Petrucci 8th ed Problem set #5: Structure Page 1 of 6
Petrucci 8th Edition problem set #5: Structure of
liquids and solids
Chapter 13
13-1 a) ∆Hvap symbolizes the molar enthalpy of vaporization, the quantity of heat
needed to convert one mole of a substance from liquid to vapor.
b) Tc symbolizes the critical temperature, that temperature above which a
gas cannot be condensed to a liquid, no matter how high the applied
pressure.
c) An instantaneous dipole is a momentary imbalance of the positive and
negative charges in an atom or a molecule, caused when the electron cloud
moves off center in a random manner.
d) Coordination number refers to the number of atoms of one type
surrounding another atom at the same distance, often those touching the
central atom.
e) A unit cell is the smallest portion of a crystal that will replicate the crystal
through simple translations.
13-3 a) Adhesive forces are those between different types of substances, usually
forces between two condensed phases. Cohesive forces are those within a
substance, again usually referring to liquids or solids.
b) Vaporization is the physical process of transforming a liquid into a vapor.
Condensation is the reverse process: vapor to liquid.
c) A triple point of a substance is the point where three phases coexist,
often solid, liquid and vapor. The critical point is that temperature and
pressure above which liquid and vapor cannot be distinguished; just a fluid
exists, called a supercritical fluid.
d) Face-centered and body-centered cubic unit cells both have atoms at the
vertices of a cube. In a face-centered unit cell there also is an atom on the
center of each of the six faces of the cube. In contrast, in a body-centered
Petrucci 8th ed Problem set #5: Structure Page 2 of 6
cubic unit cell there is an atom exactly in the center of the three-dimensional
cubic cell.
e) A tetrahedral hole is one surrounded by four atoms at the vertices of a
tetrahedron. An octahedral hole is surrounded by six atoms: above and
below, in front and behind, and to the right and the left.
13-8 a) mass CHCl3 vaporized = 6.62 kJ x (1000 J / 1 kJ) x (1 g CHCl3 / 247 J)
= 26.8 g CHCl3
b) ∆Hvap = ( 247 J / 1 g CHCl3 ) x (1 kJ / 1000 J) x (119.38 g CHCl3/ 1 mol
CHCl3) = 29.5 kJ / mol CHCl3
c) heat evolved = 19.6 g CHCl3 x ( 247 J / 1 g CHCl3 ) x (1 kJ / 1000 J) =
4.84 kJ
13-11 We use the ideal gas equation, with n = moles Br2 = 0.486 g Br2 x (1 mol
Br2 / 159.8 g Br2) = 3.04 x 10-3 mol Br2
P = (n R T) / V = ((3.04 x 10-3 mol Br2 x 0.08206 L atm mol
K)/ 0.2500 L) x (760 mmHg / 1 atm) = 226 mmHg
–1
K-1 x 298
13-24 a) The atoms touch along the face diagonal, d. The length of that diagonal
thus is 4r ( r from one corner atom + 2r from the center atom + r from the
other corner atom). The Pythagorean theorem relates the length of the unit
cell, l, to the face diagonal: d2 = l2 + l2 (the cell is square; both sides are
equal) or d = (2)½ l. Two quantities equal to the face diagonal are equal to
each other: d = (2)½ l = 4 r = 4 x 128 pm.
l = (4 x 128 pm) / 1.414 = 362 pm
b) The volume of the unit cell is that of a cube: V = l3 = (362 pm)3
V = 4.74 x 107 pm3
c) In a face-centered unit cell, there are eight “corner” Cu atoms, each
shared by eight unit cells; there are six “face” Cu atoms, each shared by two
unit cells. No. Cu atoms = (8 x 1/8) + (6 x ½) = 4 Cu atoms.
Petrucci 8th ed Problem set #5: Structure Page 3 of 6
d) The ratio of the occupied volume to the unit cell volume is
fv =
volume of spheres in unit cell
volume of unit cell
If the radius of the atom is r, the volume of a sphere is (4/3) πr3, and the
volume of the unit cell is known in part b):
4
4 × πr 3
3
= 0.7413
fv =
4.74 × 10 7
since there are four complete spheres in the unit cell. Thus, 74.13% of the
unit cell is occupied and 25.87% of the unit cell is empty
e) The mass of 4 atoms is determined from the atomic mass.
mass
=
unit cell
4 Cu atoms
unit cell
×
1mol Cu atoms
×
63.55g Cu
6.022 × 10 Cu atoms 1mol Cu atoms
23
=
4.221 × 10 − 22 g Cu
unit cell
f)
3


mass
4.221× 10 g Cu  1mol Cu atoms 
=
×
density =
= 8.91g / cm 3

volume
 6.022 × 10 23 Cu atoms 
unit cell


− 22
13-32 nacetonitrile = PV/RT
nacetonitrile = (1.00 atm x 1.17 L) / (0.08206 L atm mol-1 k-1 x (273.15 +
81.6) K)
nacetonitrile = 0.0402 mol acetonitrile
∆Hvap = (1.00 kJ / 0.0402 mol) = 24.9 kJ/mol acetonitrile
13-42
As the N2 is bubbled through the liquid CCl4 it becomes saturated with CCl4(g).
Petrucci 8th ed Problem set #5: Structure Page 4 of 6
Therefore,
PTotal = PN2 + PCCl4 = 742 mmHg + 261 mmHg = 1003 mmHg
This gas mixture is confined to the 7.53 L of gas that passed through the
CCl4(l). If the final pressure of this mixture becomes 742 mmHg,
Vf = (1003 mmHg/742 mmHg) X 7.53 L = 10.2 L
(i.e. the moles of gas and the temperature remain constant)
13-45
 760mmHg  25.5 X 10 3 J mol −1  1
1

 =
−
In

−1
−1 
 375mmHg  8.3145 J mol K  T 329.35 K 
1
1
−
= 2.30325 X 10 − 4
T 329.35
1
= 3.2666 X 10 −3
T
T = 306 K or 33°C
13-52 a) The upper-right region of the phase diagram is the liquid region, while the
lower-right region is the region of gas. One way to figure this out is to
imagine moving from left to right (toward higher temperatures) at constant
pressure. (Assume 45 atm for this case.) From experience, we know that the
progression of states is solid (low temperature ) → liquid (intermediate
temperature ) → gas (high temperature).
b) Melting means that we convert the solid to a liquid. As the phase diagram
shows, the lowest pressure at which liquid exists is at the triple point
pressure, 43 atm. 1.00 atm is far below 43 atm; liquid cannot exist at this
temperature.
c) As we move from point A to point B by lowering the pressure, initially
nothing happens. At a certain pressure, the solid liquefies. The pressure
continues to drop, with the entire sample being liquid while it does, until
another, lower pressure is reached. At this pressure the entire sample
vaporizes. The pressure then continues to drop, with the gas becoming less
dense as it does so, until point B is reached.
Petrucci 8th ed Problem set #5: Structure Page 5 of 6
13-61 a) HCl is not a very heavy diatomic molecule; London forces are expected to
only moderately important. Hydrogen bonding is weak in the case of H-Cl
bonds; Cl is not one of the three atoms (F, O, N) that from strong hydrogen
bonds. Finally, because Cl is an electronegative atom, and H is only
moderately electronegative, dipole-dipole interactions should be relatively
strong.
b) In Br2 neither hydrogen bonds nor dipole-dipole attractions are important;
there are no H atoms in the molecule, and homonuclear molecules are
nonpolar. London forces are more important than in HCl since Br2 is heavier.
c) In ICl there are no hydrogen bonds since there are no H atoms in the
molecule. The London forces are as strong as in Br2 since the two molecules
have the same number of electrons. However, dipole-dipole interactions are
important in ICl; the molecule is polar toward Cl.
d) In HF London forces are not very important; the molecule has only 10
electrons. Hydrogen bonding is quite important and definitely overshadows
even the strong dipole-dipole interactions.
e) In CH4, H bonds are not important; the H atoms are not bonded to F, O, or
N. In addition the molecule is not polar, so there are no dipole-dipole
interactions. Finally, London forces are quite weak since the molecule
contains only 10 electrons, and this is why CH4 is a gas with low critical
temperature.
13-71 Yes, the attractive force between a pair of oppositely charged ions increases
with decreased ionic sizes, i.e. smaller ions produce larger interionic forces.
13.89
Mg+(g) → Mg2+(g) + e-
I2 = 1451 kJ/mol
L.E. = -2526 kJ/mol
Sublimation:
Mg2+(g) + 2Cl-(g) →
MgCl2
Mg(s) → Mg(g)
First ionization energy:
Mg(g) → Mg (g) + e
I1 = 738 kJ/mol
Dissociation energy:
Cl2(g) → 2Cl(g)
D.E. = (2 x 122) kJ/mol
Second ionization
energy:
Lattice energy:
+
∆Hsub = 146 kJ/mol
-
Petrucci 8th ed Problem set #5: Structure Page 6 of 6
Electron affinity:
2Cl(s) + 2 e- → 2Cl-(g)
2 x E.A. = 2(-349) kJ/mol
∆H°f = ∆Hsub + I1 + I2 + D.E. + (2 x E.A.) + L.E.
= (146 + 738 + 1451 + 244 – 698 – 2526) kJ/mol
= -645 kJ/mol
13-108
First diagram for T > than the temperature of the triple point, the
substance should remain a gas. In the diagram the substance vapor
pressure curve indicates otherwise.
Second diagram: from the triple point as P increases the line
representing the effect of pressure on the melting point (fusion curve)
should be nearly vertical since the melting temperature is essentially
unaffected by moderate pressures. At very high pressures, the fusion
curve will have a positive slope, ultimately reaching temperatures
above the critical temperature. The fusion curve has a negative slope
in the second diagram.