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CHEM 122
General Chemistry
Summer 2014
Name:
Midterm Examination 1
Multiple Choice Questions
1.
If a solution of Water (H2O) and Methyl Amine (CH3NH2) were to form, the
dominant Intermolecular Force between molecules of Water and Methyl Amine
would be:
a)
b)
c)
d)
2.
A solution prepared by mixing liquid Pentane (C5H12) and liquid Octane (C8H18) is
expected to have an Enthalpy of Solution that is:
a)
b)
c)
d)
e)
3.
large and exothermic (Hsoln << 0)
small and exothermic (Hsoln < 0)
about zero (Hsoln ~ 0) *****
small and endothermic (Hsoln > 0)
large and endothermic (Hsoln >> 0)
Hpent ~ small (LDF)
Hoctan ~ small (LDF)
Hmix ~ small (LDF)
Hsoln ~ small + small - small ~ 0
The Henry's Law Constant for O2 in Water is kO2 = 0.00128 M/atm. How many
grams of O2 can dissolve in 2.00L Water if the partial pressure of O2 over the Water
is 0.20 atm? (Hint: First calculate the solubility of the O2 in the Water.)
a)
b)
c)
d)
4.
LDF
Dipole-Dipole
H-Bonding *****
Ion-Dipole
0.123g
0.105g
0.063g
0.016g *****
SO2 = kO2 PO2 = (0.00128 M/atm) (0.20 atm)
= 2.56 x 10-4 M
#g = (2.56 x 10-4 M) (2.00 L) (32 g/mol) = 0.016g
When a diver dives to great depths, an excess of gaseous Nitrogen will dissolve in
his blood. If he surfaces too rapidly, he can suffer from the Bends. This is because
his blood is
with respect to Nitrogen.
a) saturated
b) unsaturated
c) super-saturated *****
5.
A solution is prepared by mixing 3.50g Naphthalene (moth-balls, C10H11) into 29.0g
of Benzene (a hydrocarbon liquid, C6H6). What is the mole fraction of Benzene in
this solution?
a)
b)
c)
d)
6.
0.978
0.933 *****
0.871
0.711
# mol Nap = (3.50 g) (1 mol Nap / 131g) = 0.0267 mol Nap
# mol Ben = (29.0 g) (1 mol Ben / 78g) = 0.3718 mol Ben
xBen = 0.3718 / (0.3718 + 0.0267) = 0.933
Brass is a solution of Copper and Zinc. Red Brass is approximately 20% Zinc.
What is the molality of this solid solution?
a)
b)
c)
d)
15.3 m
4.7 m
3.8 m *****
0.25 m
Assume 100 g Soln: 20g Zn & 80g Cu
# mol = 20g / (65.4 g/mol) = 0.306 mol
molality = 0.306 mol / (0.80 kg) = 3.8 m
7.
While cooking, you add a "pinch" of Table Salt, NaCl, to a quart of water
simmering in a sauce pan. Take the "pinch" of salt to be about 0.1 gram and the
amount of water to be close to 1 L; ~1 kg. What is the elevation in the boiling point
of this solution? For water, Kb = 0.521 oC/m.
a)
b)
c)
d)
1.003oC
0.1291oC
0.0214oC
0.0018oC *****
NaCl(aq)
Na+(aq) + Cl-(aq)
#mol NaCl = (0.1 g) (1 mol NaCl / 58.5g) = 0.00171 mol NaCl
m = (0.00171 mol) / (1 kg) = 0.00171 m
Tb = Kbm = (2) (0.521 oC/m) (0.00171 m) = 0.0018oC
8.
Milk is a colloidal emulsion, meaning it is a
phase.
a)
b)
c)
d)
9.
gas; liquid
liquid; solid
liquid; liquid *****
solid; liquid
dispersed into a
Liquid Fat globules dispersed in an Aqueous Sol'n
As a pan of water is heated, the concentration of the dissolved gases will:
a) increase.
b) decrease. *****
c) remain fairly constant.
10. Hydrogen reacts explosively with Oxygen to produce Water
2 H2(g) + O2(g)
2 H2O(g)
The Rate of Consumption of O2 is:
a)
b)
c)
d)
11.
one quarter that of H2.
one half that of H2. *****
twice that of H2.
four times that of H2.
Rate = - (1/2) [H2] / t = - [O2] / t
Chloroform reacts with Chlorine to form Carbon Tetrachloride:
CHCl3 + Cl2
CCl4 + HCl
The rate law for this reaction is:
Rate = k [CHCl3] [Cl2]1/2
If the concentration of the Chlorine is increased by a factor of four, the reaction rate
will increase by a factor of:
a)
b)
c)
d)
2. *****
4.
8.
16.
Rate ~ 41/2 x = 2 x
12. The decomposition of Nitramide occurs according to first-order kinetics with a rate
constant of 0.00563 min-1:
NH4NO2aq)
N2O(g) + H2O(l)
How long will it take for a solution that is 0.105M Nitramide to decompose to
0.041M?
a)
b)
c)
d)
23.4 min
97.1 min
167 min *****
245 min
[Nit] = [Nit]o exp(-kt)
(0.041 M) = (0.105 M) exp( - (0.00563 min-1) t)
t = 167 min
13. The following reaction is Second Order:
2 NO2(g)
2 NO(g) + O2(g)
with a rate constant k = 0.543M-1sec-1. If we start with [NO2]o=0.01M, how long
will it take for the concentration of the reactant to be cut in half?
a)
b)
c)
d)
2892 sec
593 sec
184 sec *****
34 sec
1 / [N2O5] = 1 / [N2O5]o - kt
1 / (0.005 M) = 1 / (0.01 M) - (0.543 M-1sec-1) t
t = 184 sec
14. The rearrangement of Methy Isonitrile has been extensively studied:
CH3-NC(g)
CH3-CN(g)
Kinetic data at 189.7oC is plotted below.
The Rate Constant k for this reaction is:
a)
b)
c)
d)
0.0002 min-1
0.0015 min-1 *****
0.1133 min-1
0.3341 min-1
k = - slope
15. The bromination of nitric oxide:
2 NO + Br2
2 NOBr
occurs according to the following rate law:
Rate = k [NO]2 [Br2]
The overall Order of this reaction is:
a)
b)
c)
d)
1st
2nd
3rd *****
4th
order = 2 + 1 = 3
Short Answer
1.
Initial rate data is provided for the following reaction:
2 NO + H2
[NO]o (M)
6.4 x 10-3
12.8 x 10-3
6.4 x 10-3
Trial #1
Trial #2
Trial #3
N2O + H2O
[H2]o (M)
2.2 x 10-3
2.2 x 10-3
4.4 x 10-3
rateo (M/sec)
2.60 x 10-5
10.4 x 10-5
5.20 x 10-5
Determine the Reaction Order with respect to NO and H2. Write the resulting Rate
Law for this reaction. (Show your work.)
Assume
Rate = k [NO]m [H2]n
Then
Rate2 / Rate1 = ([NO]2 / [NO]1)m
(10.4 / 2.60) = (12.8 / 6.4)m
4 = 2m
m = 2
Rate3 / Rate1 = ([H2]3 / [H2]1)n
(5.20 / 2.60) = (4.4 / 2.2)n
2 = 2n
n = 1
Finally,
Rate = k [NO]2 [H2]
2.
For the decomposition of N2O5, we have:
2 N2O5
4 NO2 + O2
Time (sec)
0
600
1200
1800
[N2O5] (M)
1.65 x 10-2
1.24 x 10-2
0.93 x 10-2
0.71 x 10-2
Calculate the reaction rate during the first and last time intervals. Comment on your
results.
Rate = - (1/2) [N2O5] / t
1st Interval
Rate = - (1/2) ((1.24 - 1.65) x 10-2 M) / (600 - 0 sec)
= 3.42 x 10-6 M/sec
Last Interval
Rate = - (1/2) ((0.71 - 0.93) x 10-2 M) / (1800 - 1200 sec)
= 1.83 x 10-6 M/sec
The rate has decreased as expected when the concentration of the
reactants decreases.
3.
Amnesic Shellfish Poisoning is a result of eating shell which contain high levels of
Domoic Acid. How were the harvesters of Blue Muscles on Prince Edward Island
advised to avoid raising shellfish with high levels of Domoic Acid and why does
this remedy work?
Sol'n: Wait a sufficient amount of time after a Red Tide before
harvesting the shellfish.
Domic Acid is very hydrophillic. So it will not absorb into the fleshy
parts of the muscle. It will instead be expelled from the digestive
tract of the muscle with the water.
4.
Polymers are high molecular weight molecules. 6.053g of poly(Vinyl Alcohol) in a
100.0mL solution has an Osmotic Pressure of 0.272 atm at 25oC.
a)
What is the Molarity of this solution?
C =  / RT = (0.272 atm) / (0.08206 L atm/K mol) (298.15 K)
= 0.011 M
b) What is the number of mole poly(Vinyl Alcohol) dissolved in this solution?
# mol = (0.0111 M) (0.1000 L) = 0.00111 mol
c) What is the MW of the polymer?
MW = 6.053 g / 0.00111 mol = 5.4 x 103 g/mol
5.
Why do we expect the Rate of a chemical reaction to decrease as the reaction
proceeds? Why might this not happen?
As the molecules get further apart, because they are more dilute, they collide
less frequently and therefore should react more slowly.
This may not happen if the reaction relies on some hidden variable or if
the molecules collide and react in a way not explained by the overall
reaction stoichiometry.
Appendix
Constants
NA
= 6.022045 x 1023 entities/mole
kB
= 1.380662 x 10-23 J/K
c
= 2.99792458 x 108 m/sec
h
= 6.626176 x 10-34 J sec
R
= 8.314 J/K mol
= 0.08206 L atm/K mol
1 amu = 1.6605 x 10-24 g