1.
Explain
the structure of a hydrogen bond.
The electrons of an electronegative atom are attracted
to a partially positively charged H which is bonded to another
electronegative atom.

2.
Write
the Lewis structures and give the shapes of each of the following molecules;
predict which substance of each pair has the higher boiling point. Explain your reasoning.
a)
Ethanol,
CH3CH2OH or dimethyl ether, CH3-O-CH3?
Ethanol can hydrogen bond therefore the
intermolecular forces are stronger and it has a higher boiling point.
b)
Butane,
C4H10 or octanol, C8H18?
Octanol has a higher molecular
weight and therefore is has stronger dispersion forces (
c)
PF3
or PCl3?
Because F is more electronegative than Cl, PF3
is more polar and has stronger intermolecular forces therefore it has a
higher boiling point.

d)
SO2
or CO2?
e)
See
question below. Polar SO2 has
stronger intermolecular forces than nonpolar CO2.
f)
Extra
question: Explain why sulfur dioxide, SO2,
has a dipole moment of 1.63D, but that of CO2 is zero.
SO2 is bent therefore it has a net dipole
moment towards the oxygens. CO2
is linear and the two bond dipoles cancel.
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3.
The
phase diagram of an unknown substance is shown below:

a)
Label
the solid liquid and gas regions of the phase diagram.
b)
Label
the triple point and the critical point.
c)
Label
the normal freezing and boiling points.
There is not a normal boiling point
or freezing point as the picture is drawn.
The normal sublimation point is the point where there is a transition
between the solid and the gas phase at 1 atm pressure.
d)
Which
phase is more dense the liquid or the solid?
Explain your reasoning.
The liquid is more dense.
4.
Explain
why CH3OH is completely soluble in water but not completely soluble
in toluene, C6H5CH3.
CH3OH is polar so it
dissolves easily in polar solvent water but not so easily in nonpolar toluene.
5.
A
room in which the humidity has been lowered feels cooler. Yet the dehumidifier has not “cooled” the
room. Why is that?
As the humidity is lowered, more water can evaporate before the air is
again saturated with water. The
evaporation of water is an endothermic process which means that the heat is
drawn out of the room or the room is cooled as the water evaporates.
6.
How
might one go about establishing whether a solution of sodium sulfate in water
is saturated, unsaturated, or supersaturated?
Add a little more sodium sulfate.
If the solution is unsaturated, the salt will dissolve, if it is
saturated the salt will settle on the bottom of the solution, and if the
solution is supersaturated the addition of additional salt will cause the
excess salt to precipitate out of solution.
7.
A
10.7 m solution of NaOH has a density of 1.33 g/mL at 20oC. Calculate
a)
The
mole fraction of NaOH

b)
The
mass percentage of NaOH

c)
The
molarity of the solution

8.
A
solution of glucose (C6H12O6) is prepared by
dissolving 100.0 g of glucose in 1000. g of water. The density of the resultant solution is
1.050 g/mL. Kb for water is 0.52 oC/m
and kf for water is –1.86 oC/m.

a)
What
is the vapor pressure of the solution at 100.0oC?
Vapor pressure = (Xwater)(Pwater) =
(0.9901)(760 torr) = 752 torr
b)
What
is the boiling point of the solution?
rTb = kb•m = (0.52 oC/m)(0.5549
m) = 0.29oC
Boiling point = 100.29oC
c)
What
is the osmotic pressure of the solution at 25oC?
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9.
A
compound of carbon, hydrogen, and oxygen was burned in oxygen, and 1.000 g of
the compound produced 1.434 g CO2 and 0.783 g H2O. In another experiment, 0.1107 g of the
compound was dissolved in 25.0 g of water.
This solution had a freezing point of -0.0894oC. What is the molecular formula of the
compound?

10.
The
osmotic pressure of blood at 37oC is 7.7 atm. A solution that is given intravenously must
have the same osmotic pressure as the blood.
What should be the molarity of a glucose solution to give an osmotic
pressure of 7.7 atm at 37oC?

11. For the equilibrium C(s)
+ 2 H2(g) <==>
CH4(g) + heat
a)
Write
the equilibrium constant expression as Kc. ![]()
b) What are the units of the
equilibrium constant Kp?
(pressure) -1
c) How is Kc related to Kp (specify
quantitatively)
Kc = Kp(RT)
d) For each of the following changes to
the system at equilibrium, predict the direction of the shift and explain why
it occurs:
|
Change |
Shift (left or right) |
Reason |
|
The volume of the reaction vessel is doubled. |
left |
volume change causes P to
drop. Shift left makes more mols of
gas and thereforea higher pressure |
|
The temperature is increased. |
left |
shifts left to use up heat |
|
The pressure of H2(g) is increased. |
right |
shifts to the right to decrease
the moles of H2. Reaction
rate to right increases, to the left stays the same so there is a net shift
tothe right. |
|
C(s) is added to the system. |
no change |
The concentration of C stays the
same. |
|
Adding a catalyst |
no change |
a catalyst does not affect the
positin of the equilibrium. (So then,
just what does it do?) |
12. The equilibrium constant for the
reaction N2O4(g)
<==> 2 NO2(g) is 0.212 mol/L at 100o C. What is the value of Kc at 100oC
for:
a) 2 NO2(g) <==>
N2O4(g)
reaction is reversed
so Krxn = 1/(Kc) = 1/.212 = 4.72
b) NO2(g) <==>
1/2 N2O4(g)
reaction is halved and
reversed so Krxn =
= 2.17
13. For the reaction NO(g) + NO2(g) + H2O(g) <==>
2 HNO2(g), occurring at 28oC, [NO]i = [NO2]i = 44.1 torr and [H2O]i = 17.5 torr. If the total
pressure at equilibrium is 95.6 torr.
a) What are the equilibrium pressures
of all species?
NO(g)
+ NO2(g) + H2O(g) <==> 2 HNO2(g) total pressure
D -x
-x -x +2x -10.1 torr
final pressure = P(NO)
+ P(NO2) + P(H2O) +
P(HNO2)
95.6 torr
= (44.1-x)torr + (44.1-x)torr + (17.5-x)torr + (2x)torr
x = 10.1 torr
pressure change = -10.1 torr = -x + -x + -x + +2x = -x
x = 10.1 torr
P(NO) = 44.1 torr - 10.1 torr = 34.0 torr
P(NO2)
= 44.1 torr - 10.1 torr = 34.0 torr
P(H2O) =
17.5 torr - 10.1 torr = 7.4 torr
P(HNO2) =
2(10.1 torr) = 20.2 torr
b)
Calculate
Kp for the reaction?

14. At some temperature the system 2 SO2(g) + O2(g) <==>
2 SO3(g) is at
equilibrium when [SO2] =
0.0100 M, [O2] =
0.200 M and [SO3] = 0.100 M.
What is the value of Kc at this temperature?
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Challenge
If at the same temperature, 3.00 mol
of SO3 is added to an empty 1.00 L vessel, what will be the value of
[O2] at equilibrium? (hint:
check your answer - iterate to within 5%)
2 SO2(g)
+ O2(g) <==>
2 SO3(g)
Initial 0 0 3.00 M
D +2x
+x -2x
Equilibrium 2x x 3.00-2x
1st
iteration

x = 0.165 M
check answer: ![]()
the
assumption that 2x << 3.00 is incorrect.
2nd
iteration

x = 0.153 M
3rd iteration

x = 0.154 M
4th iteration

x = 0.154 M
x’s have converged we are finished
[SO2] =
2(0.15 M) = 0.30 M
[O2] = 0.15 M
[SO3]
= {3.00 - 2(0.15)}M = 2.70 M
15. The reaction 2 NO(g)
+ Br2(g) <==>
2 NOBr(g) has a Kp =
1.17 atm-1 at 25oC. If 1.10 atm of NOBr, 0.100 atm of NO, and
0.0100 atm of Br2 are mixed at 25oC, what reaction will
occur? Explain.

a) When 5.00 atm of NOBr is allowed to
equilibrate at 50oC, the equilibrium pressure of NOBr is measured to
be 4.30 atm. What is the value of Kp at
50oC? Compare with the value
of Kp at 25oC and explain.
2
NO(g) +
Br2(g) <==> 2 NOBr(g)
Initial 0 atm 0 atm 5.00 atm
D +2x +x
-2x
Equil 2x x 4.30 atm x = 0.35 atm
= 0.70 atm =
0.35 atm
![]()
The Kp increases as the temperature
is increased. This means the reaction
shifts to the right when heat is added so it is an endothermic reaction.
16. Assume that you place a fresh water
plant into a salt solution and examine it under a microscope. What happens to the plant cells? What if you placed a salt-water plant in pure
water? Explain. Draw pictures to illustrate your explanation.
Assume that you place a fresh water plant into a salt solution and examine it
under a microscope. What happens to the plant cells? What if you placed a salt-water plant in pure
water? Explain. Draw pictures to illustrate your explanation.
A fresh water plant in salt water
will lose water from the plant cells to dilute the salt solution and all of the
cells will shrivel.
A salt water plant in pure water
will draw water into the cells to dilute the salt in them. The cells will
burst.
17. Pure
iodine (100.0 g) is dissolved in 300.0 g of CCl4 at 65oC. Given that the vapor pressure of CCl4
at this temperature is 504 mm Hg, what is the vapor pressure of the CCl4/I2
solution at 65oC? (Assume
that I2 does not contribute to the vapor pressure.)
