Chemistry 2000 (Spring 2010)
Problem Set #4: Intermolecular Forces and States of Matter
Solutions
Answers to Questions in Petrucci (only those w/out answers at the back of the book)
10.70. Polar: HCN, SO3, OCS, SOCl2, POF3 (these molecules each have a net dipole)
Nonpolar: CS2, SiF4 (while each of these molecules contain polar bonds, they have no
net dipole due to their symmetry)
If asked this question on a test, you would be expected to draw the Lewis structure as part
of your justification.
10.72.
(a)
.. ..
....
F N O
..
..
(b)
bent (~120° angles):
(c)
..
....
F
+1
N
...-1
O.
..
.
..O.
....
F
..
+1
N
...
O.
.. -1
O ..
..
..
....
F
.. N
.
..O.
..
....
F
.. N
.
..O.
net dipole
FNO2 is trigonal planar, and all three bonds are polar. Since the electrons are pulled
away from the nitrogen in each bond, the net dipole is small.
FNO is bent, and will have a strong dipole pointing from the nitrogen out between the
fluorine and oxygen atoms:
12.48 Benzene (C6H6) is nonpolar, so it can only experience induced dipole-induced dipole
attractions. The other three molecules are all polar so they experience stronger
intermolecular forces (dipole-dipole attractions, in particular) and therefore have higher
boiling points.
Phenol (C6H5OH) can experience hydrogen bonding, one of the strongest dipole-dipole
attractions, so it has the highest boiling point.
Bromobenzene (C6H5Br) has a higher boiling point than chlorobenzene (C6H5Cl) because
bromine is a larger atom therefore bromobenzene is more polarizable and experiences
stronger dipole-induced dipole forces (and stronger induced dipole-induced dipole forces)
than chlorobenzene does.
12.52
(a)
There can be no hydrogen bonding involving butane (CH3CH2CH2CH3) because it has no
partially positive hydrogen atoms.
(b)
Intramolecular hydrogen bonding may be important in hexanedioic acid
(HOOCCH2CH2CH2CH2COOH) if it is in a dilute solution and cannot hydrogen bond
with the solvent it’s dissolved in. Otherwise, intermolecular hydrogen bonding is likely
to be more important.
(c)
There cannot be intramolecular hydrogen bonding in acetic acid (CH3COOH) because the
partially positive hydrogen atom cannot reach an oxygen atom for hydrogen bonding.
Intermolecular hydrogen bonding will be important for this compound, though.
(d)
Intramolecular hydrogen bonding will be important in ortho-phthalic acid (see below)
because the partially positive hydrogen atoms are very close in space to partially negative
oxygen atoms.
6.68
3RT
M
(v rms )2 = 3RT
M
2
M (v rms )
T=
3R
v rms =
Therefore, in order to double vrms, it is necessary to quadruple the temperature (in
Kelvin). So, vrms for H2 at 273 K is half of vrms for H2 at 1092 K.
6.78
(a)
PV = nRT
⎛
Pa ⋅ m 3 ⎞
⎟(298 K )
(1.50mol )⎜⎜ 8.3145
mol ⋅ K ⎟⎠
nRT
1000 L
1kPa
⎝
×
×
= 37.2kPa
P=
=
3
(100.0 L )
V
1000 Pa
1m
⎛
n2
⎜⎜ P + a 2
V
⎝
⎛
n2
⎜⎜ P + a 2
V
⎝
⎞
⎟⎟(V − bn ) = nRT
⎠
⎞
nRT
⎟⎟ =
⎠ (V − bn )
nRT
n2
−a
=
P=
(V − bn ) V 2
⎡⎛
⎤
⎞
⎛
Pa ⋅ m 3 ⎞
⎜
⎟
⎟⎟(298 K )
⎢ (1.50mol )⎜⎜ 8.3145
⎥
2
mol ⋅ K ⎠
⎜
L2 ⋅ atm ⎞ (1.50mol ) 101325 Pa ⎞⎥ 1kPa
1000 L ⎟ ⎛ ⎛
⎝
⎢
⎟ ×
⎟×
×
− ⎜⎜ ⎜⎜ 6.71
×
P= ⎜
3 ⎟
2 ⎟
2
⎟⎥ 1000 Pa
⎢ ⎛
atm
1
m
mol
1
⎞
L
(
)
L
100
.
0
⎛
⎞
⎝
⎠
⎠⎥
⎟ ⎝
(
⎢⎜ ⎜⎜100.0 L − ⎜ 0.0564
1.50mol )⎟⎟
⎟
⎜
⎟
mol
⎝
⎠
⎢⎣⎝ ⎝
⎥⎦
⎠
⎠
P = 37.0kPa
(b)
Calculated in the same way as part (a) (but with 50.0 L instead of 100.0 L).
P = 74.3 kPa when the ideal gas law is used and P = 73.8 kPa when the van der Waals
equation is used.
(c)
Calculated in the same way as part (a) (but with 20.0 L instead of 100.0 L).
P = 186 kPa when the ideal gas law is used and P = 183 kPa when the van der Waals
equation is used.
(d)
Calculated in the same way as part (a) (but with 10.0 L instead of 100.0 L).
P = 372 kPa when the ideal gas law is used and P = 359 kPa when the van der Waals
equation is used.
Additional Practice Problems
1.
Calculate the average kinetic energy of a molecule in an ideal monatomic gas at 25oC.
E k = 32 RT =
3
2
(8.3145 molJ⋅K )(298 K ) = 3.72 × 10 3 molJ
E k = 3.72 × 10 3
J
mol
×
1mol
= 6.17 × 10 − 21
23
6.02214 × 10 molecules
J
molecule
The average kinetic energy of a molecule of an ideal monatomic gas at 25 °C is
6.17×10-21 J
2.
List all the intermolecular forces, in decreasing order of strength, which operate in an
aqueous sodium chloride solution. Between which solution components, do each of these
forces operate.
STRONGEST
• ion-dipole forces between ions (Na+ or Cl−) and water;
• hydrogen bonding between water molecules;
• ion-ion forces between Na+ ions, between Cl− ions and between Na+ and Cl− ions
(these forces are weaker than might be anticipated because the ions are solvated by
water molecules so the distance between them is large);*
• ion-induced dipole forces between ions and water (not discussed in class, could be
left out of your answer);*
• dipole-dipole forces between water molecules;*
• dipole-induced dipole forces between water molecules;
• induced dipole-induced dipole forces (London dispersion forces) between water
molecules
WEAKEST
* These three types of IMF are comparable in strength in this situation. It is not
important which order you had them in relative to each other.
3.
For each of the compounds below, indicate whether it would be volatile* (to an
appreciable extent) at room temperature.
a. potassium iodide
b. tartaric acid (structure below)
HO
O
H
HO
C
C
C
O
C
H
OH
OH
c. bromine
*volatile = “evaporates readily”; you should be familiar with this term.
Only bromine (Br2) is volatile under these conditions. The intermolecular forces in
potassium iodide (KI) and tartaric acid are too strong.
4.
The diffusion coefficient of an enzyme, ribonuclease, from bovine pancreas, is
1.31×10−10 m2/s at 20 ˚C. Estimate the radius of this enzyme. The viscosity coefficient
of water is 1.00×10−3 Pa·s at this temperature.
k T
D= B
6πrη
⎛
− 23 J ⎞
⎟(293K )
⎜1.38065 × 10
k BT
K⎠
⎝
r=
=
2
6πηD
⎛
−3
−10 m
6π 1.00 × 10 Pa ⋅ s ⎜⎜1.31 × 10
s
⎝
(
5.
)
⎞
⎟⎟
⎠
= 1.64 × 10 −9 m = 1.64nm
Compare the intermolecular forces present in pure CCl4 and in pure CI4. One of these
compound is a liquid under standard conditions; the other is a solid. Which is which?
Justify your answer.
• Both tetrahalides are tetrahedral, non- polar compounds, so the only intermolecular
forces present are induced dipole-induced dipole forces
• CCl4 is smaller than CI4. Therefore, CI4 has a much higher polarizibility and has
much stronger induced dipole-induced dipole forces. As a consequence, CI4 has a
lower vapour pressure and a higher boiling point than CCl4.
• The difference in strengths of the intermolecular forces results in different states of
matter for these two compounds. Carbon tetrachloride is a liquid and carbon
tetraiodide is a solid.
6.
For each pair of compounds, identify the compound with the higher boiling point. Justify
each answer.
(a)
octane (CH3CH2CH2CH2CH2CH2CH3) or hexane (CH3CH2CH2CH2CH3)
Octane
Both molecules are nonpolar. Since octane is a larger molecule, it has stronger induced
dipole-induced dipole forces. Hence, octane has a higher boiling point.
(b)
helium or argon
Argon
Both atoms are non-polar. Since argon is a larger atom, it has stronger induced dipoleinduced dipole forces. Hence, argon has a higher boiling point.
(c)
SO2 or CO2
SO2
Sulfur dioxide is a polar molecule (bent geometry). On the other hand, carbon dioxide is
a nonpolar molecule (linear geometry). The dipole-dipole forces between SO2 molecules
are stronger than the induced dipole-induced dipole forces between CO2 molecules.
Hence, sulfur dioxide has a higher boiling point.
(d)
water or ethanol (CH3CH2OH)
Water and ethanol can form hydrogen bonds. Water can form two hydrogen bonds per
molecule, while ethanol has a non-polar carbon group attached to the O. The
intermolecular forces in water are stronger and, hence, the boiling point of water is higher
than that of ethanol.
7.
Calculate the root-mean-square speeds of helium and of xenon at 0.00 ˚C.
For helium:
v rms =
3RT
=
M
3(8.3145 molJ ⋅K )(273.15 K ) 1000 g 1 s 2
×
×
g
1kg
1J
4.0026 mol
= 1304.7 ms
3(8.3145 molJ ⋅K )(273.15 K ) 1000 g 1 s 2
×
×
g
1kg
1J
131.29 mol
= 227.81 ms
kg ⋅m 2
(
)
For xenon:
v rms =
8.
(a)
3RT
=
M
kg ⋅m 2
(
)
What properties of a gas would allow it to be described as “ideal”?
For a gas to be ideal two assumptions are made:
• The gas particles do not take up a significant volume, and
• The gas particles do not experience intermolecular forces.
(b)
Which quantities describing a nonideal gas (p, V, n, or T) have a correction term in the
van der Waals equation?
Volume and pressure have correction terms in the van der Waals equation. (# moles of
gas is involved in each correction term) There is no correction for temperature.
(c)
Describe the purpose of each correction factor.
Correction factor a corrects for the fact that nonideal gas particles experience
intermolecular forces of attraction. These IMF reduce the overall pressure since they
attract the gas particles toward the bulk of the sample. Polar gases tend to have larger
correction factor a values.
Correction factor b corrects for the fact that nonideal gas particles have a volume. As
such, the available ‘empty’ volume is less than the volume of the container. Large gases
tend to have larger correction factor b values.
9.
List the intermolecular forces present in each of the following compounds:
In order to answer this question you have to draw the Lewis structures and determine the
molecular geometries.
(a)
NaF
This compound consists of Na+ cations and F- anions. Neither ion has a dipole moment.
Intermolecular forces:
Ion-Ion forces
Ion-induced dipole forces
Induced dipole-induced dipole forces
(b)
H2S
This molecule has a bent molecular geometry and is polar.
Intermolecular forces:
Dipole-dipole forces
Dipole-induced dipole forces
Induced dipole-induced dipole forces
(c)
SF6
This molecule has an octahedral molecular geometry and is non-polar.
Intermolecular forces:
Induced dipole-induced dipole forces
(d)
KClO4
This compound consists of K+ cations and ClO4- anions. Neither ion has a dipole moment
(perchlorate is tetrahedral).
Intermolecular forces:
Ion-Ion forces
Ion-induced dipole forces
Induced dipole-induced dipole forces
(e)
NO2F
This molecule has a trigonal pyramidal molecular geometry and is polar.
Intermolecular forces:
Dipole-dipole interactions
Dipole-induced dipole interactions
Induced dipole-induced dipole forces
(f)
SeF4
This molecule has a seesaw molecular geometry and is polar.
Intermolecular forces:
Dipole-dipole interactions
Dipole-induced dipole interactions
Induced dipole-induced dipole forces
(g)
OF2
This molecule has a bent molecular geometry and is polar.
Intermolecular forces:
Dipole-dipole interactions
Dipole-induced dipole interactions
Induced dipole-induced dipole forces
10.
(a)
(b)
(c)
(d)
11.
(a)
(b)
(c)
Identify whether each of the statements below is correct or incorrect. If incorrect, what’s
wrong with it?
All gas molecules have the same temperature.
Incorrect. A molecule can’t have a temperature. Temperature is a macroscopic property.
All gas molecules travel with the same speed.
Incorrect. Within a sample, the speeds of the gas molecules are distributed according to a
Maxwell-Boltzmann distribution.
The temperatures of the gas molecules have a Maxwell-Boltzmann distribution.
Incorrect. A molecule can’t have a temperature. Temperature is a macroscopic property.
The speeds of the gas molecules have a Maxwell-Boltzmann distribution.
Correct.
For each pair of gases, indicate which has particles with a higher root-mean-square speed
at the same temperature?
Ar and Xe
Ar. It has a smaller molar mass.
CH4 (methane) and C3H8 (propane)
CH4. It has a smaller molar mass.
Ar and O2
O2. It has a smaller molar mass.