Answer Key, Problem Set 12 – Full - Week #1, January 19-23.

Chemistry 121 Mines

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Answer Key, Problem Set 12 – Full 1. NT1; 2. NT2; 3. NT3; 4. NT4; 5. NT5; 6. NT6; 7. NT7; 8. NT8; 9. NT9; 10. 10.34; 11. NT10; 12. NT11; 13. NT12; 14. NT13; 15. NT14; 16. NT15; 17. NT16

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Molecular Polarity (How to Determine if a Molecule is Polar)

[Mastering Only]

Identification of Interparticle Forces, and Properties that Correlate with Them

1. NT1. (a) Draw a picture illustrating a hydrogen bond between two molecules of ammonia (NH3), with the hydrogen bond

clearly indicated. (b) Describe how the interaction in your picture is consistent with the definition of a hydrogen bond. (c) State why the same picture using PH3 in place of NH3 would not illustrate a hydrogen bond. Specifically address why a P atom in place of an N atom does not result in there being a hydrogen bond.

Answers:

(a)

(b) This is consistent with the definition of a hydrogen bond in that it shows a H atom that is

covalently bonded to a N (which is an N, O, or F) on one molecule interacting with a partially negative N (which is an N, O, or F) on a different molecule.

(c) If there were a P in place of the N in each molecule above, although one could draw the two

molecules next to one another as above, there would be no hydrogen bond because P is neither electronegative enough nor small enough for a hydrogen bond to form.

Hydrogen bonding is a specialized case of dipole-dipole in which a hydrogen atom bonded to a small and highly electronegative atom (e.g., N, O, or F) [and which is therefore partially positive] interacts with a small, partially negative small atom (e.g., N, O, or F) on another molecule. The only bonds with H as one of the two atoms that are polar enough and have a small enough other atom to form H-bonds are H-O, H-N, or H-F bonds.

2. NT2. How does the strength of hydrogen bonds compare to the strength of covalent bonds? How about to ionic

bonding?

Answer: H-bonds (and all IM forces!) are much weaker than covalent bonds or ionic bonding. How do you know? The fact that most molecules stay intact upon melting or boiling (even if their molecules are attracted to one another via H-bonding) indicates that the energies needed to break covalent bonds are much greater than the energies needed to separate things that are held together by H-bonds. For example, water boils at 100C. That means the intact water molecules separate from one another, overcoming the collective IM forces between the molecules (including H-bonding + London + dipole-dipole). So the hydrogen bonds clearly “break” at this temperature, while the O-H bonds within water molecules do not break.

Separately, although your text does not mention it, the maximum values for “energy per mole” associated with H-bonds is

approximately 40 kJ/mol. This is less than one-third of the value of even the weakest covalent bond (142 kJ/mol—See Table

10.3 in Tro), and most covalent bonds are much greater than that (the triple bond in CO is over 1000 kJ/mol!). For comparing H-bonding to ionic bonding, consider the melting points of molecular substances the have H-bonding compared to those of typical ionic compounds. The melting points of ionic compounds (typically > 300 C—see Exp. 17 in lab manual) are generally well above the melting points (and often the

H

H

N H

H

H

N H

A hydrogen bond (between two molecules)

A covalent bond (within a molecule)

Answer Key, Problem Set 12

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boiling points!) of most (small) molecular compounds that contain H-bonds (mp’s typically <300 C—see Exp. 17), which indicates that the forces holding ions together are much stronger than those holding molecules together. Since H-bonding is only a subset of the (collective) IM forces between molecules, clearly ionic bonding forces are stronger than H-bonding forces.

Note also that the smallest (in magnitude) lattice energies are approximately -650 kJ/mol (see Tro, Section 10.4), which

means that it takes at least 650 kJ to break apart (completely) the ions in a mole of an ionic compound. As such, the order-

of-magnitude strength of ionic bonding is (again) well above that of even the strongest hydrogen bonding interactions (40

kJ/mol max).

3. NT3. The nonpolar hydrocarbon C25H52 is a solid at room temperature. Its (normal) melting point is over 50°C and its

(normal) boiling point is greater than 400°C. Which has the stronger intermolecular forces, C25H52 or H2O? Explain your

answer.

Answer: C25H52 has stronger IM forces than H2O.

Reasoning: It is reasonable to generalize as follows: the greater the melting and boiling point of a substance (at a given Pexternal), the greater the forces of attraction between its basic units (“interparticle forces”), because higher temperatures are associated with greater average kinetic energy of particles, and kinetic energy is, effectively, “needed” in order to overcome forces of attraction. Since C25H52 has greater (normal) melting and boiling points than H2O (50°C > 0°C, and 400°C > 100°C), it has the stronger IM forces between its molecules.

Think of it this way: at 300°C (and P = 1 atm), the molecules of H2O cannot be held together by their IM forces—the average kinetic energy is apparently too great, leading to “bulk” escape of molecules, and water exists as a gas. But at the same T (and P), the molecules of C25H52 are held together by their IM forces—though the average kinetic energy of particles is the same as in H2O, it is apparently not great enough to lead to “bulk” escape (since C25H52 is a liquid at this T). The only reasonable conclusion is that the IM forces are greater in C25H52.

4. NT4. Tro states on p. 495: “Hydrogen bonds are … the strongest of the three intermolecular forces we have discussed so far.

Substances composed of molecules that form hydrogen bonds have higher melting and boiling points than comparable substances composed of molecules that do not form hydrogen bonds.” What do you suppose “comparable” means in this context? Why does this qualification make the main premise of the statement invalid or meaningless? Use the information in (and answer to) question NT3 (prior problem) to invalidate/critique Tro’s statement. (Hint: “Are London forces one of the “three intermolecular forces”? And are they the same in all substances?)

By “comparable”, the only thing that makes sense to me is that he’s saying “if all other IM forces are similar in strength”. That is, if all other IM forces are similar in strength, then having hydrogen bonding, in addition, will make the mp and bp higher. This makes total sense and I mentioned this in class (like a dozen times, right?!). But this invalidates the premise of the first statement (that H-bonds are the “strongest”)! London forces are one of the 3 IM forces discussed, yes? So how can you try to justify a statement that H-bonding is stronger than London forces by saying that “if London forces are comparable”, H-bonding will make the mp and bp higher? It’s an invalid argument! It’s what I ranted about in class! I cannot understand how educated chemists and educators can justify stating that H-bonding is the “strongest of the 3 IM forces we’ve discussed” given the examples such as those in NT3, where the mp and bp are much higher for a nonpolar compound (like C25H52) than for water, which has H-bonding. I think they must be trying to emphasize that H-bonding forces are stronger “per atom” or “per electron” than any other IM force. If they qualified it that way, it would be valid, because you do need a lot of electrons or a lot of atoms in order to get the London forces (collectively) to be stronger. But that isn’t what they say, and I think it is, frankly, ridiculous that it keeps getting mentioned (without qualification) in general chemistry textbooks. (and it’s not like I’ve not written the author to make this point; he simply ignores my argument for some reason [he doesn’t provide his reasoning] Maybe I’m missing something, but I don’t think so…). Why not just say that London forces, collectively, can vary greatly in strength, so much so that they can be stronger than H-bonding if the molecules get large enough. And that given that H-bonding is between only two small atoms, it is remarkably strong for an IM force. Why is there a need to say that H-bonding is the “strongest” (without qualification)? I just don’t get it… Okay, I’ll stop now.

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5. NT5. Identify the types of interparticle forces present in the solids (or liquids!) of each of the following substances.

Remember: All molecules (or unbounded atoms) attract other molecules (or unbounded atoms) via London forces.

(a) NH4Cl: ion-ion forces

Substance is ionic (all symbols are nonmetals, but the “NH4” at the beginning indicates that the substance contains NH4

+ ions, and thus “Cl” here represents Cl- ions)

(b) C (diamond): covalent bonds

This is a covalent network solid, so each atom is covalently bonded to every other atom. Even in SiO2 (a compound rather than an element), the atoms of one FU are bonded to atoms in neighboring FUs covalently (there is no difference between the intraformula unit “bonding” and the interformula unit bonding.)

(c) n-octane, CH3(CH2)6CH3: London forces

Substance is molecular (binary with two nonmetals), so forces are IM forces of some kind. Substance is nonpolar (remember, C-H bonds are considered nonpolar, and C-C bonds are obviously nonpolar), so the only IM forces present are London forces.

(d) CHCl3: London forces and dipole-dipole forces

Substance is molecular (not ionic because all symbols are nonmetals, and the first part of the formula doesn’t involve “NH4”—remember, NH4

+ is the only common polyatomic cation). Substance is polar (CHCl3 has one center (carbon) surrounded by 3 Cl’s and 1 H. Since not all the surrounding atoms are

the same, and C-Cl bonds are polar (EN ~ 0.5), the bond dipoles do not completely cancel out, making the molecule polar.

(e) NH3: London forces, dipole-dipole, and H-bonding

Substance is molecular (binary with two nonmetals) and polar (a NH3 molecule has one center (N) surrounded by three H’s and a lone pair [LDS not shown, but you should be able to draw it!]. The lone pair breaks up the symmetry so the N-H bond dipoles do not completely cancel out. Substance’s molecules contain N-H bonds, so each molecule can make H-bonds to other NH3 molecules.

(f) NO: London forces, and dipole-dipole forces

Substance is molecular (binary with two nonmetals) and polar (only one bond, and that bond is polar

[EN ~ 0.5])

(g) BF3: only London forces

Substance is molecular (binary with two nonmetals) and nonpolar (each BF3 molecule has three F atoms surrounding it, but no lone pairs—exception to octet rule; see text)

6. NT6. Consider the substances Cl2, HCl, F2, NaF, and HF. Which one has a (normal) boiling point closest to that of argon?

Explain.

NOTE: Since noble gas samples contain “basic units” that are individual, non-covalently nor metallically bonded atoms, the forces between the atoms are the same as the forces between nonpolar molecules in a molecular sample--London forces. Thus we typically treat noble gases as “nonpolar molecular” substances even though they are obviously not actually “molecular”. Thus the term “IM forces” is often applied to noble gases despite the fact that it isn’t strictly appropriate.

Answer: F2

Strategy:

1) Recognize that the most similar boiling points (or mp, or Hvap, or Hfus or surface tension…) are predicted to be in the two substances that have the most similar interparticle forces.

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2) Recognize that typical ionic compounds have much stronger interparticle forces than molecular substances (or noble gases), because ion-ion forces are typically greater than any set of IM forces (see NT1, above).

3) When comparing molecular substances:

a) First assess London forces in all substances by counting the number of electrons per molecule (or noble gas atom) (OR molar mass) in each and compare. (Remember that the magnitude of London forces is roughly proportional to the number of electrons per molecule/atom (OR molar mass).)

b) For substances with similar London forces, look to see if which, if any, molecules are polar and/or can interact via H-bonding. Being polar adds forces of attraction “on top of” London because of dipole-dipole forces. Being able to form H-bonds adds additional forces of attraction.

Execution of Strategy:

Ar Cl2 HCl F2 NaF HF

ionic? Y

# electrons per unit* (London)

18 34 18

18 N/A 10

polar? N N Y N N/A Y

H-bonding? N N N N N/A Y

* Or ~MM’s: 40 g/mol 71 36 38 20

Clearly, the substance with the most similar IM forces as Ar is F2. Each one’s “basic units” are nonpolar, and have 18 electrons (or MM’s close to 40 g/mol). So the forces are similar. Note that HCl has similar London forces, but since its molecules are polar, there is an additional dipole-dipole force of attraction between its molecules, so you’d predict stronger IM forces overall and a higher boiling point for HCl vs. Ar. Cl2 should have stronger IM forces than Ar or F2 since the London forces will be stronger (more electrons/molecule or greater MM). HF should have a higher bp due to the dip-dip and H-bonding. NaF should have a very high boiling point due to the ion-ion forces holding its “units” together.

7. NT7. Consider the following compounds and formulas. (Note: The formulas are written in such a way as to give you an idea

of the structure.)

ethanol: CH3CH2OH; dimethyl ether: CH3OCH3; propane: CH3CH2CH3

The boiling points of these compounds are (in no particular order) -42.1°C, -23°C, and 78.5°C. Match the boiling points to the correct compounds.

Answers: -42.1°C is propane, -23°C is dimethyl ether, and 78.5°C is ethanol

Reasoning:

All substances have similar London forces since the number of electrons per molecule values are identical in the three (26 e-) (the molar masses are also similar: 44-46 g/mol). Propane is nonpolar because it contains only C’s and H’s, and C-C and C-H bonds are nonpolar. Thus the only IM forces between its molecules are the London forces. The ether molecule is polar (it contains a C – O – C bond, which makes the geometry around the O bent [O will have two lone pairs!]), so its IM forces and thus bp should be larger than those for propane. Ethanol has an O – H bond, so in addition to being polar, its molecules can attract one another via H-bonding, which makes its IM forces and bp the highest of the three.

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8. NT8. (a) Order from weakest to strongest intermolecular forces and explain your reasoning: CH4; He; CH3OH; NaF;

CH3F; CH3CH3 (b) Which do you think would have the highest boiling point at 1 atm pressure? (c) Which do you think would have the highest vapor pressure (at some hypothetical T where all of them were liquids)? (d)

Which to you think would have the highest Hvap? Explain.

Answer:

(a) weakest forces He < CH4 < CH3CH3 < CH3F < CH3OH < NaF strongest forces 2 e- 10 e- 18 e- 18 e- (&polar) 18 e- (w/H-bonding) ionic

Reasoning.

(NOTE: You may wish to see the strategy outlined in Problem 6 (NT6) above before reading this answer, since that is essentially what is applied here.)

First identify any ionic compounds. Here, NaF is the only one. Thus, it has the strongest interparticle forces (technically, these are not “intermolecular” so the question was slightly “flawed” in using that word here), since ion-ion forces are stronger than any of the intermolecular forces (London, dipole-dipole, or H-bonding). Next, assess the London forces in the molecular substances by counting the total number of electrons or calculating molar mass—London forces get stronger as the number of electrons (or molecular weight, which correlates with #electrons) gets larger. CH3OH, CH3F, and CH3CH3 all have the same number of electrons (and a similar molecular weight), so their London forces are similar to one another, but larger than CH4 and He, which have only 10 and 2 e-‘s respectively. Next, consider if any of the substances have forces in addition to London—i.e., either dipole-dipole or H-bonding. He, CH4, and CH3CH3, being nonpolar (these all have either no bonds [He] or only nonpolar bonds), don’t have any additional forces. Thus their order is as indicated above. But CH3F, with exactly one polar bond (C-F) is polar, and thus has dipole-dipole forces in addition to London forces, making the total (overall) IM forces greater in CH3F than in CH3CH3. Molecules of CH3OH, having an O-H bond, not only are polar, but can make hydrogen bonds with one another. So CH3OH will have even stronger IM forces (in total) than CH3F (H-bonding is “added on” to similar “other” IM forces).

(b) NaF should have the highest boiling point at 1 atm pressure because it has the greatest

intermolecular (interparticle) forces (see part a). See also the answer to some later problems (below) for a more detailed analysis of the relationship between IP forces and bp.

(c) Helium (He) should have the highest vapor pressure at a given temperature. Having weaker

forces of attraction means that more molecules (actually “atoms” here!) can escape the liquid at any given temperature, and “more gaseous molecules/atoms” equates to “greater (vapor) pressure”. Again, see the answers below for a more detailed analysis.

(d) NaF should have the highest Hvaporization because if the interparticle forces are stronger, it should take more energy to separate the particles, which is what occurs during vaporization (liquid turning

into gas means particles are separated from one another). Recall that Hvaporization refers to the change in enthalpy that accompanies the vaporization of one mole of a liquid; since this occurs at a constant temperature, the enthalpy change is essentially equal to the increase in the potential energy of the system as the particles are separated from one another).

Interparticle Forces and Different Types of Crystalline Solids

9. NT9. (a) If covalent bonds are present in both molecular solids and covalent network solids, why are the melting points

so low for molecular solids and so high for covalent network solids? (b) Why do ionic compounds tend to have higher melting points than molecular compounds with similar molar masses? (Hint: What forces hold the basic units together?)

(a) Think about and visualize the following carefully: you do not break any covalent bonds when

melting a molecular solid! Since the molecules of a molecular substance remain intact upon melting (they just separate from one another), only the much weaker intermolecular forces of attraction need to be overcome. That means that less energy is required, and generally speaking that leads to a very low melting point for molecular solids [see Exp. 17]. On the other hand, in order

Answer Key, Problem Set 12

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A B

C

Pvap

T

to melt a covalent network solid, the covalent bonds between atoms do have to be broken, and so it is not surprising that the melting points of these solids are very high.

(b) The stronger the attractive interparticle forces in a solid, the higher the melting point. Molecules are

neutral and ions are charged. This leads to fundamentally different magnitudes of electrostatic forces of attraction between “basic units”. Ions attract one another more strongly because the units are “fully” charged ions, and electrostatic forces are stronger the greater the magnitude of the charge(s) (Coulomb’s Law). Molecules are neutral, but can still attract one another electrostatically either via dipole-dipole interactions or via instantaneous and induced dipole interaction. However, in both of these cases, the “charges” that lead to the electrostatic forces are necessarily “partial”--either “partial and permanent” (if molecules are polar) or “partial and temporary/induced” (if nonpolar). Ultimately, the “partial charged” nature of IM forces are what make them intrinsically less strong than ion-ion forces. It’s all about Coulomb’s Law in the end!

10. 10.34. How does the electron sea model explain the conductivity of metals? The malleability and ductility of metals?

Answer: The electron sea model views the valence electrons as being held so loosely by the metal atoms that they flow freely and continuously between the positive “cores” of any atoms—they are “delocalized” and they are a kind of flowing, negatively charged “glue” that holds all of the cores to one another. This explains conductivity fairly obviously—if the valence electrons can flow freely between the positive “cores”, then they can flow freely through the solid, which is basically the definition of electrical conductivity (movement of charge carriers). It explains malleability by suggesting that in response to being struck by, say, a hammer, the cores will move with respect to one another, but the electron “sea” will just flow around the cores (in whatever deformed arrangement that might be generated by the “blow”), holding them together to one another. The idea is that the electrons move so fast that they can respond essentially immediately and keep the cores together (rather than having the solid fracture as an ionic solid does in response to a sharp blow). It explains ductility similarly—drawing metals into wires is a type of deformation, and as was just stated, the ability of the “sea of electrons” to flow quickly around the cores would allow for the deformation to occur while keeping the atoms bonded to one another.

Vapor Pressure, Vapor Pressure Curves, and Relation to Boiling and Boiling Point

11. NT10. Consider the following vapor pressure versus temperature plot for three different

substances A, B, and C.

If the three substances are CH4, SiH4, and NH3, match each curve to the correct substance.

Answer: A is CH4, B is SiH4, and C is NH3

Reasoning:

1) First of all, determine who has the strongest/weakest IM forces:

C has the strongest IM forces and A has the weakest, because if you look at the T needed to generate a given vapor pressure (horizontal dashed line added), you see that C requires the highest T and A requires the least. That means the forces are strongest in C and weakest in A. (You could also look at a vertical line and look at who has the higher VP at a given T—C would be lowest indicating stronger IM forces).

2) Now, figure out the order of strength of IM forces in CH4, SiH4, and NH3:

CH4 and SiH4 are both nonpolar (4 electron clouds with no lone pairs gives a tetrahedral AG, with all outer atoms the same. Actually, all bond are pretty much nonpolar anyway, so the geometry isn’t critical here), but SiH4 molecules have more electrons and thus stronger IM forces exist between SiH4 molecules.

NH3 has the same number of electrons as CH4, but it is polar (4 clouds but one lone pair) and has hydrogen bonding forces acting between its molecules, so it clearly has greater IM forces than CH4. Since the number of electrons in SiH4 isn’t that much bigger than the number in NH3 (18 vs 8), NH3 is predicted to have stronger IM forces than SiH4. NOTE: based on the I2 vs. H2O

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example, you need to have something like 50-100 electrons per molecule to get London forces the same order of magnitude as H-bonding. So NH3 has greatest IM forces and is C; CH4 has the weakest and is A.

12. NT11. (a) What is a kinetic energy distribution curve? (What are the

axes of the plot? What does the maximum of the curve represent?) (b) Assume the distribution curve at the right is for a liquid X, and that Eescape is the energy needed to escape the liquid. (i) Approximately what fraction of the molecules in a sample of liquid X at temperature T1 have enough energy to escape? (ii) Approximately how many times larger is the fraction that can escape at T2 vs T1 (i.e., how many times larger is the area under the curve to the right of Eescape in the curve of T2 compared to that in the curve of T1)? (iii) Approximately how many times larger is the average KE at the higher T than at the lower T? (iv) Are the answers to (ii) and (iii) significantly different? Comment.

Answers:

(a) A kinetic energy distribution curve shows the distribution of KE values that different particles possess at any given time in a sample. In other words, it shows how many particles in the sample (at a given moment in time) have each given KE (shown on the x-axis). As such, the axes on the plot are “number of particles” (on the y-axis) and “KE” (on the x-axis). The maximum of this plot, therefore, occurs at the single KE value that is possessed by more particles than any other in the sample. It is called the “most probable KE”. Although this is not equal to the average kinetic energy of all particles (because the curve is asymmetrical), it is close to the value of the average (slightly less, actually) and will track with the average—as the average gets larger (i.e., the peak shifts to the right), the maximum will also shift to the right.

(b)(i) Given the plot in this problem, I’d estimate the fraction of molecules with KE Eescape to be around 1/20th (~5%). I looked at the roughly triangular area represented by the area under the T1 curve past Eescape and compared it to the rest of the area under that curve (to the left of Eescape). This is only a rough estimate, so anything from 2 – 6 % is probably reasonable (I think it’s pretty clearly less than 10% though).

(ii) Looking at those two areas on the plot, it looks to me like the second one is at least 3 times as large, perhaps 4 times. That means that the fraction of molecules that can now escape is 3-4 times as large as the fraction that can escape at the lower T (maybe 15-20% vs ~5%). Note, this is 300% - 400% of the value at the lower T (since 2x larger means 200%, etc.)

(iii) I dropped dashed lines down from the maxima of the two curves, and then added some “tick” marks on the x-axis to estimate how much larger the most probable KE is at T2 vs T1. It is about 5 to 4, or 5/4 = 1.25 (which means about 25% larger).

(iv) YES THEY ARE!! When the avg KE is made (only) about 1.25 times larger (through a T increase), the fraction of molecules that have enough energy to escape became 3-4 times larger! This is a huge effect (25% increase vs 300% increase)! The fraction that can escape (which is proportional to the vapor pressure) is definitely not “proportional” to absolute T—it increases “exponentially” rather than “linearly”. See next problem!

13. NT12. Give a molecular-level explanation of why the vapor pressure of a liquid is so sensitive to changes in

temperature (exponential vs. linear dependence). (Hint: Consider the distribution of kinetic energies of particles

[distribution curve] and the concept of “escape energy”.)

Shortest Answer: A small increase in T leads to shift in the distribution curve of kinetic energies to higher energies. This results in a significant increase in the number of particles with high enough KE to escape (because of the shape and nature of a distribution curve at high KE values). This makes the concentration of gas particles significantly greater, increasing the (vapor) pressure by more than would be the case by a T increase (of a fixed amount of gas) alone. In other words, looking at P = (n/V) x R x T, when T of a liquid increases a small amount, n/V of the gas above the liquid increases quite a bit (because of the “escaping” issue), raising the vapor pressure (of the liquid) by a lot more than “just” the (small) T increase. It is not the change in T itself that causes the (vast majority of the) increase in P. It is mostly due to that increase in n/V. Reread the answers to Parts (b)(i)-(b)(iv) of the prior problem to see a numerical example of this effect.

KE

# par t i c les

Eescape T1

T2

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Full Answer: Vapor forms over a liquid in an enclosed container because there is always some fraction of the molecules in the liquid that have enough energy at any given time to overcome the forces of attraction and "escape" into the gas phase. Since the average kinetic energy of the sample depends on temperature, the higher the temperature, the greater will be the average kinetic energy—the curve shifts to “greater KE energies”, so more particles can escape. Moreover, a closer look at the way the curve shifts lead one to the following observation: it is the nature of probability distribution curves that a small change in the average has a profound effect at the end of the curve. Thus, a small increase in T, which leads to a small increase in the average kinetic energy, leads to a significant (percentage) increase in the fraction of molecules that have enough energy to overcome a fixed “escape energy” barrier. This leads to a significantly larger number of gas molecules (per L) in the gas phase before dynamic equilibrium is established. Since pressure is directly proportional to "n" (the number of moles of gas particles) the (vapor) pressure will thus increase significantly. Note: as commented on in class, the predominant factor here is the increased number of gas molecules that escape (↔ higher concentration of gas particles) rather than the temperature increase of the gas particles themselves. A very small contribution to the increased pressure will be from the T increase, but it is essentially negligible compared to the increase in pressure due to the increase in “n”. Make sure you understand how KE distribution curves relate to the argument above (NOTE: The example below is the same exact example as used in the prior problem!!):

14. NT13. (a) Define "boiling" and "boiling point" and then (b) describe how boiling relates to vapor pressure,

temperature and atmospheric (external) pressure. In other words, why does a substance boil at its boiling point or

above, but not below it?

(a) boiling: the process in which bulk liquid molecules in a sample of a liquid are converted into a gas, forming bubbles. It will occur only if energy is added to a liquid sample at a temperature equal to or above the substance's boiling point.

(Note: evaporation is the process in which surface molecules in a sample of liquid are

converted into a gas. It can happen at any temperature, and without external energy being added.)

boiling point: the temperature at which the vapor pressure of a liquid is equal to the external

pressure; it is the temperature at (or above) which a liquid can undergo boiling.

(NOTE: the boiling point of a substance depends on the external pressure, so there is no single boiling point for a given substance. There is a different boiling point for every external pressure.)

(b) In order for boiling to occur, gas must be able to form within the liquid sample, forming bubbles.

This means that the gas that is formed in the bulk liquid must be able to "push away" the bulk

# par t i c les

Eescape

T1 T2

Increased fraction of molecules that have the energy needed to escape (after T increase) [~400% larger area than before the T increase, whereas the average KE only increased by ~25%]

Original fraction of molecules that have the energy needed to escape

KE

Region where KEmolecule > Eescape

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liquid (and thus the atmosphere) to make space for the "bubble". The maximum pressure a given gas (in equilibrium with its liquid) can achieve at a given temperature is called its "vapor pressure". Thus, boiling cannot occur until the vapor pressure of the liquid is equal to or greater than the external pressure (pushing on the liquid). The temperature at which the vapor pressure becomes equal to the external pressure thus becomes the boiling point!

Since vapor pressure increases with T, if the T is less than the boiling point, the vapor pressure will be too small to push away the atmosphere and the liquid will not boil. Conversely, if the T is equal to or greater than the boiling point, the vapor pressure will be great enough to push away the atmosphere and the liquid will boil.

NOTE: If external pressure increases, the boiling point raises. If external pressure

decreases, the boiling point decreases. 15. NT14

Vapor Pressure of Substances in mmHg at Different Temperatures

0 C 20 C 50 C 80 C 100 C

water 4.6 17.6 92 354.9 760

turpentine 2.1 4.4 17.0 61.3 131.1

(a) Which choice best indicates the degree of correctness of this statement? “Water at 50 C would

boil if the opposing pressure were reduced to 90 mmHg by means of a vacuum pump.” Explain your answer.

(i) The statement is true (ii) The statement is probably true; additional data would be needed for a final decision. (iii) It is impossible to judge the statement because the data are insufficient. (iv) The statement is probably false; additional data would be needed for a final decision.

(v) The statement is false.

Reasoning. By definition, a liquid will boil if the vapor pressure of the liquid is greater than or equal to the external pressure. The chart indicates that at 50 C, the vapor pressure of water is 92 mmHg. Thus if the opposing (i.e., external) pressure were reduced to 90 mmHg, the vapor pressure would be greater than the external pressure (92 mmHg is greater than 90 mmHg) and so the water would boil.

(b) Which has the higher boiling point at 1 atm pressure, water or turpentine?

Answer: turpentine. Reasoning: You can look at this several ways. 1) Most specific

answer/view: water will boil at 100 C if the external pressure is 760 mmHg because water’s vapor

pressure is equal to 760 mmHg at this temperature (see chart). On the other hand, at 100 C, the vapor pressure of turpentine is only 131 mmHg, which is much smaller than the external pressure

(760 mmHg) and so it clearly will not boil at 100 C. That means that its boiling point must be

higher than 100 C, which means it is higher than water’s boiling point. 2) Similar view, but more general: Boiling occurs when the vapor pressure of a liquid equals or exceeds the external pressure. If the vapor pressure of water is higher than turpentine at any temperature, surely its vapor pressure will become equal to a given external pressure at a lower temperature than will the vapor pressure of turpentine. Thus water will boil at a lower temperature, or in other words, it has a lower boiling point [so turpentine has the higher boiling point]. 3) Third view (also general): The data in the chart indicate that at any temperature listed, the vapor pressure of water is greater than that of turpentine. That means that the intermolecular forces must be weaker in water than in turpentine because a higher vapor pressure at a given temperature means that more molecules are able to escape (so the forces holding them together must be weaker). If the forces in water are weaker, it should boil at a lower temperature (the molecules will need a lower average kinetic energy in order to separate from the liquid and boil, and lower kinetic energy is associated with a lower temperature). Thus, turpentine will boil at the higher temperature (its molecules are held together by stronger forces).

Answer Key, Problem Set 12

PS12-10

Rhombic

Monoclinic

P

T

Phase Diagrams

16. NT15. Analyze the figure for problem 12.77 in Tro to answer the following question: Which has a greater density,

the Rhombic or Monoclinic phase of sulfur? You must provide detailed reasoning from the figure (no numerical values should be looked up or given here!)

Answer: Rhombic

Reasoning: If you look at the line segment separating Rhombic from Monoclinic, you can see that it has a positive slope (see right, highlighted segment). That means that if you start below the line (in Monoclinic) and increase pressure at constant temperature, the system will convert to Rhombic at some point (see arrow at right and intersection with the boundary line). Since systems convert to their more dense phase at higher pressure (high pressure “compresses” the sample, making V smaller and thus density larger), Rhombic must be the more dense phase of sulfur (between the two).

Heating Curves and Related Calculations [in Mastering only]

Unusual Properties of Water

17. NT16. (a) What is odd about the relative densities of ice and water? (b) What is the basis for this property of

water? (c) Why is it important for life on this planet that water display this property?

(a) Most substances are more dense in the solid state than in the liquid state because they are more “compact” in the solid state at the molecular level. That is, solids often represent the most efficient way to “pack” atoms or molecules, and so if that is true, then clearly packing more atoms into a given volume will lead to a greater density (more mass in a given volume). Water is odd in that solid ice is less dense than liquid water.

(b) This occurs because in the solid state, water molecules pack together in a manner that maximizes the number of hydrogen bonds between molecules. It turns out that this packing leads to a more “open” structure than the more “random” arrangements found in liquid water. So there ends up being fewer water molecules in a given unit of volume and so the density actually decreases a bit upon freezing.

(c) One important consequence is that lakes do not completely freeze in the winter. When the surface water freezes into ice, the ice floats rather than sinks. If it were to sink, then ultimately the entire lake would freeze over, but since it floats, the ice layer actually insulates the liquid water beneath it so that it remains a liquid. That allows fish (and any other creatures/organisms that live in the water) to survive the winter! (This is discussed in your text on p. 530.) There are many other important consequences of water’s odd liquid/solid density properties. Sea level would be much higher, for example, if icebergs were to sink rather than float, and thus many islands and coasts would be under water right now! (This is a separate issue from the one about icebergs melting. Interestingly, an iceberg melting would not change sea level since it is already floating in the ocean, but a mass of ice on land like a glacier would cause an increase in sea level since the liquid water runoff would end up in the ocean as extra water.). Mountains degrade faster by weathering due to the fact that liquid water fills cracks in rock, and then when it freezes, it expands, thereby applying tremendous stress forces on the rock and making it crack and crumble. Cells rupture upon freezing (unless flash frozen—see text), so that impacts food storage, etc.…I’ll stop now….

=============================== FORMAL END OF SET ===============================

Answer Key, Problem Set 12

PS12-11

T

P

A B

C

D

E

F

G

H

1.0 atm

Normal fp Normal bp

=============================== FORMAL END OF SET ===============================

A couple of practice problems (with explanations) for you!

Phase Diagrams

(Practice Problem). Consider the phase diagram to

the right. What phases are present at points A

through H? Identify the triple point, normal

boiling point, normal freezing point, and critical

point. Which phase is denser, solid or liquid?

Answers:

A: solid; B: liquid; C: gas; D: solid and gas;

E: solid, liquid, and gas (this is the triple point)

F: liquid and gas; G: liquid and gas (this is the critical point)

H: supercritical fluid

The normal freezing point can be found by dropping a vertical line down from the intersection of the solid-liquid curve and the 1.0-atm horizontal line (left blue arrow shown) and seeing the T on the x-axis for this intersection point. The normal boiling point can similarly be found by dropping a vertical line down from the intersection of the 1.0-atm line with the liquid-gas curve.

The solid is more dense in this substance. You determine this by recognizing that any substance in equilibrium between its solid and liquid phases will turn into the more dense phase when pressure is increased (“squeezing” the sample converts it to its more dense [compact] state). If you start at any point on the solid-liquid equilibrium line on this phase diagram and increase the pressure, the substance turns into a solid (see added arrow). Thus the solid must be more dense.

Heating Curves

(Practice Problem). Consider a 75.0-g sample of H2O(g) at 125°C and 1 atm of pressure. What phase or phases are present

when 215 kJ of energy is removed from this sample? Cwater vapor = 2.0 J/g°C; Cwater, liguid = 4.2 J/g°C; Cice = 2.1 J/g°C; Hvap =

40.7 kJ/mol; Hfus = 6.02 kJ/mol

Answer: some solid and some liquid will be present in dynamic equilibrium (at 0°C)

Reasoning/Work:

General Approach: Realize that when heat is removed from a sample, there are only two possibilities for what occurs: either T will decrease (if the T is not at the boiling point or melting point of the substance), or a phase change will occur (if the T is at the bp or mp of the substance). So in this problem, one must reason out / realize that the sequence of events that would occur if heat were continuously and completely (first consider the hypothetical, then the “actual”) removed from a sample of gaseous water vapor above its bp are as follows:

a) gas would first cool (lower its T) until the bp is reached (q = Cgas x m x T)

A B

C

D

E

F

G

H

1.0 atm

P

T

Answer Key, Problem Set 12

PS12-12

b) gas would then condense to liquid at its bp (q = Hcondensation = -Hvaporization) (n.bp for H2O = 100.°C)

c) when all gas has been converted to liquid, the liquid would cool until the melting point is reached

(q = Cliquid x m x T)

d) liquid would then freeze at its mp/fp (q = Hfreezing = -Hfusion)

e) when all liquid has been converted to solid, the solid would then cool until it reaches absolute zero (again, assuming heat were completely removed)! (q = Csolid x m x T)

So the question in this problem is: Where do you end up with water when you don’t “completely” remove energy, but specifically remove 215 kJ’ worth of energy. Let’s find out:

1) Calculate the energy needed to be removed to reach the bp (condensation point):

q = 2.0 J/g°C x 75.0 g x (100.°C - 125°C) = -3750 J 3.8 kJ needed to be removed

Since 3.8 kJ is less than the 215 kJ we are supposed to remove, we are not done. Go to next step.

2) Calculate the (additional) energy needed to be removed to condense the entire 75.0 g sample:

q = -75.0 g/(18.0 g/mol) x 40.7 kJ/mol = -169.6 kJ 169.6 kJ (more) needed to be removed

[Total Removed Thus Far = 3.8 + 169.6 = 173.4 kJ

Since 173 kJ is less than the 215 kJ we are supposed to remove, we are not done. Go to next step.

3) Calculate the (additional) energy needed to be removed to reach the freezing point:

q = 4.2J/g°C x 75.0 g x (0.°C – 100.°C) = -31500 J 31.5 kJ needed to be removed

[Total Removed Thus Far = 173.4 kJ + 31.5 = 204.9 kJ

Since 205 kJ is less than the 215 kJ we are supposed to remove, we are not done. Go to next step.

4) Calculate the (additional) energy needed to be removed to freeze the entire 75.0 g sample:

q = -75.0 g/(18.0 g/mol) x 6.02 kJ/mol = -25.1 kJ 25.1 kJ needed to be removed

[Total Removed Thus Far = 204.9 kJ + 25.1 = 230 kJ

Since 230 kJ is greater than the 215 kJ we are supposed to remove, the system would not “make it” this

far. That is, the entire sample would not freeze—part of it would freeze, leaving the system with some

solid and some liquid water at 0°C)

Thus, if you remove 215 kJ from this sample, you will end up with some liquid and some solid at 0°C.