Chem 1A Practice Problems
4/28/16
(X-track)
1. How can you know whether London forces are stronger in one molecule or another? a) Are
they stronger in CCl4 or CH4? Why? b) Are they different in 1H2, 2H2, or 3H2? c) Are they
different in NH3 and water?
London forces increase with the total number of electrons in the molecule. Also, in general,
London forces increase when electrons are in higher shells, farther from the nucleus and held
more loosely. a) Carbon tetrachloride has stronger London forces than methane, because Cl
has more electrons than H. b) Remember that nA is a way to represent the isotope of A, where n
is the mass number (number of protons + neutrons). Although tritium (3H) is heavier than regular
hydrogen, it has the same number of electrons, so the London forces should be equal. c)
Although water and ammonia have different MW, the different is due to neutrons, not electrons.
We can’t tell if they have different London forces.
2. How can you tell if dipole-dipole forces are present in a molecule? Decide if there are
dipole-dipole forces in each of the following: C2Cl2, N2H2, PH3, BH3,
The key is to draw Lewis structures and predict the shape. C2Cl2 is linear, Cl-C-C-Cl, so it is
non-polar, no dipoles. N2H2 has a N=N bond, and the N have lone pairs, so they are bent, and it
is probably polar, depending on the twist. PH3 is trigonal pyramidal, polar. BH3 is trigonal planar,
non-polar.
3. What is required for a molecule to hydrogen bond? Can each of the following hydrogen
bond? H2CO, H4CO, (CH3)3N, (CH3)2NH
H-bonds form between H bonded to N/O/F and N/O/F lone pair.
Again, draw Lewis structures, being careful to make sure you get the right arrangement with all
atoms having octet! The first can’t, because H is not bonded to O, the second can, the third
can’t (same reason as first), the last can.
4. Hydrogen bonds can form between two different molecules. The molecule that provides the
H is called the donor and the molecule that provides the lone pair is called the acceptor.
What is required to donate a hydrogen bond? What is required to accept one? Can SH2
donate or accept? Can HCl? Can F2? Can H2CO?
To donate an H-bond, there must be an H with a large partial positive charge, which comes from
being bonded to N/O/F. To accept, there must be N/O/F lone pair on N/O/F with large partial
negative charge, which comes from being bonded to something less electronegative, like C or
H.
Hydrogen sulfide can’t, because the H isn’t bonded to N/O/F, and there isn’t N/O/F. Same for
HCl. Fluorine can’t either: F bonded to F has no partial charge. The last can accept: the O is
sufficiently negative, but the H is not sufficiently positive.
5. a) Organize CH2Cl2, CCl4, CF4 and CH4 by boiling point. Say why you chose that order. b)
CCl4 has a higher boiling point that CH2Cl2. Is this what you expected? Why or why not?
a) First, determine what types of intermolecular forces are present. All but CH2Cl2 are non-polar
and have only London forces; these increase bp with molecular weight. Probably the boiling
point increases in this order: CH4 CF4 CH2Cl2 CCl4
b) You might have expected that carbon tetrachloride has a lower boiling point than
dichloromethane, because carbon tetrachloride has only London forces and dichloromethane is
polar. Apparently the London forces are enough bigger in carbon tetrachloride that they are
stronger than the dipole-dipole forces.
6. How much heat is required to convert 10g of ice at 0 °C to steam at 105 °C? The enthalpy
of fusion of water is 1.435 kcal/mol, the enthalpy of vaporization is 9.713 kcal/mol, the
specific heat of water is 1 cal/g°C, and the specific heat of water vapor is 0.45 cal/g°C.
You have to do this by thinking about a heating diagram, in several parts.
step 1: convert 10g of ice at 0 °C to 10g of water at 0 °C: we know that the enthalpy of fusion is
1.435 kcal/mol. We convert 10 g water to moles: 0.56 mol water. The energy needed to melt the
ice is (1.435 kcal/mol)(0.56 mol) = 0.80 kcal.
step 2: heat the liquid water from 0 °C to 100 °C: q = mCΔT = (10 g)(1 cal/g°C)(100 °C) = 1 kcal
step 3: vaporize the water at 100 °C: the enthalpy of vaporization is 9.713 kcal/mol, so
q = (0.56 mol)(9.713 kcal/mol) = 5.4 kcal
step 4: heat the steam from 100 °C to 105 °C: q = mCΔT = (10 g)(0.45 cal/g°C)(5 °C) = 23 cal
Finally, add all the pieces together! Because all processes are endothermic, all heats are
positive.
0.80 kcal + 1 kcal + 5.4 kcal + 0.023 kcal = 7.2 kcal
7. Enthalpies of vaporization are usually larger than enthalpies of fusion for the same
substance. Why does this make sense?
During fusion, the material goes from solid to liquid. The density usually doesn’t change much,
so the molecules are still very close together, and the intermolecular forces haven’t broken that
much.
During vaporization, the intermolecular forces are broken completely, and the molecules
separate completely. This requires more energy.
8. How much heat is required to convert 25 g of ice at -15 °C to steam at 150 °C?
You have to do this by thinking about a heating diagram, in several parts.
step 1: heat 25 g of ice from -15 °C to 0 °C.
q = mCΔT = (25 g)(1 cal/g°C)(15 °C) = 0.375 kcal
step 2: convert 25 g of ice at 0 °C to 25 g of water at 0 °C: we know that the enthalpy of fusion is
1.435 kcal/mol. We convert 25 g water to moles: 1.39 mol water. The energy needed to melt the
ice is (1.435 kcal/mol)(1.39 mol) = 1.99 kcal.
step 3: heat the liquid water from 0 °C to 100 °C: q = mCΔT = (25 g)(1 cal/g°C)(100 °C) = 2.5
kcal
step 4: vaporize the water at 100 °C: the enthalpy of vaporization is 9.713 kcal/mol, so
q = (1.39 mol)(9.713 kcal/mol) = 13.5 kcal
step 5: heat the steam from 100 °C to 150 °C: q = mCΔT = (25 g)(0.45 cal/g°C)(50 °C) = 0.563
kcal
Finally, add all the pieces together! Because all processes are endothermic, all heats are
positive.
0.375 kcal + 1.99 kcal + 2.5 kcal + 13.5 kcal + 0.563 kcal = 18.9 kcal
9. A fancy calorimeter with heat capacity 41.7 cal/°C holds 200 mL of water (specific heat 1
cal/g°C). Then 21.8 g of solid C7H7I is added. The crystals sink and do not dissolve in the
water. The calorimeter is heated slowly using electric current, with a constant current (I) of
0.820 A and resistance (R) of 9.05 Ω, raising the temperature from RT to 34.0 °C. a) At 34.0
°C the solid begins to melt. The electrical heater continues to add heat at a slow constant
rate. What happens to the temperature of the water in the calorimeter as the solid melts?
What happens to the temperature of the water when the solid finishes melting? b) Find the
energy in joules added to the system by the heater in 4 min 41 seconds, if q = I2Rt. (Use the
current and resistance given.) Convert this to calories, using 4.18 J/cal. c) What is the molar
heat of fusion of the solid if 4 min 41s is the time it takes to melt the solid?
a) The temperature is constant while the solids melts, because the heat that is added goes to
melting the solid, not increasing the temperature. After it finishes melting, then the
temperature starts to rise again.
b) The time in s is 4 x 60 + 41 = 281 s. We are using SI units, so with seconds everything will
cancel correctly to give us J. q = I2Rt = (0.820 A)2(9.05 Ω)(281 s) = 1.71 kJ, which is 409 cal
c) nΔH = qadded = 409 cal. n = (21.8 g)(1 mol/218.03 g) = 0.100 mol, so 409 cal/0.1 mol = 4.09
kcal/mol. We don’t actually need the heat capacity of the calorimeter and water for this,
since they don’t change temperature while the solid melts.
10. When heated gently, NH4NO3 forms N2O and water. a) Write a balanced reaction. What type
of reaction is this? b) At atmospheric pressure and 25 °C, what volume of N2O do you
expect to be produced from reaction of 12 g of NH4NO3? c) You try this reaction, collecting
the gas over water. The final volume of gas is 3.8 L. Why is this different from your
prediction? d) The vapor pressure of water at 25 °C is 23.8 torr. How many moles of N2O
did you collect?
a) NH4NO3 → 2H2O + N2O, decomposition
b) (12 g)(1 mol/80 g) = 0.15 mol, produces 0.15 mol N2O, so V = nRT/P = (0.15 mol)(62.36
torrL/molK)(298 K)/(760 torr) = 3.7 L
c) You collected the gas over water. There is water vapor mixed into the gas you collected.
d) n = PV/RT = (760 -23.8 torr)(3.8 L)/(62.36 torrL/molK)(298 K) = 0.15 mol
Note: this problem should probably have more sig figs.
11. In the lab, and sometimes in hi-tech kitchens, people use low pressure to remove solvents
like water or alcohol and concentrate solutions. How and why does this work?
The way we get low pressure is by using a vacuum pump. At low pressure, the liquid will boil at
a low pressure. The vapor that is produced will be removed by the vacuum pump, so it won’t
come to equilibrium with the liquid, but will continue to evaporate. However, the solution will
rapidly get cold, because the molecules with the most kinetic energy are being removed. So
usually there is a water bath with a heater to keep the solution warm so it continues to
evaporate.
12. Find the formula (A dark, B light), and the coordination number and geometry of A and B in
the unit cells on the next page.
top left (A red, B gray):
A: 8 x 1/4(edges) + 8 x 1/2 (faces) + 2 = 8
B: 8 x 1/8(corners) + 4 x 1/2(faces) + 1 = 4
A2B
A: distorted T-shape, 3 coordinate
B: distorted octahedral, 6 coordinate
top middle (purple A, green B)
A: 2, B 1, A2B
A: 3, trigonal pyramidal, unless you count the A-A interaction
B: 6, octahedral
top right:
A, blue: 8 x 1/8 (corners) + 1 = 2, 4-coordinate, tetrahedral
B, red: 4, 2 coordinate, linear
AB2
lower left:
yellow: 2,
gray: 1 + 4 x 1/4 = 2
AB
yellow and gray: 4, tetrahedral
lower right:
A, white: 8 x 1/8 + 6 x 1/2 = 4
B, yellow: 8
AB2
white: 8-coordinate
yellow: tetrahedral, 4
(Y-track)
13. Hydrogen bonds are usually linear (X—H - - - Y are linear). Explain why.
We can explain this like any other geometry question. The H makes a bond with X, and also
interacts with a lone pair on Y. Thus, it has 2 electron domains. Since it has 2 electron domains,
it prefers to be linear.
14. If you wanted to make an ionic liquid (an ionic material that is liquid at normal temperatures)
how would you do it? Why? Would an ionic liquid have a high or low boiling point?
An ionic liquid would have a lower boiling point than most ionic substances which are solids.
Thus, it would need weaker forces than most ionic compounds, which means small charges at
big distances. This is usually done by using big ions with low charges, like tetrabutylammonium
[(C4H9)4N+].
15. Some critical point data (K, atm): He (5.19, 2.24); H2 (33.2, 12.8); F2 (144.3, 51.5); CO2
(304.2, 72.8); H2SO4 (927, 45.4); NH3 (405.5, 111.3). a) What does the critical temperature
seem to depend on? Why does this make sense? b) What do the critical pressures depend
on? Why does this make sense?
a) The critical temperature seems to depend on the strength of the intermolecular forces. He
and H2 have very low critical temperatures, because the forces are weak. Sulfuric acid and
ammonia, which have hydrogen bonding, have very high critical temperatures. This makes
sense because above the critical temperature, it’s impossible to make a liquid, because
there’s too much kinetic energy; greater intermolecular forces mean it’s easier to overcome
the kinetic forces and make a liquid.
b) Generally, if the critical temperature is lower, the critical pressure is also lower. However,
sulfuric acid has a low critical pressure, because it is much bigger: doesn’t take as a high a
pressure to force the molecules close together. Likewise, H2 has a higher critical pressure
than He. The have the same number of electrons, and thus similar London forces, but H2 is
bigger.
16. When people make delicate solids, like foams or very thin wafers, by precipitation from
solution, they often break when the solvent evaporates as a result of surface tension. a)
Explain why fragile solids break when water dries on them. b) People often use supercritical
solvents to avoid this problem. Why?
a) The solvent will evaporate until there is only a little left. At this point, cohesive forces in the
solvent might pull it into round droplets. If there are also strong adhesive forces, the
combination of cohesive and adhesive forces might break the solid by pulling it around the
droplet.
b) There aren’t strong cohesive forces in supercritical fluids, because the molecules have too
much kinetic energy to really stick together. Thus, the supercritical fluid can evaporate
without pulling on the solid.
17. Normally hydrogen bonds are asymmetrical (the H appears to have 1 short covalent bond
and 1 dipole-dipole interaction. a) Draw an example of this normal hydrogen bond between
2 molecules. b) Sometimes hydrogen bonds are symmetrical, such as inside the ion
HOOCCHCHCOO—. Try to draw a Lewis structure (the first H is hydrogen-bonded to the
last O—). Draw a resonance structure showing that both O-H distances are the same.
Note: I drew the second using “organic” drawing style, in which the bent points in the lines that
aren’t labelled are carbon, with enough H to have octet. The drawings are above.
18. a) We can measure the size of atoms by looking at distances between atoms in different
types of structures, like ionic or covalent radii. You can also find lists of van der Waals radii,
from the distance between atoms in van der Waals contact. Do you think VDW radii are
bigger or smaller than covalent radii? Why? b) We don’t have VDW radius data for all
elements. Why do you think that is? c) How would you measure VDW radius for Ar? d) How
would you measure VDW radius for Fe?
a) Usually VDW radii are bigger, because covalent bonds pull atoms together harder than VDW
forces, which are weaker. b) We don’t always have good ways to look at particular atoms
making particular bonds. c) This is easy: Ar doesn’t really make any other type of bond. Just
make solid argon, and see how far apart the atoms are (this is not hard to do, by modern
standards). d) This is hard! In Fe metal, there is metallic bonding. There’s no easy way to look at
Fe atoms that are making only VDW bonds, not metallic or covalent or ionic bonds, because
they almost never do this.
19. It seems like geckos (little lizards) climb smooth walls and ceilings using van der Waals
forces. How can the forces be strong enough to hold the gecko up? What must be true
about the geckos’ feet to make this work?
Note: van der Waals forces means both London forces and dipole-dipole forces (all
intermolecular forces). Van der Waals forces aren’t that strong. In order to climb using them, the
geckos feet must be able to maximize the number and strength of the forces, by having very
good contact with the surface. Since the forces are stronger at shorter distances, I guess the
feet are soft enough to completely squish into any roughness on the surface. Probably the feet
also have polar molecules on their skin, because most surfaces are polar also. But it would be
interesting to see how well geckos climb on a surface like Teflon, which is all negatively charged
F atoms!
20. a) Draw a diagram showing the temperature (y-axis) vs the heat added (x-axis) for a sample
of water from -10 °C to 110 °C. b) Draw a diagram showing the heat added (y-axis) vs the
temperature (x-axis) for a sample of water from -10 °C to 110 °C. c) Draw a diagram
showing the heat capacity of water (y-axis) vs temperature (x-axis) a sample of water from
-10 °C to 110 °C.
a) Same standard heating diagram
b) rotate it around the line y=x, so now there are upward diagonal parts and vertical parts.
c) This is the derivative (slope) of the previous graph. Each diagonal part turns into a horizontal
line, because the slope is constant; at the phase changes, the heat capacity jumps to a new
value, with a spike in between. When the phase changes, the slope is undefined (infinity)
because we can add a lot of heat without changing the temperature.