1
Chemistry 1011
Midterm Test #1
October 22, 2009
Name:_____________________________ MUN # ____________________________
Section A – Shorter Problems (19 marks total)
[4] A1.
A solution at 25.0 °C containing 0.8330 g of a polymer of unknown structure in
170.0 mL of an organic solvent was found to have an osmotic pressure of 5.20
mmHg. What is the molar mass of the polymer?
⎞⎟
5.20 mmHg⎛⎜1 atm
760
mmHg
Π
⎝
⎠
=
= 2.797 x 10-4 mol ⋅ L−1
Π = MRT so M =
0.08206 L ⋅ atm ⋅ K -1 ⋅ mol-1 (298.15 K )
RT
(
M=
)
(
)
moles
so moles = MV = 2.797 x 10 -4 x 10 -4 mol ⋅ L−1 (0.1700 L ) = 4.754 x 10 -5 mol
volume
molar mass =
mass
0.8330 g
=
= 1.75 x 10 4 g ⋅ mol−1
-5
moles 4.754 x 10 mol
The molar mass of the polymer is 1.75 x 104 g⋅mol-1.
[4] A2.
Arrange the following in order of decreasing boiling point: RbF, CO2, CH3OH,
and CH3Br. Explain your reasoning.
RbF is an ionic solid where there are very strong electrostatic attractions between the
Rb+ ions and the F- ions. To see boiling, we must overcome these very strong
interactions, which will only occur at very high temperatures. This compound has the
highest boiling point.
The other three compounds are molecular compounds, who’s boiling points are
determined by the strength of the intermolecular forces. CH3OH has hydrogen bonding,
dipole-dipole forces, and dispersion forces. CH3Br has dipole-dipole forces, and
dispersion forces. Finally, CO2 is non-polar and only has dispersion forces. Since
hydrogen bonding tends to be stronger than dipole-dipole forces, which themselves are
stronger than dispersion forces, the boiling points of these compounds decrease in the
order given above.
[3] A3.
The strength of London dispersion forces tends to depend on two separate factors.
Name these factors and briefly describe how these factors affect the strength of
the dispersion forces.
London dispersion force strengths essentially depend on how easily a molecule can
create a temporary induced dipole, which is related to how easily the electrons can move
to form an asymmetric distribution. The first factor that affects this is the polarizability
of the molecule or species. A more polarizable molecule can more easily redistribute the
electrons. Larger molecules (more electrons) tend to be more polarizable because more
electrons are found further from the nuclei and therefore and more easily move.
The second factor that affects dispersion forces is the shape of a molecule. More
compact symmetrical molecules are less likely to have contact with each other compared
to long chain-like molecules. More contact between molecules translates into stronger
dispersion forces
2
A4.
Complete the following table. (8 marks)
Species
“Best” Lewis dot
structure
(show all formal charges
and show all resonance
structures, if any exist)
Name for electron group
distribution
ICl4-
N3-
-1 formal charge on the I
octahedral
linear
Name of molecular
geometry
square planar
linear
Hybridization of the
central atom
sp3d2
sp
Sketch of molecular
geometry
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Section B – Longer Problems – (14 marks total) – Write legibly and show all work
for all steps in the problem. For mathematical relationships show both the
equations used and substitution of values into the equation. Maintain correct units
throughout your calculations and report your answers with the correct significant
figures.
B1.
[4]
The reaction 2 NO (g) + 2 H2 (g) → N2 (g) + 2 H2O (g) was studied at 904 °C, and
the following rate of appearance of nitrogen data were obtained:
(a)
[NO], M
[H2], M
Rate of appearance of N2, M⋅s-1
0.420
0.122
0.136
0.210
0.122
0.0340
0.210
0.244
0.0680
0.105
0.488
0.0340
Determine the order of reaction for each reactant and the overall reaction
order.
We see that experiments 1 and 2 have only a halving of the [NO] which results in
a rate that is approximately ¼ which means the reaction is second order in NO.
Experiments 2 and three see a doubling of [H2] which shows a doubling of the
rate, so the reaction is first order with respect to H2. The reaction is third order
overall.
[1]
(b)
Write the rate law for the reaction.
rate = k [NO]2 [H2]
[2]
(c)
Calculate the rate constant of the reaction at 904 °C.
Using the data from experiment 1
k = rate / ([NO]2 [H2])
k = 0.136 M⋅s-1 / ([0.420 M]2 [0.122 M])
k = 6.32 M-2⋅s-1
[1]
(d)
Calculate the rate of appearance of N2 when [NO] = 0.350 M and [H2] =
0.205 M.
rate = k [NO]2 [H2] = (6.32 M-2⋅s-1) ([0.350 M]2 [0.205 M]) = 0.159 M⋅s-1
4
[6] B2.
Balance the following redox skeleton equation in basic solution:
H2S (g) + NO3- (aq) → NO2 (g) + S8 (s)
The first step is assigning oxidation numbers
H 2 S (g) + N O3- (aq) ⎯
⎯→ N O 2 (g) + S8 (s)
+1 -2
+5
-2
+4
-2
0
The oxidation and reduction skeleton half-reactions are
⎯→ N O 2 (g)
oxidation H 2 S (g) ⎯
⎯→ S8 (s) and reduction N O3- (aq) ⎯
+5
+4
−2
0
14444244
443
144424443
each N gains 1 electron
each S loses 2 electrons
The reduction half-reaction is already balanced in terms of nitrogen, but the oxidation
reaction isn’t balanced for S. To balance for S
oxidation 8 H 2S (g) ⎯
⎯→ S8 (s)
We can now balance both half-reactions for oxygen by adding water
8 H 2S (g) ⎯
⎯→ S8 (s) and NO 3- (aq) ⎯
⎯→ NO 2 (g) + H 2 O (l)
We can now balance for hydrogen by adding H+
8 H 2S (g) ⎯
⎯→ S8 (s) + 16 H + (aq) and NO 3- (aq) + 2 H + (aq) ⎯
⎯→ NO 2 (g) + H 2 O (l)
We can now balance for charge by adding e8 H 2S (g) ⎯
⎯→ S8 (s) + 16 H + (aq) + 16 e - and
NO 3- (aq) + 2 H + (aq) + 1 e - ⎯
⎯→ NO 2 (g) + H 2 O (l)
Make the number of e- match by multiplication
8 H 2S (g) ⎯
⎯→ S8 (s) + 16 H + (aq) + 16 e - and
16 NO 3- (aq) + 32 H + (aq) + 16 e - ⎯
⎯→ 16 NO 2 (g) + 16 H 2 O (l)
Add the half-reactions together
16 NO 3- (aq) + 32 H + (aq) + 16 e - + 8 H 2S (g) ⎯
⎯→ 16 NO 2 (g) + 16 H 2 O (l) + S8 (s) + 16 H + (aq) + 16 e -
Simplify by cancellation and collection
16 NO3- (aq) + 16 H + (aq) + 8 H 2S (g) ⎯
⎯→ 16 NO 2 (g) + 16 H 2 O (l) + S8 (s)
To further balance for a basic solution we add enough OH- to both sides to convert H+ to
water
16 NO 3- (aq) + 16 H + (aq) + 16 OH - (aq) + 8 H 2S (g) ⎯
⎯→ 16 NO 2 (g) + 16 H 2 O (l) + S8 (s) + 16 OH - (aq)
1444424444
3
becomes 16 H 2O (l)
The 16 H2O on both sides cancel each other out, so the correct balanced equation is
16 NO 3- (aq) + 8 H 2S (g) ⎯
⎯→ 16 NO 2 (g) + S8 (s) + 16 OH - (aq)
The End