key III

CHEMISTRY 102B
Practice Hour Exam III
Spring 2015
Dr. D. DeCoste
Name ______________________________
Signature ___________________________
T.A. _______________________________
This exam contains 22 questions on 8 numbered pages. Check now to make sure you have a
complete exam. You have one hour and thirty minutes to complete the exam. Determine the
best answer to the first 20 questions and enter these on the special answer sheet. Also, circle
your responses in this exam booklet.
Show all of your work and provide complete answers to questions 21 and 22.
1-20
(40 pts.)
_________
21
(20 pts.)
_________
22
(20 pts.)
_________
(80 pts)
_________
Total
Useful Information:
CHEMISTRY 102B
Practice Hour Exam III
1.
Spring 2015
Page No. 1
Consider the carbon monoxide molecule pictured below. Note that the bond may be a
single, double, or triple bond. What can best be said about the electrons involved in the
bond?
C
Box A
O
Box B
a) There is an equal chance of a bonding electron being in the area marked by either box.
This is because carbon and oxygen are both in the same row of the periodic table.
b) There is an equal chance of a bonding electron being in the area marked by either box.
This is because the molecule is linear.
c) A bonding electron is more likely to exist in the area marked by Box A because carbon
needs more electrons to get to a noble gas electron configuration.
d) A bonding electron is more likely to exist in the area marked by Box B because the
oxygen nucleus has a stronger attraction for electrons than does the carbon nucleus.
e) Because we never know where the electron is, we cannot say one way or the other
where a bonding electron is more likely to be.
2.
Which of the given statements best describes the following data?
First ionization energy for Na: 495 kJ/mol
First ionization energy for Cl: 1255 kJ/mol
Heat of formation of NaCl: -525 kJ/mol
Electron affinity for Na: -50 kJ/mol
Electron affinity for Cl: -349 kJ/mol
a) Energy is released when a sodium ion is formed from a sodium atom and when a
chlorine ion is formed from a chlorine atom. This makes the formation of sodium
chloride exothermic.
b) Energy is required to remove an electron from sodium, but more energy is released
when chlorine takes an electron. This makes the formation of sodium chloride
exothermic.
c) Energy is required to form both the sodium ion and the chloride ion, so the reaction to
form sodium chloride is endothermic.
d) Energy is required to remove an electron from sodium and energy is released when
chlorine takes an electron. Less energy is released than required for these two
processes, though, so overall, the formation of sodium chloride is endothermic.
e) Energy is required to remove an electron from sodium, and energy is released when
chlorine takes an electron. Less energy is released than required for these two
processes, but overall the process is exothermic.
3.
In which case is the bond polarity incorrect?
a)
b)
c)
d)
e)
δ+
H–Fδδ+
S–Oδδ+
Mg–Hδδ+
Cl–Iδδ+
Si–Sδ-
CHEMISTRY 102B
Practice Hour Exam III
4.
Spring 2015
Page No. 2
Given the following information:
Li(s) → Li(g)
HCl(g) → H(g) + Cl(g)
Li(g) → Li+(g) + e–
Cl(g) + e– → Cl–(g)
Li+(g) + Cl–(g) → LiCl(s)
H2(g) → 2H(g)
Heat of sublimation of Li(s) = 166 kJ/mol
Bond energy of HCl = 427 kJ/mol
Ionization energy of Li(g) = 520. kJ/mol
Electron affinity of Cl(g) = –349 kJ/mol
Lattice energy of LiCl(s) = –829 kJ/mol
Bond energy of H2 = 432 kJ/mol
Calculate the net change in energy for the reaction 2Li(s) + 2HCl(g) → 2LiCl(s) + H2(g).
a) –562 kJ
5.
b) 562 kJ
c) –497 kJ
d) 497 kJ
e) –994 kJ
Use the bond energy values from Table 8.4 to estimate ΔH for the following reaction.
Note: you will need to draw Lewis structures for the reactants and products to determine
the possible presence of multiple bonds.
HCN(g) + 2H2(g) → CH3NH2(g)
a) –158 kJ
6.
b) 158 kJ
c) –744kJ
d) 744 kJ
e) 1913 kJ
Which of the following has a Lewis structure most like that of the carbonate ion?
a)
b)
c)
d)
e)
carbon dioxide
the sulfate ion
the nitrate ion
ozone (O3)
nitrogen dioxide
CHEMISTRY 102B
Practice Hour Exam III
7.
Spring 2015
Page No. 3
How many of the following statements concerning resonance structures is/are correct?
I. The concept of resonance is used because the Lewis structure model is incomplete in
describing bonding in a molecule.
II. For a species having three resonance structures, it is best to think of the species to exist
as each of these structures one-third of the time.
III. All charged molecules have resonance structures.
IV. The octet rule must not be violated in writing resonance structures.
a) 0
8.
c) 2
d) 3
e) 4
Which of the following best describes BF3 and NF3?
a)
b)
c)
d)
e)
9.
b) 1
They each have variable geometries and shapes due to potential resonance structures.
They have the same geometry and different shapes.
They have the same geometry and the same shape.
They have different geometries and the same shape.
They have different geometries and different shapes.
Which of the following correctly labels the molecules as polar or non-polar?
BH3
CO2
SO2
a)
polar
polar
polar
b)
non-polar
non-polar
non-polar
c)
polar
non-polar
polar
d)
polar
non-polar
non-polar
e)
non-polar
polar
polar
-----------------------------------------------------------------------------------------------------------------10-12: Determine the shape (molecular structure) of the following molecules/ions and the
hybridization around the central atom
10.
[XeF5]+
a)
b)
c)
d)
e)
11.
trigonal bipyramid, d2sp3
see-saw, dsp3
trigonal bipyramid, dsp3
square pyramid, d2sp3
None of these
[ICl2]a)
b)
c)
d)
e)
bent, sp3
bent, dsp3
linear, dsp3
linear, sp3
None of these
CHEMISTRY 102B
Practice Hour Exam III
12.
Spring 2015
Page No. 4
SF4
a) tetrahedral, sp3
b) see-saw, dsp3
c) square planar, d2sp3
d) trigonal bipyramid, sp3
e) None of these
---------------------------------------------------------------------------------------------------------------13. Identify the strongest intermolecular force for the given molecule
a)
b)
c)
d)
e)
C2H6
LDF
dipole-dipole
H-bond
LDF
LDF
NH3
H-bond
LDF
H-bond
dipole-dipole
H-bond
CH3OCH3
H-bond
dipole-dipole
dipole-dipole
dipole-dipole
dipole-dipole
14.
Which of the following decreases as the strength of intermolecular forces increases?
a) boiling point
b) melting point
c) vapor pressure
d) surface tension
e) All of the above increase with as the strength of intermolecular forces increases.
15.
You mix 500.0 mL of 4.00 M Fe(NO3)3 with 500.0 mL 6.00 M KSCN. They react according to the
following equation:
Fe3+(aq) + SCN-(aq)
FeSCN2+(aq)
Since FeSCN2+ in solution appears red (and the reactants are colorless) you use a
spectrophotometer and determine the concentration of FeSCN2+ at equilibrium to be 1.00 M.
Determine the value of the equilibrium constant, K, for the reaction above.
a) 0.0417
16.
b) 0.0667
c) 0.100
d) 0.167
e) 0.500
Consider an acidic chromate-dichromate system at equilibrium in which the color is an
orange-yellow. The reaction is represented by the equation
Cr2O72-(aq) + H2O(l)
orange
CrO42-(aq) + H+(aq)
yellow
Which of the following best describes what happens if a strong base is added to the system
at equilibrium?
a)
b)
c)
d)
e)
Both the color of the solution and the value for K remain the same.
The solution turns more yellow and the value for K increases.
The solution turns more orange and the value for K decreases.
The solution turns more orange and the value for K does not change.
The solution turns more yellow and the value for K does not change.
CHEMISTRY 102B
Practice Hour Exam III
17.
Consider a chemical system at equilibrium. The reaction is exothermic as written and the
temperature of the system is raised. Which of the following is true?
a)
b)
c)
d)
e)
18.
Equilibrium shifts to the left and the value of K increases.
Equilibrium shifts to the right and the value of K increases.
Equilibrium shifts to the right and the value of K decreases.
Equilibrium shifts to the left and the value of K decreases.
Equilibrium shifts, but value of K stays constant.
You mix two aqueous ionic solutions together and they do not react. To this you add
(dropwise) a third aqueous ionic solution. At first a white solid forms. As you continue to
add the third solution, a yellow solid begins to form. Which of the following can you say
with confidence?
I.
II.
III.
IV.
The solubility of the white solid is greater than the solubility of the yellow solid.
The solubility of the yellow solid is greater than the solubility of the white solid.
The Ksp value for the white solid is greater than the Ksp value for of the yellow solid.
The Ksp value for the yellow solid is greater than the Ksp value for of the white solid.
a) I and III
19.
b) II and IV
c) I only
d) II only
e) I and IV
The Ksp of AgI is 1.5 x 10-16. Calculate the solubility of AgI in a 0.30 M NaI solution.
a)
b)
c)
d)
e)
20.
Spring 2015
Page No. 5
1.7 x 10-8 M
0.30 M
2.6 x 10-8 M
8.5 x 1017 M
5.0 x 10-16 M
Titrating 30.00 mL of a saturated calcium iodate solution requires 28.91 mL of a 0.092 M
solution of Na2S2O3 according to the equation. Calculate Ksp for Ca(IO3)2.
IO3- + 6S2O32- + 6H+ → I- + 3S4O62- + 3H2O
Calculate the value of the solubility product constant (Ksp) for Ca(IO3)2.
a)
b)
c)
b)
e)
1.09 x 10-4
7.14 x 10-8
6.97 x 10-4
2.79 x 10-3
1.61 x 10-6
CHEMISTRY 102B
Practice Hour Exam III
21.
Spring 2015
Page No. 6
For each of the following questions, fully explain your answer, using Lewis structures and
VSEPR theory when appropriate.
a.
The periodic table is an amazing achievement in human intellectual progress.
Developed before atomic theory was widely accepted, it helps explain a variety of
observations. For example, being in the same column, chlorine, bromine, and iodine
are all diatomic in their standard states. It turns out, though that at room conditions,
one is a gas, one is a liquid, and one is a solid. Match each element in its standard
state with its phase and explain your answer. [6 points]
• Use Lewis structures to show these are all linear and non-polar.
• Since they are non-polar, the strongest IMFs are LDFs for each of these.
• With more electrons, the LDFs are stronger, thus the IMFs are strongest for I2
and weakest for Cl2.
• The stronger the IMF the more energy required to separate the molecules, thus
the higher the boiling point or melting point.
• Thus, I2 has the highest boiling and melting points and Cl2 has the lowest
boiling and melting points.
• Thus, Cl2 is the gas, Br2 is the liquid, and I2 is the solid.
b.
You and your lab partner accidentally and independently synthesize a compound with
the formula XeCl2F2. Interestingly, your compound is a gas and your partner’s is a
liquid, each at lab conditions. Explain how these compounds can have the same
formulas yet exist in different phases at the same temperature and pressure. [6 points]
• Draw Lewis structures and use VSEPR to show that these are square planar.
• The chlorine atoms can either be 90° or 180° from each other. If 90°, the
molecule is polar; if 180°, the molecule is non-polar.
• The polar molecule has stronger IMFs, meaning more energy required to
separate the molecules, meaning a higher boiling point
• The polar molecule is the liquid, the non-polar molecule is the gas.
CHEMISTRY 102B
Practice Hour Exam III
21.
Spring 2015
Page No. 7
(con’t)
c.
The formula of the reactant, N2O, doesn’t tell us if the skeletal structure is N-O-N or
N-N-O, nor does it tell us if there is a best Lewis structure. Your goal is to find the
best Lewis structure for the compound, using formal charge as the primary basis.
Define formal charge in your answer. Justify your answer completely. [8 pts.]
Formal charge: (# valence electrons in free atom) – (# valence electrons assigned to the
atom in the molecule)
Five possible Lewis structures (note: lone pairs are not included):
I
N=N=O
II
N≡N−O
III
N−N≡O
Formal charges for I: -1, +1, 0
Formal charges for II: 0, +1, -1
Formal charges for III: -2, +1, +1
Formal charges for IV: -1, +2, -1
Formal charges for V: 0, +2, -2
IV
N=O=N
V
N≡O−N
(N: 5-6 = -1, N: 5-4 = +1, O: 6-6 = 0)
(N: 5-5 = 0, N: 5-4 = +1, O: 6-7 = -1)
(N: 5-7 = -2, N: 5-4 = +1, O: 6-5 = +1)
(N: 5-6 = -1, O: 6-4 = +2, N: 5-6 = -1)
(N: 5-5 = 0, O: 6-4 = +2, O: 5-7 = -2)
We want to minimize formal charge, so structures I and II are the best. Since oxygen is
more electronegative than nitrogen, oxygen should have the negative formal charge
(since it has more attraction for a shared electron).
So, structure II is better than structure I when taking formal charge into account.
CHEMISTRY 102B
Practice Hour Exam III
22.
Spring 2015
Page No. 8
Gaseous hydrazine (N2H4) decomposes to nitrogen gas and hydrogen gas. At a certain
temperature, Kp = 3.50 x 104 for this reaction balanced in standard form. Pure gaseous
hydrazine is placed in an otherwise empty rigid vessel and the pressure is noted. The
reaction is allowed to reach equilibrium at this temperature and it is observed that the
equilibrium pressure is double its initial value.
a.
Determine the initial pressure of the hydrazine (in atm). Define any variables. [11 pts]
Pressure directly related to number of moles (constant volume and
temperature), so we can use pressures in the ICE chart.
I
C
E
N2H4(g)
X
-Y
X-Y
N2(g) + 2H2(g)
0
0
+Y
+2Y
Y
2Y
X = initial pressure (pressure of hydrazine only)
Y = change in pressure of hydrazine
(X-Y) + Y + 2Y = X+2Y = final pressure
(X+2Y)/X = 2 or X =2Y;
3.5 x 104 =
(2Y) 2 (Y )
(2Y) 2 (Y )
=
; Y = 93.5
(Y)
( 2Y − Y )
Initial pressure of hydrazine = X = 2Y = 187 atm
b.
What percent (in terms of moles) of hydrazine decomposed to reach equilibrium?
Explain your answer. [7 pts]
Initial pressure = 2Y
Change in pressure = Y
Thus Y/2Y or 50% decomposed (pressure)  related to moles (constant
volume and temperature), so 50% moles decomposed.
c.
Determine the equilibrium pressure of hydrogen (in atm). [2 pts]
Equilibrium pressure of hydrogen gas = 2Y = 187 atm