Silberberg7e solution manual ch 12

Boiling point increases with increasing strength of intermolecular forces; hydrogen bonds are stronger than dipole-dipole forces, which are stronger than dispersion forces.. 12.39 Plan:

CHAPTER 12 INTERMOLECULAR FORCES: LIQUIDS, SOLIDS, AND PHASE CHANGES

FOLLOW–UP PROBLEMS

12.1A Plan: This is a three step process: warming the ice to 0.0°C, melting the ice, and warming the liquid water to

16.0°C Use the molar heat capacities of ice and water and the relationship q = n x C x T to calculate the heat use for the warming of the ice and of the water; use the heat of fusion and the relationship q = nHfus to calculate the heat required to melt the ice

Solution:

The total heat required is the sum of three processes:

1) Warming the ice from –7.00°C to 0.00°C

q1 = n x Cice x T = (2.25 mol)(37.6 J/mol•°C) [0.0 – (–7.00)]°C = 592.2 J = 0.592 kJ 2) Phase change of ice at 0.00°C to water at 0.00°C

q3 = n x Cwater x T = (2.25 mol)(75.4 J/mol•°C) [16.0 – (0.0)]°C = 2714.4 J = 2.71 kJ The three heats are positive because each process takes heat from the surroundings (endothermic) The phase

change requires much more energy than the two temperature change processes The total heat is q1 + q2 + q3 =

(0.592 kJ + 13.5 kJ + 2.71 kJ) = 16.802 = 16.8 kJ

12.1B Plan: This is a three step process: cooling the bromine vapor to 59.5°C, condensing the bromine vapor, and

cooling the liquid bromine to 23.8°C Use the molar heat capacities of bromine vapor and liquid bromine and the

relationship q = n x C x T to calculate the heat released by the cooling of the bromine vapor and of the liquid bromine; use the heat of vaporization and the relationship q = n(–Hvap) to calculate the heat released when bromine vapor condenses

Solution:

Amount (mol) of bromine = (47.94 g Br2) 1 mol Br2

159.8 g Br 2 = 0.3000 mol Br2The total heat released is the sum of three processes:

1) Cooling the bromine vapor from 73.5°C to 59.5°C

q1 = n x Cgas x T = (0.3000 mol)(36.0 J/mol°C) [59.5 – 73.5]°C = –151.2 J = –0.151 kJ 2) Phase change of bromine vapor at 59.5°C to liquid bromine at 59.5°C

q2 = n(–Hvap) = (0.3000 mol) –29.6 kJ

mol = –8.88 kJ 3) Cooling the liquid from 59.5°C to 23.8°C

q3 = n x Cliquid x T = (0.3000 mol)(75.7 J/mol°C) [23.8 – 59.5]°C = –810.7 J = –0.811 kJ The three heats are negative because each process releases heat to the surroundings (exothermic) The phase

change releases much more energy than the two temperature change processes The total heat is q1 + q2 + q3 =

(–0.151 kJ + –8.88 kJ + –0.811 kJ) = –9.842 = –9.84 kJ

12.2A Plan: The variables Hvap, P1, T1, and T2 are given, so substitute them into the Clausius-Clapeyron equation and

solve for P2 Convert both temperatures from °C to K Convert Hvap to J so that the units cancel with R

Solution:

T1 = 34.1 + 273.15 = 307.2 K T2 = 85.5 + 273.15 = 358.6 K P1 = 40.1 torr

2 1

ln40.1 torr

In the subtraction, the number of significant figures decreased from 4 to 3

In a log term (like the 2.28 above, since ln (P2/P1) = 2.28), only the numbers after the decimal point are

significant Thus, in 2.28, there are 2 significant figures that get carried over through the next calculation (9.8, the non-log term, has 2 significant figures)

The temperature increased, so the vapor pressure should be higher

12.2B Plan: The variables Hvap, P1, P2, and T1 are given, so substitute them into the Clausius-Clapeyron equation and

solve for T2 Convert T1 from °C to K Convert Hvap to J so that the units cancel with R After solving for T2, convert the temperature from K to oC

Solution:

T1 = 20.2 + 273.15 = 293.4 K P1 = 24.5 kPa P2 = 10.0 kPa

2 1

103 J

1 kJ

2.40403 x 10–4 K = 1

T2 – 293.4 K1 3.64872 x 10–3 K = 1

T2

T2 = 274.1 K – 273.15 = 0.95°C

The pressure decreased, so the temperature should be lower, as it is

12.3A Plan: Refer to the phase diagram to describe the specified changes In this problem, carbon is heated at a constant

12.4A Plan: Hydrogen bonding occurs in compounds in which H is directly bonded to O, N, or F In hydrogen bonding,

there is a –A:… H–B– sequence in which A and B are O, N, or F

Solution:

a) The –A:… H–B– sequence is present in both of the following structures:

O

CO

O

H

HH

HH

H

HH

CO

O

CO

H

b) The –A:… H–B– sequence can only be achieved in one arrangement

CO

CO

H

H

H

The hydrogens attached to the carbons cannot form H bonds

c) Hydrogen bonding is not possible because there are no O–H, N–H, or F–H bonds

12.4B Plan: Hydrogen bonding occurs in compounds in which H is directly bonded to O, N, or F In hydrogen bonding,

there is a –A:… H–B– sequence in which A and B are O, N, or F

Solution:

a) Hydrogen bonding is not possible because there are no O–H, N–H, or F–H bonds

b) The –A:… H–B– sequence is present in the following structure (one of the possible hydrogen bonding

structures):

HH

NH

H

c) The –A:… H–B– sequence is present in the following structure (one of the possible hydrogen bonding

structures):

The hydrogens attached to the carbons cannot form H bonds

12.5A Plan: Hydrogen bonding occurs in compounds in which hydrogen is directly bonded to O, N, or F Dipole-dipole

forces occur in compounds that are polar Dispersion forces are the dominant forces in nonpolar compounds Boiling point increases with increasing strength of intermolecular forces; hydrogen bonds are stronger than

dipole-dipole forces, which are stronger than dispersion forces Forces generally increase in strength as molar mass increases

Solution:

a) Both CH3Br and CH3F are polar molecules that experience dipole-dipole and dispersion intermolecular

interactions Because CH3Br (M = 94.9 g/mol) is ~3 times larger than CH3F (M = 34.0 g/mol), dispersion forces

result in a higher boiling point for CH 3 Br

b) CH3CH2CH2OH, n–propanol, forms hydrogen bonds and has dispersion forces whereas CH3CH2OCH3, ethyl

methyl ether, has only dipole-dipole and dispersion attractions CH 3 CH 2 CH 2 OH has the higher boiling point

c) Both C2H6 and C3H8 are nonpolar and experience dispersion forces only C 3 H 8, with the higher molar mass, experiences greater dispersion forces and has the higher boiling point

12.5B Plan: Hydrogen bonding occurs in compounds in which hydrogen is directly bonded to O, N, or F Dipole-dipole

forces occur in compounds that are polar Dispersion forces are the dominant forces in nonpolar compounds Boiling point increases with increasing strength of intermolecular forces; hydrogen bonds are stronger than dipole-dipole forces, which are stronger than dispersion forces Forces generally increase in strength as molar mass increases

Solution:

a) Both CH3CHO and CH3CH2OH are polar molecules that experience dipole-dipole and dispersion

intermolecular interactions Because of its O–H bond, CH3CH2OH also experiences hydrogen bonding Because

CH 3 CHO does not experience the stronger hydrogen bonds, it has a lower boiling point than CH3CH2OH b) Both CHCl3 and CHI3 are polar molecules that experience dipole-dipole and dispersion intermolecular

interactions Because CHCl3 (M = 119.37 g/mol) is about a third the size of CHI3 (M = 393.7 g/mol), it will have

weaker dispersion forces, resulting in a lower boiling point for CHCl 3

c) Both H2NCOCH2CH3 and (CH3)2NCHO are polar molecules that experience dipole-dipole and dispersion intermolecular interactions Because of its N–H bonds, H2NCOCH2CH3 also experiences hydrogen bonding

Because (CH 3 ) 2 NCHO does not experience the stronger hydrogen bonds, it has a lower boiling point than

H2NCOCH2CH3

12.6A Plan: To determine the number of atoms or ions in each unit cell, count the number of atoms/ions at the corners,

faces, and center of the unit cell Atoms at the eight corners are shared by eight cells for a total of 8 atoms x 1/8 atom per cell = 1 atom; atoms in the body of a cell are in that cell only; atoms at the faces are shared by two cells: 6 atoms x 1/2 atom per cell = 3 atoms Count the number of nearest neighboring particles to obtain the coordination number

Solution:

a) Looking at the sulfide ions, there is one ion at each corner and one ion on each face The total number of sulfide

ions is 1/8 (8 corner ions) + 1/2 (6 face ions) = 4 S 2–

ions There are also 4 Pb 2+ ions due to the 1:1 ratio of S 2–

ions to Pb2+ ions Each ion has 6 nearest neighbor particles, so the coordination number for both ions is 6

b) There is one atom at each corner and one atom in the center of the unit cell The total number of atoms is 1/8 (8

corner atoms) + 1 (1 body atom) = 2 W atoms Each atom has 8 nearest neighbor particles, so the coordination number for W is 8

c) There is one atom at each corner and one atom on each face of the unit cell The total number of atoms is 1/8 (8

corner atoms) + 1/2 (6 face atoms) = 4 Al atoms Each atom has 12 nearest neighbor particles, so the coordination number for Al is 12

12.6B Plan: To determine the number of atoms or ions in each unit cell, count the number of atoms/ions at the corners,

faces, and center of the unit cell Atoms at the eight corners are shared by eight cells for a total of 8 atoms x 1/8 atom per cell = 1 atom; atoms in the body of a cell are in that cell only; atoms at the faces are shared by two cells: 6 atoms x 1/2 atom per cell = 3 atoms Count the number of nearest neighboring particles to obtain the coordination number

Solution:

a) There is one atom at each corner and one atom in the center of the unit cell The total number of atoms is 1/8 (8

corner atoms) + 1 (1 body atom) = 2 K atoms Each atom has 8 nearest neighbor particles, so the coordination number for K is 8

b) There is one atom at each corner and one atom on each face of the unit cell The total number of atoms is 1/8 (8

corner atoms) + 1/2 (6 face atoms) = 4 Pt atoms Each atom has 12 nearest neighbor particles, so the coordination number for Pt is 12

c) Looking at the chloride ion, there is one ion in the center of the unit cell, so there is 1 Cl – ion in the unit cell There is one Cs+ ion at each corner, so there is 1/8 (8 corner ions) = 1 Cs +

ion Each ion has 8 nearest neighbor

particles, so the coordination number for both ions is 8

12.7A Plan: Use the radius of the Co atom to find the volume of one Co atom Use Avogadro’s number to find the

volume of one mole of Co atoms, and convert from pm3 to cm3 Use the volume of one mole of Co atoms and the packing efficiency of the solid (0.74 for face centered cubic) to find the volume of one mole of Co metal

Calculate the density of Co metal by dividing the molar mass of Co by the volume of one mole of Co metal Solution:

1 mol Co

1 mol Co 6.66 cm 3 = 8.85 g/cm 3

12.7B Plan: Use the volume of the Fe atom to find the radius of the atom; use the radius to find the edge length of the

body-centered cubic cell Find the volume of the cell by cubing the value of the edge length Use the fact that there are two Fe atoms in the cell The molar mass and density of iron are then used to find the number of Fe atoms in a mole (Avogadro’s number)

Solution:

34

7.874 g mol2.46407x10 cm

r = 0.405 nm/√8 = (0.143 nm) 1000 pm1 nm = 143 pm

TOOLS OF THE LABORATORY BOXED READING PROBLEMS

B12.1 Plan: The Bragg equation gives the relationship between the angle of incoming light, θ, the

wavelength of the light, λ, and the distance between layers in a crystal, d

B12.2 Plan: The Bragg equation gives the relationship between the angle of incoming light, θ, the

wavelength of the light, λ, and the distance between layers in a crystal, d

Solution:

a) d, the distance between the layers in the NaCl crystal, must be found The edge length of the NaCl unit cell is

equal to the sum of the diameters of the two ions:

Edge length (pm) = 204 pm (Na+) + 362 pm (Cl–) = 566 pm The spacing between the layers is

half the distance of the edge length: 566/2 = 283 pm

θ = 33.20958 = 33.2°

END–OF–CHAPTER PROBLEMS

12.1 The energy of attraction is a potential energy and denoted Ep The energy of motion is kinetic energy and denoted

Ek The relative strength of Ep vs Ek determines the phase of the substance In the gas phase, Ep << Ek because the gas particles experience little attraction for one another and the particles are moving very fast In the solid phase,

Ep >> Ek because the particles are very close together and are only vibrating in place

Two properties that differ between a gas and a solid are the volume and density The volume of a gas expands to fill the container it is in while the volume of a solid is constant no matter what container holds the solid Density

of a gas is much less than the density of a solid The density of a gas also varies significantly with temperature and pressure changes The density of a solid is only slightly altered by changes in temperature and pressure

Compressibility and ability to flow are other properties that differ between gases and solids

12.2 a) Gases are more easily compressed than liquids because the distance between particles is much greater in a gas

than in a liquid Liquids have very little free space between particles and thus can be compressed (crowded together) only very slightly

b) Liquids have a greater ability to flow because the interparticle forces are weaker in the liquid phase than in the solid phase The stronger interparticle forces in the solid phase fix the particles in place Liquid particles have enough kinetic energy to move around

12.3 a) intermolecular b) intermolecular c) intermolecular d) intramolecular

12.4 a) Heat of fusion refers to the change between the solid and the liquid states and heat of vaporization refers to the

change between liquid and gas states In the change from solid to liquid, the kinetic energy of the molecules must

increase only enough to partially offset the intermolecular attractions between molecules In the change from liquid to gas, the kinetic energy of the molecules must increase enough to overcome the intermolecular forces The energy to overcome the intermolecular forces for the molecules to move freely in the gaseous state is much greater than the amount of energy needed to allow the molecules to move more easily past each other but still stay very close together

b) The net force holding molecules together in the solid state is greater than that in the liquid state Thus, to change solid molecules to gaseous molecules in sublimation requires more energy than to change liquid molecules

to gaseous molecules in vaporization

c) At a given temperature and pressure, the magnitude of vap is the same as the magnitude of cond The only difference is in the sign: vap = –cond.

12.5 Plan: Intermolecular forces (nonbonding forces) are the forces that exist between molecules that attract the

molecules to each other; these forces influence the physical properties of substances Intramolecular forces (bonding forces) exist within a molecule and are the forces holding the atoms together in the molecule; these forces influence the chemical properties of substances

Solution:

a) Intermolecular — Oil evaporates when individual oil molecules can escape the attraction of other oil

molecules in the liquid phase

b) Intermolecular — The process of butter (fat) melting involves a breakdown in the rigid, solid structure of fat

molecules to an amorphous, less ordered system The attractions between the fat molecules are weakened, but the bonds within the fat molecules are not broken

c) Intramolecular — A process called oxidation tarnishes pure silver Oxidation is a chemical change and

involves the breaking of bonds and formation of new bonds

d) Intramolecular — The decomposition of O2 molecules into O atoms requires the breaking of chemical bonds,

i.e., the force that holds the two O atoms together in an O2 molecule

Both a) and b) are physical changes, whereas c) and d) are chemical changes In other words, intermolecular forces are involved in physical changes while intramolecular forces are involved in chemical changes

12.6 a) intermolecular b) intramolecular c) intermolecular d) intermolecular

12.7 a) Condensation The water vapor in the air condenses to liquid when the temperature drops during the

night

b) Fusion (melting) Solid ice melts to liquid water

12.8 a) deposition b) sublimation c) crystallization (freezing)

12.9 The propane gas molecules slow down as the gas is compressed Therefore, much of the kinetic energy lost by

the propane molecules is released to the surroundings upon liquefaction

12.10 Sublimation and deposition

12.11 The gaseous PCl3 molecules are moving faster and are farther apart than the liquid molecules As they

condense, the kinetic energy of the molecules is changed into potential energy stored in the dipole-dipole

interactions between the molecules

12.12 The two processes are the formation of solid from liquid and the formation of liquid from solid (at the macroscopic

level) At the molecular level, the two processes are the removal of kinetic energy from the liquid molecules as they solidify and the overcoming of the dispersion forces between the molecules as they turn to liquid

12.13 In closed containers, two processes, evaporation and condensation, occur simultaneously Initially there are few

molecules in the vapor phase, so more liquid molecules evaporate than gas molecules condense Thus, the

number of molecules in the gas phase increases, causing the vapor pressure of hexane to increase Eventually, the number of molecules in the gas phase reaches a maximum where the number of liquid molecules evaporating

equals the number of gas molecules condensing In other words, the evaporation rate equals the condensation rate

At this point, there is no further change in the vapor pressure

12.14 a) At the critical temperature, the molecules are moving so fast that they can no longer be condensed This

temperature decreases with weaker intermolecular forces because the forces are not strong enough to overcome

molecular motion Alternatively, as intermolecular forces increase, the critical temperature increases

b) As intermolecular forces increase, the boiling point increases because it becomes more difficult and takes

more energy to separate molecules from the liquid phase

c) As intermolecular forces increase, the vapor pressure decreases for the same reason given in b) At any

given temperature, strong intermolecular forces prevent molecules from easily going into the vapor phase and thus vapor pressure is decreased

d) As intermolecular forces increase, the heat of vaporization increases because more energy is needed to

separate molecules from the liquid phase

12.15 Point 1 is depicted by C This is the equilibrium between melting and freezing

Point 2 is depicted by A This is the equilibrium between vaporization and condensation

Point 3 is depicted by D This is the equilibrium between sublimation and deposition

12.16 a) The final pressure will be the same, since the vapor pressure is constant as long as some liquid is present

b) The final pressure will be lower, according to Boyle’s law

12.17 If the solid is more dense than the liquid, the solid-liquid line slopes to the right; if less dense, to the left

12.18 When water at 100°C touches skin, the heat released is from the lowering of the temperature of the water The

specific heat of water is approximately 75 J/mol•K When steam at 100°C touches skin, the heat released is from

the condensation of the gas with a heat of condensation of approximately 41 kJ/mol Thus, the amount of heat

released from gaseous water condensing will be greater than the heat from hot liquid water cooling and the burn from the steam will be worse than that from hot water

12.19 Plan: The total heat required is the sum of three processes: warming the ice to 0.00°C, the melting point; melting

the ice to liquid water; warming the water to 0.500°C The equation q = c x mass x T is used to calculate the

heat involved in changing the temperature of the ice and of the water; the heat of fusion is used to calculate the heat involved in the phase change of ice to water

12.20 0.333 mol x 46.07 g/mol = 15.34131 g ethanol

Cooling vapor to boiling point:

q1 = c x mass x T = (1.43 J/g°C)(15.34131 g)(78.5 – 300)°C = –4859.28 J

Condensing vapor: (note Hcond = – Hvap)

q2 = nHcond   = (0.333 mol)(–40.5 kJ/mol)(103 J/kJ) = –13,486.5 J Cooling liquid to 25.0°C:

q3 = c x mass x T = (2.45 J/g°C)(15.34131 g)(25.0 – 78.5)°C = –2010.86 J

qtotal = q1 + q2 + q3 = (–4859.28 J) + (–13,486.5 J) + (–2010.86 J) = –20356.64 = – 2.04x10 4

J

12.21 Plan: The Clausius-Clapeyron equation gives the relationship between vapor pressure and temperature We are

given Hvap , P1, T1, and T2; these values are substituted into the equation to find the P2, the vapor pressure Solution:

P1 = 1.00 atm T1 = 122°C + 273 = 395 K

P2 = ? T2 = 113°C + 273 = 386 K Hvap = 35.5 kJ/mol

vap 2

12.23 Plan: The Clausius-Clapeyron equation gives the relationship between vapor pressure and temperature We are

given P1, P2, T1, and T2; these values are substituted into the equation to find Hvap The pressure in torr must be converted to atm

vap

H = 21,343.248 = 2x10 4

J/mol

(The significant figures in the answer are limited by the 1 atm in the problem.)

12.24 The Clausius-Clapeyron equation gives the relationship between vapor pressure and temperature

vap 2

distinction between the gas, liquid, and solid states, and the melting point T, which is located directly above the

critical T Solid ethylene is denser than liquid ethylene since the solid-liquid line slopes to the right with

12.27 Plan: Refer to the phase diagram to describe the specified changes In part a), a substance sublimes if it converts

directly from a solid to a gas without passing through a liquid phase In part b), refer to the diagram to explain the changes that occur when sulfur is heated at a constant pressure of 1 atm

Solution:

a) Rhombic sulfur will sublime when it is heated at a pressure less than 1 x 10-4 atm

b) At 90 oC and 1 atm, sulfur is in the solid (rhombic form) As it is heated at constant pressure, it passes through

the solid (monoclinic) phase, starting at 114 oC At about 120 oC, the solid melts to form a liquid At about

445 oC, the liquid evaporates and changes to the gas state

12.28 Plan: Refer to the phase diagram to describe the specified changes In part a), identify the phase at the given

conditions In part b), refer to the diagram to explain the changes that occur when xenon is compressed at a constant temperature of -115 oC

Solution:

a) At room temperature and pressure, xenon is a gas

b) At -115 oC and 0.5 atm, Xe is a gas As it is compressed at constant temperature, it passes through the liquid

phase (starting at about 0.6 atm) At about 0.75 atm, Xe becomes a solid (undergoes fusion)

12.29 This is a stepwise problem to calculate the total heat required

Melting SO2:

q1 = nHfus   = (2.500 kg)(103 g/1 kg)(1 mol SO2/64.06 g SO2)(8.619 kJ/mol)(103 J/kJ)

= 336,364.3 J Warming liquid SO2:

q2 = c x mass x T = (0.995 J/g°C)(2.500 kg)(103 g/1 kg)[–10 – (–73)]°C = 156,712.5 J Vaporizing SO2:

q3 = nHvap   = (2.500 kg)(103 g/1 kg)(1 mol SO2/64.06 g SO2)(25.73 kJ/mol)(103 J/kJ)

= 1,004,136.7 J Warming gaseous SO2:

q4 = c x mass x T = (0.622 J/g°C)(2.500 kg)(103 g/1 kg)[60 – (–10.)]°C = 108,850 J

qtotal = q1 + q2 + q3 + q4 = (336,364.3 J) + (156,712.5 J) + (1,004,136.7 J) + (108,850 J)

= 1,606,063.5 = 1.606x10 6

J

12.30 Plan: The Clausius-Clapeyron equation gives the relationship between vapor pressure and temperature We are

given Hvap , P1, T1, and T2; these values are substituted into the equation to find P2 Convert the temperatures from °C to K and Hvap from kJ/mol to J/mol to allow cancellation with the units in R

Solution:

P1 = 2.3 atm T1 = 25.0°C + 273 = 298 K

P2 = ? T2 = 135°C + 273 = 408 K Hvap = 24.3 kJ/mol

vap 2

12.31 a) At 20ºC and 40ºC, no liquid exists, only gas At –40ºC, liquid exists At –120ºC, no liquid exists, only solid b) No, at any pressure below the triple point pressure, the CO2(s) will sublime

b) dipole-dipole forces < hydrogen bonds < ion-dipole

12.34 To form hydrogen bonds, the atom bonded to hydrogen must have two characteristics: small size and high

electronegativity (so that the atom has a very high electron density) With this high electron density, the attraction for a hydrogen on another molecule is very strong Selenium is much larger than oxygen (atomic radius of 119 pm

vs 73 pm) and less electronegative than oxygen (2.4 for Se and 3.5 for O) resulting in an electron density on Se in

H2Se that is too small to form hydrogen bonds

12.35 The I–I distance within an I2 molecule is shorter than the I–I distance between adjacent molecules This is because

the I–I interaction within an I2 molecule is a true covalent bond and the I–I interaction between molecules is a dispersion force (intermolecular) which is considerably weaker

12.36 All particles (atoms and molecules) exhibit dispersion forces, but these are the weakest of intermolecular forces

The dipole-dipole forces in polar molecules dominate the dispersion forces

12.37 Polarity refers to a permanent imbalance in the distribution of electrons in the molecule Polarizability refers to the

ability of the electron distribution in a molecule to change temporarily The polarity affects dipole-dipole interactions,while the polarizability affects dispersion forces

12.38 If the electron distribution in one molecule is not symmetrical (permanent or temporary), that can induce a

temporary dipole in an adjacent molecule by causing the electrons in that molecule to shift for some (often short) time 12.39 Plan: Dispersion forces are the only forces between nonpolar substances; dipole-dipole forces exist between polar

substances Hydrogen bonds only occur in substances in which hydrogen is directly bonded to either oxygen, nitrogen, or fluorine

Solution:

a) Hydrogen bonding will be the strongest force between methanol molecules since they contain O–H bonds

Dipole-dipole and dispersion forces also exist

b) Dispersion forces are the only forces between nonpolar carbon tetrachloride molecules and, thus, are the

c) Dispersion forces are the only forces between nonpolar chlorine molecules and, thus, are the strongest forces 12.40 a) Hydrogen bonding b) Dipole-dipole c) Ionic bonds

12.41 Plan: Dispersion forces are the only forces between nonpolar substances; dipole-dipole forces exist between polar

substances Hydrogen bonds only occur in substances in which hydrogen is directly bonded to either oxygen, nitrogen, or fluorine

Solution:

a) Dipole-dipole interactions will be the strongest forces between methyl chloride molecules because the C–Cl

bond has a dipole moment

b) Dispersion forces dominate because CH3CH3 (ethane) is a symmetrical nonpolar molecule

c) Hydrogen bonding dominates because hydrogen is bonded to nitrogen, which is one of the three atoms (N, O,

or F) that participate in hydrogen bonding

12.42 a) Dispersion b) Dipole-dipole c) Hydrogen bonding

12.43 Plan: Hydrogen bonds are formed when a hydrogen atom is bonded to N, O, or F

Solution:

a) The presence of an OH group leads to the formation of hydrogen bonds in CH 3 CH(OH)CH 3

There are no hydrogen bonds in CH3SCH3

CO

CO

FHF

12.44 a) The presence of H directly attached to the N in (CH 3 ) 2 NH leads to hydrogen bonding More than one

arrangement is possible

N

HC

CH

HH

HH

H

N

CH

HH

HHH

b) Each of the hydrogen atoms directly attached to oxygen atoms in HOCH 2 CH 2 OH leads to hydrogen bonding More than one arrangement is possible In FCH2CH2F, the H atoms are bonded to C so there is no hydrogen bonding

12.45 Plan: In the vaporization process, intermolecular forces between particles in the liquid phase must be broken as

the particles enter the vapor phase In other words, the question is asking for the strongest interparticle force that must be broken to vaporize the liquid Dispersion forces are the only forces between nonpolar substances; dipole-dipole forces exist between polar substances Hydrogen bonds only occur in substances in which hydrogen

is directly bonded to either oxygen, nitrogen, or fluorine

Solution:

a) Dispersion, because hexane, C6H14, is a nonpolar molecule

b) Hydrogen bonding; hydrogen is bonded to oxygen in water A single water molecule can engage in as many

as four hydrogen bonds

c) Dispersion, although the individual Si–Cl bonds are polar, the molecule has a symmetrical, tetrahedral shape

and is therefore nonpolar

12.46 a) Dispersion b) Dipole-dipole c) Hydrogen bonding

12.47 Plan: Polarizability increases down a group and decreases from left to right because as atomic size

increases, polarizability increases

Solution:

a) Iodide ion has greater polarizability than the bromide ion because the iodide ion is larger The electrons can be

polarized over a larger volume in a larger atom or ion

b) Ethene (CH 2 =CH 2 ) has greater polarizability than ethane (CH3CH3) because the electrons involved in  bonds

are more easily polarized than electrons involved in  bonds

c) H 2 Se has greater polarizability than water because the selenium atom is larger than the oxygen atom

12.48 a) Ca b) CH 3 CH 2 CH 3 c) CCl 4

In all cases, the larger molecule (i.e., the one with more electrons) has the higher polarizability

12.49 Plan: Weaker attractive forces result in a higher vapor pressure because the molecules have a smaller energy

barrier in order to escape the liquid and go into the gas phase Decide which of the two substances in each pair has the weaker interparticle force Dispersion forces are weaker than dipole-dipole forces, which are weaker than hydrogen bonds

Solution:

a) C 2 H 6 C2H6 is a smaller molecule exhibiting weaker dispersion forces than C4H10

b) CH 3 CH 2 F CH3CH2F has no H–F bonds (F is bonded to C, not to H), so it only exhibits dipole-dipole forces,

which are weaker than the hydrogen bonding in CH3CH2OH

c) PH 3 PH3 has weaker intermolecular forces (dipole-dipole) than NH3 (hydrogen bonding)

12.50 a) HOCH 2 CH 2 OH has a stronger intermolecular force, because there are more OH groups to hydrogen bond

b) CH 3 COOH has a stronger intermolecular force, because hydrogen bonding is stronger than dipole-dipole forces

c) HF has a stronger intermolecular force, because hydrogen bonding is stronger than dipole-dipole forces 12.51 Plan: The weaker the interparticle forces, the lower the boiling point Decide which of the two substances in each

pair has the weaker interparticle force Dispersion forces are weaker than dipole-dipole forces, which are weaker than hydrogen bonds, which are weaker than ionic forces

Solution:

a) HCl would have a lower boiling point than LiCl because the dipole-dipole intermolecular forces between

hydrogen chloride molecules in the liquid phase are weaker than the significantly stronger ionic forces holding the ions in lithium chloride together

b) PH 3 would have a lower boiling point than NH3 because the intermolecular forces in PH3 are weaker than those

in NH3 Hydrogen bonding exists between NH3 molecules but weaker dipole-dipole forces hold PH3 molecules together

c) Xe would have a lower boiling point than iodine Both are nonpolar with dispersion forces, but the forces

between xenon atoms would be weaker than those between iodine molecules since the iodine molecules are more polarizable because of their larger size

12.52 a) CH 3 CH 2 OH, hydrogen bonding (CH3CH2OH) vs dispersion (CH3CH2CH3)

b) NO, dipole-dipole (NO) vs dispersion (N2)

c) H 2 Te, the larger molecule has larger dispersion forces

12.53 Plan: The weaker the intermolecular forces, the lower the boiling point Decide which of the two substances

in each pair has the weaker intermolecular force Dispersion forces are weaker than dipole-dipole forces, which are weaker than hydrogen bonds, which are weaker than ionic forces

c) HBr, the dipole-dipole forces of hydrogen bromide are weaker than the hydrogen bonding forces of water

12.54 a) CH 3 OH, hydrogen bonding (CH3OH) vs dispersion forces (CH3CH3)

b) FNO, greater polarity in FNO vs ClNO

c)

FCH

CF

H

This molecule has dipole-dipole forces since the two C–F bonds do not cancel and the molecule is polar The other molecule has only dispersion forces since the two C–F bonds do cancel, so that the molecule is nonpolar 12.55 The trend in both atomic size and electronegativity predicts that the trend in increasing strength of

hydrogen bonds is N–H < O–H < F–H As the atomic size decreases and electronegativity increases, the electron density of the atom increases High electron density strengthens the attraction to a hydrogen atom on another

molecule Fluorine is the smallest of the three and the most electronegative, so its hydrogen bonds would be the strongest Oxygen is smaller and more electronegative than nitrogen, so hydrogen bonds for water would be

stronger than hydrogen bonds for ammonia

12.56 The molecules of motor oil are long chains of CH2 units The high molar mass results in stronger dispersions forces

and leads to a high boiling point In addition, these chains can become tangled in one another and restrict each other’s motions and ease of vaporization

12.57 The ethylene glycol molecules have two sites (two OH groups) which can hydrogen bond; the propanol has only

one OH group

12.58 The molecules at the surface are attracted to one another and to those molecules in the bulk of the liquid Since this

force is directed downwards and sideways, it tends to “tighten the skin.”

12.59 The shape of the drop depends upon the competing cohesive forces (attraction of molecules within the drop

itself) and adhesive forces (attraction between molecules in the drop and the molecules of the waxed floor) If the cohesive forces are strong and outweigh the adhesive forces, the drop will be as spherical as gravity will allow If,

on the other hand, the adhesive forces are significant, the drop will spread out Both water (hydrogen bonding) and mercury (metallic bonds) have strong cohesive forces, whereas cohesive forces in oil (dispersion) are

relatively weak Neither water nor mercury will have significant adhesive forces to the nonpolar wax molecules,

so these drops will remain nearly spherical The adhesive forces between the oil and wax can compete with the weak, cohesive forces of the oil (dispersion) and so the oil drop spreads out

12.60 The presence of the ethanol molecules breaks up some of the hydrogen bonding interactions present between the water

molecules, lowering the surface tension

12.61 Surface tension is defined as the energy needed to increase the surface area by a given amount, so units of energy

(J) per surface area (m2) describe this property

12.62 The strength of the intermolecular forces does not change when the liquid is heated, but the molecules have

greater kinetic energy and can overcome these forces more easily as they are heated The molecules have more energy at higher temperatures, so they can break the intermolecular forces and can move more easily past their neighbors; thus, viscosity decreases

12.63 Plan: The stronger the intermolecular force, the greater the surface tension Decide which of the substances has

the weakest intermolecular force and which has the strongest Dispersion forces are weaker than dipole-dipole forces, which are weaker than hydrogen bonds, which are weaker than ionic forces

Solution:

All three molecules exhibit hydrogen bonding (H is bonded to O), but the extent of hydrogen bonding increases with the number of O–H bonds present in each molecule HOCH2CH(OH)CH2OH with three O–H groups can form more hydrogen bonds than HOCH2CH2OH with two O–H groups, which in turn can form morehydrogen bonds than CH3CH2CH2OHwith only one O–H group The greater the number of hydrogen bonds, the stronger the intermolecular forces, and the higher the surface tension

CH 3 CH 2 CH 2 OH < HOCH 2 CH 2 OH < HOCH 2 CH(OH)CH 2 OH

12.64 CH 3 OH > H 2 CO > CH 3 CH 3

The intermolecular forces would decrease as shown (hydrogen bonding > dipole-dipole > dispersion), as would the

12.65 Plan: Viscosity is a measure of the resistance of a liquid to flow, and is greater for molecules with stronger

intermolecular forces The stronger the force attracting the molecules to each other, the harder it is for one

molecule to move past another Thus, the substance will not flow easily if the intermolecular force is strong Decide which of the substances has the weakest intermolecular force and which has the strongest Dispersion forces are weaker than dipole-dipole forces, which are weaker than hydrogen bonds, which are weaker than ionic forces

Solution:

The ranking of decreasing viscosity is the opposite of that for increasing surface tension (Problem 12.61)

HOCH 2 CH(OH)CH 2 OH > HOCH 2 CH 2 OH > CH 3 CH 2 CH 2 OH

The greater the number of hydrogen bonds, the stronger the intermolecular forces, and the higher the viscosity 12.66 Viscosity and surface tension both increase with increasing strength of intermolecular forces

16.04 g CH 1 mol CH

Mercury takes more energy

b) Calculate the energies involved using the heats of vaporization

16.04 g CH 1 mol CH

Methane takes more energy

c) Mercury involves metallic bonding and methane involves dispersion forces

12.69 The pentanol has stronger intermolecular forces (hydrogen bonds) than the hexane (dispersion forces)

12.70 Water is a good solvent for polar and ionic substances and a poor solvent for nonpolar substances Water is a polar

molecule and dissolves polar substances because their intermolecular forces are of similar strength Water is also