Chemistry 2000
Problem Set #3: Chapter 13
Answers to Practice Problems
1.
In lab, we often rinse wet glassware with acetone to remove the water then use a stream
of air to evaporate off the acetone. The structure of acetone and vapour pressure curves
for acetone and water are shown below.
Vapour Pressure Curves for Acetone and Water
Equilibrium Vapour Pressure (mmHg)
800
700
600
500
Acetone
Water
400
300
200
P20'C(acetone) = ~178 mmHg
100
0
-80
-60
-40
-20
P20'C(H2O) = 17 mmHg
(a)
0
20
40
o
Temperature ( C)
60
80
100
120
Tobp(acetone) = 56.5oC Tobp(H2O) = 100oC
What type(s) of intermolecular forces are responsible for water’s solubility in acetone?
dipole-dipole interactions (including hydrogen bonding between H of water and O of
acetone)
(b)
What are the normal boiling points of acetone and water? see graph above
(c)
What are the vapour pressures of acetone and water at room temperature (20 ˚C)? see
graph above
(d)
Briefly, justify the relative boiling points and vapour pressures of water and acetone.
Water molecules can hydrogen bond with each other. Acetone has no Hδ+ so acetone
molecules cannot hydrogen bond with each other. As such, the intermolecular forces
between water molecules are stronger than the intermolecular forces between acetone
molecules. If the intermolecular forces are stronger, fewer molecules will have enough
energy to break through the surface tension and escape the liquid phase. Thus, the
vapour pressure of water is lower. On the same note, the molecules of water will require
more kinetic energy (higher temperature) to have enough evaporate that the vapour
pressure is equal to atmospheric pressure. i.e. The boiling point of water is higher.
(e)
Why does blowing a stream of air over acetone-wet glassware accelerate evaporation?
The liquid acetone and acetone vapour exist in equilibrium. The stream of air reduces the
vapour pressure of acetone, so more acetone must evaporate to reach the equilibrium
vapour pressure. In other words, the air pushes the acetone vapour away, so more
acetone has to evaporate to restore the equilibrium.
(f)
Why is blowing a stream of air over water-wet glassware much less effective?
Water has a much lower equilibrium vapour pressure than acetone. As such, there are
fewer water vapour molecules to push away and fewer water molecules need to evaporate
to restore the equilibrium. This method would eventually work but it might take hours!
(g)
If you need dry glassware for the following week’s lab, is this method of drying
necessary? Why or why not?
No, it isn’t necessary. As long as the wet glassware is not stored in a sealed environment,
the water (or acetone) will slowly evaporate until the glassware is dry. (See lecture notes
for an explanation of this phenomenon.)
2.
Another common lab technique is to vacuum filter using a Büchner
funnel. A good aspirator is able to reduce the pressure in the receiving
flask to ~20 mmHg if the apparatus is sealed properly. In the first
experiment in Chemistry 2000, you filtered yellow crystals out of water,
rinsed with water then acetone and left the vacuum on for ~5 minutes.
When you took apart your apparatus, you may have noticed that the
receiving flask felt a bit cold. (You may want to use data from the
vapour pressure curve on the previous page in your answers.)
(a)
In what phase is acetone at 20 ˚C and 20 mmHg?
gas
(b)
Why would the receiving flask feel cold?
Evaporation is an endothermic process. As the higher-energy molecules evaporate, the
average kinetic energy of the remaining molecules is reduced. This reduces the
temperature of the filtrate, making it feel cold.
(c)
Why did your lab instructor allow you to dispose of the filtrate down the sink rather than
into the acetone waste?
After 5 minutes under vacuum, most of the acetone would have evaporated. This would
leave an aqueous solution which was environmentally safe to put down the sink.
3.
Iodine is the only naturally occurring halogen that is a solid at room temperature.
(a)
What type of solid is iodine?
I2 is a molecular solid.
(b)
Why does iodine have a higher melting point than bromine?
I2 is more polarizable than Br2. (i.e. It is easier to make an induced dipole in I2 than Br2.)
This is because the iodine atoms are larger so the valence electrons are more diffuse
(“spread out”) and more shielded from the positive charge of the nucleus. Thus, it is
easier to ‘shift’ their position, generating a temporary dipole.
4.
What conditions would be necessary to make supercritical water? (see Table 13.5 in text)
To make supercritical water,
1. heat water above the critical temperature (374 ˚C) and, at the same time,
2. compress it to a pressure higher than the critical pressure (217.7 atm).
5.
The enthalpy of vaporization of butane is 22.4 kJ/mol. If the normal boiling point of
propane is –0.5 ˚C, calculate the vapour pressure of butane at room temperature (20 ˚C).
Step 1: Collect Data
∆H˚vap = 22.4 kJ/mol
If normal boiling point = -0.5 ˚C then vapour pressure at -0.5 ˚C is 760 mmHg
P1 = 760 mmHg
T1 = -0.5 ˚C = 272.6 K (keep 272.65 K in calculator)
Looking to find vapour pressure at 20 ˚C
P2 = ???
T2 = 20 ˚C = 293 K (keep 293.15 K in calculator)
Step 2: Solve for ln(P2/P1)
ln(P2/P1) = ∆H˚vap ( 1 – 1 )
R
T1 T2
= ∆H˚vap ( 1 – 1 )
R
T1 T2
= ∆H˚vap ( 1 – 1 )
R
T1 T2
=
(22.4 kJ/mol)
× (
1
1
) × 1000 J
-1 -1
8.3145 J mol K
272.6 K
293 K
1 kJ
ln(P2/P1) = 0.691
***3 sig .fig.***
Step 3: Solve for P2
ln(P2/P1) = 0.691
recall that eln(x) = x
P2/P1 = e0.691
P2 = P1 × e0.691
= (760 mmHg) × e0.691
P2 = 1.52 × 103 mmHg
***3 sig .fig.***
Note that it is not necessary to do steps 2 and 3 separately. I have done so here for clarity.