FORM B
Name KEY
CHEM 1474
Test #1
Spring 2007
Buckley
Answer the first twenty questions on your Scantron sheet. Be sure to include your name
on both the Scantron and this test paper. Include your form letter (A or B) on the
Scantron. My answers are in bold-faced italic underline.
1.
Which intermolecular forces would be operative in CH3OCH3?
(12 of 31 correct)
a.
b.
c.
d.
e.
2.
Only dispersion forces
Only dispersion forces and hydrogen bonding
Only dipole-dipole and hydrogen bonding
Only dispersion forces and dipole-dipole
(Dispersion forces as always, the molecule is polar, but the H is not
attached to N, O, or F so no hydrogen bonding.)
Dispersion, dipole-dipole, and hydrogen bonding forces
Which of the following molecules would have a linear molecular geometry?
(10 of 31 correct)
CO2
a.
b.
c.
d.
e.
3.
SO2
HCN
HF
CO2 only
CO2 and SO2 and HF only
CO2 and HF only
HCN and HF and SO2 only
CO2 and HCN and HF only
(SO2 has nonbonded pairs, causing an angular molecular geometry.)
Which of the molecules in question 2 are nonpolar?
(14 of 31 correct)
a.
Only CO2
b.
Only HF and HCN
c.
Only SO2 and CO2
d.
Only CO2 and HF
e.
Only HF and SO2
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4.
Of the following compounds, which would you expect to have the highest boiling
point? (25 of 31 correct)
CH3CH2CH3
a.
b.
c.
d.
e.
5.
CH3CH2CH2CH3
CH3CH2OH
CH3CH2CH3
CH3CH2CH2CH3
CH3CH2OH
(First and second only have dispersion – third has dipole-dipole and
hydrogen bonding.)
CH3CH2OH and CH3CH2CH3 would be tied for the high boiling point
It is impossible to predict from the information given.
Consider the following four aqueous solutions. Which would have the lowest
freezing point? (12 of 31 correct)
0.10 m BaCl2, 0.20 m NaCl, 0.15 m Ba(NO3)2, 0.10 m C12H22O11 (sugar)
(Consider molality of particles, in order: 0.30 m, 0.40 m, 0.45 m, 0.10 m)
a.
b.
c.
d.
e.
6.
As the temperature of a solution rises,
(16 of 31 correct)
a.
b.
c.
d.
e.
7.
0.10 m BaCl2
0.20 m NaCl
0.15 m Ba(NO3)2
0.10 m C12H22O11
0.10 m BaCl2 and 0.15 m Ba(NO3)2 would be tied for the lowest
the solubility of a solid typically decreases and that of a gas increases
the solubility of a solid typically increases and that of a gas decreases
the solubility of solids and gases typically increases
the solubility of solids and gases typically decreases
the solubility of solids typically increases but that of gases is unaffected
A hydrophobic colloid is a colloidal particle that is:
(31 of 31 correct)
a.
repelled by water
b.
attracted to water
c.
repelled by whatever solvent it is in
d.
attracted by whatever solvent it is in
e.
too small to dissolve in the solvent
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8.
A colloid is coagulated to:
(23 of 31 correct)
a.
make it more soluble in the solvent
b.
decrease the particle size
c.
increase the particle size
d.
make it water soluble
e.
decrease the vapor pressure of a solution
9. An enzyme is suspected of having a molecular weight on the order of 16000 amu.
Which of the colligative properties would be most appropriate for accurately
determining the molecular weight of the enzyme?
(12 of 31 correct)
a.
freezing point depression determination
b.
boiling point elevation determination
c.
measurement of vapor pressure lowering
d.
osmotic pressure determination
(It is difficult to get high enough concentrations with high molecular
weight solutes to attain measurable changes in colligative properties
except for osmotic pressure.)
e.
they would all be equally appropriate
Consider the phase diagram below for questions 10 and 11.
1.5
b
a
P (atm)
1.0
c
z
0.5
-30
-20
-10
0
10
20
Temperature (EC)
30
40
10. The region in the phase diagram that corresponds to the liquid phase is:
(23 of 31 correct)
a.
a
b.
b
c.
c
d.
z
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11. The normal boiling point of the substance depicted in the phase diagram is:
(5 of 31 correct)
a.
-30 EC
b.
-10 EC
c.
-7 EC
d.
0 EC
e.
16EC
(The normal boiling point occurs when the vapor pressure – indicated
along the vapor-liquid line between b and c – equals 1 atmosphere.)
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12. Which of the following substances is more likely to dissolve in methanol, CH3OH?
(21 of 31 correct)
a.
CCl4
b.
Kr
c.
N2
d.
CH3CH2OH
(Like dissolves like. Methanol and CH3CH2OH
are both polar and have hydrogen-bonding.)
e.
H2
13. The magnitudes of Kf and Kb depend on the identity of the __________
(12 of 31 correct)
a.
solute
b.
solvent
c.
solution
d.
solute and solvent
e.
solute and temperature
14.
The definition of molarity is:
(26 of 31 correct)
a. moles solute/L solvent
b. moles solute/moles solvent
c. moles solute/moles solution
d. moles solute/L solution
e. moles solute/kg solvent
15.
The definition of molality is:
(30 of 31 correct)
a.
moles solute/L solvent
b.
moles solute/moles solvent
c.
moles solute/moles solution
d.
moles solute/L solution
e.
moles solute/kg solvent
16.
A reaction is found to be first order in A and first order overall. An increase in
the A concentration from 1.0 M to 1.5 M will result in an increase in rate from 2.0
x 10-2 M s-1 to __________. (26 of 31 correct)
a.
b.
c.
d.
e.
3.0 x 10-2 M s-1
([A] increases by a factor of 1.5 and, since it is first order in A, the rate
will also increase by a factor of 1.5 and become 1.5 x 2.0 x 10-2 Ms-1.
4.0 x 10-2 M s-1
0.75 x 10-2 M s-1
1.0 x 10-2 M s-1
4.5 x 10-2 M s-1
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17.
Consider the reaction:
4 NH3 +
7 O2
Æ
4 NO2 +
6 H2O
The rate of the reaction is found to be 2.0 x 10-2 M min-1. Which of the following
statements is true? (2 of 31 correct)
a.
b.
c.
d.
e.
18.
The rate of disappearance of NH3 is 0.5 x 10-2 M min-1.
The rate of formation of NO2 is 8.0 x 10-2 M min-1.
The rate of disappearance of oxygen is twice that of NH3.
The rates of change of concentration of all reactants and products is equal
to 2.0 x 10-2 M min-1.
The rate of appearance of H2O is greater than the rate of disappearance of
O2.
The rate law for a reaction is
rate = k [A] [B]2
Which of the following statements is false? (19 of 31 correct)
a.
b.
c.
d.
e.
19.
The reaction is first order in A.
The reaction is second order in B.
Possible units for the rate constant are M-2 s-1.
The reaction is second order overall.
(Actually is third order overall.)
If the concentration of A is tripled, the rate will increase by a factor of
three.
One difference between first- and second- order reactions is that __________.
(21 of 31 correct)
a.
b.
c.
d.
e.
the half-life of a first-order reaction does not depend on [A]o; the halflife of a second-order reaction does depend on [A]o
the rate of a first-order reaction does not depend on reactant
concentrations; the rate of a second-order reaction does depend on
reactant concentrations
the rate of a first-order reaction depends on reactant concentrations; the
rate of a second-order reaction does not depend on reactant concentrations
a first-order reaction can be catalyzed; a second-order reaction cannot be
catalyzed
the half-life of a first-order reaction depends on [A]o; the half-life of a
second-order reaction does not depend on [A]o
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20.
For reaction that is first-order in A,
(14 of 31 correct)
a.
b.
c.
d.
e.
A plot of [A] vs. time will be a straight line with positive slope.
A plot of [A] vs. time will be a straight line with negative slope.
A plot of 1/[A] vs. time will be a straight line with positive slope.
A plot of ln [A] vs. time will be a straight line with positive slope.
A plot of ln [A] vs. time will be a straight line with negative slope.
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Problems (10 points each). SHOW YOUR WORK.
21.
15.0-g of benzene, C6H6, are dissolved in 250-g of carbon tetrachloride, CCl4.
The density of the resulting solution is 1.42 g/mL. Find the freezing point,
boiling point, osmotic pressure at 25 EC and vapor pressure at 25 EC of the
solution. Information regarding CCl4 and C6H6 is below:
For CCl4:
Normal boiling point:
Normal freezing point:
Boiling point elevation constant (Kb):
Freezing point depression constant (Kf):
Vapor pressure at 25 EC:
Molar masses:
CCl4:
C6H6:
76.8 EC
-22.3 EC
5.02 EC/m
29.8 EC/m
130 torr
154
78
Moles benzene (solute) = 15.0g/78g/mol = 0.1923 mol benzene
Moles carbon tetrachloride (solvent) = 250g/154 g/mol = 1.623 mol carbon tetrachloride
Volume of solution = mass of solution/density = 265.0 g/1.42 g/mL = 186.6 mL
Kg solvent = 0.250 kg
For concentrations:
Molarity = moles benzene/L solution = 0.1923 mol benzene/0.1866 L = 1.031 M
Molality = moles benzene/kg carbon tet = 0.1923 mol benzene/0.250kg = 0.7692 m
Xcarbon tet = moles carbon tet/(moles carbon tet + moles benzene) = 1.623 mol/(1.623 +
0.1923) = 0.8941
Benzene is a nonelectrolyte – 1 particle per molecule.
∆Tf = Kf x m = 29.8 °C/m x 0.7692 m = 22.92 °C so Tf = -22.3 – 22.92 = -45.2 °C = Tf
∆Tb = Kb x m = 5.02 °C/m x 0.7692 m = 3.86 °C so Tb = 76.8 + 3.86 = 80.7 °C = Tb
Π = MRT = 1.031 M x 0.08206 Latm/molK x 298 K = 25.2 atm
P = XCCl4 x P°CCl4 = 0.8941 x 130 torr = 116 torr
8 of 11
22.
Initial rate experiments were conducted to determine the rate law for the reaction:
A + B Æ
Products
The results are given in the table below:
Experiment #
1
2
3
Initial [A] (/M)
0.30
0.30
0.15
Initial [B] (/M)
0.20
0.05
0.20
Initial Rate (M/min)
3.45 x 10-4
8.6 x 10-5
8.6 x 10-5
a.
Write the rate law for this reaction based on the data above.
Compare experiments 1 and 2. Going from 2 to 1, [B] quadruples, initial rate goes up
by a factor of four so it must be first order in B. Compare experiments 1 and 3,
going from 3 to 1 A doubles and the rate quadruples. This must mean it is
second order in A.
Rate = k [A]2[B]
b.
Determine the value of the rate constant being sure to include the correct
units.
k = rate/{[A]2[B]} so use any experiment to evaluate k. Using experiment 1:
k = 3.45 x 10-4 Mmin-1/{(0.30 M)2(0.20 M)} = 1.92 x 10-2 M-2 min-1
c.
What concentrations of [A] and [B] could be mixed together to provide a
reaction mixture with an initial rate of 2.76 x 10-3 M min-1? Note there are
an infinite number of answers to this question – you don’t have to supply
them all. Just one example of [A] and [B] that would result in an initial
rate of 2.76 x 10-3 M min-1 is sufficient.
Any combination of [A] and [B] such that:
Rate = k [A]2[B]
so
2.76 x 10-3 M min-1 = 1.92 x 10-2 M-2 min-1 [A]2[B]
So [A]2[B] = 2.76 x 10-3 M min-1 /1.92 x 10-2 M-2 min-1 = 0.144 M3
9 of 11
23.
a.
A reaction is first order in A and first order overall. If the rate constant for
the reaction 7.45 x 10-4 s-1 and the initial concentration of A is 0.345 M,
how long will it take for the concentration to drop to 0.100 M?
[ A]o
0.345M
= kt ⇒ ln
= (7.45 x10− 3 s −1 )t ⇒ 1.2384 = (7.45 x10− 3 s −1 )t ⇒
0.100M
[ A]
1.2384
t=
= 166s
7.45 x10− 3 s −1
ln
b.
What is the half-life for this first order reaction?
If you don’t happen to remember the half-life equation (ln2 = kt1/2), just find out how
long it takes this first order reaction to drop from 0.345 M to 0.345/2 M. The half-life
from any concentration would be the same.
t1 =
2
ln 2
0.693
=
= 93.0 s
k
7.45 x10− 3 s −1
c.
After three half-lives, how much of the starting concentration of A would
be left?
The simplest way to think of this is that after three half-lives the remaining A is:
1 1 1 1
or 1/8 x 0.345 M = 0.0431 M
x x =
2 2 2 8
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Potentially useful information:
ln
[ A]o
= kt
[ A]
1
1
−
= kt
[ A] [ A]o
R = 0.08206
L • atm
mol • K
11 of 11