1
Chapter 11 - Homework Solutions
11.2) What types of Intermolecular Attractive Forces (IAF) are shown in the following?
a) HF molecules
H-F ------ H-F
(This dashed line represents a H-bond.)
What is being shown is a hydrogen bond. This occurs between a lone-pair of
electrons on the F atom of one molecule and the H atom of another molecule.
Remember, H bonding is a special type of very strong dipole-dipole interaction. It
occurs when a H atom bonded directly to a an atom of high electronegativity (N, O
or F) forms a strong dipole-dipole attraction to an unshared electron pair on a
nearby small electronegative atom or ion (N, O or F). You need a H covalently
bonded (directly connected) to a N, O or F in a molecule in order to have hydrogen
bonding and a lone pair on a N, O or F.
You will also have London Forces (LF) and Dipole-Dipole AF (DD) between HF
molecules. Not all HF molecules in the liquid will be H-bonded to each other. The
molecules have to line up just right to form H bonds. Thus, when you have liquid
HF the molecules are attracted to each other by all three types of AF (LF, DD and
H-bonding). In solid HF (if it is a prefect crystalline solid) the H-bonding will be
maximized and on average the HF molecules form 2 H-bonds per molecule. In the
liquid the HF molecules form between 1-2 H-bonds per molecule.
Polar molecules with a H bonded directly to a N, O or F atom can form H-bonds in
addition to LF and DD. Thus, polar molecules are attracted to each other by LF,
DD and H-bonds.
b) F2 molecules
F-F --- F-F
(This dashed line represents LF.)
This molecule is NONpolar. All molecules have LF between them. NONpolar
molecules have ONLY LF between them.
2
11.2) (cont.)
c) Na+ and H2O
Na+ *** H2O
(This represents a Ion-Dipole AF.)
Ion-Dipole AF (I-D) occur between an ion and a polar molecule. This is a very
strong IAF. It occurs in solutions and we will discuss it in detail and use it in Ch.
13.
In this particular case, the negative end of the polar H2O (the oxygen atom) is
attracted to the positive Na+ cation. This AF is weaker than the Ion-Ion AF forces
in ionic compounds but much stronger than H-bonding.
d) SO2 molecules
O=S
----\
O
O=S
(This represents LF and DD AF)
\
O
This molecule is polar. The S-O bonds are polar and the molecule is bent (with a
bond angle of about 120°). The S atom has a partial positive charge (*+) on it and
the O atoms have a partial negative charge (*&). Thus, the partially negatively
charged O atom of one molecule forms an attraction to the positively charged S
atom of another molecule. This is in addition to LF that occur between these
molecules. Thus, SO2 molecules are attracted to each other by both LF and DD
AF.
Polar molecules have BOTH LF AND DD AF between them.
The strengths of the AF we are talking about in chapters 11 and 13 increase in the
following order (increasing in strength from left to right).
LF < DD < H-bonds < I-D < Ionic & Covalent
Increasing AF
3
11.5) Attractive forces between particles effect properties of solids and liquids.
Remember:
Nonpolar
Polar
H-bonds
===>
===>
===>
In General:
Stronger Att. Forces
|
London Forces (LF) only
LF & Dipole-Dipole (DD)
Molecules with a H covalently bonded to a N, O, F
will form H-bonds to an N, O, F with a lone pair of
e! (lpe) on another molecule.
Lower Vapor Pressure
Slower rater of evaporation
Higher b.p., m.p., sublimation pt.
)Hvap, )Hfus, )Hsub,
critical temp., surface tension, viscosity
In this question we are comparing the boiling points between isomers with the
molecular formula C3H8O, propanol (CH3CH2CH2-OH) and ethyl methyl ether
(CH3 - O - CH2CH3). The stronger the AF the greater the average KE needed in
order to break these attractive forces. Thus, we have to go to a higher temperature
to increase the KE when there are stronger AF and we get a higher b.p.
The propanol, CH3CH2CH2-OH, can form LF, DD and H-bonds between two such
molecules (due to the H bonded to the O atom of the -OH group). The ethyl methyl
ether can NOT form H bonds and has only LF and DD between two of these
molecules. Since the molecules have the same molar mass (mol. wt.) the LF are
approximately the same. They are both polar, with the propanol being slightly
more polar (because the O-H covalent bond is more polar than an O-C covalent
bond). However, it is the H-bonding that forms between propanol molecules that is
the largest contributor to the AF between propanol molecules. Thus, propanol has
greater IAF and thus a higher b.p.
4
11.6)
S - Solid
a)
L - Liquid
G - Gas
The points labeled on the phase diagram are:
A:
Sublimation point (equilibrium between solid and gas). The curve AB
is the vapor-pressure curve of the solid. Every point on this line
represents an equilibrium between the solid and gas, at a particular
pressure and temperature.
B:
Triple point (equilibrium between solid, liquid and gas). At this point,
all three states of mater are in equilibrium at the same time. You can
have more than one triple point on a phase diagram. The best example
is water in which there are many different types of solid water (normal
ice and other types of ice that are actually more dense than liquid
water) giving rise to several triple points.
C:
Normal melting point (equilibrium between solid and liquid at 1 atm).
The curve BC is the melting-point curve. Every point on this line
represents an equilibrium between the solid and liquid (a melting point,
m.p., at a particular barometric pressure). The direction it slopes
depends on the relative densities of the solid and liquid. The phase
diagram above shows this line with a positive slope (leans to the right
like CO2). This occurs when the solid is more dense than the liquid,
as occurs in most substances. In this case, the m.p. increases as
barometric pressure increases. If the m.p. curve has a negative slope
(leans to the left like H2O) this means the solid is less dense than the
liquid and the m.p. decreases as barometric pressure increases.
For this substance the n.m.p. (n.f.p.) is ~ 260 K.
5
11.6) (cont.)
D:
Normal boiling point (equilibrium between liquid and gas at 1 atm).
The curve BD is the vapor-pressure curve of the liquid. Every point on
this line represents an equilibrium between the liquid and gas, at a
particular pressure and temperature (a boiling point at a particular
barometric pressure). This curve ends at the critical point (not
shown).
The critical point is the point above which there is no longer a liquid
and gas phase in equilibrium and only one supercritical fluid phase
exists. The critical temp. is the temp. above which the gas can no
longer be liquefied by applying pressure. It is the last point in which
the liquid and gas are in equilibrium.
- can NOT liquefy compound above the critical temperature no
matter what pressure is applied
For this substance the n.b.p. is ~ 360 K.
b)
The physical states at the various P and T conditions are given below and
shown on the phase diagram.
(i)
c)
T = 150 K, P = 0.2 atm
gas (just below curve AB - although a
little tough to tell)
(i i) T = 100 K, P = 0.8 atm
solid
(i i i) T = 300 K, P = 1.0 atm
liquid
See part (a) above. Labeled as point B on the phase diagram.
6
11.7)
Niobium(II) oxide crystallizes in a cubic unit cell such that the Nb2+ ions are
located on the faces of the cube with the O2& ions on the edges of the cube as shown
below.
a) How many Nb2+ and O2& ions are within the unit cell?
Nb2+ ions are located on the faces. A face is shared by one other unit cell in
the whole crystal. In a single unit cell there is 1/2 of a particle on a face and
6 faces per cube. Thus, if there is a particle on every face there are a total of
3 particles (although they aren’t 3 whole particles - just 6 pieces that add up
to 3 particles in total).
1/2 particle/face, 6 faces : 6 x 1/2 = 3 particles
Face particles contribute 3 particles to the u.c. (if there is a particle on every
face).
Thus, there are 3 Nb2+ in the u.c.
O2& ions are located on the edges. An edge is shared by 4 other unit cells in
the whole crystal. In a single unit cell there is 1/4 of a particle on an edge
and 12 edges per cube. Thus, if there is a particle on every edge there are a
total of 3 particles (although they aren’t 3 whole particles - just 12 pieces that
add up to 3 particles in total).
1/4 particle/edge, 12 edges : 12 x 1/4 = 3 particles
Edge particles contribute 3 particles to the u.c. (if there is a particle on every
edge).
Thus, there are 3 O2& in the u.c.
7
11.7) (cont.)
b) What is the empirical formula of niobium(II) oxide?
Based on part (a) there are 3 Nb2+ and 3 O2& in the unit cell. We could write the
formula as Nb3O3. However, the formulas for ionic compounds are empirical
formulas which give the smallest whole number ratios of the ions. These ions are
in a one-to-one ratio and the formula is:
NbO
(There are 3 NbO f.u./uc)
This makes sense based on the charges and what you learned in 121. When you
wrote the empirical formulas for ionic compounds in 121 you did so by balancing
the charges on the cations and anions. Since the charges on the cation and anion
are the same they will be present in a one-to-one ratio and you get the same formula
as above. This of course should be the case since the unit cell is a representation of
the compound and should give us the same properties as the compound (formula,
density, etc.).
c) Is this a molecular, covalent-network or ionic solid?
Since this compound is composed of a metal and nonmetal you would expect it to
be an ionic compound. Although there are cases when the bonding in compounds
of metals and nonmetals are more covalent than ionic and they are molecular
compounds, it doesn’t occur that often (probably less than 1% of the time).
Sometimes compounds of metals and nonmetals are ionic in the solid state and
form ions when liquefied (of dissolved in solution) but can form molecular
compounds in the gas phase (such as AlCl3 and Al2Cl6).
8
11.8)
The picture in the book is of a stack of oranges. These are pretty spherical objects.
Spherical objects can pack in several ways. Three examples are the 3 cubic unit
cells, simple cubic (sc) or primitive cubic, body-centered cubic (bcc) and facecentered cubic (fcc).
When you pack perfect spheres there are two ways to pack them to give the
minimum amount of empty space. These are called close-packed structures.
Hexagonal close packed (hcp): ABABABCCC
Cubic close packed (ccp):
ABCABCCCC
(same as fcc)
In both cases, there is a minimum amount of empty space (about 26%) and a
coordination number (number of nearest neighbors - # particles each particle
contacts in the interior of the crystal) of 12.
a)
You can see in the picture that the arrangement is close-packed. However,
since you can’t see the interior of the structure you really can’t tell if it’s hcp
or ccp.
b)
As given above, the coordination number in both hcp and ccp is 12. In the
interior of the structures each particle touches 12 other particles. In a
particular layer a single particle will touch 6 other particles, one particle in
the middle surrounded by 6 others. The particle in the middle will also touch
3 particles in the layer above and 3 particles in the layer below.
c)
If the oranges represent Ar atoms the substance will be molecular. The Ar
atoms are “nonpolar” and Ar is a gas at room temperature (low b.p.). Thus,
there are only LF forces between Ar atoms. These are weak AF. Molecular
substances often form one of the close-packed structures (ccp or hcp).
NOTE:
the table below gives things discussed in the book and/or presented in
class. You need to know this.
sc (pc)
bcc
fcc(ccp)
hcp
# part.
1
2
4
X
coord #
6
8
12
12
% empty
space
48
32
26
26
4r
-------(2)1/2
X
R =
4r
-------(4)1/2
4r
-------(3)1/2
9
11.10)
a) In solids, particles are vibrating about essentially in fixed positions. The KE is
much less than the AF between the particles (KE << AF). In liquids, the particles
are still close together but there is a little more empty space and disorder. The KE
and AF are on the same order of magnitude (KE . AF). In gases, particles are far
apart and in constant, random motion. In gases the KE is much greater than the AF
(KE >> AF) and the AF are ineffective (they are still present just not effective).
See part (b) below.
b)
c) Assuming a constant temperature, the average KE of the particles is constant.
Compressing the gas cause the particles to get closer together. As the volume
decrease and the molecules have less room to move there are more collisions and
the AF become more important and effective. The closer the particles the more
effective the AF between them. As the total particle-to-particle AF increase the
particles liquefy.
10
11.13)
11
11.15)
a)
I2 is a nonpolar linear molecule so it has only London Forces (LF) that must
be overcome in order to convert it from the solid to the gas.
b)
CH3CH2OH is a polar molecule. It is bent around the O atom like water.
Since it is polar it has LF and Dipole-Dipole (DD) AF. Furthermore, since
there is a H atom attached directly to the O atom and the O atom also has 2
lone pairs of electrons there are also H-bonding forces between two such
molecules. All three must be overcome to convert from the liquid to gas.
(There must be a H attached directly to a N, O or F atom, which must also
have at least one lone pair of electrons).
c)
H2Se is a polar molecules and is bent around the Se atom (looks like H2O).
Since it is polar it has LF and Dipole-Dipole (DD) AF. These must be
overcome to go to the gas phase. H2Se molecules can not form H bonds
(there must be a H attached directly to a N, O or F atom, which must also
have at least one lone pair of electrons).
11.16)
12
13
11.17)
a)
b)
Polarizability: The ease with which the charge distribution (e- cloud) in a
molecule (atom or ion) can be distorted to produce a transient (temporary)
dipole. This leads to London Forces (London Dispersion or Dispersion).
Sb is the most polarizable. It is the largest & its valence e! are farthest from
the nucleus and less tightly held - the e! cloud can be distorted (“pulled”)
more easily.
LF inc. w. inc. polarizability (which generally inc. with size and thus At. Wt.
or Mol. Wt.)
N < P < As < Sb
Increasing size & AW, Inc. Polarizability, Inc. LF
(Larger/Higher b.p., m.p., sub. pt., )Hv, )Hfus, )Hsub,
viscosity, surface tension, critical temp.
Smaller/ Lower Vapor pressure, slower rate of evap.
c)
Polarizability generally inc. w. molecular size (and thus Mol. Wt.)
MW
CH4 < SiH4 < SiCl4 < GeCl4 < GeBr4
16
32
170
214
392
Increasing size & AW, Inc. Polarizability, Inc. LF
All of these molecules are tetrahedral and NONpolar and thus have only LF
d)
The magnitude of LF and thus b.p. of molecules inc. as polarizability inc.
The polarizability inc. as the size (# e! & MW) inc. (although shape does
play a role). Thus, generally, LF inc. as MW (AW for atoms) inc. So the
order of inc. b.p. is the order of inc. LF (inc. MW or AW) for nonpolar
molecules (or polar molecules with about the same DD AF). Thus the order
of increasing b.p. is the same as the order of inc. polarizability (the order
given in part (c) above).
For molecules with similar structures, the strength of the London Forces
(dispersion forces) increases with molecular size (Mol. Wt. and number of
electrons in the molecules).
LF inc. with inc MW
14
11.20)
Which member of the following pairs has the stronger intermolecular dispersion
(London) forces (LF)?
For molecules with similar structures, the strength of the London Forces
(dispersion forces) increases with molecular size (Mol. Wt. and number of
electrons in the molecules).
LF inc. with inc MW
a) Br2 or O2
Both Br2 and O2 are linear and nonpolar and thus have only LF. The MWs of
Br2 and O2 are 159.81 amu and 32.00 amu, respectively. Since Br2 is larger it
will have the larger LF.
b) CH3CH2CH2CH2SH or CH3CH2CH2CH2CH2SH
Both are straight chains with the same types of atoms (C, H and S). The
larger one, CH3CH2CH2CH2CH2SH, will have the larger LF.
c) CH3CH2CH2Cl or (CH3)2CHCl
CH3 - CH2 - CH2 - Cl
CH3
|
H - C - Cl
|
CH3
These molecules are isomers (have the same number and kind of atoms and
thus the same molecular formula, C3H7Cl). The Mol. Wts. are the same (78.5
amu). In this case, the shape of the molecules determines which has the
stronger LF. The cylindrical (straight chain) molecule has more points of
contact compared to the branched (more compact) molecule. Thus the
straight chain molecule, CH3CH2CH2Cl, will have stronger LF. See Figure
11.6 that shows pentane (C5H12, straight chain) and neopentane (compact,
spherically shaped branched isomer). The straight chain has the stronger LF
and higher b.p. I used this example as a demo in class.
15
11.23)
11.26)
Remember the following:
Nonpolar
Polar
H-bonds
===>
===>
===>
Stronger Att. Forces
|
London Forces (LF) only
LF & Dipole-Dipole (DD)
Molecules with a H covalently bonded to a N, O, F
will form H-bonds to an N, O, F with a lone pair of
e! (lpe) on another molecule.
Lower Vapor Pressure
Slower rater of evaporation
Higher b.p., m.p., sublimation pt.
)Hvap, )Hfus, )Hsub,
critical temp., surface tension, viscosity
16
11.26) (cont.)
a)
C6H14
MW
86
nonpolar
LF
or
<
C8H18
114
nonpolar
LF
Molecules containing only C & H are nonpolar and thus have only LF between
them. C8H18 is larger and will have greater LF and it should have the higher b.p.
b)
C3H8
MW
44
nonpolar
LF
or
.
CH3-O-CH3
46
polar
LF
DD
Both molecules have LF, which are approximately equal (same size and shape).
CH3-O-CH3 is bent around the O atom (as in H2O) and slightly polar so it also has
DD AF. Thus the AF for CH3-O-CH3 are larger so it should have the higher b.p.
Note: CH3-O-CH3 can NOT form H-bonds to another one since the H’s are on the
C’s, not the O atom (no H-bonding as a pure substance).
c)
HOOH
MW
34
polar
LF
DD
H-bonds
or
<
>
HSSH
66
polar
LF
DD
H2S2 is larger and has larger LF. Both are polar, but H2O2 is more polar since O is
more eletronegative (OH bond is more polar than SH bond). H2O2 molecules form
H-bonds to each other (H2S2 can not). This H-bonding is very important and makes
a bigger difference then the LF and DD, which sort of “balance” each other out (see
figure 11.7 for the effect of H-bonding on the b.p. of H2O and H2S). Thus H2O2
has stronger IAF and a higher b.p.
17
11.26) (cont.)
d)
NH2NH2
MW
32
polar
LF
DD
H-bonds
or
.
CH3CH3
30
nonpolar
LF
NH2NH2 has much stronger AF since it has DD AF (polar) and especially because
it has H-bonding between molecules. Thus, it should have the higher b.p.
11.27)
H2O
MW
18
polar
LF
DD
H-bonds
nbp (°C) 100.00
nmp (°C)
0.00
or
<
>
H2S
34
polar
LF
DD
-60.7
-85.5
Water is a fairly small molecule and has small LF compared to the other larger
similar molecules in group 6A (and other molecules in other groups). However, is
is very polar, much more so than the other group 6A molecules, and has fairly large
DD AF. H2S is polar but not nearly as polar as H2O.
In addition, H2O can form H-bonds between the molecules while H2S (and other
group 6A molecules) can not. Also, because H2O has a nice balance of H atoms
and lpe (2 of each) it can form many H-bonds per molecule (on average from 2-4
H-bonds per molecule in the liquid, 4 H-bonds per molecule in the solid). Thus,
H2O has very strong IAF for such a small molecule.
This is born out by the n.b.p. and n.m.p. for H2O compared to H2O. H2O has a very
high n.b.p. (and n.m.p.) for such a small molecule.
18
11.27) (cont.)
a) Why is ice less dense than liquid water while H2S(s) is more dense than H2S(R)?
As stated above, H2O molecules form H-bonds between them. This occurs in both
water and ice. In water, the molecules are in constant motion due to the fact that
the KE .AF. Hydrogen bonds are constantly being broken and formed. Not all
water molecules are H-bonded to each other and the number varies over time. Each
H2O molecule forms between 1-4 H-bonds to other H2O molecules. Since not all
H2O molecules have the maximum number of H-bonds they can actually get closer
together than in ice. In ice, the H-bonding between the H2O molecules is
maximized. Ice adopts a very ordered structure with the maximum number of Hbonds to each molecule (each interior H2O forms 4 H-bonds to 4 other H2O’s
oriented tetrahedrally to each other) and they simply vibrate around these fixed
positions. When molecules form H-bonds they can get only so close. This causes a
network structure (figure 11.10) with more open space between the H2O molecules
than in the liquid. This empty space causes a larger volume for the same mass of
H2O and thus a lower density for the solid (ice) compared to the liquid (water).
In H2S, since there is no H-bonding between molecules, it would be expected that
the molecules can pack closer together, especially in the solid, than they can in
H2O. If the H2S molecules can get closer together in the solid than H2O molecules
can, the H2S solid will occupy a smaller volume than the liquid and thus be more
dense (smaller vol for the same mass means larger density).
b) Why does water have a high specific heat?
H2O has a relatively high specific heat compared to other substances its size.
Remember, specific heat is the heat required to increase the temperature of 1 gram
of a substance by 1°C. Temperature is a measure of the average KE. In order to
increase the KE of the H2O molecules the strong H-bonding AF must be broken. A
large amount of heat must be added in order to do this and increase the temp. by
1°C.
19
11.29)
Viscosity and surface tension are properties of liquids and are largely governed by
the intermolecular attractive forces (IAF) between the particles and for viscosity,
structural features that cause the molecules to become entangled are also important.
Viscosity: resistance to flow
Surface Tension: energy required to increase the surface area of a liquid by a
given amount. Think of it as the energy necessary to pull a
molecule off the surface of the liquid.
a) As stated above, these properties are a result of the IAF between the particles in
the liquid. As temperature increases the KE increases and more molecules will
have sufficient KE to overcome the AF. This will cause the viscosity and surface
tension to decrease.
Viscosity & Surf. Ten.
DEC as Temp INC
b) Both Viscosity and surface tension are related to the strength of the IAF between
the particles. The same AF that cause a high surface tension (make it difficult to
separate the surface molecules) also are responsible for holding the molecules
together in the interior of the liquid causing them to resist moving relative to one
another (resistance to flow). Therefore, liquids with high surface tensions will also
tend to have high viscosities.
Viscosity & Surf. Ten.
INC as IAF INC
INC as IAF INC (Stronger AF => higher viscosity & surface tension)
DEC as Temp INC (KE increase causing some AF to break)
Remember:
Stronger Att. Forces
|
Lower Vapor Pressure
Slower rater of evaporation
Higher b.p., m.p., sublimation pt.
)Hvap, )Hfus, )Hsub,
critical temp., surface tension, viscosity
20
21
11.35)
When the solid is melted (s ----> R) the increase in KE is not enough to completely
separate the molecules. Not all the AF are broken when going from the solid to the
liquid. The AF are still fairly important in the liquid (KE .AF) and the molecules
remain close together. However, when liquid is vaporized (R ----> g) the molecules
are essentially completely separated (for most substances). All (or nearly all) the
AF have to be overcome. This requires more energy. Thus, generally,
)Hv > )Hfus
You can see this in a heat curve (Figure 11.19 in the textbook). The horizontal line
representing boiling is longer than that for melting.
Note: The heat of sublimation is approximately equal to the sum of the heats of
vaporization and fusion (melting). This makes sense since the phase change is
solid to gas (remember Hess’s Law).
)Hsub = )Hv + )Hfus
11.38)
Do 11.39 first.
For heat transfer problems remember the following:
Heat gained or lost within a state (no phase change):
q = C C )T
q = m C s C )T
q = n C c C )T
C = heat capacity (J/°C) , )T = temp change (Tf - Ti)
s = specific heat (J/gC°C), m = mass
c = molar heat capacity (J/molC°C), n = moles
Heat transfer (gained or lost) for a phase change (using vaporization):
qv = n C )Hv
qv = m C )Hv
n = moles )Hv (kJ/mol, normally)
m = mass )Hv (kJ/g or J/g)
Heat transferred between substances:
qgained = & qlost
(heatgained = & heatlost)
(or ggained = |qlost| or qin = |qout|)
(qlost is neg., qlost < 0; qgained is pos., qgained > 0)
22
11.38) (cont.)
What mass of CCl2F2 must evaporate (gain heat) to freeze (heat lost) 200 g (assume
3 sig. fig.) of H2O initially at 15°C?
Calculate heat lost by H2O (qlost) when it is cooled from 15°C to 0°C and frozen.
It’s easier to do the problem if you draw a heating curve (Fig. 11.19). We are
starting at point A (15°C) and cooling (going from right to left). (Could draw a
“cooling diagram” - put heat lost on the x-axis and go from left to right. See 10th
edition solnutions.)
qlost = qlost by cooling liquid + qlost by freezing
= (mR C sR C )TR) + (mR C )Hfreezing)
= (mR C sR C )TR) + (mR C !)Hfus)
)Hfreezing = !)Hfus = !334 J/g
= [200 g C 4.18 J/g-°C C (0 - 15)°C] + (200 g C !334 J/g)
= !12540 J + !66800 J = !79340 J = !79.3 kJ
Total heat released is 79.3 kJ.
qgained by CCl2F2 = !qlost by H2O = !(!79340 J) = 79340 J
Also,
qgained by CCl2F2 = mCCl2F2 C )Hv
mCCl2F2 = qgained by CCl2F2/)Hv = 79340 J/(289 J/g)
mCCl2F2 = 274.53 g = 2.7 x 102 g
23
11.39)
See question 38 for some definitions.
b) Want to determine the amount of heat necessary to convert 25.0 g of ethanol,
C2H5OH, at 25°C (liquid) to the vapor phase at 78°C (gas). The best thing to do is
to draw a heating curve (figure 11.19 in the textbook). You determine the heat
needed in each region. Since heat (enthalpy) is a state function we can add the heat
required in each individual step to get the total heat required for the whole process.
Given the following data:
b.p. 78°C
)Hv = 38.56 kJ/mol
sR = sp. ht. of liquid = 2.3 J/g C°C (same as J/gCK)
Heat req. or released w/in a state: q = m C s C )T
Heat req. or released for a phase transition:
n C )H (mass C )H, if in J/g)
24
11.39) (cont.)
1 mol
? mol C2H6O = 25.0 g x ------------ = 0.5427 mol C2H6O
46.06 g
q = qheat liquid + qvap
= (mR C sR C )TR) + (nR C )Hv)
= [25.0 g C 2.3 J/g-°C C (78 & (25))°C] + (0.5427 mol C 38560 J/mol)
= 3047.5 J + 20929.2 J
(2 s.f.)
= 23976.7 J
(3 s.f.)
(Can only report to the 100's place since each step is known only to
the 100's place)
= 2.39767 x 104 J
= 24.0 kJ
You might ask why each of the above steps has the sig. fig. listed. In heating the
liquid (step 1), the specific heat has only 2 s.f., as does the )T, so the result has
only 2 s.f. and is known only to the 100's place. The vaporization step (step 2) has
3 sig. fig. because the moles and )Hv each have 3 s.f. and is known only to the
100's place. The last place common to both numbers is the 100's place so that’s the
last place in the answer, which gives 3 s.f.
25
11.39) (cont.)
b) Want to determine the amount of heat necessary to convert 5.00 L of ethanol,
C2H5OH, at -140°C (solid) to the vapor phase at 78°C (gas). The best thing to do is
to draw a heating curve (figure 11.19 in the textbook). You determine the heat
needed in each region. Since heat (enthalpy) is a state function we can add the heat
required in each individual step to get the total heat required for the whole process.
Given the following data:
density = 0.789 g/mL
m.p. -114°C
b.p. 78°C
)Hfus = 5.02 kJ/mol
)Hv = 38.56 kJ/mol
ss = sp. ht. of solid = 0.97 J/gC°C (same as J/gCK)
sR = sp. ht. of liquid = 2.3 J/g C°C (same as J/gCK)
Heat req. or released w/in a state: q = m C s C )T
Heat req. or released for a phase transition:
n C )H (mass C )H, if in J/g)
26
11.39) (cont.)
103 mL 0.789 g
? mol C2H6O = 5.00 L x ---------- x ----------- = 3945 g C2H6O
1L
1 mL
1 mol
? mol C2H6O = 3945 g x ----------- = 85.649 mol C2H6O
46.06 g
Heat the solid from -140°C to -114°C, then melt the solid at -114°C, then heat the
liquid from -114°C to 78°C and finally vaporize (boil) at 78°C.
q = qheat solid + qfusion + qheat liquid + qvap
= (ms C ss C )Ts) + (ns C )Hfus) + (mR C sR C )TR) + (nR C )Hv)
= [3945 g C 0.97 J/g-°C C (-114 & (-140))°C] + (85.649 mol C 5020 J/mol)
+ [3945 g C 2.3 J/g-°C C (78 & (-114))°C] + (85.649 mol C 38560 J/mol)
= 99492.9 J + 429958.74 J + 1742112 J + 3302631.35 J
(2 s.f.)
= 5574194.99 J
(3 s.f)
(2 s.f.)
(3 s.f.)
(Can only report to the 100000's place, 6th place to left of
decimal, since step 3 is known only to the 100000's place)
= 5.57419499 x 106 J
= 5.57419499 x 103 kJ = 5.6 x 103 kJ
You might ask why each of the above steps has the sig. fig. listed. Both the
melting (fusion) and vaporization steps have 3 sig. fig. because the moles has 3 s.f.
and each )H has 3 or 4 s.f. In heating the liquid the specific heat has 2 s.f. so the
result has 2 s.f. (to the 100000's place). In heating the solid the specific heat has 2
s.f. so the result has 2 s.f (100's place). Since the 3rd step is only known to the
100000's place (6th digit to the left of the decimal) the answer can only be reported
to that place and winds up with only 2 s.f.
27
11.41)
a) The critical pressure is the pressure required to cause liquefication at the critical
temp.
b) As the AF between molecules inc., the critical temp. of the compounds inc.
c) The temp. of N2 (R) is 77 K (-196 °C). All the gases in Table 11.5 (p. 452) have
critical temp. higher than 77 K, so all can be liquefied at this temp by applying
sufficient pressure.
The critical temp. is the highest temp at which a gas can be liquefied, regardless of
pressure. Above this temp. a substance can not be liquefied by applying pressure.
Above the critical temp. and press. only one fluid phase exists, the supercritical
fluid phase. It has properties similar to both liquids and gases.
In order to liquefy a gas by applying pressure the temperature of a substance must
be at or below its critical temperature, Tc.
11.42)
Tc (K)
Pc (atm)
CCl3F
471
43.5
CCl2F2
385
40.6
CClF3
302
38.2
CF4
227
37.0
Tc = critical temperature
Pc = critical pressure
a)
CF4 is nonpolar and has only LF. CCl3F, CCl2F2 and CClF3 are polar and
have both LF and DD. The polarity of the polar molecules isn’t going to vary
much but the size does. As each F atom is replaced with a Cl atom the
molecules size increases and the LF inc. The trends for the above molecules
are mostly due to these inc. LF.
28
11.42) (cont.)
b)
According to the solution to 11.41(b), the higher the critical temp, Tc, the
stronger the AF. Therefore the strength of the AF increases moving right to
left as the molecules (and MW) get bigger,
CF4 < CClF3 < CCl2F2 < CCl3F
c)
Looking at the molecules in the table one sees that as they get larger the Tc
and Pc get larger. From this one would expect Tc and Pc to be larger for CCl4
than for CCl3F. Looking at the numbers for the polar molecules the values
for Tc increase by 83 K going from CClF3 to CCl2F2 and 86 K going from
CCl2F2 to CCl3F. Tc increases a little less, 75 K, going from CF4 to CClF3
indicating the fact the CClF3 is polar has little to no effect and the dominating
AF is LF. Based on the inc. for the polar molecules one might expect an inc.
of about 89 K going from CCl3F to CCl4. Again, looking at Pc for the three
polar molecules one sees an increase of 2.4 atm and 2.9 atm going from
CClF3 to CCl3F. Thus and increase of about 3.2 atm might be expected.
Property
Tc (K)
Pc (atm)
CCl3F
471
43.5
CCl4 (predicted)
560
46.7
CCl4 (CRC)
556.6
44.6
The predicted values for CCl4 are in good agreement with the CRC
(experimental) values. Since CCl4 is nonpolar, this indicates the polarity of
or the molecules plays only a small role and the LF (rather than the DD)
dominate the physical properties for these molecules.
Remember, just because one molecule is polar (has LF and DD) and the other
is nonpolar (has only LF) doesn’t mean the polar molecule will always have
the stronger overall AF. If the nonpolar molecule is big enough its AF (due
to LF) can be bigger than those of a smaller polar molecule (especially if the
molecule isn’t very polar).
29
11.43)
a)
No effect (vol. of liquid does not effect VP).
b)
No effect (as the surface area inc. so does vol. above the liquid so there is no
effect on VP).
c)
Vapor pressure would dec. with inc. intermolecular attractive forces (IAF)
because fewer molecules would have sufficient KE to overcome the AF and
escape to the vapor phase. IAF inc. => VP dec.
d)
VP inc. with inc. temp. because the average KE of the molecules inc. (there
are more molecules w. KE greater than the energy required for the molecules
to escape the liquid).
e)
As density increases VP decreases. Density (m/V) increases as MW
increases (for the same volume of liquid). As MW increases (bigger
molecule) the strength of the LF inc. At a given temp., the larger the AF the
lower the VP (see part (c) above for the reason).
30
11.45)
a)
CBr4 < CHBr3< CH2Br2< CH2Cl2 < CH3Cl < CH4
The weaker the IAF, the higher the VP, the more volatile the compound. The
order of inc. volatility is the same as the order of dec. strength of the IAF.
Even though four of the molecules (CHBr3, CH2Br2, CH2Cl2 and CH3Cl) are
polar the AF are dominated by the LF. This is analogous to the boiling
points of H2S, H2Se and H2Te (Section 11.2 in the text and Figure 11.5 on
page 443 of the text).
The polarity of the polar molecules doesn’t vary to a great degree so the trend
is dominated by the inc. in LF due to the inc. in size of the Cl compared to H
and the Br compared to Cl (and thus a large inc. in size and MW of the
molecules as each Cl or Br is added). CBr4 is considerably larger than CHBr3
so its LF are much greater than the LF and DD of CHBr3. Generally going
down a group with similar molecules the trend in AF winds up being
dominated by the LF due to the inc. in size of the atoms going down the
group.
b)
CH4 < CH3Cl < CH2Cl2 < CH2Br2 < CHBr3 < CBr4
Boiling point inc as the strength of AF inc, so the order of b.p. is the order of
the inc AF. The lower the VP the higher the b.p. (order is opp that in part a).