Chem 121 Problem set VIII SOLUTIONS - 1
Problem Set VIII Liquids, Solids, Intermolecular Forces and Phase Diagrams
1a) this is a point on the vapour pressure curve
1b) gas
1c) gas to liquid
Water
CO2
2a) solid to vapour or sublimes, 2b) 5.2 atm at the triple point,
2c) The fusion curve has a positive slope. A verticle line drawn in the direction of increasing pressure from
any point on the fusion curve takes carbon dioxide into the solid state; that is an increase in pressure will
cause liquid carbon dioxide to solidify. The solid carbon dioxide must therefore be more compact than the
liquid and have a greater density.
3a) liquid, b) liquid
4a) liquid to solid to vapor, b) liquid to vapour
5a) no, b) yes
Chem 121 Problem set VIII SOLUTIONS - 2
6. A solid
B liquid
C gas
D solid & gas
E solid, liquid & gas
F liquid & gas
7c) the pure compound changes from a solid to a liquid.
8. Both molecules will have a similar dispersion force. However carbon monoxide is polar, nitrogen is
nonpolar and so CO will also have a dipole-dipole force giving it the higher melting point. The melting points
of CO and N2 are −199 and −210ºC respectively.
Chem 121 Problem set VIII SOLUTIONS - 3
9. Again both molecules will have a similar dispersion force but because H2S has the larger dipole moment, it
will have the larger dipole-dipole force, giving it the higher boiling point and heat of vapourization.
10. Hydrogen bonding only occurs with the small electronegative elements, N, O and F attached to a
hydrogen. NH3 (N-H bond), CH3OH (O-H bond) and CH3NH2 (N-H bond) show hydrogen bonding. HBr, PH3
and CH3F (only has a C-F bond) do not.
11a) Chlorine, Cl2 is larger and has more electrons than oxygen, O2 and so has the larger dispersion force
giving it the higher melting point. The normal melting points of Cl2 and O2 are −101ºC and −218ºC
respectively.
11b) Butane is the larger molecule and so has more electrons and thus has the stronger dispersion force and
the higher melting point. The normal melting points of propane and butane are −190ºC and −138ºC
respectively.
11c) Again SiH4 is larger and has more electrons then CH4 and thus has the stronger dispersion force and the
higher melting point. The normal melting points of CH4 and SiH4 are −170ºC and −110ºC respectively.
12. Largest vapor pressure at 100 K: SiH4 CH4 GeH4 SnH4 ; all four molecules are hydrides of group IV
elements and are liquids at 100K, since all molecules are tetrahedral, the bond dipoles will cancel and all are
nonpolar so the only intermolecular force is dispersion. Smallest molecule, weakest intermolecular forces and
highest vapour pressure and this is CH4
Highest boiling point: H2S H2Se H2Te H2O ; all four molecules are hyrides of group VI elements and are
bent in shape with two lone pairs of electrons. All will exhibit the dispersion force, with H2Te being the
strongest with the most electrons. H2S, H2Se and H2Te exhibit dipole-dipole intermolecular forces while H2O
exhibits hydrogen bonding. In this case the hydrogen bonding of water is stronger than the dispersion of
H2Te. Since H2O shows the strongest intermolecular force it will require the higher temperature for its vapour
pressure to reach 1 atm and so will have the highest boiling point.
13a) Hg is a metallic solid, SnO2 is an ionic solid, CO2 and butane are molecular solids.
CO2 is a nonpolar molecule (vectors cancel), so it only has the dispersion force (22 electrons) and has the
weakest attractive force of the four (bp −78°C).
Hg is a metallic solid. The metallic force can be a strong force, but all metals are solids except Hg, and it has
the weakest attractive force of the metals (bp 357°C).
SnO2 is an ionic solid (∆EN = 1.7), with a strong force of attraction and has strongest attractive force of the
four (bp 1630°C).
C4H10 is a nonpolar hydrocarbon molecule so has the dispersion force (42 electrons) and has a stronger force
of attraction than CO2 (bp −0.5°C).
The substance with the highest boiling point will have the strongest force of attraction and this is SnO.
13b) SiO2 is a network covalent solid and the other three are molecular solids.
SiO2 is a network covalent solid (quartz) and has a very strong force, the strongest of the four (bp 2950°C).
PH3 has the dispersion force (18 electrons) and the dipole-dipole force (dipole-dipole force: ∆EN (P-H) = 2.12.1 = 0 so thus the vector is due to the lone pair and is relatively strong). Both forces are relatively weak
compared to (bp −88°C).
NH3 has the dispersion force (10 electrons) and hydrogen bonding which is a strong force of attraction. The
dipole moment of NH3 is the larger than PH3 and PCl3 (∆EN (N-H) = 3.0 - 2.1 = 0.9 and is in the direction of
the lone pair so the vectors add). Hydrogen bonding is the significant attractive force in ammonia and the
attractive force is stronger than PH3. (bp −33°C).
PCl3 has the dispersion force (66 electrons) and the dipole-dipole force.(dipole-dipole force: ∆EN (P-Cl) = 3.02.1 = 0.9 however this vector is in the opposite direction to the lone pair and so this molecule has the smallest
dipole moment). However the dispersion force is significant with 66 electrons and the overall force of
attraction will be larger than NH3 or PH3 (bp 76°C).
Chem 121 Problem set VIII SOLUTIONS - 4
P
H
P
N
H
H
H
Cl
H
Cl
Cl
H
The substance with the largest vapour pressure will have the weakest force of attraction and this is PH3.
13c) All four molecules are hydrides of the Group V elements, and are pyramidal in shape with a lone pair.
SbH3 has a large dispersion force (54 electrons) and a small dipole-dipole force (∆EN (H-Sb) = 2.1-1.9 = 0.2
and this vector is in the opposite direction to the lone pair). The dispersion force is the largest of the four
molecules (bp −17°C).
NH3 has a weak dispersion force (10 electrons) and a strong hydrogen bonding force of attraction. The dipole
moment of NH3 is the larger than PH3 (∆EN (N-H) = 3.0 - 2.1 = 0.9) and is in the direction of the lone pair so
the vectors add (bp −33°C).
AsH3 has a relatively large dispersion force (36 electrons) and a weak dipole-dipole force (∆EN (H-As) = 2.12.0 = 0.1 and this vector is in the opposite direction to the lone pair). The dipole-dipole force is slightly smaller
than SbH3, however the difference is insignificant. The dispersion force is also smaller than SbH3 (bp −63°C).
PH3 has a weak dispersion force (18 electrons) and a relatively strong dipole-dipole force (∆EN (P-H) = 2.12.1 = 0, so the vector is entirely due to the lone pair and the vector is larger than SbH3 and AsH3). PH3 has
the weakest force of attraction of the four molecules (bp −88°C).
Sb
H
H
As
N
H
H
H
H
H
H
H
P
H
H
H
The order of increasing strength for the dipole-dipole interaction is SbH3 < AsH3 < PH3 < NH3.
However the dispersion force is always stronger than the dipole-dipole force (except for hydrogen bonding)
and the order of increasing strength for the dispersion interaction is NH3 < PH3 < AsH3 < SbH3 (the reverse
order of the dipole-dipole force). Ammonia has the strongest intermolecular force of all four molecules as it
has hydrogen bonding. The molecule with the lowest boiling point is that with the weakest force of attraction
or the weakest dispersion force (with the fewest electrons), and that is PH3. See the figure (Bp of Hydrides)
on page 42.
13d) All four molecules are molecular solids.
CH4 is a nonpolar molecule and has a weak dispersion force (10 electrons), (bp −161°C).
CH3CH3 is also nonpolar and has a slightly stronger dispersion force (18 electrons), (bp −89°C ).
CH3OH has the hydrogen bonding force of attraaction and the dispersion force is starting to become noticable
(18 electrons). The hydrogen bonding force of attraction is significantly stronger than the dispersion force in
CH4 or CH3CH3 as both of these molecules are gases and CH3OH is a liquid (bp 65°C).
CH3CH2OH also has the hydrogen bonding force and the dispersion force (26 electrons). It has a stronger
overall force of attraction compared to CH3OH due to its larger dispersion component (bp 78°C).
The substance with the smallest vapour pressure will have the strongest force of attraction and that is
CH3CH2OH.
13e) KCl is an ionic solid and the other three are molecular solids.
KCl is an ionic solid with a very strong force of attraction (∆EN = 2.2), the strongest of the four entities (bp
771°C).
H2O has the dispersion force (10 electrons) and the hydrogen bonding force, which is a strong van der Waals
force, but weak compared to the ionic, metallic and network covalent forces of attraction. (bp 100°C)
H2S has the dispersion force (18 electrons) and the dipole-dipole force which is weak compared to the
hydrogen bonding and dispersion forces (∆EN (S-H) = 2.6-2.1 = 0.5 and this vector is in the same direction as
the lone pairs). It has a weaker attractive force than water (bp −60°C).
CH4 is a nonpolar molecule and only has the dispersion force with 10 electrons and so has the weakest force
of attraction of the four entities (bp−161°C).
Chem 121 Problem set VIII SOLUTIONS - 5
The substance with the highest boiling point will have the strongest force of attraction and that is KCl.
CH3
N
H
CH3
CH2
CH2
CH2
N
CH3
H
N
H
CH3
CH3
CH3
N
CH3
H
H
13f CH3NHCH3 has the hydrogen bonding force of attraction and the dispersion force (26 electrons), (bp
17°C).
CH3CH2CH2CH2NH2 has the hydrogen bonding force of attraction with two N-H bonds. It is also linear and so
can pack close together strengthening the dispersion force (42 electrons), giving this molecule the strongest
force of attraction (bp 77°C).
CH3N(CH3)CH3 has no hydrogen bonding force of attraction, instead it has a weak dipole-dipole force (∆EN
(N-C) = 3.0-2.1 = 0.9 and this vector is in the same direction as the lone pair). It’s dispersion force (34
electrons) will be relatively weak as it is spherical and the molecules cannot pack close together. Thus the
overall force of attraction will be weak (bp 3°C).
CH3NH2 has hydrogen bonding with two N-H bonds and a small dispersion component (18 electrons)
compared to the other three molecules. Note that this molecule has the lowest bp even though it can form two
hydrogen bonds; in this case the dispersion force dominates (bp −6°C).
The substance with the smallest vapour pressure will have the strongest force of attraction and that is
CH3CH2CH2CH2NH2.
14a)The electronegativity vectors for CO2 will cancel however there will be a net vector toward oxygen in
OCS.
∆EN(O-C) = 3.5 - 2.5 = 1.0, ∆EN(S-C) = 2.5 - 2.5 = 0.
OCS has the strongest dipole-dipole intermolecular force.
O
C
O
O
C
S
b) The electronegativity vectors for SO3 will cancel, while SO2 has a net vector toward the lone pair.
∆EN(O-S) = 3.5 - 2.5 = 1.0
SO2 has the strongest dipole-dipole intermolecular force.
O
S
O
S
O
O
O
c) The electronegativity vectors for XeCl2 will cancel, while OCl2 has a net vector between the lone pairs.
∆EN(O-Cl) = 3.5 - 3.0 = 0.5, ∆EN(Cl-Xe) = 3.0 - 2.6 = 0.4.
OCl2 has the strongest dipole-dipole intermolecular force.
Cl
Cl
Xe
Cl
Cl
O
Cl
Cl
d) The electronegativity vectors for SiF4 will cancel, (∆EN(S-Si) = 2.5 - 1.8 = 0.7), the molecule will not have a
dipole moment and will have the weakest dipole-dipole intermolecular forces.
SF4 is not symmetrical and will have a net electronegativity vector. We will take the lone pair electronegativity
vectors as approximately 1.5 (since ∆EN(lp-N) = lp - 3.0 ≈ 1, from lecture notes, then lp ≈ 4 and thus ∆EN(lpS) = 4 - 2.5 ≈ 1.5). As ∆EN(F-S) = 4.0 - 2.5 = 1.5, all the electronegativity vectors are approximately the
same and will come close to cancelling. The molecule will have a net electronegativity vector, which will be
small but it will have a diploe moment and a small dipole-dipole intermolecular force.
SF4 has the stronger dipole-dipole intermolecular force.
Chem 121 Problem set VIII SOLUTIONS - 6
F
Si
F
F
F
F
F
F
F
Si
F
F
F
S
F
F
F
F
F
15. The vapour pressure of a liquid is only dependant on temperature and independant of volume. Thus if
there is liquid water in the container when the volume is reduced by three its vapour pressure will be 12 torr
(the temperature does not change). Since the gas is saturated with water vapour when the volume is V, and
when we reduce the volume to one third the original volume, one third of the number of moles of water are
necessary to maintain a pressure of 12 torr, the other two thirds will form liquid water.
However the pressure of the gas will increase by a factor of three when we reduce the volume by
three. Thus we first need to find the partial pressure of the gas.
Ptotal = Pgas + Pwater vapour
Pgas = 284 − 12 = 272 torr
Let the initial volume of the container be V so the final pressure of the gas is:
V
V
Pfinal = Pinitial × initial = 272 torr ×
= 816torr
1
Vfinal
3V
this is the pressure of the gas and we need to add to this the vapour pressure of water so
Ptotal = Pgas + Pwater vapour = 816 torr + 12 torr = 828 torr.
16.
We use the Clausius Clapeyron equation and first we need to find ∆Hvap for mercury from the data
given.
P1 = 17.287 torr, T1 = 473.15 K and P2 = 0.2729 torr, T2 = 373.15 K so
∆H vap ⎛
1
1 ⎞
⎛ 0.2729 torr ⎞
−
ln⎜
⎜
⎟
⎟ =
J
⎝ 17.287 torr ⎠ 8.314 mol.K ⎝ 47315
. K 37315
. K⎠
and ∆Hvap = 6.0896 x 104 J/mol
now the vapour pressure at the normal boiling point of Hg is 1 atm and thus we need to find the
temperature when the vapour pressure is 1760 torr. Using the Clausius Clapeyron equation with P3 = 760 torr
and T3 = ?
4
1 ⎞
⎛ 0.2729 torr ⎞ 6.0896 × 10 J mol ⎛ 1
ln⎜
=
−
⎜
⎟
⎟
⎝ 760torr ⎠
. K⎠
8.314 J mol.K ⎝ T3 37315
T3 = 626.19 K or 353.0°C
17.
Again we use the Clausius Clapeyron equation and initially we need to find ∆Hvap for first liquid from
the data given: P1 = 0.50 atm, T1 = 273.15 K and P2 = 1.00 atm, T2 = 323.15 K so
∆H vap ⎛
1
1 ⎞
⎛ 1.00atm ⎞
ln⎜
−
⎟ =
⎜
⎟
J
⎝ 0.50atm ⎠ 8.314 mol.K ⎝ 27315
. K 32315
. K⎠
and ∆Hvap = 1.017 x 104 J/mol.
Since ∆Hvap(1) = 2∆Hvap(2), ∆Hvap(2) =5.0816 x 103 J/mol,
and using the Clausius Clapeyron equation to find the normal boiling point of the second liquid with
P1 = 0.50 atm, T1 = 273.15 K and P2 = 1.00 atm, T2 = ? K and ∆Hvap(2) =5.0816 x 103 J/mol we have
3
1
1 ⎞
⎛ 1.00atm ⎞ 5.0816 × 10 J mol ⎛
⎜⎜
⎟
−
ln⎜
⎟=
J
8.314 mol.K ⎝ 273.15K T2 ⎟⎠
⎝ 0.50atm ⎠
T2 = 395.02 K or the normal boiling point of the second liquid is 122.5°C.
18. Read text on types of crystaline solids, their bonding and properties.
Gold is a metal, sugar is a molecular solid, quartz is a covalent network solid and fluorite is an ionic solid.
Chem 121 Problem set VIII SOLUTIONS - 7
19.
SiO2
KBr
covalent network solid
(quartz)
ionic solid
PF3
molecular solid
C
H2O
covalent network solid
(diamond, graphite)
molecular solid
U
metallic solid
contain no molecules and the entire solid is held together with strong covalent
bonds (mp 1713°C cristobalite).
strong ionic bonds. The particles that make up the crystal lattice are positive
and negative ions. Each ion is surrounded by neighbours of the opposite charge
and there are no molecules (mp 734°C)
dispersion and Dipole-dipole forces. The dispersion force is an instantaneous
dipole and is dependant on the number of electrons in the molecule. PF3 has 42
electrons and thus this force is relatively significant. The dipole-dipole force or
permanent dipole is dependant on the sum of the electronegativity vectors
(∆EN(F-P)=1.8 and ∆EN(lp-P)≈1.8). The VSEPR shape of the molecule is
tetrahedral and the four vectors will come close to canceling and we would
expect the dipole-dipole force to not be very large. The dispersion force would
be similar to butane (also 42 electrons) and the mp’s are also similar (mp
−138°C, butane; −152°C, PF3).
contain no molecules and the entire solid is held together with strong covalent
bonds (mp 4440°C, diamond; 4492°C, graphite).
dispersion and the Hydrogen bonding attractive forces. The dispersion is
relatively small (10 electrons) and the hydrogen bonding force is the largest of
all molecules. Hydrogen bonding is an extreme form of the permanent dipole,
where a highly electronegative atom strips the electrons from a hydrogen,
leaving an exposed proton or positive charge which will interact with another
electronegaive atom (mp 0°C)
Metallic attractive force. The metal atoms occupy regular lattice sites, however
the valence electrons occupy molecular orbitals that extend throughout the
entire solid. For two atomic orbitals interacting (two metal atoms coming
together) two rather widely spaced energy levels result. As more atomic orbitals
become available to form molecular orbitals, the resulting energy levels become
more closely spaced., finally producing a band of very closely spaced orbitals
that extend through the entire solid. Since the force of attraction is not localised,
the metal atoms can slide from one lattice site to another, making metals
maleable and ductile (mp 1135°C)