Chem 1515 Section 2
Problem Set #1
Spring 1998
Name__________________________
TA Name ______________________
Lab Section #______
ALL work must be shown to receive full credit. Due in lecture at 1:30 p.m.
Friday, January 23rd
PS1.1. Indicate all the various types of intermolecular attractive forces that may operate
in each of the following:
a) C2H6(l);
London dispersion only
b) H2S(l);
Dipole-dipole and London dispersion
c) CH3NH2(l); Hydrogen-bonding and London dispersion
d) MgCl2(s)
Ionic
PS1.2. What are the required structural features for a substantial hydrogen-bonding
contribution to the intermolecular attractive forces between two identical or
different substances?
XÐH ..... :Y where X can be oxygen, nitrogen or fluorine and :Y is
oxygen, nitrogen, fluorine, sulfur or chlorine.
PS1.3. What is the strongest intermolecular attraction, or bond, that must be broken
when each of the following substances is melted?
a) nitrogen monoxide Dipole-dipole and London dispersions forces :
since nitrogen and oxygen are bond second period weÕll assume
the dipole-dipole is more important than the dispersion forces.
b) boron trifluoride
London dispersion forces is the only force in
this case so it is the most important by default.
c) ammonium chloride Ionic interaction and London dispersion
forces : but LDF are extremely small compared to ionic bonds.
d) bromine
London dispersion forces are the only forces.
e) propane
London dispersion forces are the only forces.
January 26, 1998
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Spring 1998
PS1.4 For each of the following pairs of substances predict which will have the
higher boiling point and indicate why:
a) CO2 or OCS
b) CH3CH2OH or HOCH2CH2OH
OCS will have the higher boiling
point for two reasons; it is a polar
compound while CO 2 does not, so
OCS has some dipole-dipole
forces, and since S will have a
higher polarizability compared to O
we would expect a greater
contribution from the dispersion
forces.
Ethylene glycol will have the
higher boiling point compared to
ethanol, because it has one more
OH functional group than ethanol.
The additional OH means more
hydrogen bonding is possible.
c) HCl or KCl
d) N2 or Ba
KCl will have the higher boiling
point because it is an ionic
compound and the interparticle
attractive forces are amoung the
strongest-ionic bonds. HCl is
polar covalent and the dipoledipole and dispersion forces are
very weak compared to the ionic
bond in KCl.
Ba will have the higher boiling
point due to the strongere metallic
bonds that occur in the Ba solid.
N 2 has only dispersion forces
which are very weak compared to
metallic bonds.
January 26, 1998
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Spring 1998
PS1.6. Define the term equilibrium vapor pressure.
the pressure due to particles of a substance in the vapor phase
above its liquid in a closed container at a given temperature.
b) Use a vapor-pressure table (in your text or some other reference book) to
look up the equilibrium vapor pressure of a sample of water at 95 ûC and
at 83 ûC.
The vapor pressure of water at 95 ûC is 633.9 mmHg and at 83
ûC the vapor pressure is 400.6 mmHg.
c) Consider two closed containers each partially filled with liquid water one
at 95 ûC and the other at 83 ûC. Can the pressure of water vapor in the
gas phase in either container ever exceed the equilibrium vapor pressure at
the particular temperature? Explain why or why not.
No. At a given temperature we cannot have a pressure due to
the vapor above a liquid greater than the equilibrium vapor
pressure. If we attempt to add additional water, in the vapor
phase, to a system already at equilibrium, the rate of
condensation increases until the vapor pressure re-establishes
equilibrium. The net result is there is no change in the vapor
pressure.
PS1.7. A sample of water vapor in a flask of constant volume exerts a pressure of
635 mm Hg at 100 ûC. The flask is slowly cooled.
a) Assuming no condensation, use the Ideal Gas Law to calculate the
pressure of the vapor at 95 ûC; at 83 ûC.
P áT
635mmHg ¥ 368K
@95ûC P 2 = 1 2 =
= 626 mmHg
T1
373K
@83ûC P 2 =
P 1 áT 2
635mmHg ¥ 356K
=
T1
373K
= 606 mmHg
b) Will condensation occur at 95 ûC; 83 ûC?
Since the calculated pressure of the sample in the vapor at 95
ûC is less than the equilibrium vapor pressure no
condensation occurs. Condensation occurs at 83ûC because
the calculate pressure exerted by the vapor is greater than the
equilibrium vapor pressure at that temperature.
c) On the basis of your answers in a) and b), predict the pressure exerted by
the water vapor at 95 ûC; at 83 ûC.
The pressure due to the vapor at 95ûC is 626 mmHg, at 83ûC
the vapor pressure is 400.6 mmHg.
Can you determine the volume of water which has condensed at
83ûC?
January 26, 1998
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Spring 1998
PS1.8. Consider the following data for the vapor pressure:
Lithium
T (K)
Pv (mm Hg)
750
1
890
10
1080
100
1240
400
1310
760
Magnesium
T (K)
620
740
900
1040
1110
1
T
for each metal and use your graph to determine the slope of the best line
through the data. The heat of vaporization of a liquid can be obtained
from such a plot. The relationship is given as,
∆Hûvap
slope = Ð
J
8.314 moláK
Calculate the heat of vaporization for lithium and magnesium. (Note: Be
sure to clearly label the graph.)
Slope = Ð11566 K Ð 1
Vapor Pressure Data for Lithiu
∆Hû vap
slope = Ð
J
8 . 3 1 4 moláK
8
y = -11566x + 15.363
J
Ð11566 K Ð1 á Ð8.314 mo
6
∆Hû vap = 96.2 kJ
4
ln(vp)
ln (vapor
pressure)
a) Use Cricket Graph (or any other graphing software) to plot ln (Pv) vs.
Linear (ln(vp)
2
0
0.0005
-2
0.001
0.0015
1/Temperature
ln
(vapor
pressure)
Vapor
(K)
Pressure
of
Magnesium
7
y = -9314.3x + 14.968
6
5
4
ln(vp)
Linear (ln(vp)
3
Slope = Ð9314 KÐ 1
∆Hû vap
slope = Ð
J
8 . 3 1 4 moláK
J
Ð9314 K Ð1 á Ð8.314 mol
∆Hû vap = 77.4 kJ
2
1
0
-0.0005
1
0.001
0.0015
1/Temperature
January 26, 1998
0.002
(K)
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Spring 1998
PS1.8. (Continued)
b) In which metal is the bonding stronger? Lithium has stronger bonding
than magnesium on the basis of the ∆Hûvap. That lithium has the
higher ∆Hûvap suggests the stronger bonding. From data in
another reference the ∆Hûvap for lithium is 148 kJ and 127 kJ
for magnesium. Either the data that I used is funky or such
1
data for metals does not fit the ln (Pv ) vs.
T relationship as
well.
c) Determine the temperature of a sample of lithium and of magnesium when
the vapor pressure is 350 mmHg.
Lithium
1
y = Ð1.16 x 104 x + 15.4 or
ln(vp) = Ð1.16 x 104 æ Tö +
è ø
15.4
1
ln(350) = Ð1.16 x 10 4 æ Tö + 15.4
è ø
1
5.858 = Ð1.16 x 104 æ Tö + 15.4
è ø
1
1.16 x 104 æ Tö = 15.4 Ð 5.858
è ø
1
9.54
=4 8.22 x 10Ð 4
T = 1.22 x 103 K
T = 1.16 x 10
Magnesium
1
ln(vp) = Ð9.31 x 103 æ Tö +
y = Ð9.31 x 103 x + 15.0 or
è ø
15.0
1
ln(350) = Ð9.31 x 10 3 æ Tö + 15.4
è ø
1
5.858 = Ð9.31 x 103 æ Tö + 15.4
è ø
1
9.31 x 103 æ Tö = 15.4 Ð 5.858
è ø
1
9.54
=3 1.02 x 10Ð 3
T = 9.76 x 102 K
T = 9.31 x 10
January 26, 1998
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Spring 1998
PS1.8. (Continued)
d) Determine the vapor pressure of a sample of lithium and of magnesium at
800. ûC.
Lithium
y = Ð1.16 x 104 x + 15.4
1
ln(vp) = Ð1.16 x 104 æ Tö +
è ø
or
15.4
1
ln(vp) = Ð1.16 x 10 4 æ 1 0 7 3ö + 15.4
è
ø
ln(vp) = 4.59
e ln(vp) = e4 . 5 9
vp = 98.4 mmHg
Magnesium
y = Ð9.31 x 103 x + 15.0
1
ln(vp) = Ð9.31 x 103 æ Tö +
è ø
or
15.0
1
ln(vp) = Ð9.31 x 10 3 æ 1 0 7 3ö + 15.0
è
ø
ln(vp) = 6.32
e ln(vp) = e6 . 3 2
vp = 557 mmHg
PS1.9. The normal boiling point of acetone, (CH3)2CO is 56.2 ûC and its DHûvap =
kJ
32.0 mol . Draw a Lewis structure for acetone and calculate the temperature
at which acetone has a vapor pressure of 570. mmHg.
P
ln P 1
2
-DH
R
( T11
(
(
)
)
(
)
(
January 26, 1998
Lewis structure
)
1
- T
2
J
- 3 2 0 0 0 mol
1
570
1
ln 7 6 0 =
T1 - 3 2 9 . 2 K
J
8.314 mol ¥ K
1
1
ln (0.750) = -3848.9 T
- 329.2
1
1
1
-0.288 = -3848.9 T
- 329.2
1
1
1
7.48 x 10 -5 = T
- 329.2
1
1
-3
T 1 = 321 K
T 1 = 3.11 x 10
=
)
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Spring 1998
PS1.10. When a volatile liquid vaporizes into the atmosphere from an uninsulated
container, the temperature of the liquid remains equal to the temperature of the
surrounding. If the same liquid is placed into an insulated container the
temperature of the liquid falls below the temperature of the surrounding.
Explain these observations.
Evaporation is an endothermic process. When it occurs in a
container in thermo contact with the surrounding heat flows from
the surrounding into the system to keep the temperature of the
system constant. In an container which is thermally insulated heat
is not transferred to the system from the surrounding and the
temperature drops.
Evaporation is an endothermic process because only those particles
on the surface with sufficient energy to overcome the
intermolecular attractive forces can escape into the vapor phase.
Those are generally the particles with the greater kinetic energy.
When they leave the system the average kinetic energy of the
system drops, therefore, the temperature decreases.
January 26, 1998
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Spring 1998