chapter 11 intermolecular forces, liquids, and solids Flashcards
(19 cards)
Which of the following intermolecular forces would be present in liquid CH3CH2CH2CH2F?
I. hydrogen bonding II. dipole-dipole forces III. London dispersion forces
a. I only
b. I and II
c. II and III
d. II only
e. I, II, and III
c. II and III
I got it wrong because I did think that there was hydrogen bonding due to the CH2F but… What I don’t understand is, can you know just by looking at ‘CH2F’ that the F is connected to the H or the C
This molecule is polar and dipole-dipole forces will be present but the fluorine atom is bonded to carbon rather than hydrogen (no F-H bonds) and therefore hydrogen bonding is not possible. It will also have London dispersion forces as all molecules do.
Which of the following would have the greatest London dispersion forces?
a. H2O
b. H2S
c. HCl
d. HF
e. H2Se
e.
Yay!
Greater London dispersion forces are associated with higher molecular weight and greater surface area. These answer choices differ greatly in molecular weight and as H2Se has the highest molecular weight it also has the greatest London dispersion forces.
Which of the following molecules would be capable of hydrogen bonding when mixed with water?
a. CH3CH2OH
b. CH3CH2OCH2CH3
c. CH3COCH3
d. (CH3)3N
e. All of the above
e.
I am completely failing this.
To be capable of hydrogen bonding as a pure substance a molecule must have an F-H, O-H, or N-H bond in its structure. The requirement is different when mixed with water. Water can already act as a hydrogen bond donor (has the hydrogen atoms bonded to oxygen). All that is needed from the compound mixing with water is a hydrogen bond acceptor (F, O, or N with at least one lone pair of nonbonding electrons available). As all of these molecules contain either oxygen or nitrogen they would all be capable of hydrogen bonding when mixed with water.
Which of the following is true about the relative boiling points of compounds A and B shown below?
There are pictures here but it’s not going to copy over. The first one (A) has a large surface area, 5 carbon atoms, and 12 hydrogen atoms. The second one (B) has a smaller surface area, 5 carbon atoms, and 12 hydrogen atoms. Both appear to be non-polar.
a. Compounds A and B will have the same boiling point because they have the same molecular weight.
b. Compounds A and B will have the same boiling point because they have the same surface area.
c. Compound A has a higher boiling point because it is more polar.
d. Compound B has a higher boiling point because it is more polar.
e. Compound A has a higher boiling point because it has greater London dispersion forces.
e.
Compounds A and B are isomers and have the same molecular weight. They are also nonpolar and therefore London dispersion forces are the only intermolecular forces present for both compounds. Even though they have the same molecular weight, compound B has a more compact structure and therefore compound A has a greater surface area, greater London dispersion forces, and therefore a higher boiling point.
Which of the following is true about the relative boiling points of HBr and HI?
a. HBr has a higher boiling point because it is more polar.
b. HI has a higher boiling point because it is more polar.
c. HBr has a higher boiling point due to hydrogen bonding.
d. HI has a higher boiling point due to greater London dispersion forces.
e. HBr has a higher boiling point due to greater London dispersion forces.
d.
I guessed and I guessed wrong. But I did learn something that hopefully sticks. I picked HI because I thought it was more polar but that’s actually not true. I is actually less electronegative than Br so there is less of a difference between it and the hydrogen atom. In fact, the electronegativities between I and Br are very similar.
HI does have the higher boiling point but it’s because it has significantly higher molecular weight therefore stronger London-dispersion forces.
Which of the following is true about the relative boiling points of CH3CH2CH2CH2CH3 and C(CH3)4?
a. CH3CH2CH2CH2CH3 has a higher boiling point due to a greater dipole moment.
b. C(CH3)4 has a higher boiling point due to a greater dipole moment.
c. C(CH3)4 has a higher boiling point due to greater London dispersion forces.
d. CH3CH2CH2CH2CH3 has a higher boiling point due to a greater surface area.
e. C(CH3)4 has a higher boiling point due to a greater surface area.
d.
Finally got one right.
Just try to draw the 2nd molecule out and you’ll see that it covers less surface area.
They both are pretty similar in molecular weight. And neither seems very polar.
Rank the following in order of decreasing boiling point:
CH3Cl, CH2Cl2, CBr4, CHCl3
a. CHCl3 > CH3Cl > CBr4 > CH2Cl2
b. CBr4 > CH3Cl > CHCl3 > CH2Cl2
c. CBr4 > CHCl3 > CH2Cl2 > CH3Cl
d. CH2Cl2 > CHCl3 > CBr4 > CH3Cl
e. CH3Cl > CH2Cl2 > CHCl3 > CBr4
c.
I put a lot of thought into this but it still felt like a wild type guess. After a while, I figured out that none have hydrogen bonding because there’s no F-H, O-H, or N-H. But I knew that London-dispersion forces were going to come into play and some of these are polar and some aren’t. But anyway, here is the full description:
None of these are capable of hydrogen bonding but some are polar and dipole-dipole forces and London dispersion forces will both be present and one is nonpolar (CBr4) and only London dispersion forces will be present. When molecules are similar in molecular weight the more polar one will typically have the higher boiling point due to the greater dipole-dipole forces. But if there are significant differences in molecular weight (as is the case here) then the heaviest molecule will have the greatest London dispersion forces and will have the highest boiling point even if it is not the most polar. The difference in London dispersion forces will typically more than make up for any difference in dipole-dipole forces. As CBr4 has the highest molecular weight by a significant amount it will have significantly greater London dispersion forces and the highest boiling point.
What is the vapor pressure of water at 100 degrees C?
a. 0
b. 0.25 atm
c. 0.5 atm
d. 0.75 atm
e. 1.0 atm
e. 1.0 atm
Idk. I feel like this question is so obvious that it feels like a trick question. Basically, if water is boiling at 100 degrees C then you know you’re at sea level (1 atm).
I guess just note that as you atm changes, the boiling point changes too. e.g. at 5,000 feet, water boils at 95 degrees C
Which of the following could be the vapor pressure of water at 97⁰C?
a. 740 torr
b. 760 torr
c. 800 torr
d. 1480 torr
e. none of these
a. 740 torr
Which of the following will affect the vapor pressure of a liquid?
I. surface area II. temperature III. volume of the liquid
a. I only
b. II only
c. I and II
d. II and III
e. I, II, and III
b.
II. temperature only
Yay!
Which of the following has the highest viscosity?
a. CH3CH2F
b. CH3CH2OH
c. CH3F
d. HOCH2CH2OH
e. HOCH2CH(OH)CH2OH
e. phew!
I analyzed them all and it just seemed like the best choice. Here’s what the book says:
The greater the intermolecular forces the greater the viscosity. These are all molecular compounds and will be distinguished by comparing the presence and degree of hydrogen bonding, dipole-dipole forces, and London dispersion forces in each. Hydrogen bonding is the strongest of the intermolecular forces and the following three choices are capable of hydrogen bonding as they each have an O-H bond:
CH3CH2OH
HOCH2CH2OH
HOCH2CH(OH)CH2OH
As the last answer has the greatest number of O-H bonds it is capable of the greatest degree of hydrogen bonding and will have the highest viscosity.
The fluorine atoms in the remaining two compounds are bonded to carbon rather than hydrogen (no F-H bonds present) and hydrogen bonding will not be possible.
Which of the following molecules has the highest surface tension?
a. CH3CH3
b. H2NNH2
c. HCl
d. H3COCH3
e. CH3F
b.
Yay!
I almost thought it might be c. because it’s very polar but all the hydrogen bonds in b. win out
Which of the following is exothermic?
I. vaporization II. fusion III. condensation
a. I. only
b. II. only
c. III. only
d. I. and II.
e. I. II. and III.
c. III.
I might be ashamed to say that I really had to think about this one. And condensation is a hard one to even think about.
But I do know that the other ones are taking on heat which makes them endothermic.
Which of the following makes the phase diagram of water unique from most other substances?
a. An unusually high triple point
b. An unusually low critical point
c. A negative slope for the solid/liquid line of equilibrium
d. A positive slope for the solid/liquid line of equilibrium
e. A positive slope for the liquid/gas line of equilibrium
c.
A positive slope for the liquid/gas line of equilibrium
What makes water unique is the negative slope for the solid/liquid line of equilibrium. Most substances have a positive slope for this line. This correlates to the fact that liquid water is more dense than solid water and is due to the significant degree of hydrogen bonding present in water.
Water’s triple point isn’t unusually high and its critical point isn’t unusually low.
Water’s liquid/gas line of equilibrium has a positive slope but so do most other compounds.
Which of the following is true about the critical point?
a. It is the temperature above which a compound explodes.
b. It is the temperature above which a compound decomposes.
c. It is the temperature above which spontaneous combustion occurs.
d. It is the temperature above which a compound cannot exist in the liquid phase.
e. It is the pressure above which a compound decomposes.
d.
I don’t completely understand this. I just know that the other ones don’t makes sense.
The critical point is the temperature and pressure above which a compound cannot exist in the liquid phase. At temperatures exceeding the critical point the average kinetic energy of the molecules is greater than the intermolecular forces, no matter the pressure, and the gas to liquid phase change does not occur.
What is the phase change indicated in the following phase diagram?
An arrow is moving from the liquid area to the solid area. It moving straight up which means pressure is increasing.
a. fusion
b. sublimation
c. deposition
d. vaporization
e. freezing
e.
I guess it doesn’t matter if the pressure is increasing or if the temperature is decreasing, if it’s going from a liquid to a solid, it’s freezing.
What is the phase change indicated in the following phase diagram?
An arrow is moving from the gas area to the solid area
a. fusion
b. deposition
c. sublimation
d. vaporization
e. decomposition
b.
The phase change in the diagram is the conversion of a gas to a solid which is deposition (or vapor deposition).
If the substance corresponding to the following diagram initially at 1.2atm and 0˚C is heated isobarically to 60˚C, what phase change occurs?
Let’s just say it’s going from a solid to a liquid. FYI: ‘isobarically’ means ‘at one constant pressure’
a. fusion
b. condensation
c. sublimation
d. vaporization
e. deposition
a.
What are the normal melting and boiling points of the following compound?
There’s a picture so I’m just going to re-phrase this question.
How about: define what makes a boiling point and melting point ‘normal’
a. 10˚C and 92˚C
b. 20˚C and 160˚C
c. 30˚C and 92˚C
d. 40˚C and 160˚C
e. -20˚C and 100˚C
It’s the melting and boiling points at 1 atm
