Chemistry 111 Sect. 2
Exam #3
May 4, 2006
Name: ________________________________________________
KEY
Student #: ______________________________________________
Please put your name and student number on all pages in case they become separated. The last
page of the exam has some useful information. The exam lasts 60 minutes. There are a total of
100 points.
MULTIPLE CHOICE QUESTIONS
Place an X in the box corresponding to the correct answer.
1. (5 pts) In the molecule formaldehyde (H2CO, Lewis structure shown below), what is the
hybridization of the carbon atom?
†
X
†
†
†
†
sp
sp2
sp3
sp3d
sp3d2
:O:
H
C
H
There are three sigma bonds and no lone pairs on the carbon, so
one s and two p atomic orbitals are needed to hybridize.
2. (5 pts) A 2.00 gram sample of the rocket fuel hydrazine, N2H4 (molar mass = 32.0 g/mol), is
burned in a constant pressure bomb calorimeter. The temperature rises from 24.6 °C to 31.7 °C.
The total heat capacity of the calorimeter (including the water) is 5860 J/K. Calculate the molar
heat of combustion of hydrazine, in kJ/mol.
X
†
†
†
†
†
-666
-41.6
-20.8
41.6
666
The temperature rise is ∆T = 31.7 °C – 24.6 °C = 7.1 °C = 7.1 K. The heat q
needed to warm the calorimeter is q = Ctot ∆T = (5860 J/K) (7.1 K) = 41.6 kJ.
This heat came from the combustion reaction, so qcomb = ∆Hcomb = -q = -41.6
kJ. The number of moles of hydrazine in the calorimeter was n = 2.00 g ÷
32.0 g/mol = 6.25 × 10-2 mol. Thus, the heat of combustion per mole is
∆Hcomb = -41.6 kJ ÷ 6.25 × 10-2 mol = -666 kJ/mol.
3. (5 pts) Which of the following gas species has the fastest rms speed at 0 °C and 1 atm pressure?
†
X
†
†
†
†
Ar
CH4
Cl2
N2
Ne
Because urms = √3RT/M, the lowest molar or atomic mass gas species will
have the fastest urms. Of the species listed, CH4 has the smallest mass, 16
g/mol.
Page 1
Chemistry 111 Sect. 2
Exam #3
May 4, 2006
Name: ________________________________________________
KEY
Student #: ______________________________________________
4. (5 pts) One liter of an unknown hydrocarbon (containing only H and C) gas is found to weigh
1.23 g at STP. Which of the following hydrocarbons is the unknown?
†
†
†
X
†
†
CH3
C2H2
C2H4
C2H6
C3H8
Density relates to pressure, temperature and molar mass as d = PM/RT.
Standard temperature is 298 K and standard pressure is 1 atm. Thus, solving
the equation for M gives M = dRT/P. Plugging in the given values leaves us
with M = (1.23 g/ 1 L)(0.082 L⋅atm/mol⋅K)(298 K)/(1 atm) = 30.1 g/mol. The
molecular weights of the species listed, in order, are 15, 26, 28, 30 and 44, all
in g/mol. The unknown is ethane, C2H6.
5. (5 pts) How many atoms of Ar are there in 1 L of Ar gas at STP? (You may assume that Ar
behaves as an ideal gas.)
†
†
X
†
†
†
For 1 L of gas att 1 atm pressure and 298 K temperature, the ideal gas law
says that
4.1 × 10-2
2.4 × 101
2.5 × 1022
6.0 × 1023
1.5 × 1025
n = PV/RT = (1 atm)(1 L)/(0.082 L⋅atm/mol⋅K)(298 K) = 4.1 × 10-2 mol.
Multiplying by Avogadro’s number gives the number of atoms: (4.1 × 10-2
mol)(6.02 × 1023 atoms/mol) = 2.5× 1022 atoms.
6. (5 pts) The enthalpy change in the reaction of SO2 to make SO3, shown below, at STP is
SO2(g) + ½ O2(g) → SO3(g)
†
†
†
X
†
†
-692.6 kJ/mol
-198.0 kJ/mol
-148.4 kJ/mol
-99.0 kJ/mol
+148.4 kJ/mol
Look up the ∆Hof for SO2 (g) and SO3 (g) in the table on the last page. The
enthalpy change in a reaction equals the sum of enthalpies of formation of
reactants minus those of products, so
∆Horxn = {∆Hof for SO3 (g)} - {∆Hof for SO2 (g) + ½ ∆Hof for SO2 (g)}
= {-395.8 kJ/mol} –{-296.8 kJ/mol + ½ (0 kJ/mol)} = -99.0 kJ/mol.
Page 2
Chemistry 111 Sect. 2
Exam #3
May 4, 2006
Name: ________________________________________________
KEY
Student #: ______________________________________________
7. (5 pts) For which of the following is ∆Hof not equal to 0 kJ/mol?
†
X
†
†
†
†
He (g)
O (g)
Na (s)
C (s, graphite)
S (s, rhombic)
All of these are elements in their standard states except O (g). The standard
state form of oxygen is O2 (g).
8. (5 pts) The specific heat capacity of copper is 0.385 J/g⋅K. What quantity of heat is required
to heat 200 g of copper from -30 °C to +120 °C?
†
†
†
†
X
†
The amount of heat required to change the temperature of a substance is
0.3 kJ
2.3 kJ
4.6 kJ
9.2 kJ
11.6 kJ
q = m C ∆T.
We are given m = 200 g and C = 0.385 J/g⋅K. We can easily compute ∆T =
120 °C – (-30 °C) = 150 °C = 150 K. Plug into the formula to get
q = (200 g)(0.385 J/g⋅K)(150 K) = 11.6 kJ.
MULTIPLE MULTIPLE CHOICE QUESTIONS
Place an X in the boxes corresponding to the correct answers. Any number of answers may be
correct, including none of them.
9. (10 pts) In class, a chunk of solid CO2 (dry ice) was placed in a sealed bag. The bag inflated, lifting a
book about 3 cm off the table top. Which of the following is TRUE of this reaction?
X
†
X
†
†
†
†
the enthalpy H of the CO2 changed during the reaction It went up
work was done by the CO2 during the reaction The CO2 lifted the book, and did work
heat flowed out of the CO2 during the reaction Heat flowed in, work came out
the number of moles of CO2 in the bag changed during the reaction The bag was sealed
heat entering the CO2 equaled the work done on the book during the reaction
Heat entering equaled the work done plus the change in
internal energy (which is greater since the solid has
become a gas, and that takes energy)
Page 3
Chemistry 111 Sect. 2
Exam #3
May 4, 2006
Name: ________________________________________________
KEY
Student #: ______________________________________________
10. (10 pts) Four equal volume tires on a car are each filled with a different gas to a pressure of 3
atm at T = 300 K. Tire 1 is filled with Ar, tire 2 with N2, tire 3 with He and tire 4 with an
unknown gas. Assume all gases are ideal. Which of the following is true?
Since P, V and T are the same for all tires,
n must also be the same. That means the #
† the mass of gas added is the same for each tire
of grams are different since each gas has a
X
†
the number of moles of gas in each tire is equal
different molar (or atomic) mass.
X
†
the rms speed of gas molecules in tire 2 is greater than that in tire 1 M of He is < M of N2, so urms
of He is faster
† the unknown gas in tire 4 is H2 No way to know
† the rate at which tire 3 will deflate by effusion is slower than that for tire 1
M of He is < M of Ar, so urms of He is faster, meaning that effusion of He is faster
11. (10 pts) Carbon monoxide and hydrogen react to produce methanol with ∆Horxn = -90.0 kJ/mol
according to CO (g) + 2 H2 (g) → CH3OH (g). From this and data on the last page you can
determine that
X
†
X
†
†
X
†
†
∆Hof = -200.5 kJ/mol for CH3OH (g) ∆Hof,prod = ∆Horxn + ∆Hof,reactants = (-90) + (-110.5)) = -200.5
this reaction is occurring at 25 °C The superscript 0 means 25 °C and 1 atm pressure
this reaction is endothermic ∆Horxn is negative, so heat is produced – it’s exothermic
if the reaction goes to completion in a sealed vessel, the total pressure will drop All gases here. # of
moles goes from 3 to
CH3OH (g) has a more negative ∆Hof than CH3CH2OH
1, so the pressure
Need more information to determine this
must drop
SHORT ANSWER QUESTIONS
Answer in the spaces indicated.
12. (10 pts) The reaction of carbon monoxide and hydrogen can produce methanol according to
the following equation: CO (g) + 2 H2 (g) → CH3OH (g), Suppose you start with 2.8 g CO
and 2.0 g H2 in a 10 L flask at 227 °C.
The initial numbers of moles are CO: n = (2.8 g)/(28 g/mol) = 0.1 mol and H2: n = (2 g)/(2 g/mol) = 1 mol.
The partial pressures of CO and H2 at the start are, then PCO = nRT/V = (0.1 mol)(0.082 L⋅atm/mol⋅K)(500
K)/(10 L) = 0.41 atm and PH2 = (1 mol)(0.082 L⋅atm/mol⋅K)(500 K)/(10 L) = 4.1 atm. Add them together
to get the total pressure.
4.5 atm
What is the pressure (in atm) in the flask before a reaction occurs?
____________
When the reaction goes to completion, the number of moles of ethanol will be equal to the number of
moles of CO with which we started due to the reaction stoichiometry. Thus, the final number of moles of
CH3OH are n = 0.1 mol. The reaction will have consumed twice as many moles of H2, or 0.2 mol. That
will leave 0.8 mol of H2. The partial pressure of H3COH will be 0.41 atm and PH2 = (0.8 mol)(0.082
L⋅atm/mol⋅K)(500 K)/(10 L) = 3.3 atm. Add them together to get the total pressure.
3.7 atm
What is the pressure (in atm) in the flask after the reaction goes to completion? _____________
Page 4
Chemistry 111 Sect. 2
Exam #3
May 4, 2006
Name: ________________________________________________
KEY
Student #: ______________________________________________
13. (10 pts) The Lewis structure of carbon monoxide (CO) is shown at right.
:C
O:
Indicate which of the original atomic orbitals types of carbon (indicated by the
letters A-F) are used to form CO (some orbitals may be rearranged into molecular orbitals). All
letters A-F must be used. Some can be used multiple times. Some lines will have multiple letters.
A. 1s
B, C
_______________
used to form the CO sigma bond
B. 2s
C
_______________
used to form the first of the CO pi bonds
C. 2p
C
_______________
used to form the second of the CO pi bonds
D. 3s
B, C
_______________
used to house the carbon lone pair of electrons
A, D, E, F
_______________
not used
The valence electrons are used in bonding. To make the hybrid sp
F. 3d
orbitals for the lone pair and the sigma bonding electrons, one 2s and
one 2 p orbital hybridize. Another of the 2 p orbitals forms the first
pi bond. The remaining 2p orbital forms the second pi bond.
14. (10 pts) A lecture demonstration used a glass tube,
NH4Cl
pictured at the right, involving diffusing NH3 and HCl
HCl
NH3
gases. Where these gases meet, they react to form a cloud
of NH4Cl. The experiment was done at T = 25 ºC.
E. 3p
If the temperature of the experiment was increased to T = 100 ºC, would you expect the NH4Cl
to form at the same place, to the right or to the left of the location where it formed at T = 25 ºC?
Diffusion rate is proportional to urms = √3RT/M.. The position of the
NH4Cl will be at a point where the diffusing NH3 and HCl meet. If T
increases, both the NH3 and HCl will diffuse faster, and both will go
faster by the same factor. Thus the NH4Cl will form more quickly, but
will be at exactly the same location.
same place
_________________
NH3 also forms a visible compound if the HCl is replaced with HF. Would you expect the NH4F
to form at the same place, to the right or to the left of the location where the NH4Cl formed?
The molar mass of HF is less than that of HCl, so its urms will be faster
than that of HCl. Thus, the cloud of NH4F will form farther from the HF
end than the cloud of NH4Cl forms from the HCl end. That is, the cloud
of NH4F will form to the left of the position where the cloud of NH4Cl
formed.
to the left
_________________
Page 5
Chemistry 111 Sect. 2
Exam #3
May 4, 2006
Name: ________________________________________________
KEY
Student #: ______________________________________________
Useful information
NA = 6.02 × 1023 molecules/mol
c = 3.0 × 108 m/sec
h = 6.6 × 10-34 J sec
Rhc = 2.2 × 10-18 J
me = 9.1 × 10-28 g
a0 = 0.0529 nm
R = 0.082 L atm/mol K
2-linear, 3-trigonal planar, 4-tetrahedral
5-trigonal bipyramidal, 6-octahedral
q = C m ∆T
q = Lv m or q = Lf m
∆H = qp
d = m/V = PM/RT
E = -Rhc/n2
E = hν
c = λν
λ = h/(m v)
Z* = Z - ninner electons
∆Hrxn = Σ(bond ener)react - Σ(bond ener)prod
∆H0net = Σ (∆H0f)prod - Σ (∆H0f)react
∆H0net = Σ (∆H0rxn)
PV = nRT
urms =
3RT
M
Some useful thermodynamic data
Material
∆H0f (kJ/mol)
∆H0f (kJ/mol)
-74.8
H2O (l)
-285.8
-110.5
H2O (g)
-241.8
-393.5
H2O2 (l)
-187.8
279.0
N (g)
472.7
-296.8
NH3 (g)
-46.1
-395.8
NO (g)
90.9
Material
CH4 (g)
CO (g)
CO2 (g)
S (g)
SO2 (g)
SO3 (g)
PERIODIC TABLE OF THE ELEMENTS
IA
IIA
IIIB
IVB
VB
VIB
VIIB
VIIIB
IB
IIB
IIIA
IVA
VA
VIA
VIIA VIIIA
1
2
H
He
1.008
4.003
3
4
5
Li
Be
B
C
N
O
F
Ne
6.939
9.012
10.81
12.01
14.01
16.00
19.00
20.18
11
12
13
14
15
16
17
18
Na
Mg
Al
Si
P
S
Cl
Ar
22.99
24.31
26.98
28.09
30.97
32.07
35.45
39.95
19
20
21
22
K
Ca
Sc
Ti
39.10
40.08
44.96
37
38
39
Rb
Sr
85.47
55
23
6
7
8
9
10
24
25
26
27
28
29
30
31
32
33
34
35
36
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
47.90
50.94
52.00
54.94
55.85
58.93
58.71
63.55
65.39
69.72
72.61
74.92
78.96
79.90
83.80
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
87.62
88.91
91.22
92.91
95.94
(99)
101.1
102.9
106.4
107.9
112.4
114.8
118.7
121.8
127.6
126.9
131.3
56
57
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
Cs
Ba
La
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
132.9
137.3
138.9
178.5
181.0
183.8
186.2
190.2
192.2
195.1
197.0
200.6
204.4
207.2
209.0
(209)
(210)
(222)
87
88
89
Fr
Ra
Ac
(223)
226.0
227.0
Page 6
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Xe