Homework Questions – Chapter 11 – Intermolecular Forces
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Explain why you can skate on ice, but not on solid CO2 (dry ice).
What would you expect to see in a container holding iodine at its triple point?
What is a supercritical fluid? (hint: it’s not the fluid’s mother)
At what T (C) and P (mmHg) is the triple point of water.
(a) What is the practical difference between attractive forces and bonds?
(b) What is the actual difference between attractive forces and bonds?
Name the main types of intermolecular forces that exist in a pure substance (not a mixture) and briefly describe
what is necessary for each one to be present between two molecules.
When predicting the relative boiling points of two compounds (i.e. which is greater), we consider the following;
molar mass, molecule polarity, H-bonding, molecular shape. Rank these in order of their effect on the boiling
point, listing first the one that has the biggest effect.
List three examples of H-bonding in nature.
What type of intermolecular force is stronger than a H-bond? Give an example of where this type of force is
found.
(a) Stronger intermolecular forces result in; (i) higher or lower boiling point? (ii) higher or lower vapor pressure?
(iii) higher or lower surface tension? (iv) higher of lower viscosity?
(b) Explain your answer to part (i).
Explain why capillarity causes water to rise higher in a narrow glass tube than in a wider glass tube.
Why does heating usually decrease the viscosity of a liquid?
For butane, Hvap = 320. kJ/kg, and its boiling point is -0.5 C. Estimate the vapor pressure of butane at 257 K.
Ammonia boils at -33.34 C and has a vapor pressure of 11.57 psi at 235.00 K. What is the heat of vaporization
of ammonia, in J/g?
For compound X, Hvap = 5571 cal/mol, and its vapor pressure is 352 torr at 225 K. What is the boiling point of
compound X?
(a) When the temperature of compound Q decreases from 77.9 C to 28.4 C, its vapor pressure drops to exactly
one eighth of its original value. What is the molar heat of vaporization for compound Q?
(b) Follow up question of the same type as part (a), to be done for practice after working through the solution on
D2L: When the temperature of compound M increases from 20.0 C to 30.0 C, its vapor pressure exactly
doubles. What is the molar heat of vaporization for compound M?
Which of the following compounds would you expect to have the higher melting point, and explain why;
H
18. Which of the following compounds would you expect to have the higher vapor pressure, and explain why;
19. Which of the following compounds would you expect to have the lower viscosity, and explain why;
20. List every type of force and/or bond in a solid sample of each of the following;
(a) HF (b) HBr (c) NaBr (d) H2O (e) NH3 (f) Octane (g) C (diamond) (h) I2
21. State which compound/element you would expect to have the lower boiling point in each of the following pairs,
and explain why; (a) HF and HBr (b) I2 and NaBr (c) H2O and octane (d) NaBr and C (diamond)
22. Give definitions for the following terms;
(a) supercritical fluid (b) phase diagram (c) capillarity (d) temporary dipole (e) physical bond
23. Use wikipedia.org to look up the boiling points of; CH3OH (methanol), CH3CH2CH2CH2CH2CH3 (hexane),
HOCH2CH2OH (ethylene glycol), and CH3CH2CH2CH2CH2CH2CH2CH2OH (1-octanol). Use these values to
figure out approximately how much extra molar mass is required so that the LDF are as strong as one –OH
hydrogen bonding group for this type of organic compound.
Answers
1. Ice is frozen water. Water is one of a very small number of materials that has a backwards sloping solid/liquid
equilibrium line in its phase diagram. This allows solid water (ice) to change phase into liquid water when the
pressure is increased (unlike CO2). Since the blades of skates are sharp, they create very high pressure on the ice,
causing it to melt at the point of contact. The skate (and skater) can then glide along on the thin layer of water,
which immediately refreezes after the skate passes by. Since CO2 does not melt under increased pressure, the
skate would feel a great deal of friction against the solid CO2 surface and skating would be impossible. (note: the
melting of the ice due to pressure is actually only a part of the explanation, rather than the whole reason, but it is
sufficient for this class)
2. Solid, liquid, and gaseous iodine.
3. A fluid above its critical point. It has characteristics of both a liquid and a gas.
4. 0.0098 C, 5 mmHg.
5. (a) Bonds are strong attractions between particles and are not easily broken under normal circumstances (such as
ionic and covalent bonds). Attractive forces are weaker attractions between particles that are more easily broken
under normal circumstances (such as London dispersion forces, or hydrogen bonds).
(b) There is no actual difference between attractive forces and bonds – any distinction is arbitrary.
6. London dispersion forces (LDF) – anything with electron clouds will experience LDF (all atoms/ions/molecules);
Dipole-dipole forces – a polar molecule is required (uneven distribution of electrons); Hydrogen bonds – a
hydrogen atom covalently bonded to an oxygen, nitrogen, or fluorine atom is required for H-bonding.
7. H-bonding, molar mass, molecule polarity, molecular shape.
8. H-bonds are present between water molecules (without it water would be a gas at room temperature and life as we
know it wouldn’t exist); H-bonds are present in DNA molecules (they hold the two strands together and help them
to twist into the famous double helix); H-bonds are present in proteins (they pin together different parts of the
long protein chain, creating a specific shape for the protein that is the key to its biological functionality).
9. An ion-dipole force is stronger than a hydrogen bond. It is found between solvated ions and the water molecules
its solvation shell.
10. (a) (i) higher boiling point (ii) lower vapor pressure (iii) higher surface tension (iv) higher viscosity
(b) To boil a (molecular) liquid, all of the attractive forces between the molecules must be broken. The stronger
these forces are, the more energy is needed to break them. If more energy is needed, the boiling point will be
higher.
11. Water rises up a narrow glass tube because of the attractive forces between the water molecules and the glass
walls of the tube. To maximize these forces, the water tries to coat as much of the glass surface as possible,
traveling up the walls to do so. This movement is limited by gravity, which pulls down the water molecules in the
center of the tube (these molecules are not in contact with the wall), which in turn hold back the water molecules
that are creeping up the glass walls. In a narrower tube, fewer water molecules are in the center, and with fewer
molecules holding them back, the water molecules in contact with the wall can spread further upwards, making
the water rise higher.
12. When molecules are heated, they gain kinetic energy (they move faster). Molecules that are moving faster tend to
spread out to fill more space and are therefore further apart, on average. Since the strength of the attractive forces
depends partly on the separation of the molecules, heating results in the molecules being less strongly attracted to
each other. Viscosity is a measure of the ‘stickiness’ of particles, and this ‘stickiness’ is greater when the forces
of attraction are stronger. Heating a liquid decreases the strength of the attractive forces between molecules and
therefore decreases the viscosity.
13. 0.61 atm (0.6067 atm)
14. 1370 J/g (1369.5 J/g)
15. -33 °C or 240. K (-33.34 °C or 239.81 K)
16. (a) 37.0 kJ/mol (b) 51.21 kJ/mol
17. Both compounds have two hydrogen bonding sites per molecule, so they have the same hydrogen bonding
strength, and both have the same molar mass, so they have the same London dispersion forces. However, each
molecule of (a) has a greater surface area than each molecule of (b), allowing there to be more London dispersion
interactions between adjacent molecules. This results in a stronger intermolecular force of attraction between (a)
molecules. Melting requires the intermolecular forces between molecules to be broken, so more energy would be
needed to break molecules of (a) apart. Therefore (a) would have the higher melting point.
18. Compound (a) has a greater molar mass than (b), so it will have stronger London dispersion forces. However,
compound (a) is a non-polar molecule, whereas compound (b) is a polar molecule. When only dipole-dipole and
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London dispersion forces are present, the LDF usually dominate, so in this case, compound (a) will have stronger
intermolecular forces. Vapor pressure is the result of liquid molecules escaping into the vapor phase, and the
stronger the intermolecular forces, the harder it is for molecules to escape from the liquid. Therefore, stronger
intermolecular forces lead to a lower vapor pressure. Compound (b) has weaker intermolecular forces than (a),
and will therefore have the higher vapor pressure.
Both compounds (a) and (b) have polar bonds that cancel out to leave a non-polar molecule. The only
intermolecular forces present are therefore London dispersion forces. CS2 has a larger molar mass than CO2, so it
will have stronger LDF. Liquid viscosity tells us how easily the molecules ‘roll over’ each other as the liquid
flows (i.e. how ‘runny’ it is). Stronger intermolecular forces prevent the molecules from easily ‘rolling over’ each
other, thus reducing the flow (the liquid is ‘stickier’) – the liquid has a higher viscosity. Since compound (b) has
stronger intermolecular forces, it should also have the higher viscosity.
(a) Covalent bond (between H and F), LDF, dipole-dipole, H-bond (b) Covalent bond (between H and Br), LDF,
dipole-dipole (c) Ionic bond (between Na+ and Br), LDF (d) Covalent bond (between H and O), LDF, dipoledipole, H-bond (e) Covalent bond (between H and N), LDF, dipole-dipole, H-bond (f) Covalent bond (between
H and C, and C and C), LDF (g) Covalent bond (between C and C), LDF (h) Covalent bond (between I and I),
LDF.
When a liquid boils, all the forces of attraction between the molecules (molecular compound), ions (ionic
compound), or atoms (macromolecular compound) must be overcome to completely separate the particles. The
weaker the forces, the lower the temperature needs to be to supply sufficient energy to overcome the forces,
resulting in a lower boiling point. (a) HBr and HF are molecular, and both have LDF and dipole-dipole type
intermolecular forces. However, HF also has hydrogen bonding forces due to the F-atom attached directly to an
H-atom. Since H-bonds are much stronger than LDF or dipole-dipole forces, HF has much stronger
intermolecular forces than HBr. Therefore HBr has the lower boiling point. (b) I2 is molecular and non-polar. I2
molecules are held together by weak London dispersion forces alone. NaBr is an ionic compound, and its ions are
held together by extremely strong ionic bonds (much stronger than LDF). The much weaker forces in I2 cause it
to have the lower boiling point. (c) H2O is molecular, but in addition to LDF and dipole-dipole forces, its
molecules are also held together by strong hydrogen bonds. In contrast, octane is a non-polar molecular
compound. Although octane has a much greater molar mass than water, and hence much stronger LD forces, the
H-bonds between water molecules are even stronger than these LD forces. Octane therefore has weaker total
intermolecular forces, and a lower boiling point. (d) NaBr is an ionic compound, and its ions are held together by
extremely strong ionic bonds. Diamond is a macromolecular compound in which all the carbon atoms are held
together by extremely strong covalent bonds (even stronger than ionic bonds). To boil a macromolecular
material, all of the covalent bonds must be broken (not just the intermolecular forces between molecules, like the
molecular covalent compounds). As the ionic bonds in NaBr are weaker than the covalent bonds in diamond,
NaBr will have the lower boiling point.
Give definitions for the following terms;
(a) Fluid in the region of the phase diagram beyond the critical point. In this region, there is no distinction (phase
transition) between liquid and gas. (b) A diagram showing the state of a substance under varying conditions of
temperature and pressure. (c) Capillarity is when a liquid is drawn up inside a narrow tube (or other narrow
space), against the force of gravity, because the attraction between the liquid molecules and the walls of the tube
is greater than the attraction between the liquid molecules themselves. (d) A dipole is an uneven distribution of
electrons (more accurately, electron density) within an atom, bond, or molecule. A temporary dipole is a dipole
that only exists due to the temporary movement of electrons that causes an uneven distribution. (e) A generic
name for all types of intermolecular forces (typically H-bonds, dipole-dipole forces, and London dispersion
forces).
Approximately 50-70 g/mol extra molar mass produces the same intermolecular forces as one hydrogen bonding
site (-OH).