Chem 107 - Hughbanks Exam 3, Solutions

Chem 107 - Hughbanks
Exam 3, Solutions
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Key
UIN #
Section 504
Exam 3, Version B - Solutions
On the last page of this exam, you’ve been given a periodic table and some physical constants. You’ll
probably want to tear that page off the to use during the exam – you don’t need to turn it in with the rest
of the exam.
The exam contains 10 problems, with 5 numbered pages. You have the full 50 minutes to complete the
exam. Please show ALL your work as clearly as possible – this will help us award you partial credit
if appropriate. Even correct answers without supporting work may not receive credit. You may use an
approved calculator for the exam, one without extensive programmable capabilities or the ability to store
alphanumeric information. Print your name above, provide your UIN number, and sign the honor code
statement below.
On my honor as an Aggie, I will neither give nor receive unauthorized assistance on this exam.
SIGNATURE:
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Version B, Solutions
(For grading)
Scores
(1) (6 pts) Which of the following molecules possess a dipole moment? Circle all
the correct answers.
(a) NH3
(b) H2O
H
F
F
(d)
(c) BF3
H
(e)
F
H
H
H
F
F
(f)
F
H
(2) (8 pts) For each of the following pairs of elements or compounds, circle the
choice with the higher boiling point.
(a) O2 or N2
(b) SiH4 or GeH4
(c) SO2 or CO2
(d) HF or HBr
1
2
3
4
5
6
7
8
9
10
Tot.
/6
/8
/5
/5
/12
/7
/12
/10
/20
/15
/100
(3) (5 pts) Doping a small amount of arsenic (As) into a crystal of pure germanium
(Ge) produces a
(I) metal
(II) p-type semiconductor
(III) n-type semiconductor
(IV) semiconductor with a small number of holes in the bonding valence band
(V) semiconductor with a small number of electrons in the antibonding conduction band
(A) only III is correct
(D) II and IV are correct
Ans. 3
(4)
(B) only II is correct
(E) I, IV, and V are correct
(C) III and V are correct
c
(5 pts) For which one of the following processes/reactions is the difference between ∆H and ∆E
largest? (Put the letter of the correct answer in the blank provided.)
(A) C(s, diamond) → C(s, graphite)
(B) HNC(g) → HCN(g)
(C) CaO(s) + SiO2(s) → CaSiO3(s)
(D) CaCO3(s) → CaO(s) + CO2(g)
(E) H2O(s) → H2O(ℓ)
Ans. 4
D
1
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Version B, Solutions
(5) (12 points) Use the vapor pressure curves
illustrated here to answer the questions that
follow. Put answers in the blanks provided.
(a) (2 pts) What is the vapor pressure of
ethanol (C2H5OH) 70˚C?
~ 550 mm Hg
(b) (2 pts) Considering only carbon disulfide
(CS2) and ethanol, which has the stronger
intermolecular forces in the liquid state?
Ethanol
(c) (2 pts) At what temperature does heptane (C7H16) have a vapor pressure of 400 mm Hg?
_______
(d) (3 pts) What are the approximate normal boiling points of each of the three substances?
Ethanol:
78 ˚C
Carbon disulfide:
Heptane:
46 ˚C
98 ˚C
(e) (3 pts) At a pressure of 400 mm Hg and a temperature of 70˚C, for each substance indicate
whether it is entirely a liquid, entirely a gas, or a mixture of liquid and vapor?
Ethanol: ___gas___
Carbon disulfide: ___gas___
Heptane: ___mixture___
(6) (7 points) The reaction shown below is involved in the chemistry of the stratosphere, and has been
studied extensively due to concern over depletion of the ozone layer.
ClO2 (g) + O (g) → ClO (g) + O2 (g)
ΔHf° (kJ/mol)
ClO (g)
ClO2 (g)
O (g)
O2 (g)
O3 (g)
109
102.5
249.17
142.7
Find ΔH° for the reaction.
109 + 0 – 102.5 – 249.17 kJ = – 243 kJ
2
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Version B, Solutions
Questions 7-10 – show all work (59 pts)
(7) (12 points) The molecule shown at right is vanillin (C8H8O3), and is
responsible for the taste and smell of vanilla. (Be careful here, as some of
these are trickier than they seem. Realize that not all of the atoms are shown
in the drawing! Lone pairs are not shown here, but the structure obeys the
octet rule.) Put answers in the blanks provided.
(a) (2 pts) What type of orbital hybridization would be expected for the
oxygen atom labeled as #2?
sp3
(b) (2 pts) What type of orbital hybridization would be expected for the carbon atom labeled as #1?
sp2
(c) (2 pts) The molecule contains carbon-oxygen bonds of 2 different lengths: 122 pm and 143 pm.
How many of the shorter bonds are present? __1___
(d) (3 pts) How many sigma (σ) bonds are present in the molecule? __19___
(e) (3 pts) How many pi (π) bonds are present in the molecule? __4___
(8) (10 points) Nitroglycerin, C3H5(NO3)3(ℓ), is an explosive most
often used in mine and quarry blasting. It is a powerful explosive
because four gases (N2, O2, CO2, and steam) are formed when
nitroglycerin is detonated (combustion, burning with O2, is not
occurring). In addition, 6.26 kJ of heat is given off per gram of
nitroglycerin detonated.
(a) (3 pts) Write a balanced thermochemical equation for the reaction.
C3H5(NO3)3 → 3 CO2(g) + 3/2 N2(g) + 5/2 H2O(g) + ¼ O2(g)
1 mole reactant → 7¼ moles of products
(b) (7 pts) Calculate ∆H when 8.25 mol of products is formed.
6.26 kJ
227 g 1 mol C3H 5 ( NO3 )3
×
×
× 8.25 mol products
7.25 mol products
g C3H 5 ( NO3 )3 mol
= 1617 kJ
3
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(9)
Version B, Solutions
(20 points) Nitrogen can combine with oxygen and fluorine to form either nitrosyl fluoride (NOF)
or nitryl fluoride (NO2F). In both of these compounds, nitrogen is the central atom, with the
oxygen and fluorine atoms bound to it.
(a) (7 pts) Draw one or more complete Lewis structures and specify the nitrogen atom
hybridization for the NOF molecule. Draw an accurate diagram showing the shape; give an
approximate value for the O-N-F angle.
sp2
F
N
113 pm
110˚
O
(b) (7 pts) Draw one or more complete Lewis structures and specify the nitrogen atom
hybridization for the NO2F molecule. Draw an accurate diagram showing the shape; give
approximate and consistent values for the O-N-F angles and the O-N-O angle.
123 pm
O
N
F
118˚
O
–
O
N
F
–
sp2
O
(c) (6 pts) Using your results, predict which molecule has the shortest N–O bond. (To receive
credit, your answer must also include a sufficient justification for your choice.)
The N–O bond in NOF is a double bond and those in NO2F have bond orders of
1½, so we expect the N–O bond in NOF to be shorter.
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Version B, Solutions
(10) (15 points) Molecular orbitals can be used to describe the bonding in any molecule, not just the
homonuclear diatomics. For nitric oxide (NO), the MO treatment is much better than the Lewis
model for handling the odd number of electrons.
(a) (9 pts) Draw a molecular orbital energy level diagram for the NO molecule. Include all the
orbitals formed from the valence atomic orbitals (i.e., 2s and 2p). Label the diagram completely,
identifying all of the molecular orbitals, and use arrows to indicate the electron configuration.
(HINT: The molecular orbitals formed are the same ones as we've seen for O2. So you can draw
the orbital diagram for O2, and then just use the correct number of electrons for NO.).
σ*2p
π*2p
π2s
σ2s
σ*2s
σ2s
(b) (3 pts) Compare neutral NO to the NO+ ion. Which will have the longer bond length and how do
you know? (HINT: use your diagram from part (a) to decide.)
In forming the NO+ ion, NO loses its π* electron. Thus, the bond shortens in the NO+ ion.
(c) (3 pts) Rank the following molecules in order of increasing bond strength: N2, O2, NO. Give a
BRIEF explanation of your choice.
(strongest bond)
bond orders:
N2
3
<
NO
2.5
<
O2
2
(weakest bond)
The molecules have, in order, 0 π* electrons, 1 π* electron, and 2 π* electrons – hence the bond orders.
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