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NAME:_________________________
Fall 2009
INSTRUCTIONS:
Section:_____
Student Number:________________
Chemistry 1000 Midterm #2A
____/ 55 marks
1) Please read over the test carefully before beginning. You should have 7
pages of questions, a blank page that can be used if you run out of space
on any question, and 2 pages of data/formula/periodic table sheet.
2) If you use the “overflow” page, indicate this next to the question and
clearly number your work on the “overflow” page.
3) If your work is not legible, it will be given a mark of zero.
4) Marks will be deducted for incorrect information added to an otherwise
correct answer.
5) Marks will be deducted for improper use of significant figures and for
missing or incorrect units.
6) Show your work for all calculations. Answers without supporting
calculations will not be given full credit.
7) You may use a calculator.
8) You have 90 minutes to complete this test.
Confidentiality Agreement:
I agree not to discuss (or in any other way divulge) the contents of this exam until after 5pm
Mountain Time on Tuesday, November 17th, 2009. I understand that, if I were to break this
agreement, I would be choosing to commit academic misconduct and that is a serious offense which
will be punished. The minimum punishment would be a mark of 0/55 on this exam and removal of
the “overwrite midterm mark with final exam mark” option for my grade in this course; the
maximum punishment would include expulsion from this university.
Signature: ___________________________
Course: CHEM 1000 (General Chemistry I)
Semester: Fall 2009
The University of Lethbridge
Date: _____________________________
Question Breakdown
Q1
Q2
Q3
Q4
Q5
Q6
Q7
Q8
Q9
Q10
/ 16
/3
/5
/6
/3
/3
/6
/2
/7
/4
Total
/ 55
NAME:_________________________
Section:_____
Student Number:________________
1.
Answer the following questions by filling in the blanks.
[16 marks]
(a)
In lab, you constructed face centered cubic unit cells from white foam spheres. If you were
to construct a body centered cubic lattice from the same white foam spheres, would it have a
higher or lower density than the face-centered cubic lattice?
The density of the body-centered cubic lattice would be __lower__.
(b)
Lithium has very similar properties to one of the alkaline earth metals. That alkaline earth
metal is _magnesium (Mg)__.
(c)
If there are no lone pairs on the phosphorus atom in PBr5 then the molecular geometry is
_trigonal bipyramidal_.
(d)
Sodium metal is produced through the electrolysis of sodium chloride in what state of
matter? _liquid_
(e)
Aluminium metal reacts with some acids but not others. An example of an acid with which
aluminium metal will not react is _nitric acid (HNO3)_.
or any other oxidizing acid
or any weak acid
(f)
The alkali metal that reacts most vigorously with water is _francium (Fr)_.
(g)
Give an example of a metal oxide that reacts with both acid and base. _beryllium oxide
(BeO) or aluminium oxide (Al2O3)_
or any other oxide of a group 13 element
(h)
The correct name for CrCl3 is _chromium(III) chloride_.
(i)
The alkaline earth metal that produces salts giving green flame tests is _barium (Ba)_.
(j)
The gas produced when magnesium metal is added to hydrochloric acid is _hydrogen (H2)_.
(k)
Zinc(II) nitride has the molecular formula _Zn3N2_.
(l)
The two cations that contribute the most to the hardness of water are _Mg2+_ and _Ca2+_.
(Credit will only be given if both symbol and charge are correct!)
(m)
An atom which has octahedral electron group geometry (two lone pairs + the appropriate
number of bonded atoms) has _square planar_molecular geometry.
(n)
A face-centered cubic lattice has two types of holes. The _octahedral_ holes are larger than
the _tetrahedral_ holes.
NAME:_________________________
2.
Section:_____
Student Number:________________
Given that Ba2+(aq) is highly toxic, which of the compounds below would make a more
effective rat poison? Briefly, justify your answer. Your justification should include at least
one balanced chemical equation.
[3 marks]
BaCO3 or BaSO4?
BaCO3 and BaSO4 are both insoluble in water; however, BaCO3 reacts with stomach acid,
releasing toxic Ba2+(aq) cations. BaSO4 does not react with acid.
2+
+
BaCO 3(s) + 2H (aq)
→ Ba (aq)
+ H 2 O (l) + CO 2(g)
3.
In the gas phase, BeCl2 can exist as discrete molecules with polar covalent bonds. Under
standard conditions, MgCl2 exists as an ionic lattice.
[5 marks]
(a)
Draw the best Lewis structure for BeCl2.
..
.. Cl
..
Be
[1 mark]
.. .
Cl
.. .
This is the Lewis structure that minimizes formal charge. It is preferable to a structure
which has Be=Cl double bonds since those would give each Cl a formal charge of +1 and
the much less electronegative Be a formal charge of -2. Group 2 and Group 13 elements
can have less than a full octet (4 electrons for Group 2; 6 electrons for Group 13).
(b)
What is the molecular geometry for BeCl2?
[1 mark]
linear
(c)
What is the Cl-Be-Cl bond angle in BeCl2?
[1 mark]
180°
(d)
Why would BeCl2 be a molecular compound when all of the other alkaline earth metals react
with chlorine gas to give ionic compounds?
[2 marks]
The Be2+ cation is very small. Therefore, its charge density is much higher than any other
M2+ cation formed from an alkaline earth metal. This makes Be2+ a much less stable cation
than other M2+ cations.
For this reason, beryllium atoms are more stable when part of molecular compounds (in
which they usually have a neutral formal charge).
NAME:_________________________
Section:_____
Student Number:________________
4.
(a)
[6 marks]
Draw all valid resonance structures for the sulfite ion ( SO 32- ). Include any non-zero formal
[4 marks]
charges on the appropriate atom(s).
..
.. ..O
-1
..
S
..
..O
.. . -1
O
.. .
.. . -1
O
.. .
S
.. ..O ..
.. O ..
(b)
..
-1
What is the average S-O bond order in the sulfite ion?
.. ..O
..
-1
..
..
S
.. O..
..
O
..
-1
[1 mark]
1⅓
(b)
Draw one valid Lewis structure for sulfurous acid ( H 2SO 3 ). Include any non-zero formal
charges on the appropriate atom(s).
[1 mark]
H
..
..O
..
S
..
O
..
H
.. O..
Note that the hydrogen atoms MUST be bonded to oxygen – not sulfur. It is logical that the
H+ would attach themselves to the O- of the SO32- anion.
NAME:_________________________
Section:_____
Student Number:________________
5.
Write a balanced chemical equation for each of the following reactions. If no reaction
[3 marks]
occurs, instead write “NO REACTION”.
Include all states of matter.
(a)
Bromine is poured over sodium.
[1 mark]
2 Na (s) + Br2(l) → 2 NaBr(s)
(b)
Sodium is exposed to nitrogen.
[1 mark]
no reaction
(c)
Barium is dropped into water.
[1 mark]
2+
Ba (s) + 2 H 2 O (l) → Ba (aq)
+ 2 OH -(aq) + H 2(g)
It is acceptable to combine the barium and hydroxide ions, writing Ba(OH)2(aq)
6.
Potassium metal exists as a body centered cubic lattice. Given that the radius of a potassium
atom is 227.2 pm, what is the size of the unit cell? Express your answer as the length of one
edge of the unit cell in pm.
[3 marks]
The relationships between the three sides of the triangle drawn at
the right were derived in class using the Pythagorean theorem.
x
3
As can be seen in the figure at the right, the distance between
opposite corners of the BCC unit cell is equal to four atomic radii.
x
2
x
3x = 4r
4r
x=
3
4(227.2pm )
x=
3
x = 524.7pm
NAME:_________________________
7.
(a)
Section:_____
Student Number:________________
g
at 25 °C. The metal exists as a cubic closest
cm 3
[6 marks]
packed lattice, and the edge of a unit cell is 392 pm long.
An unknown metal has a density of 21.5
Calculate the atomic mass of this metal. Report your answer in g/mol.
[5 marks]
cubic closest packed = face-centered cubic therefore, one unit cell contains 4 atoms.
1m
100 cm
x = 392 pm × 12
×
= 3.92 × 10 −8 cm
10 pm
1m
(
)
3
V = x 3 = 3.92 × 10 −8 cm = 6.02 × 10 −23 cm 3
d=
m
V
(
)
g ⎞
⎛
mcell = d ⋅ V = ⎜ 21.5 3 ⎟ 6.02 × 10 − 23 cm 3 = 1.30 × 10 − 21 g
cm ⎠
⎝
mcell = 4matom
matom =
mcell 1.30 × 10 −21 g
=
= 3.24 × 10 −22 g
4
4
M = matom ⋅ N A = 3.24 × 10
− 22
g
g
6.02214 × 10 23 atoms
×
= 195
atom
mol
1mol
(b)
Identify the metal by referring to the periodic table given on the data sheet.
You do not need to name the metal; the symbol is sufficient.
platinum (Pt)
8.
Which of the metal oxides below would you expect to have the highest melting point?
Circle your choice and briefly justify your answer.
[2 marks]
Li2O
MgO
Al2O3
[1 mark]
CO2
Eliminate CO2 as it is not a metal oxide.
Of the remaining choices, compare the charge density of the ions. The lattice whose ions
have the highest charge density would be expected to have the highest enthalpy of lattice
formation and therefore the highest melting point.
Li2O consists of Li+ and O2-. MgO consists of Mg2+ and O2-. Al2O3 consists of Al3+ and O2-.
As such, Al2O3 would be expected to have the highest melting point since M3+ can be
expected to have a higher charge density than M2+ or M+.
NAME:_________________________
Section:_____
Student Number:________________
9.
We cannot measure the second enthalpy of electronic attraction (∆HEA2) of oxygen directly.
It can, however, exist in the solid state where the high energy requirement for its formation
[7 marks]
is offset by the large lattice energies of ionic oxides.
(a)
Sketch a Born-Haber cycle diagram for MgO. Clearly label the enthalpy change involved
[4 marks]
with each step. (i.e. give the name or symbol for each enthalpy change)
Steps to a Born-Haber cycle:
- Write overall chemical equation for formation of ionic compound from its
constituent elements (“enthalpy of formation”)
- Convert each element into single gaseous atoms
- Make the appropriate ions from those atoms (“ionization energy” to make
cations; “enthalpy of electronic attraction” to make anions)
- Bring ions together to form a lattice (“enthalpy of lattice formation”)
Mg2+(g)
+
I2(Mg)
∆EA2H(O)
Mg+(g)
O-(g)
∆EA1H(O)
I1(Mg)
Mg(g)
(b)
∆LFH (MgO)
O(g)
∆sublH(Mg)
Mg(s)
O2-(g)
½ ∆BDH(O2)
+
½ O2(g)
∆fH° (MgO)
MgO(s)
Use your Born-Haber cycle and the information from the data sheet to calculate ∆EA2H for
[2 marks]
oxygen.
∆ f H °( MgO ) = ∆ subl H ( Mg ) + I1 ( Mg ) + I 2 ( Mg ) + 12 ∆ BD H (O2 ) + ∆ EA1 H (O ) + ∆ EA 2 H (O ) + ∆ LF H ( MgO )
∆ EA 2 H (O ) = ∆ f H °( MgO ) − [∆ subl H ( Mg ) + I1 ( Mg ) + I 2 ( Mg ) + 12 ∆ BD H (O2 ) + ∆ EA1 H (O ) + ∆ LF H ( MgO ) ]
kJ
kJ
kJ
kJ
) − [(146 mol
) + (737 .7 mol
) + (1451 mol
)+
= (− 601 .7 mol
1
2
kJ
kJ
kJ
(498 mol
) + (− 141.0 mol
) + (− 3925 mol
)]
kJ
∆ EA 2 H (O ) = +881 mol
(c)
Values for ∆EA1H are usually negative. Why do you think the value for ∆EA2H is positive?
[1 mark]
Adding an electron to an anion is not as favourable as adding an electron to a neutral atom
due to increased charge repulsion (between the negative ion and the negative electron).
NAME:_________________________
Section:_____
Student Number:________________
10.
Aluminium metal is prepared by electrolysis of aluminium oxide using a carbon (C(s)) anode.
[4 marks]
This process produces carbon dioxide as a byproduct.
(a)
Write a balanced chemical equation (including states of matter) for this reaction.
[1 mark]
2Al 2 O 3(l) + 3C (s) → 4Al (l) + 3CO 2(g)
Except for states of matter, all reactants and products were listed in the sentence at the top
of the page.
(b)
Calculate the mass of carbon dioxide produced for every kilogram of aluminium prepared
according to this method. (This does not, of course, include carbon dioxide generated by
[3 marks]
burning fuels in order to generate the necessary energy for this reaction.)
Give your answer with three significant figures.
1 kg = 1000 g
n Al = 1000 g Al ×
1 mol
= 37.1 mol Al
26.9815 g
n CO 2 = 37.1 mol Al ×
3 mol CO 2
= 27.8 mol CO 2
4 mol Al
m CO 2 = 27.8 mol CO 2 ×
44.0098 g
= 1223 g CO 2 = 1.22 × 10 3 g CO 2 = 1.22 kg CO 2
1 mol
Every stoichiometry problem should be approached with the same three-step process:
1. Find moles of known species (in this case, Al)
2. Use mole ratio to find moles of species of interest (in this case, CO2)
3. Answer question. (In this case, find mass of CO2.)
You CANNOT use the mole ratio directly on masses!
NAME:_________________________
Section:_____
Student Number:________________
Some Useful Constants and Formulae
Fundamental Constants and Conversion Factors
Atomic mass unit (u)
1.6605 × 10-27 kg
Avogadro's number
6.02214 × 1023 mol–1
Bohr radius (a0)
5.29177 × 10-11 m
9
2 -2
Coulomb constant (1/(4πε0)) 8.988 × 10 N·m ·C
Electron charge (e)
1.6022 × 10-19 C
Electron mass
5.4688 × 10-4 u
Ideal gas constant (R)
8.3145 J·mol-1·K-1
8.3145 m3·Pa·mol-1·K-1
Planck's constant
Proton mass
Neutron mass
Rydberg Constant (RH)
Speed of light in vacuum
Standard atmospheric pressure
6.626 × 10-34 J·Hz-1
1.0072765 u
1.0086649 u
2.179 × 10-18 J
2.9979 × 108 m·s-1
1 bar = 100 kPa
Formulae
c = λυ
rn = a0
E = hυ
n2
Z
E n = − RH
λ=
p = mv
Z2
n2
Ek =
1 ( z + e)( z − e)
F=
4πε
d2
1 2
mv
2
h
p
∆x ⋅ ∆p >
h
4π
PV = nRT
1 ( z + e)( z − e)
E=
4πε
d
Some Thermodynamic Data
∆ sub H ( Mg )
I 1 ( Mg )
I 2 ( Mg )
∆ EA1 H ( Mg )
kJ
mol
kJ
+ 737.7
mol
kJ
+ 1451
mol
kJ
0
mol
+ 146
∆ BD H (O = O)
I1 (O )
I 2 (O )
∆ EA1 H (O )
kJ
mol
kJ
+ 1314
mol
kJ
+ 3388
mol
kJ
- 141.0
mol
+ 498
∆ LF H ( MgO)
∆ f H °(MgO)
kJ
mol
kJ
- 601.7
mol
- 3925
NAME:_________________________
1
Section:_____
Student Number:________________
CHEM 1000 Standard Periodic Table
18
1.0079
4.0026
H
He
2
13
14
15
16
17
6.941
9.0122
10.811
12.011
14.0067
15.9994
18.9984
Li
Be
B
C
N
O
F
Ne
3
22.9898
4
24.3050
5
26.9815
6
28.0855
7
30.9738
8
32.066
9
35.4527
10
39.948
1
2
20.1797
Na
Mg
11
39.0983
12
40.078
3
4
5
6
7
8
9
10
11
12
44.9559
47.88
50.9415
51.9961
54.9380
55.847
58.9332
58.693
63.546
65.39
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
19
85.4678
20
87.62
21
88.9059
22
91.224
23
92.9064
24
95.94
26
101.07
27
102.906
28
106.42
29
107.868
30
112.411
31
114.82
32
118.710
33
121.757
34
127.60
35
126.905
36
131.29
Rb
Sr
37
132.905
38
137.327
Cs
Ba
55
(223)
56
226.025
Fr
87
Ra
Y
39
La-Lu
Ac-Lr
88
P
S
Cl
Ar
15
74.9216
16
78.96
17
79.904
18
83.80
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
41
180.948
42
183.85
43
186.207
44
190.2
45
192.22
46
195.08
47
196.967
48
200.59
49
204.383
50
207.19
51
208.980
52
(210)
53
(210)
54
(222)
Hf
Ta
W
Re
Os
Ir
Pt
Au
72
(261)
73
(262)
74
(263)
75
(262)
76
(265)
77
(266)
78
(281)
79
(283)
Rf
Db
Sg
105
106
138.906
140.115
140.908
144.24
La
Ce
Pr
Nd
57
227.028
58
232.038
59
231.036
60
238.029
Ac
Si
14
72.61
40
178.49
104
89
25
(98)
Al
13
69.723
Th
90
Pa
91
U
92
Bh
107
Hs
Mt
Dt
Hg
Tl
Pb
Bi
Po
At
80
81
82
83
84
85
174.967
Rg
108
109
110
111
(145)
150.36
151.965
157.25
158.925
162.50
164.930
167.26
168.934
173.04
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
61
237.048
62
(240)
63
(243)
64
(247)
65
(247)
66
(251)
67
(252)
68
(257)
69
(258)
70
(259)
71
(260)
Np
93
Pu
94
Am
95
Cm
96
Rn
86
Bk
97
Cf
98
Es
99
Fm
100
Md
101
No
102
Lr
103
Developed by Prof. R. T. Boeré