CHEM 1023: Study Group Problem Answer Keys

Chapter 12

1. Which forces are INTRAmolecular and which are INTERmolecular?

a. Those preventing oil from evaporating at room temperature

INTERmolecular: oil evaporates when individual oil molecules can escape the

attraction of other oil molecules in the liquid phase.

b. Those allowing silver to tarnish

INTRAmolecular: Oxidation tarnishes pure silver. Oxidation involves the

breaking of bonds and formation of new bonds.

c. Those preventing butter from melting in a refrigerator

INTERmolecular: The process of butter (fat) melting involves a break down in

the rigid, solid structure of fat molecules to an amorphous, less-ordered system.

The attractions between the molecules are weakened but the bonds within the

molecules are not broken.

2. Name the phase change for each of these events.

a. Dew appears on the lawn in the morning

Condensation: the water vapor in the air condenses to liquid when the temp.

drops during the night.

b. Icicles change into liquid water.

Fusion (melting): solid ice melts to liquid water

c. Wet clothes dry on a summer day.

Evaporation; liquid water on clothes evaporates to water vapor

3. Liquid propane, a widely used fuel, is produced by compressing gaseous propane at

20°C. During the process, ~15 kJ of energy is released for each mole of gas liquefied.

From where does this energy come?

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.

4. From the data below, calculate the total heat (in J) needed to convert 12.00 g of ice at

minus (-) 5.00°C to liquid water at 0.500°C. (HINT: Three step process, include phase

change) (Hint: notice the units (grams and moles) and perform your math accordingly)

a. Melting point (mp) at 1 atm: 0.0°C

b. cliquid: 4.21 J/g°C

c. csolid: 2.09 J/g°C

d. Hfus: 6.02 kJ/mol

Step 1: Warming the ice from -5.00°C to 0.500°C.

q1= Cm T = (2.09 J/g°C)(12.00 g)(0.0- (5.00))°C = 125.4 (unrounded)

Step 2: Phase change of ice at 0.00°C to water at 0.00°C

q2= n Hfus = (12.0 g (1 mol/18.02 g))(6.02 kJ/mol)(1000J/1 kJ) = 4008.879 J

(unrounded)

Step 3: Warming the liquid from 0.00°C to 0.500°C

q3= Cm T = (4.21 J/g°C)(12.00 g)(0.500- (0.0))°C = 25.26 J (unrounded)

The three heats are positive because each process takes heat from its surroundings

(endothermic). The phase change requires much more energy than the two

temperature change processes. The total heat is q1 + q2 + q3 = (125.4 J + 4008.879 J +

25.26 J) = 4160 J = 4.16x103 J

5. A liquid has a Hvap of 35.5 kJ/mol and a boiling point of 122°C at 1.00 atm. What is its

vapor pressure at 109°C?

The Clausius-Clapeyron equation gives the relation between vapor pressure and

temperature. Boiling point is defined as the temperature when vapor pressure of the

liquid equals atmospheric pressures, usually assumed to be exactly 1 atm. In the

calculations below, 1.00 atm is used to emphasize the additional significant figures.

ln P2 = -35.5kJ/mol ( 1 - 1________) (1000J) = -0.3678758

1.00 atm 8.314 J/mol*K (273+109)K (273+122)K (1 kJ)

Then take the inverse of ln (by using e)…to learn more about ln and e, go to

http://en.wikipedia.org/wiki/Natural_logarithm

P2 = (0.692203)

1.0 atm

P2 = (0.692203) (1.00 atm) = 0.692 atm

6. What is the strongest interparticle force in each substance?

a. CH3OH

Hydrogen bonding will be the strongest force between methanol molecules since

they contain O-H bonds. Dipole-dipole and dispersions also exist.

b. CCl4

Dispersion forces are the only forces between nonpolar carbon tetrachloride

molecules and thus, are the strongest forces.

c. CH3Br

Dipole-dipole interactions will be the strongest forces between methyl bromide

molecules because the C-Br bond has a dipole moment.

7. Which forces oppose vaporization of each substance?

a. hexane

Dispersion because hexane (C4H14) is a nonpolar molecule.

b. Water

Hydrogen bonding, a single water molecule, can engage in as many as four

hydrogen bonds.

8. Which substance has the higher boiling point?

a. LiCl or HCl

Lithium chloride would have a higher boiling point than HCl because the ions in

lithium chloride are held together by ionic forces, which are stronger than the

dipole-dipole intermolecular forces between HCl molecules in the liquid phase.

b. NH3 or PH3

Ammonia would have a higher boiling point than phosphine (PH3) because the

intermolecular forces in ammonia are stronger than those in phosphine are.

Hydrogen bonding exists between ammonia molecules but weaker dipole-dipole

forces hold phosphine molecules together.

c. Xe or I2

Iodine would have a higher boiling point than xenon. Both are nonpolar with

dispersion forces, but the forces between iodine molecules would be stronger

than between xenon atoms are since the iodine molecules are more polarizable

because of their larger size.

9. What type of crystal lattice does each metal form? (The number of atoms per unit cell is

given in parentheses.)

a. Ni (4)

Ni is face-centered cubic since there are 4 atoms/unit cell.

b. Cr (2)

Cr is body-centered cubic since there are 2 atoms/unit cell.

c. Ca (4)

Ca is face-centered cubic since there are 4 atoms/unit cell.

10. Of the five major types of crystalline solids, which does each of the following form?

a. Sn

You may be familiar with tin, Sn, as a companent of tin cans. Tin is a metal

(malleable, excellent electrical conductor) that forms metallic bond.

b. Si

Silicon is in the same group as carbon, so it exhibits similar bonding properties.

Since diamond and graphite are both network covalent solids, it makes sense

that Si forms the same type of bonds.

c. Xe

Xenon, Xe, is a noble gas and is monatomic. Xe is an atomic solid.

d. F2

Fluorine forms a molecular solid since dispersion forces hold the F2 molecules

together.

e. CH3OH

Methanol forms a molecular solid since the CH3OH molecules are held together

by hydrogen bonds.

11. Zinc selenide (ZnSe) crystallizes in the zinc blende structure and has a density of 5.42

g/cm3.

a. How many Zn and Se ions in each unit cell?

To determine the number of Zn2+

ions and Se2-

ions in each unit, picture the

lattice (Figure 12.32) and count the number of ions at the corners, faces, and

center of unit cell. Looking at selenide ions, there is one ion at 4 Se2-

ions. Three

aer also 4Zn2+

ions due to the 1:1 ratio of Se ions to Zn ions.

b. What is the mass of a unit cell?

Mass of unit cell = (4)(mass of Zn atom) + (4)(mass of Se atom) = (4)(65.39 amu)

+ (4)(78.96 amu) = 577.40 amu