Chemistry 121
Mines
Answer Key, Problem Set 12 – Full
1. MP; 2. MP; 3. MP; 4. MP; 5. MP; 6. WO1; 7. WO2; 8. WO3; 9. WO4; 10. WO5; 11. WO6; 12. WO7; 13. WO8; 14. MP; 15.
WO9; 16. WO10; 17. WO11; 18. WO12; 19. WO13; 20. WO14; 21. WO15; 22. MP; 23. MP(11.73); 24. MP(11.74); 25. MP; 26.
WO16; 27. MP; 28. MP; 29. WO17
---------------------------Molecular Polarity (How to Determine if a Molecule is Polar)
1. MP. No answer in this key to Mastering Problem.
2. MP. No answer in this key to Mastering Problem.
Identification of Interparticle Forces, and Properties that Correlate with Them
3. MP. No answer in this key to Mastering Problem.
4. MP. No answer in this key to Mastering Problem.
5. MP. No answer in this key to Mastering Problem.
6. WO1.
(a) Draw a picture illustrating a hydrogen bond between two molecules of ammonia (NH3), with the hydrogen bond
clearly indicated. (b) Describe how the interaction in your picture is consistent with the definition of a hydrogen bond.
(c) State why the same picture using PH3 in place of NH3 would not illustrate a hydrogen bond. Specifically address
why a P atom in place of an N atom does not result in there being a hydrogen bond.
Answers:
(a)
A covalent bond
(within a molecule)
H
H
N
H
H
N
A hydrogen bond
(between two molecules)
H
H
(b) This is consistent with the definition of a hydrogen bond in that it shows a H atom that is
covalently bonded to a N (which is an N, O, or F) on one molecule interacting with a partially
negative N (which is an N, O, or F) on a different molecule.
(c) If there were a P in place of the N in each molecule above, although one could draw the two
molecules next to one another as above, there would be no hydrogen bond because P is
neither electronegative enough nor small enough for a hydrogen bond to form.
Hydrogen bonding is a specialized case of dipole-dipole in which a hydrogen atom bonded to a
small and highly electronegative atom (e.g., N, O, or F) [and which is therefore partially
positive] interacts with a small, partially negative small atom (e.g., N, O, or F) on another
molecule. The only bonds with H as one of the two atoms that are polar enough and have a
small enough other atom to form H-bonds are H-O, H-N, or H-F bonds.
7. WO2.
How does the strength of hydrogen bonds compare to the strength of covalent bonds? How about to ionic
bonding?
Answer: H-bonds (and all IM forces!) are much weaker than covalent bonds or ionic bonding.
How do you know? The fact that most molecules stay intact upon melting or boiling (even if their
molecules are attracted to one another via H-bonding) indicates that the energies needed to break
covalent bonds are much greater than the energies needed to separate things that are held together by
H-bonds. For example, water boils at 100C. That means the intact water molecules separate from one
another, overcoming the collective IM forces between the molecules (including H-bonding + London +
dipole-dipole). So the hydrogen bonds clearly “break” at this temperature, while the O-H bonds within
water molecules do not break.
PS12-1
Answer Key, Problem Set 12
Separately, although your text does not mention it, the maximum values for “energy per mole” associated with H-bonds is
approximately 40 kJ/mol. This is less than one-third of the value of even the weakest covalent bond (142 kJ/mol—See Table
9.3 in Tro), and most covalent bonds are much greater than that (the triple bond in CO is over 1000 kJ/mol!).
For comparing H-bonding to ionic bonding, consider the melting points of molecular substances the
have H-bonding compared to those of typical ionic compounds. The melting points of ionic compounds
(typically > 300 C—see Exp. 17 in lab manual) are generally well above the melting points (and often
the boiling points!) of most (small) molecular compounds that contain H-bonds (mps typically <300 C—
see Exp. 17), which indicates that the forces holding ions together are much stronger than those holding
molecules together. Since H-bonding is only a subset of the (collective) IM forces between molecules,
clearly ionic bonding forces are stronger than H-bonding forces.
Note also that the weakest lattice energies (see Tro, Section 9.4) are approximately -700 kJ/mol, which means that it takes at
least 700 kJ to break apart (completely) the ions in a mole of an ionic compound, and so the order-of-magnitude strength of
ionic bonding is (again) well above that of even the strongest hydrogen bonding interactions (40 kJ/mol max).
8. WO3.
The nonpolar hydrocarbon C25H52 is a solid at room temperature. Its (normal) melting point is over 50°C and its
(normal) boiling point is greater than 400°C. Which has the stronger intermolecular forces, C25H52 or H2O? Explain your
answer.
Answer: C25H52 has stronger IM forces than H2O.
Reasoning: It is reasonable to generalize as follows: the greater the melting and boiling point of a
substance (at a given Pexternal), the greater the forces of attraction between its basic units (“interparticle
forces”), because higher temperatures are associated with greater average kinetic energy of particles, and
kinetic energy is, effectively, “needed” in order to overcome forces of attraction. Since C25H52 has greater
(normal) melting and boiling points than H2O (50°C > 0°C, and 400°C > 100°C), it has the stronger IM
forces between its molecules.
Think of it this way: at 300°C (and P = 1 atm), the molecules of H2O cannot be held together by their IM
forces—the average kinetic energy is apparently too great, leading to “bulk” escape of molecules, and
water exists as a gas. But at the same T (and P), the molecules of C25H52 are held together by their IM
forces—though the average kinetic energy of particles is the same as in H2O, it is apparently not great
enough to lead to “bulk” escape (since C25H52 is a liquid at this T). The only reasonable conclusion is that
the IM forces are greater in C25H52.
9. WO4.
Tro states on p. 495: “Hydrogen bonds are … the strongest of the three intermolecular forces we have discussed so far.
Substances composed of molecules that form hydrogen bonds have higher melting and boiling points than comparable
substances composed of molecules that do not form hydrogen bonds.” What do you suppose “comparable” means in this
context? Why does this qualification make the main premise of the statement invalid or meaningless? Use the
information in (and answer to) question WO3 (prior problem) to invalidate/critique Tro’s statement. (Hint: “Are London
forces one of the “three intermolecular forces”? And are they the same in all substances?)
By “comparable”, the only thing that makes sense to me is that he’s saying “if all other IM forces are
similar in strength”. That is, if all other IM forces are similar in strength, then having hydrogen bonding,
in addition, will make the mp and bp higher. This makes total sense and I mentioned this in class (like a
dozen times, right?!). But this invalidates the premise of the first statement (that H-bonds are the
“strongest”)! London forces are one of the 3 IM forces discussed, yes? So how can you try to justify a
statement that H-bonding is stronger than London forces by saying that “if London forces are
comparable”, H-bonding will make the mp and bp higher? It’s an invalid argument! It’s what I ranted
about in class! I cannot understand how educated chemists and educators can justify stating that Hbonding is the “strongest of the 3 IM forces we’ve discussed” given the examples such as those in WO3,
where the mp and bp are much higher for a nonpolar compound (like C25H52) than for water, which has
H-bonding. I think they must be trying to emphasize that H-bonding forces are stronger “per atom” or
“per electron” than any other IM force. If they qualified it that way, it would be valid, because you do
need a lot of electrons or a lot of atoms in order to get the London forces (collectively) to be stronger.
But that isn’t what they say, and I think it is, frankly, ridiculous that it keeps getting mentioned (without
qualification) in general chemistry textbooks. (and it’s not like I’ve not written the author to make this
point; he simply ignores my argument for some reason [he doesn’t provide his reasoning] Maybe I’m
missing something, but I don’t think so…). Why not just say that London forces, collectively, can vary
PS12-2
Answer Key, Problem Set 12
greatly in strength, so much so that they can be stronger than H-bonding if the molecules get large
enough. And that given that H-bonding is between only two small atoms, it is remarkably strong for an
IM force. Why is there a need to say that H-bonding is the “strongest” (without qualification)? I just don’t
get it… Okay, I’ll stop now.
10. WO5.
Identify the types of interparticle forces present in the solids (or liquids!) of each of the following substances.
Remember: All molecules (or unbounded atoms) attract other molecules (or unbounded atoms) via London
forces.
(a) NH4Cl: ion-ion forces
Substance is ionic (all symbols are nonmetals, but the “NH4” at the beginning indicates that the
substance contains NH4+ ions, and thus “Cl” here represents Cl- ions)
(b) C (diamond): covalent bonds
This is a covalent network solid, so each atom is covalently bonded to every other atom. Even in SiO2
(a compound rather than an element), the atoms of one FU are bonded to atoms in neighboring FUs
covalently (there is no difference between the intraformula unit “bonding” and the interformula unit
bonding.)
(c) n-octane, CH3(CH2)6CH3: London forces
Substance is molecular (binary with two nonmetals), so forces are IM forces of some kind. Substance
is nonpolar (remember, C-H bonds are considered nonpolar, and C-C bonds are obviously nonpolar),
so the only IM forces present are London forces.
(d) CHCl3: London forces and dipole-dipole forces
Substance is molecular (not ionic because all symbols are nonmetals, and the first part of the formula
doesn’t involve “NH4”—remember, NH4+ is the only common polyatomic cation). Substance is polar
(CHCl3 has one center (carbon) surrounded by 3 Cl’s and 1 H. Since not all the surrounding atoms are
the same, and C-Cl bonds are polar (EN ~ 0.5), the bond dipoles do not completely cancel out,
making the molecule polar.
(e) NH3: London forces, dipole-dipole, and H-bonding
Substance is molecular (binary with two nonmetals) and polar (a NH3 molecule has one center (N)
surrounded by three H’s and a lone pair [LDS not shown, but you should be able to draw it!]. The lone
pair breaks up the symmetry so the N-H bond dipoles do not completely cancel out. Substance’s
molecules contain N-H bonds, so each molecule can make H-bonds to other NH3 molecules.
(f) NO: London forces, and dipole-dipole forces
Substance is molecular (binary with two nonmetals) and polar (only one bond, and that bond is polar
[EN ~ 0.5])
(g) BF3: only London forces
Substance is molecular (binary with two nonmetals) and nonpolar (each BF3 molecule has three F
atoms surrounding it, but no lone pairs—exception to octet rule; see text)
11. WO6.
Consider the substances Cl2, HCl, F2, NaF, and HF. Which one has a (normal) boiling point closest to that of argon?
Explain.
NOTE: Since noble gas samples contain “basic units” that are individual, non-covalently nor metallically
bonded atoms, the forces between the atoms are the same as the forces between nonpolar molecules in a
molecular sample--London forces. Thus we typically treat noble gases as “nonpolar molecular” substances
even though they are obviously not actually “molecular”. Thus the term “IM forces” is often applied to noble
gases despite the fact that it isn’t strictly appropriate.
Answer: F2
PS12-3
Answer Key, Problem Set 12
Strategy:
1) Recognize that the most similar boiling points (or mp, or Hvap, or Hfus or surface tension…) are
predicted to be in the two substances that have the most similar interparticle forces.
2) Recognize that typical ionic compounds have much stronger interparticle forces than molecular
substances (or noble gases), because ion-ion forces are typically greater than any set of IM forces
(see WO1, above).
3) When comparing molecular substances:
a) First assess London forces in all substances by counting the number of electrons per molecule (or
noble gas atom) (OR molar mass) in each and compare. (Remember that the magnitude of
London forces is roughly proportional to the number of electrons per molecule/atom (OR molar
mass).)
b) For substances with similar London forces, look to see if which, if any, molecules are polar and/or
can interact via H-bonding. Being polar adds forces of attraction “on top of” London because of
dipole-dipole forces. Being able to form H-bonds adds additional forces of attraction.
Execution of Strategy:
Ar
Cl2
HCl
F2
ionic?
# electrons
per unit*
(London)
polar?
H-bonding?
* Or ~MM’s:
NaF
HF
Y
18
34
18
18
N/A
10
N
N
Y
N
N/A
Y
N
N
N
N
N/A
Y
40 g/mol
71
36
38
20
Clearly, the substance with the most similar IM forces as Ar is F2. Each one’s “basic units” are nonpolar,
and have 18 electrons (or MM’s close to 40 g/mol). So the forces are similar. Note that HCl has similar
London forces, but since its molecules are polar, there is an additional dipole-dipole force of attraction
between its molecules, so you’d predict stronger IM forces overall and a higher boiling point for HCl vs.
Ar. Cl2 should have stronger IM forces than Ar or F2 since the London forces will be stronger (more
electrons/molecule or greater MM). HF should have a higher bp due to the dip-dip and H-bonding. NaF
should have a very high boiling point due to the ion-ion forces holding its “units” together.
12. WO7.
Consider the following compounds and formulas. (Note: The formulas are written in such a way as to give you an
idea of the structure.)
ethanol: CH3CH2OH;
dimethyl ether: CH3OCH3;
propane: CH3CH2CH3
The boiling points of these compounds are (in no particular order) -42.1°C, -23°C, and 78.5°C. Match the boiling points to the
correct compounds.
Answers: -42.1°C is propane, -23°C is dimethyl ether, and 78.5°C is ethanol
Reasoning:
All substances have similar London forces since the number of electrons per molecule values are identical
in the three (26 e-) (the molar masses are also similar: 44-46 g/mol). Propane is nonpolar because it
contains only C’s and H’s, and C-C and C-H bonds are nonpolar. Thus the only IM forces between its
molecules are the London forces. The ether molecule is polar (it contains a C – O – C bond, which makes
the geometry around the O bent [O will have two lone pairs!]), so its IM forces and thus bp should be larger
than those for propane. Ethanol has an O – H bond, so in addition to being polar, its molecules can attract
one another via H-bonding, which makes its IM forces and bp the highest of the three.
PS12-4
Answer Key, Problem Set 12
13. WO8. (a) Order from weakest to strongest intermolecular forces and explain your reasoning:
CH4; He; CH3OH;
CH3F; CH3CH3 (b) Which do you think would have the highest boiling point at 1 atm pressure? (c) Which do
you think would have the highest vapor pressure (at some hypothetical T where all of them were liquids)? (d)
Which to you think would have the highest Hvap? Explain.
Answer:
(a) weakest forces He < CH4 < CH3CH3 < CH3F <
2 e-
10 e-
18 e-
18 e- (&p)
CH3OH
strongest forces
18 e- (w/H-bonding)
Reasoning.
(NOTE: You may wish to see the strategy outlined in Problem 11 (WO6) above before
reading this answer, since that is essentially what is applied here.)
The only forces of attraction in nonpolar substances are London forces, which get stronger as the
number of electrons (or molecular weight, which correlates with #electrons) gets larger. Of the
nonpolar substances, He has the fewest (2), then CH4 (10), then CH3CH3 (18) (and the molecular
weights also increase steadily). So that explains the ordering of these three. CH3F has the same
number of electrons (and a similar molecular weight) as CH3CH3 but is polar. So it has similar
London forces but has, in addition, dipole-dipole forces. So the overall (or “total”) IM forces are
expected to be greater in CH3F than CH3CH3. CH3OH molecules have the same number of
electrons as CH3F (and a similar molecular weight), are polar, and can interact with one another
via hydrogen bonding. So it will have even stronger IM forces (in total) than CH3F (H-bonding is
“added on” to similar “other” IM forces).
(b) CH3OH should have the highest boiling point at 1 atm pressure because it has the greatest
intermolecular forces (see part a). See also the answer to WO15 below for a more detailed
analysis of the relationship between IM forces and bp.
(c) Helium (He) should have the highest vapor pressure at a given temperature. Having weaker
forces of attraction means that more molecules (actually “atoms” here!) can escape the liquid at
any given temperature, and “more gaseous molecules/atoms” equates to “greater (vapor)
pressure”. Again, see the answer to WO15 for a more detailed analysis.
(d) CH3OH should have the highest Hvaporization because if the intermolecular forces are stronger, it
should take more energy to separate the molecules, which is what occurs during vaporization (liquid
turning into gas means molecules are separated from one another). Recall that Hvaporization refers to
the change in enthalpy that accompanies the vaporization of one mole of a liquid; since this occurs
at a constant temperature, the enthalpy change is essentially equal to the increase in the potential
energy of the system as the molecules are separated from one another).
14. MP. No answer in this key.
Interparticle Forces and Different Types of Crystalline Solids
15. WO9. (a) If covalent bonds are present in both molecular solids and covalent network solids, why are the melting points
so low for molecular solids and so high for covalent network solids? (b) Why do ionic compounds tend to have
higher melting points than molecular compounds with similar molar masses? (Hint: What forces hold the basic
units together?)
(a) Think about and visualize the following carefully: you do not break any covalent bonds
when melting a molecular solid! Since the molecules of a molecular substance remain intact
upon melting (they just separate from one another), only the much weaker intermolecular forces of
attraction need to be overcome. That means that less energy is required, and generally speaking
that leads to a very low melting point for molecular solids [see Exp. 17]. On the other hand, in order
to melt a covalent network solid, the covalent bonds between atoms do have to be broken, and so it
is not surprising that the melting points of these solids are very high.
PS12-5
Answer Key, Problem Set 12
(b) The stronger the attractive interparticle forces in a solid, the higher the melting point. Molecules are
neutral and ions are charged. This leads to fundamentally different magnitudes of electrostatic
forces of attraction between “basic units”. Ions attract one another more strongly because the units
are “fully” charged ions, and electrostatic forces are stronger the greater the magnitude of the
charge(s) (Coulomb’s Law). Molecules are neutral, but can still attract one another electrostatically
either via dipole-dipole interactions or via instantaneous and induced dipole interaction. However, in
both of these cases, the “charges” that lead to the electrostatic forces are necessarily “partial”-either “partial and permanent” (if molecules are polar) or “partial and temporary/induced” (if
nonpolar). Ultimately, the “partial charged” nature of IM forces are what make them intrinsically less
strong than ion-ion forces. It’s all about Coulomb’s Law in the end!
16. WO10. 9.34 in Tro.
How does the electron sea model explain the conductivity of metals? The malleability and ductility of
metals?
Answer: The electron sea model views the valence electrons as being held so loosely by the metal
atoms that they flow freely and continuously between the positive “cores” of any atoms—they are
“delocalized” and they are a kind of flowing, negatively charged “glue” that holds all of the cores to
one another. This explains conductivity fairly obviously—if the valence electrons can flow freely
between the positive “cores”, then they can flow freely through the solid, which is basically the
definition of electrical conductivity (movement of charge carriers). It explains malleability by
suggesting that in response to being struck by, say, a hammer, the cores will move with respect to
one another, but the electron “sea” will just flow around the cores (in whatever deformed
arrangement that might be generated by the “blow”), holding them together to one another. The
idea is that the electrons move so fast that they can respond essentially immediately and keep the
cores together (rather than having the solid fracture as an ionic solid does in response to a sharp
blow). It explains ductility similarly—drawing metals into wires is a type of deformation, and as was
just stated, the ability of the “sea of electrons” to flow quickly around the cores would allow for the
deformation to occur while keeping the atoms bonded to one another.
Vapor Pressure, Vapor Pressure Curves, and Relation to Boiling and Boiling Point
17. WO11. Consider the following vapor pressure versus temperature plot for three different
substances A, B, and C.
A
If the three substances are CH4, SiH4, and NH3, match each curve to the correct substance.
B
Answer: A is CH4, B is SiH4, and C is NH3
Reasoning:
Pvap
1) First of all, determine who has the strongest/weakest IM forces:
C has the strongest IM forces and A has the weakest, because if you
look at the T needed to generate a given vapor pressure (horizontal
dashed line added), you see that C requires the highest T and A
requires the least. That means the forces are strongest in C and
weakest in A. (You could also look at a vertical line and look at who
has the higher VP at a given T—C would be lowest indicating stronger IM forces).
2) Now, figure out the order of strength of IM forces in CH4, SiH4, and NH3:
CH4 and SiH4 are both nonpolar (4 electron clouds with no lone pairs gives a tetrahedral AG, with
all outer atoms the same. Actually, all bond are pretty much nonpolar anyway, so the geometry
isn’t critical here), but SiH4 molecules have more electrons and thus stronger IM forces exist
between SiH4 molecules.
NH3 has the same number of electrons as CH4, but it is polar (4 clouds but one lone pair) and has
hydrogen bonding forces acting between its molecules, so it clearly has greater IM forces than
CH4. Since the number of electrons in SiH4 isn’t that much bigger than the number in NH3 (18 vs
8), NH3 is predicted to have stronger IM forces than SiH4. NOTE: based on the I2 vs. H2O
example, you need to have something like 50-100 electrons per molecule to get London forces
the same order of magnitude as H-bonding.
PS12-6
T
C
Answer Key, Problem Set 12
So NH3 has greatest IM forces and is C; CH4 has the weakest and is A.
18. WO12.
(a) What is a kinetic energy distribution curve? (What are the
axes of the plot? What does the maximum of the curve represent?) (b)
Assume the distribution curve at the right is for a liquid X, and that Eescape is
the energy needed to escape the liquid. (i) Approximately what fraction of
the molecules in a sample of liquid X at temperature T1 have enough
energy to escape? (ii) Approximately what fraction have enough energy
to escape at temperature T2? (iii) Approximately how many times larger
is the fraction that can escape at the higher temperature than at the
lower temperature? (iv) Approximately how many times larger is the
average KE at the higher T than at the lower T?
Answers:
#
p
a
r
t
i
c
l
e
s
T1
Eescape
T2
KE
(a) A kinetic energy distribution curve shows the distribution of
KE values that different particles possess at any given time
in a sample. In other words, it shows how many particles in the sample (at a given moment in
time) have each given KE (shown on the x-axis). As such, the axes on the plot are “number of
particles” (on the y-axis) and “KE” (on the x-axis). The maximum of this plot, therefore, occurs at
the single KE value that is possessed by more particles than any other in the sample. It is called
the “most probable KE”. Although this is not equal to the average kinetic energy of all particles
(because the curve is asymmetrical), it is close to the value of the average (slightly less, actually)
and will track with the average—as the average gets larger (i.e., the peak shifts to the right), the
maximum will also shift to the right.
(b)(i) Given the plot in this problem, I’d estimate the fraction of molecules with KE  Eescape to be
around 1/20th (~5%). I looked at the roughly triangular area represented by the area under the
T1 curve past Eescape and compared it to the rest of the area under that curve (to the left of
Eescape). This is only a rough estimate, so anything from 2 – 6 % is probably reasonable (I think
it’s pretty clearly less than 10% though).
(ii) It looks to me like the area under the T2 curve to the right of Eescape is roughly 1/8th (~12-15%).
(iii) Actually, it now looks to me like I should have just asked part (iii) without even asking part (ii)
[or perhaps before (ii), because it is a lot easier to estimate how many times bigger the area
under the curve for T2 but to the right of Eescape than to estimate what fraction of the total area it
is. Oh, well. Based on my answers to (i) and (ii), (ii) is 2-3 times larger (12%/5% ~2, 15%/5%
= 3). However, looking just at the two areas directly on the plot, it looks to me like the second
one is clearly more than 3 times as large, perhaps 4 times.
(iv) I dropped lines down from the maxima of the two curves, and then added some “tick” marks
on the x-axis to estimate how much larger the most probable KE is at T2 vs T1. It is about 5 to
4, or 5/4 = 1.25 (which means about 25% larger). NOTE that when the avg KE is made about
1.25 times larger (through a T increase), the fraction of molecules that have enough energy to
escape became 3-4 times larger! Definitely not “proportional” (or “linear”).
19. WO13.
Give a molecular-level explanation of why the vapor pressure of a liquid is so sensitive to changes in
temperature (exponential vs. linear dependence). (Hint: Consider the distribution of kinetic energies of particles
[distribution curve] and the concept of “escape energy”.)
Shortest Answer: A small increase in T leads to shift in the distribution curve of kinetic energies to
higher energies. This results in a significant increase in the number of particles with high enough KE
to escape. This makes the concentration of gas particles significantly greater, increasing the (vapor)
pressure by more than would be the case by a T increase (of a fixed amount of gas) alone. In other
words, looking at P = (n/V) x R x T, when T of a liquid increases, n/V of the gas above the liquid
increases quite a bit, raising the vapor pressure (of the liquid).
Full Answer: Vapor forms over a liquid in an enclosed container because there is always some
fraction of the molecules in the liquid that have enough energy at any given time to overcome the
forces of attraction and "escape" into the gas phase. Since the average kinetic energy of the sample
depends on TEMPERATURE, the higher the temperature, the GREATER will be the average kinetic
energy—the curve shifts to “greater KE energies”, so more particles can escape. Moreover, a closer
look at the way the curve shifts lead one to the following observation: it is the nature of distribution
curves that a small change in the average has a great effect at the end of the curve. Thus, a
PS12-7
Answer Key, Problem Set 12
small increase in T, which leads to a small increase in the average kinetic energy, leads to a great
(percentage) increase in the fraction of molecules that have enough energy to overcome a fixed
“escape energy” barrier. This leads to a significantly larger number of gas molecules (per L) in the
gas phase before dynamic equilibrium is established. Since pressure is directly proportional to "n"
(the number of moles of gas particles) the (vapor) pressure will thus increase significantly. Note: as
commented on in class, the predominant factor here is the increased number of gas molecules that
escape (↔ higher concentration of gas particles) rather than the temperature increase of the gas
particles themselves. A very small contribution to the increased pressure will be from the T increase,
but it is essentially negligible compared to the increase in pressure due to the increase in “n”. Make
sure you understand how KE distribution curves relate to the argument above (NOTE: The example
below is the same exact example as used in the prior problem!!):
T1
T2
#
p
a
r
t
i
c
l
e
Eescape
Increased fraction of
molecules that have the
energy needed to escape
(after T increase) [~400%
larger area than before the T
increase, whereas the average
KE only increased by ~20-30%]
KE
Region where KEmolecule > Eescape
Original fraction of
molecules that have
the energy needed
to escape
20. WO14.
(a) Define "boiling" and "boiling point" and then (b) describe how boiling relates to vapor pressure,
temperature and atmospheric (external) pressure. In other words, why does a substance boil at its boiling point or
above, but not below it?
(a) boiling: the process in which bulk liquid molecules in a sample of a liquid are converted into a
gas, forming bubbles. It will occur only if energy is added to a liquid sample at a
temperature equal to or above the substance's boiling point.
(Note: evaporation is the process in which surface molecules in a sample of liquid are
converted into a gas. It can happen at any temperature, and without external energy
being added.)
boiling point: the temperature at which the vapor pressure of a liquid is equal to the external
pressure; it is the temperature at (or above) which a liquid can undergo boiling.
(NOTE: the boiling point of a substance depends on the external pressure, so there is no
single boiling point for a given substance. There is a different boiling point for
every external pressure.)
(b) In order for boiling to occur, gas must be able to form within the liquid sample, forming bubbles.
This means that the gas that is formed in the bulk liquid must be able to "push away" the bulk
liquid (and thus the atmosphere) to make space for the "bubble". The maximum pressure a given
gas (in equilibrium with its liquid) can achieve at a given temperature is called its "vapor
pressure". Thus, boiling cannot occur until the vapor pressure of the liquid is equal to or greater
than the external pressure (pushing on the liquid). The temperature at which the vapor pressure
becomes equal to the external pressure thus becomes the boiling point!
Since vapor pressure increases with T, if the T is less than the boiling point, the vapor pressure
will be too small to push away the atmosphere and the liquid will not boil. Conversely, if the T is
PS12-8
Answer Key, Problem Set 12
equal to or greater than the boiling point, the vapor pressure will be great enough to push away
the atmosphere and the liquid will boil.
NOTE: If external pressure increases, the boiling point raises. If external pressure
decreases, the boiling point decreases.
21. WO15
water
turpentine
Vapor Pressure of Substances in mmHg at Different Temperatures
0 C
20 C
50 C
80 C
100 C
4.6
17.6
92
354.9
760
2.1
4.4
17.0
61.3
131.1
(a) Which choice best indicates the degree of correctness of this statement? “Water at 50 C would
boil if the opposing pressure were reduced to 90 mmHg by means of a vacuum pump.” Explain
your answer.
(i) The statement is true
(ii) The statement is probably true; additional data would be needed for a final decision.
(iii) It is impossible to judge the statement because the data are insufficient.
(iv) The statement is probably false; additional data would be needed for a final decision.
(v) The statement is false.
Reasoning. By definition, a liquid will boil if the vapor pressure of the liquid is greater than or equal
to the external pressure. The chart indicates that at 50 C, the vapor pressure of water is 92
mmHg. Thus if the opposing (i.e., external) pressure were reduced to 90 mmHg, the vapor
pressure would be greater than the external pressure (92 mmHg is greater than 90 mmHg) and
so the water would boil.
(b) Which has the higher boiling point at 1 atm pressure, water or turpentine?
Answer: turpentine. Reasoning: You can look at this several ways. 1) Most specific
answer/view: water will boil at 100 C if the external pressure is 760 mmHg because water’s
vapor pressure is equal to 760 mmHg at this temperature (see chart). On the other hand, at 100
C, the vapor pressure of turpentine is only 131 mmHg, which is much smaller than the external
pressure (760 mmHg) and so it clearly will not boil at 100 C. That means that its boiling point
must be higher than 100 C, which means it is higher than water’s boiling point. 2) Similar view,
but more general: Boiling occurs when the vapor pressure of a liquid equals or exceeds the
external pressure. If the vapor pressure of water is higher than turpentine at any temperature,
surely its vapor pressure will become equal to a given external pressure at a lower temperature
than will the vapor pressure of turpentine. Thus water will boil at a lower temperature, or in other
words, it has a lower boiling point [so turpentine has the higher boiling point]. 3) Third view (also
general): The data in the chart indicate that at any temperature listed, the vapor pressure of water
is greater than that of turpentine. That means that the intermolecular forces must be weaker in
water than in turpentine because a higher vapor pressure at a given temperature means that more
molecules are able to escape (so the forces holding them together must be weaker). If the forces
in water are weaker, it should boil at a lower temperature (the molecules will need a lower average
kinetic energy in order to separate from the liquid and boil, and lower kinetic energy is associated
with a lower temperature). Thus, turpentine will boil at the higher temperature (its molecules are
held together by stronger forces).
Phase Diagrams
22. MP.
23. MP(11.73)
No answer in this key for these problems. However, please try the very similar
problem “22-24 Practice Problem”, and consult the detailed solution, at the
end of this key, for guidance, as needed.
24. MP(11.74)
PS12-9
Answer Key, Problem Set 12
25. MP No answer in this key for this problem, but draw a simple sketch of the phase diagram for this
substance using the information provided, and then analyze it as you did for the prior problem(s).
26. WO16. Analyze the figure for problem 11.77 in Tro to answer the following question:
Which has a greater density,
the Rhombic or Monoclinic phase of sulfur? You must provide detailed reasoning from the figure (no numerical
values should be looked up or given here!)
Answer: Rhombic
P
Reasoning: If you look at the line segment separating Rhombic
from Monoclinic, you can see that it has a positive slope (see
right, highlighted segment). That means that if you start below
the line (in Monoclinic) and increase pressure at constant
temperature, the system will convert to Rhombic at some point
(see arrow at right and intersection with the boundary line).
Since systems convert to their more dense phase at higher
pressure (high pressure “compresses” the sample, making V
smaller and thus density larger), Rhombic must be the more
dense phase of sulfur (between the two).
Rhombic Monoclinic Heating Curves and Related Calculations
27. MP. No answer in this key, but see Powerpoint 26 and/or text to help you with this. We covered this
extensively during class (I believe).
28. MP. No answer in this key, but see “28 (Practice Problem)” at the end of this key for a similar
problem, with its complete solution, to help you here if needed.
Unusual Properties of Water
29. WO17.
(a) What is odd about the relative densities of ice and water? (b) What is the basis for this property of
water? (c) Why is it important for life on this planet that water display this property?
(a) Most substances are more dense in the solid state than in the liquid state because they are more
“compact” in the solid state at the molecular level. That is, solids often represent the most efficient
way to “pack” atoms or molecules, and so if that is true, then clearly packing more atoms into a
given volume will lead to a greater density (more mass in a given volume). Water is odd in that
solid ice is less dense than liquid water.
(b) This occurs because in the solid state, water molecules pack together in a manner that maximizes
the number of hydrogen bonds between molecules. It turns out that this packing leads to a more
“open” structure than the more “random” arrangements found in liquid water. So there ends up
being fewer water molecules in a given unit of volume and so the density actually decreases a bit
upon freezing.
(c) One important consequence is that lakes do not completely freeze in the winter. When the surface
water freezes into ice, the ice floats rather than sinks. If it were to sink, then ultimately the entire
lake would freeze over, but since it floats, the ice layer actually insulates the liquid water beneath it
so that it remains a liquid. That allows fish (and any other creatures/organisms that live in the water)
to survive the winter! (This is discussed in your text on p. 469.) There are many other important
consequences of water’s odd liquid/solid density properties. Sea level would be much higher, for
example, if glaciers were to sink rather than float, and thus many islands and coasts would be under
water right now! (This is one of the reasons scientists are worried about global warming melting
glaciers—sea level would rise). Mountains degrade faster by weathering due to the fact that liquid
water fills cracks in rock, and then when it freezes, it expands, thereby applying tremendous stress
forces on the rock and making it crack and crumble.…I’ll stop now….
PS12-10
T
Answer Key, Problem Set 12
=============================== FORMAL END OF SET ===============================
A couple of practice problems (with explanations) for you! 
Phase Diagrams
22-24 (Practice Problem).
H
P
Consider the phase
diagram to the right. What phases are present at
points A through H? Identify the triple point,
normal boiling point, normal freezing point, and
critical point. Which phase is denser, solid or
liquid?
A
B
G
F
1.0 atm
E
C
D
T
Answers:
A: solid; B: liquid; C: gas; D: solid and gas;
E: solid, liquid, and gas (this is the triple point)
F: liquid and gas;
point)
G: liquid and gas (this is the critical
H: supercritical fluid
H
P
A
B
F
1.0 atm
The normal freezing point can be found by dropping a
vertical line down from the intersection of the solid-liquid
curve and the 1.0-atm horizontal line (left blue arrow shown)
and seeing the T on the x-axis for this intersection point.
The normal boiling point can similarly be found by dropping
a vertical line down from the intersection of the 1.0-atm line
with the liquid-gas curve.
G
E
C
D
Normal fp
Normal bp T
The solid is more dense in this substance. You determine this by recognizing that any substance in
equilibrium between its solid and liquid phases will turn into the more dense phase when pressure is
increased (“squeezing” the sample converts it to its more dense [compact] state). If you start at any
point on the solid-liquid equilibrium line on this phase diagram and increase the pressure, the
substance turns into a solid (see added arrow). Thus the solid must be more dense.
Heating Curves
28 (Practice Problem).
Consider a 75.0-g sample of H2O(g) at 125°C and 1 atm of pressure. What phase or phases are present
when 215 kJ of energy is removed from this sample? Cwater vapor = 2.0 J/g°C; Cwater, liguid = 4.2 J/g°C; Cice = 2.1 J/g°C; Hvap = 40.7
kJ/mol; Hfus = 6.02 kJ/mol
Answer: some solid and some liquid will be present in dynamic equilibrium (at 0°C)
Reasoning/Work:
General Approach: Realize that when heat is removed from a sample, there are only two possibilities for
what occurs: either T will decrease (if the T is not at the boiling point or melting point of the
substance), or a phase change will occur (if the T is at the bp or mp of the substance). So in this
problem, one must reason out / realize that the sequence of events that would occur if heat were
continuously and completely (first consider the hypothetical, then the “actual”) removed from a
sample of gaseous water vapor above its bp are as follows:
PS12-11
Answer Key, Problem Set 12
a) gas would first cool (lower its T) until the bp is reached
(q = Cgas x m x T)
b) gas would then condense to liquid at its bp (q = Hcondensation = -Hvaporization) (n.bp for H2O =
100.°C)
c) when all gas has been converted to liquid, the liquid would cool until the melting point is reached
(q = Cliquid x m x T)
d) liquid would then freeze at its mp/fp
(q = Hfreezing = -Hfusion)
e) when all liquid has been converted to solid, the solid would then cool until it reaches absolute
zero (again, assuming heat were completely removed)! (q = Csolid x m x T)
So the question in this problem is: Where do you end up with water when you don’t “completely”
remove energy, but specifically remove 215 kJ’ worth of energy. Let’s find out:
1) Calculate the energy needed to be removed to reach the bp (condensation point):
q = 2.0 J/g°C x 75.0 g x (100.°C - 125°C) = -3750 J  3.8 kJ needed to be removed
Since 3.8 kJ is less than the 215 kJ we are supposed to remove, we are not done. Go to next step.
2) Calculate the (additional) energy needed to be removed to condense the entire 75.0 g sample:
q = -75.0 g/(18.0 g/mol) x 40.7 kJ/mol = -169.6 kJ  169.6 kJ (more) needed to be removed
[Total Removed Thus Far = 3.8 + 169.6 = 173.4 kJ
Since 173 kJ is less than the 215 kJ we are supposed to remove, we are not done. Go to next step.
3) Calculate the (additional) energy needed to be removed to reach the freezing point:
q = 4.2J/g°C x 75.0 g x (0.°C – 100.°C) = -31500 J  31.5 kJ needed to be removed
[Total Removed Thus Far = 173.4 kJ + 31.5 = 204.9 kJ
Since 205 kJ is less than the 215 kJ we are supposed to remove, we are not done. Go to next step.
4) Calculate the (additional) energy needed to be removed to freeze the entire 75.0 g sample:
q = -75.0 g/(18.0 g/mol) x 6.02 kJ/mol = -25.1 kJ  25.1 kJ needed to be removed
[Total Removed Thus Far = 204.9 kJ + 25.1 = 230 kJ
Since 230 kJ is greater than the 215 kJ we are supposed to remove, the system would not “make it” this
far. That is, the entire sample would not freeze—part of it would freeze, leaving the system with some
solid and some liquid water at 0°C)
Thus, if you remove 215 kJ from this sample, you will end up with some liquid and some solid at 0°C.
PS12-12