M
NaF
0.10 M
MgCl2
0.10 M
C2H5OH
0.10 M
CH3COOH (HC2H3O2)
Answer the following questions, which refer to the 100 mL samples of aqueous solutions at 25 ˚C in the
stoppered flasks shown above
(a) Which solution has the lowest electrical conductivity? Explain.
The lowest electrical conductivity has the fewest ions… C2H5OH. All other substances above
form ions in solution.
(b) Which solution has the lowest freezing point? Explain.
MgCl2 has he lowest freezing point. Since MgCl2 produces the most ions in solution with an I
value of 3, ∆ Tf will be the largest.
(c) Above which solution is the pressure of water vapor the greatest? Explain.
C2H5OH does not ionize, so there are the fewest particles in solution so it will have the higher
vapor pressure.
(d) Which solution has the highest pH? Explain.
NaF will have the highest pH. F- is a weak base.
CHEM 1515 EXAM V
PAGE 4
(16) 5a. Define the term 'equilibrium vapor pressure'.
(b)
If a sample of a liquid is injected into an evacuated container, describe what phase change occurs.
(Note the amount of the liquid injected is small compared to the volume of the container.)
Since the container is evacuated when some liquid is introduced, some of the liquid will
immediately vaporize until the equilibrium vapor pressure of the liquid is reached, or until all of
the liquid evaporates. So if the equilibrium vapor pressure is reached, both liquid and vapor are
present. Otherwise only vapor.
(c)
Is the phase change exothermic or endothermic? Explain?
Evaporation is endothermic, because to escape from the liquid phase to the vapor phase
intermolecular attractive forces must be over come. Although weak these forces must be broken
for particles to enter the vapor phase.
(d)
In terms of the 'pressure exerted by the vapor' and the 'equilibrium vapor pressure' how do we
determine what phase(s) are present in the container after adding the sample?
If the pressure due to the vapor is equal to the equilbirium vapor pressure both liquid and
vapor must be present. If the pressure due to the vapor is less than the equilibrium vapor
pressure only vapor is present. The pressure dur to the vapor can never exceed the
equilibrium vapor pressure.
CHEM 1515 EXAM V
PAGE 5
(18) 6. A commercial aqueous solution of ammonia is 28% NH3 by mass and has a density of 0.900 g mL-1.
(a) Calculate the molarity of the solution.
Assume 100 gram of solution
28 g of NH3 and 72 grams of water. Convert the mass of NH3 to moles, then divide the
100 gram of solution by the density to get mLs of solution. Convert mLs to liters and divide
into the moles for the molarity
Answer is 14.8 M
(b) Calculate the molality of the solution.
Use the moles determined in part a and divide by the mass of water in kilograms.
Answer is 22.9 molal
(c) Describe how you would prepare 1.00 L of a 0.400 M NH3 solution from the more concentrated
solution in part a).
Using the dilution equation Mc·Vc = Md · Vd
The volume of concentrated ammonia needed is 27.0 mL.
Add 27 mL of NH3 to about 900 mL of water, be sure all the NH3 is dissolved than add
enough water to reach a final volume of 1000 mLs.
(10) (d) Calculate the pH of the solution prepared in part c)
.
The pH of the solution is 11.43. This is a simple weak base calculation where the initial
concentration of NH3 is 0.400 M
(10) (e) Calculate the pH of the solution prepared by adding 0.250 mol of solid NH4Cl to the solution prepared
in part c).
This is a common ion/buffer solution. The pH is 9.45.
CHEM 1515 EXAM V
PAGE 6
(12) 7. For the gaseous equilibrium represented below, it is observed that greater amounts of PCl3 and Cl2 are
produced as the temperature is increased.
PCl3(g) + Cl2(g)
PCl3(g)
ä
(a)
What is the sign of ∆S˚ for the reaction? Explain.
∆ S˚ is positive. In the balanced equation there are more moles of gase in the products
compared to the reactants. So the products are more disordered.
(b)
What change, if any, will occur in ∆G˚ for the reaction as the temperature is increased? Explain
your reasoning in terms of thermodynamic principles.
The reaction is endothermic because , ", it is observed that greater amounts of PCl3 and Cl2 are
produced as the temperature is increased ". So if ∆ H˚ is positive and ∆ S˚ is positive ∆ G˚ will be
∆ S˚ means the
come more negative with increasing temperature. The equation ∆ G˚ = ∆ H˚ – T∆
∆
reaction with become more spontaneous as T increases. The T∆ S˚ term will eventually be greater
than ∆ H˚ and the reaction will be spontaneous.
(c)
If the volume of the reaction mixture is decreased at constant temperature to half the original
volume, what will happened to the number of moles of Cl2 in the reaction vessel? Explain.
If the volume is decreased the pressure will increase in the container and the reaction will
proceed in a direction to relive the pressure, so it will got from right to left and the
amount of Cl2 will decrease.
Q is inversely related to the volume, so as volume decreases Q increases. To re-establish
equilibrium the reaction must go from R to L and amount of Cl2 will decrease.
CHEM 1515 EXAM V
PAGE 7
Periodic Table of the Elements
IA
VIIIA
1
H
1
1.008
3
2
He
IIA
IIIA IVA VA VIA VIIA 4.00
4
Li Be
2
6.94 9.01
11
12
Na Mg
3
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
lnvp  = – R T – T 
density of H2O = 1.00 3
1
2
1
cm
∆H˚rxn = ∑n(∆Hf˚(products)) – ∑ m(∆Hf˚(reactants))
6.02 x 1023
∆S˚rxn = ∑ n(S˚(products)) – ∑ m(S˚(reactants))
∆G˚rxn = ∑ n(∆Gf˚(products)) – ∑ m(∆Gf˚(reactants))
∆G˚ = ∆H˚ - T∆S˚
pH = –log[H+]
∆G˚ = –RTlnK
x 1,2 =
pH + pOH = 14
Kp = Kc(RT)∆n
–b± b2 – 4ac
for ax2 + bx + c = 0
2a
pOH = –log [OH–]
Kw = 1.00 x 10–14
CHEM 1515 EXAM V
PAGE 8
1 pm = 10-12 m
Psolution = χsolventP˚solvent
∆T = ikm
˚C
kf(H2O) = 1.86 m
edge length (l) = 2r
˚C
kb(H2O) = 0.512 m
edge length (l) = 2 2· r
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
edge length (l) =
Temperature (°C)
50
55
60
65
70
75
80
85
90
95
100
4r
3
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 V
Name
PAGE 9
Formula
Ka1
Acetic
Ascorbic
HC2H3O2
Arsenic
H3AsO4
1.8 x 10– 5
8.0 x 10– 3
5.6 x 10– 3
Arsenous
Benzoic
Boric
Butyric acid
Carbonic
Cyanic
Citric
H3AsO3
HC7H5O2
H3BO3
HC4H7O2
H2CO3
HCNO
H3C6H5O7
6.0
6.5
5.8
1.5
4.3
3.5
7.4
Formic
Hydroazoic
Hydrocyanic
Hydrofluoric
Hydrogen chromate ion
Hydrogen peroxide
Hydrogen selenate ion
Hydrogen sulfate ion
Hydrogen sulfide
Hypobromous
Hypochlorous
Hypoiodus
Iodic
Lactic
Malonic
Oxalic
HCHO2
HN3
HCN
HF
HCrO4–
H2O2
HSeO4–
HSO4–
H2S
HBrO
HClO
HIO
HIO3
HC3H5O3
H2C3H2O4
H 2C 2O 4
1.8
1.9
4.9
7.2
3.0
2.4
2.2
1.2
5.7
2.0
3.0
2.0
1.7
1.4
1.5
5.9
Nitrous
Phenol
Phosphoric
Paraperiodic
Propionic
Pyrophosphoric
Selenous
Sulfuric
Sulfurous
Tartaric
HNO2
HC6H5O
H3PO4
H5IO6
HC3H5O2
H4P2O
H2SeO3
H2SO4
H2SO3
H2C4H4O6
E.2
HC6H7O6
10–10
10– 5
10–10
10– 5
10– 7
10– 4
10– 4
x 10– 4
x 10– 5
x 10–10
x 10– 4
x 10– 7
x 10–12
x 10– 2
x 10– 2
x 10– 8
x 10– 9
x 10– 8
x 10–11
x 10– 1
x 10– 4
x 10– 3
x 10– 2
x 10– 4
x
x
x
x
x
x
x
4.5
1.3 x 10–10
7.5 x 10– 3
2.8 x 10– 2
1.3 x 10– 5
3.0 x 10– 2
2.3 x 10– 3
strong acid
1.7 x 10– 2
1.0 x 10–3
Ka2
Ka3
1.0 x 10– 7
3.0 x 10–12
5.6 x 10–11
1.7 x 10– 5
4.0 x 10– 7
1.3 x 10–13
2.0 x 10– 6
6.4 x 10– 5
6.2 x 10– 8
5.3 x 10– 9
4.4
5.3
1.2
6.4
4.6
x
x
x
x
x
4.2 x 10–13
10– 3
10– 9
10– 2
10– 8
10– 5
DISSOCIATION CONSTANTS FOR BASES AT 25˚C
Name
Formula
Ammonia
Aniline
Dimethylamine
Ethylamine
Hydrazine
NH3
C6H5NH2
(CH3)2NH
C2H5NH2
H2NNH2
Kb
1.8
4.3
5.4
6.4
1.3
x
x
x
x
x
10– 5
10–10
10– 4
10– 4
10– 6
Name
Formula
Hydroxylamine
Methylamine
Pyridine
Trimethylamine
HONH2
CH3NH2
C5H5N
(CH3)3N
Kb
1.1
4.4
1.7
6.4
x
x
x
x
10– 8
10– 4
10– 9
10– 5
CHEM 1515 EXAM V
PAGE 10
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