Chapter 10:  Liquids, Solids, and Phase Changes

Chapter 13 of the supplementary notes has a great deal of the material we look at here in an alternate form with different examples.  Taking a peek at that would probably be useful!
 

Assigned Problems:

Problem 10.22 (p. 425):

Zinc sulfide, or sphalerite, crystallizes in the following cubic unit cell:

(a)  What kind of packing do the sulfide ions adopt?
This is an example of cubic close packing.  This is a face-centered cubic cell and, if you refer to p. 411 (Table 10.10), you will see that this has cubic closest-packing!
(b)  How many S2- and how many Zn2+ ions are in the unit cell?
Zinc is easy:  Here, there are exactly 4 ions completely inside the fcc unit cell.

Sulfide is a little more challenging.  Here, there a eight sulfides at corners so that we have 8 x (1/8) donated from these.  There are sulfides on each of the six faces and these are shared by two cells each so that, from these, the donation is 6 x (1/2).  So, there are four sulfides since 4 = 1 + 3.  Note that we have equal numbers of zincs and sulfides--and this must be true in order to maintain electrical neutrality!


Problem 10.24 (p. 425):

The phase diagram of a substance is shown below:

(a)  Approximately what is the normal boiling point and what is the normal melting point of the substance?
If you can read graphs, this should be easy.  The word "normal" here refers to 1 atm pressure.  For the mp, I'd read approximately 180K and for the nbp, about 300K.
(b)  What is the physical state of the substance under the following conditions?
 
(i) T = 150K P = 0.5 atm
(ii) T = 325K P = 0.9 atm
(iii) T = 450K P = 165 atm
Again, just read the diagram.  The answers should be obvious and are as follows:
(i):     Solid
(ii):    Gas (vapor)
(iii):   Gas (as a supercritical fluid since you are above the critical pt. here)
Problem 10.26 (p. 425):
Imagine a tiled floor made of square and octagonal tiles in the following pattern.  Identify the smallest repeating rectangular unit, analogous to a unit cell.

There two possibilities (cf. example in the supplementary notes with perskovite--this shows possibilities for more than one definition in three dimensions) here.  The small squares form a box of 4 which repeats as does the unit formed by joining lines at the centers of the octagons.  This is shown in the following picture stolen from the book:


Problem 10.32 (p. 426):
Methanol (CH3OH; bp = 65oC) boils nearly 230o higher than methane (CH4; bp = -164oC), but 1-decanol (C10H21OH; bp = 229oC) boils only 55o higher than decane (C10H22; bp = 174oC).  Explain.

Methanol and methane are very short molecules and, thus, do not have much in the way of dispersion forces (i.e., there are too few electrons here to make much of a difference).  On the other hand, hydrogen bonding is quite strong in methanol, relatively.  With the larger 10-carbon chains of decanol and decane, dispersion forces have become quite noticeable and the relative effects of H-bonding have diminished.  H-bonding is still important but it has had its effects "diluted" by the long carbon chains.


Problem 10.36 (p. 426):

How can you account for the fact that the dipole moment of SO2 is 1.63D but that of CO2 is zero?

Both S-O and C-O bonds are quite polar (after all, oxygen has a rather large electronegativity).  The solution to the problem is, then, only apparent after one draws the Lewis dot structures:

SO2 is bent and, thus, the bond dipole moments can add vectorially to give a total dipole moment greater than zero.  On the other hand, CO2 is linear and the two bond dipoles point in opposite directions.  Since they are equal--and opposite--their vector sum is zero and the net dipole moment is zero.  (Of course, if one were to look at quadrupole moments...!)


Problem 10.46 (p. 426):

How much energy (in kJ) is needed to heat 5.00 g of ice from -10.0oC to 30.0oC?  The heat of fusion of water is 6.01 kJ/mol and the molar heat capacity of ice is 36.6 J/(K-mol) and the molar heat capacity of liquid water is 75.3 J/(K-mol).

This is a rather good problem in that it requires putting together the following:

What we need to do here is calculate the heat for each step in the above process.  We now list and label the three steps in the process:
 
 
Step
Process
(1)
Heating ice from -10.0oC to 0.0oC.
(2)
Melting the ice (at 0.0oC)
(3)
Heating the liquid water from 0.0oC to 30.0oC.
Looking at our units and remembering the relevant formulas, the general formula for heating a solid to its melting point, melting it, and then heating the liquid is given now.  Here, T1 is the initial temperature and T2is the final temperature, with Tf being the melting point temperature.

Here, we are using our usual symbols.  The "1000's" shown are to convert from J to kJ.  Since, Tf = 0.0oC, life here is particularly easy!  We now plug in the various numbers and get the final result.

If nothing else, this was surely a lot of pHun!


Problem 10.54 (p. 427):

Carbon disulfide, CS2, has Pvap= 100 mm [Hg] at -5.1oC and a normal boiling point of 45.6oC.  What is DHvap for carbon disulfide in kJ/mol?

This is a fairly straightforward problem using the Clasius-Clapeyron equation.  To do this, however, we might want to set up the equation in a form where doing the calculations is made as easy and error-free as possible.  A form of this equation, given in the book is

To simplify this, we first note that

We use this result and do a little more algebra to get the next equation (which is a little easier to handle numerically):

This is then easily rearranged to solve for the heat of vaporization, as follows:

This form is more easily punched into a calculator!  Now, we at 273.15K to each temperature to convert each to degrees K.  Since, T2 is the normal boiling point, P2 is obviously 760 mm Hg.  Finally, note that R must be expressed kJ/(K-mol).  Putting all these things together gives the final result, which we now evaluate:

There are other ways this could be done, but this is the easiest way (and hence the best way) to do this.


Problem 10.67 (p. 427):

Which of the four kinds of packing used by metals makes the most efficient use of space, and which makes the least efficient use?

In the book, the four types of packing correspond to the simple cubic, face-centered cubic, and body-centered cubic lattices.  A fourth type of packing refers to a noncubic lattice (which happens to be the hexagonal lattice).  A rather detailed description of packing factors is given in the supplementary notes (Chapter 13:  "Packing Factors").  The fcc and hexagonal lattices use the hexagonal close packing and cubic close packing arrangements, respectively.  Both of these occupy about 74% of the total space whereas the others occupy less space (the numbers are shown both in the Packing Factor write-up and on p. 411 of McMurry-Fay).  The simple cubic arrangement takes just 52% of the space and the bcc arrangement 68%.  Thus, fcc and hexagonal are to most efficient and sc is the least efficient.


Problem 10.68 (p. 427):

Copper crystallizes in a face-centered cubic unit cell with an edge length of 362 pm.  What is the radius of a copper atom (in pm)?  What is the density of copper in g/cm3?

Details relevant to this problem are in Chapter 13 of the supplementary notes.  Here, we shall content ourselves with just using the relevant formulas.  If the side of the cubic unit cell is denoted by a, then the radius of an atom in such a cell is given by

Simple plugging in of a = 362 pm into the above formula gives

a  =  128.0 pm.

The general formula for the density of a solid (with a cubic lattice) is simply (cf. the supplementary Chapter 13 notes here)

The symbols are as usual with, in addition, nc being the number of atoms per unit cell.  For the fcc lattice this number is 4.  Remembering that 1 pm = 10-10 cm, we find that

Problem 10.80 (p. 428):
Look at the phase diagram of CO2 in Figure 10.29 and tell what phases are present under the following conditions:
 
(a)  T = -60oC, P = 0.75 atm
(b) T = -35oC, P = 18.6 atm
(c) T = -80oC, P = 5.42 atm 

Before doing the solutions, here is the figure in question:

Now, it is just a simple matter of reading the diagram for each set of conditions given.  We give the answers now (it should be simple to just find the appropriate point for T and P and then to draw the conclusion):
(a)  gas
(b)  liquid
(c)  solid
That's all there is to it!


Problem 10.82 (p. 428):

Bromine has Tt = -7.3oC, Pt = 44 mm Hg [this is the triple point], Tc = 315oC, and Pc = 102 atm [this is the critical point].  The density of the liquid is 3.1 g/cm3 and the density of the solid is 3.4 g/cm3.  Sketch a phase diagram for bromine and label all points of interest.

Before we get into the really fun things, we first note that 44 mm Hg = 0.0579 atm approximately (this comes from dividing 44 by 760).  We note that the solid is denser than the liquid also (which gives a positive slope to the P vs. T line).  With this information, we can draw the following rough graph (stolen from the back of the textbook):

Most phase diagrams we might draw have this appearance.  One point of interest, however, is how the line between the solid and liquid phases is drawn.  Do you think that there could be solid phase above the critical temperature?  The authors gloss over this point, but it should be noted that this line probably stops before the critical pressure/temperature horizontal line.  Phase diagrams come from experiment; the above comes from just a few reasonable guesses.  In any event, the phases and points of interest are labeled probably as well as they can be!
 

Problem 10.111 (p. 429):
Niobium oxide crystallizes in the following cubic unit cell:

(a) How many niobium atoms and how many oxygen atoms are in each unit cell?
If you count the Nb atoms, you see that you have 6.  Each of these is on a face and, hence is shared by two cells.  Thus, we have 6 x (1/2) = 3 Nb atoms per unit cell.

Counting the oxygen atoms gives us 12 in the picture.  Each of these is on an edge and an edge is shared by 4 cells.  Thus, we have 12 x (1/4) = 3 O atoms.

(b)  What is the formula of niobium oxide?
We have equal numbers of Nb and O atoms.  Hence, the formula is just NbO.  Note:  Writing this as "Nb3O3" would make you mentally ill or--even worse(!)--stupid!
(c)  What is the oxidation state of niobium?
Easy!  Oxygen is a "boss atom" and takes a valence of -2.  Thus, Nb is in its +2 state and we can say that we have Nb2+ cations!