095 Version Make up Test 3 laude (50995)
This print-out should have 30 questions.
Multiple-choice questions may continue on
the next column or page nd all choices
before answering.
001
6.0 points
You set up a bomb calorimetry experiment using 1 liter of water as your heat sink
and combusting a 4.409 g sample of propane
(C3 H8 ). If the initial and nal temperature are 24.90 C and 77.96 C respectively,
what is the molar enthalpy of combustion of
ethene? (Assume the calorimeter itself absorbs no heat. Assume the density of water is
1 g mL)
1. 222.0 kJ mol1
2. 22, 200 kJ mol1
3. 22.20 kJ mol1
4. 2, 220 kJ mol1 correct
5. 222, 000 kJ mol1
Explanation:
T = Tf Ti = 77.96 C 24.90 C =
53.06 C = 53.06 K
1000 mL 1.00 g
= 1000 g
L
mL
1mol
n = 4.409 g propane
= 0.1 mol
44.09 g
propane
m=1L
Hcal
mcT
Hrxn
=
=
n
n
n
4.184 J
1000 g gK 53.06 K
=
0.1 mol
= 2, 220 kJ mol
002
1
1
G if it is performed under standard conrxn
ditions.
H f
S
kJ
J
species
( mol )
( molK )
HNO3 ()
NO2 (g)
HNO3 ()
H2 (g)
285.8
33.2
174.1
0
70
240
156
131
1. 85 kJ mol1
2. 6, 200 kJ mol1
3. 85 kJ mol1 correct
4. 6, 200 kJ mol1
Explanation:
Hrxn = Hf,products Hf,reactants =
(285.8 kJ mol1
+
33.2 kJ mol1 )(174.1 kJ mol1 )=
78.5 kJ mol1
Srxn = Sm,products Sm,reactants =
(70 J mol1 K1
+
1
1
1
1
240 J mol K )(156 J mol K
+
131 J mol1 K1 )=
23 J mol1 K1 =
0.023 kJ mol1 K1
G
=
rxn
Hrxn T Srxn = 78.5 kJ mol1
(273 K)(0.023 kJ mol1 K1 )=
85 kJ mol1
003
6.0 points
What is the change in entropy for iodine
melting? The standard molar entropy for
I2 (s) and I2 () are 116.14 J mol1 K1 and
260.69 J mol1 K1 , respectively.
1. 376.83 J mol1 K1
2. 144.55 J mol1 K1
6.0 points
3. 376.83 J mol1 K1
Consider the reaction below:
HNO3 () + H2 (g) H2 O() + NO2 (g)
Using the provided table values, calculate
4. 144.55 J mol1 K1 correct
Explanation:
The change in entropy for condensation
Version 095 Make up Test 3 laude (50995)
of water can be described by the equation
I2 (s) I2 ().
Smelting = Sm,liquid Sm,solid
= 260.69 J mol1 K1
116.14 J mol1 K1
= 144.55 J mol1 K1
004
6.0 points
Ammonia (the common name for NH3 ,
the odor of windex and cat urine), has a
Hvap = 23.35 kJ mol1 and a Svap =
97.43 J mol1 K1 . What is the normal
boiling point of ammonia expressed in centigrade?
1. 273 C
2. 0.2 C
3. 238.7 C
2
2. 1, 124 kJ mol1 correct
3. 1, 124 kJ mol1
4. 562 kJ mol1
Explanation:
Hrxn = Hf,products Hf,reactants =
(2 20.63 kJ mol1 + 3 0) (2
285.83 kJ mol1 +2 296.83 kJ mol1 )=
1, 124 kJ mol1
006 6.0 points
Which of the following statements is/are true
concerning the rst law of thermodynamics?
I) The internal energy (U ) of the universe
is conserved.
II) The internal energy of a system plus that
of its surroundings is conserved.
III) The change in internal energy (U ) of a
system and its surroundings can have the
same sign.
4. 33.3 C correct
Explanation:
Because boiling is an equilibrium process,
G = 0 = Hvap T Svap .
vap
And so T Svap = Hvap
and T = Hvap /Svap
= 23, 350J mol1 /97.43 J mol1 K1
= 238.7 K
= 33.3 C
005
6.0 points
Based on enthalpy of formation data calculate Hrxn for the following:
2H2 O() + 2SO2 (g) 2H2 S(g) + 3O2 (g)
species
H f
kJ
( mol )
H2 S(g)
O2 (g)
H2 O()
SO2 (g)
20.63
0
285.83
296.83
1. 562 kJ mol1
1. I, III
2. II only
3. II, III
4. I only
5. I, II, III
6. I, II correct
7. III only
Explanation:
Statement I and II are true; the rst law
states that the internal energy of the universe
is conserved and since the system plus the
surroundings is the universe, their sum is also
conserved. Statement III is false; for example,
if both the system and its surroundings had
a positive change in internal energy, then the
internal energy of the universe would have
increased or decreasesed - in violation of the
rst law.
Version 095 Make up Test 3 laude (50995)
3
1. d < a < b < c
007
6.0 points
2. c < d < b < a
Which of the following compounds is unstable under standard conditions?
I) C8 H18 ()
II) H2 S(aq )
III) SF6 (g )
species
C8 H18 ()
H2 S(aq )
SF6 (g )
G
f
kJ
( mol )
6. 4
27.8
1105.3
1. I correct
2. I, II, III
3. I, II
4. III
5. II
3. a < c < b < d
4. b < c < d < a
5. d < b < a < c
6. c < b < d < a correct
Explanation:
Reaction c has a substantial decrease both
in total moles and especially in moles of gas;
Srxn will likely be a large negative number. Reaction b has no change in total moles,
but does form a solid from a pair of aqueous
species; Srxn will likely be a small negative
number. Reaction d has no change in total
moles or moles of gas; Srxn will be close to
zero. Reaction a produces more total moles,
and importantly evolves a gas; Srxn will
likely be a large positive number.
009
6. II, III
7. All compounds are stable
8. I, III
6.0 points
Which of the following reactions is an enthalpy of formation reaction
1. CH4 (g) Cgraphite + 2 H2(g)
Explanation:
Examine the standard free energies of formation of the compounds.
Compounds with a negative free energy of
formation are stable, whereas those with a
positive free energy are unstable with respect
to the elements that compose them.
008
6.0 points
Rank the following reactions from most
negative to most positive in terms of change
in entropy (Srxn ):
a) CaCO3 (s) CaO(s) + CO2 (g)
b) AgNO3 (aq) + NaCl(aq)
NaNO3 (aq) + AgCl(s)
c) 4Fe(s) + 3O2 (g) 2Fe2 O3 (s)
d) CH4 (g) + 2O2 (g) CO2 (g) + 2H2 O(g)
2. 2 Fe(s) +
3
O2 (g) Fe2 O3 (s) correct
2
3.
NaOH(aq) + HCl(g) H2 O() + NaCl(aq)
3
1
N2 () + H2 (g) NH3 (g)
2
2
Explanation:
Formation reactions describe production of
exactly one mole of one product from stoichiometric quantities of elements in their standard
states.
4.
010 6.0 points
Which of the following is not a denition of
enthalpy or change in enthalpy?
1. A state function that inuences the spon-
Version 095 Make up Test 3 laude (50995)
taneity of a reaction.
2. H = Hf inal Hinitial
3. H = U + P V
4. The heat transfer at constant volume.
correct
Explanation:
The heat transfer at constant volume is
equal to the change in internal energy, not the
change in enthalpy. The other choices are all
mathematical identities relating to enthalpy
or change in enthalpy or prose restatements
of these mathematical identities.
011
Ethane is C2 H6 .
below:
6.0 points
Consider the reaction
C2 H6 (g) 2Cgraphite (s) + 3H2 (g)
For a given reaction, if Hrxn is (neg
ative/positive/either) and Srxn is (negative/positive/either), then the value of G
rxn
will always be negative, regardless of the temperature.
1. either, negative
2. either, positive
3. negative, positive correct
4. negative, either
5. positive, negative
6. positive, either
Explanation:
This can be intuited from the G = H
T S formula. In order for G to always
rxn
be negative, Hrxn must always be negative
and Srxn must always be positive.
G = 7.86 kJ mol1
rxn
The (formation/decomposition) of octane is
(spontaneous/non-spontaneous) and therefore octane is (stable/unstable).
1. decomposition, non-spontaneous, stable
correct
4
013
6.0 points
Consider the reaction below and approximate the value of the work function at room
temperature.
C2 H4 (g) + 3O2 (g) 2CO2 (g) + 2H2 O()
2. formation, spontaneous, unstable
3. formation, non-spontaneous, unstable
1. 5 kJ
4. formation, non-spontaneous, stable
2. 2.5 kJ
5. decomposition, non-spontaneous, unstable
Explanation:
The reaction written above describes the
decomposition of ethane and has a positive
change in Gibbs free energy, i.e. is nonspontaneous, meaning octane must be thermodynamically stable.
012
6.0 points
3. 5 kJ correct
4. 0 kJ
5. 2.5 kJ
Explanation:
At room temperature ( 300 K), the product of the RT term in w = ngas RT is approximately 2.5 kJ. Since ngas is 2 for this
reaction, the work function is approximately
5 kJ.
Version 095 Make up Test 3 laude (50995)
014
6.0 points
If an MP3 player does 500 kJ of work and
releases 100 kJ of heat, what is the change in
internal energy for the MP3 player?
1. 400 kJ
2. 400 kJ
3. 600 kJ correct
5
When calculating the maximum positional
entropy of a system at absolute zero using
the Boltzmann formula (S = k ln W ), which
of the following statements is/are true?
I) For a given compound, a 1 mol has
a maximum positional entropy roughly
6.022 1023 times that of 1 molecule.
II) The minimum value that W can have is
1.
III) For the same number of molecules, the
experimental value of W for NO is less
than the experimental value for CO.
1. I, III
4. 600 kJ
Explanation:
U = q + w
2. I, II, III correct
3. II only
= 500 kJ + 100 kJ = 600 kJ
4. II, III
015 points
Which 6.0 of the following statements concerning
calorimetry is/are true?
I) Bomb calorimeters hold the volume of
the system constant.
II) In real calorimetry experiments, the
calorimeter itself absorbs heat.
III) H = q in a bomb calorimeter.
1. III only
2. I, II, III
3. I, III
4. I, II correct
5. II, III
6. II only
5. I, II
6. III only
7. I only
Explanation:
Statement I is true; when the exponent of
the W term is some multiple of Avogadros
number, it can be factored out and moved to
the front of the expression, serving as a coecient. Statement II true; the natural log
of a value less than one would result in a
negative value for entropy, which would violate the third law. Statement III is true;
both molecules have 2 orientations in theory,
but both have IMF which will perturb this
value toward 1, more so for NO because it has
stronger IMF.
017
7. I only
Explanation:
Statements I and II are true. Statement
II is falsewhen the volume is constant, the
change in internal energy is equal to heat, not
the change in enthalpy.
016
6.0 points
6.0 points
Rank the following compounds by molar entropy, from least to greatest:
O2 (), O2(g), O3 (g), O2(s).
1. O3 (g) < O2 () < O2 (s) < O2 (g)
2. O2 (s) < O2 () < O3 (g) < O2 (g)
Version 095 Make up Test 3 laude (50995)
3. O3 (g) < O2 (s) < O2 () < O2 (g)
4. O2 (g) < O2 () < O2 (s) < O3 (g)
5. O2 (s) < O2 () < O2 (g) < O3 (g) correct
Explanation:
The three phases of O2 can be easily ranked,
since phase changes have very predictable effects on molar entropy, typically increasing
signicantly going from solid to liquid and
again from liquid to gas. Comparing O2 (g)
to O3 (g), simply require comparing number
of bonds, with O3 (g) having greater molar
entropy because it has more bonds.
018 6.0 points
Given the data
H2 SO4 () H2 S(g) + 2 O2 (g)
H = +793 kJ mol1
H2 O(g) + SO3 (g) H2 SO4 ()
H = 176 kJ mol1
H2 O() + SO3 (g) H2 S(g) + 2 O2 (g)
H = +661 kJ mol1
calculate H for the reaction
019
6
6.0 points
If a system absorbs 545 J of heat from the
surroundings at 272 C, what is change in
entropy of the surroundings?
1. 545 J K1 correct
2. 2 J K1
3. 2 J K1
4. 545 J K1
Explanation:
Heat absorbed by something which is always near equilibrium at constant temperature will increase its entropy by an amount
equal to q/T . For most situations we consider, the surroundings can be assumed to
meet this criterion, but the systems we consider usually do not. In this case 545J/1K =
545 J K1 .
020
H2 O() H2 O(g)
6.0 points
How much energy would be associated with
the rotational motion of 0.75 moles of N2 ?
1. 214 kJ mol1
2. 1278 kJ mol1
1. 2RT
3. 1278 kJ mol1
2. 2kT
4. 41 kJ mol1
3. 0.75kT
5. 41 kJ mol1 correct
4. 0.37kT
Explanation:
The rst two reactions need to be reversed:
H2 SO4 () H2 S(g) + 2 O2 (g)
H = 793 kJ mol1
H2 O(g) + SO3 (g) H2 SO4 ()
H = +176 kJ mol1
H2 O() + SO3 (g) H2 S(g) + 2 O2 (g)
H = +661 kJ mol1
Hrxn = 44 kJ mol1
5. 0.37RT
6. 0.75RT correct
Explanation:
The energy associated with any motional
mode is 0.5kT , which when multiplied by 2
rotational modes (N2 is linear) and 0.75 moles
(i.e. 0.5 times Avogadros number), results in
0.75RT .
Version 095 Make up Test 3 laude (50995)
021
6.0 points
Assuming you have 1 mol of each of the
following, rank them by the value of the term
W in the Boltzmann equation, from greatest
to least: HCl, HBr, HF.
7
II) 3CO2 (g) + 4H2 O(g) C3 H8 (g) + 5O2
III) 3Cgraphite + 4H2 (g) C3 H8 (g)
1. I only
2. I, III
3. III only
1. HCl > HF > HBr
4. II only
2. HF > HBr > HCl
5. none of the above correct
3. HBr > HF > HCl
6. I, II
4. HCl > HBr > HF correct
Explanation:
Since the strength of the IMF determines
the extent of the perturbation of W from its
ideal value, this problem essentially requires
ranking by increasing IMF. The greater the
IMF, the closer the value will be to unity.
022 6.0 points
Which of the following is not a denition of
internal energy or change in internal energy?
1. The change in enthalpy after correction
for P V work.
2. The dierence between the nal and initial internal energy of a system.
3. The heat transferred at constant pressure.
correct
4. U = q + w
Explanation:
All of the reactions consume gas, or have
fewer mols of products than reactants, suggesting they all have a negative value for
change in entropy and thus would become less
spontaneous as the temperature increases.
024 6.0 points
Which of the following statements concerning
the second and third laws of thermodynamics
is/are true?
I) Gsystem /T = Suniverse
II) In a perfect, pure crystal at absolute zero,
entropy would be a small negative value.
III) The entropy of the universe is always
increasing.
1. I only
2. II only
Explanation:
Internal energy is not equal to heat at constant pressure. The other statements are
all mathematical identities describing internal energy or prose restatements of the same.
023
7. II, III
6.0 points
Which of the following reactions will likely
become more spontaneous as you raise temperature?
I) CO(g) + H2 (g) CH2 O()
3. III only correct
4. I, III
5. I, II
6. I, II, III
7. II, III
Explanation:
Statement I is false;
Gsystem/T
=
Version 095 Make up Test 3 laude (50995)
Suniverse . Statement II is false; the third
law states that as the temperature of a system approaches absolute zero, the entropy of
thee system approaches its minimum, which
in the case of a perfect crystalline solid, is zero
- moreover, entropy can never have a negative
value. Statement III is true; the second law
of thermodynamics states that the entropy of
the universe is always increasing.
025 6.0 points
Which if the following phrases corresponds to
a positive change in a state function?
I) non-spontaneous
II) exothermic
III) an increase in disorder
8
Explanation:
Work quantitatively describes a transition
between states and is thus a process quantity
(path function), not a state function.
027
6.0 points
Calculate Suniverse for the condensation
of heavy water (D2 O) at 95 C if the Sm for
D2 O(g) and D2 O() are 198 J mol1 K1
and 76 J mol1 K1 , respectively.
Assume the standard boiling point of heavy
water is 101 C and that Hsystem =
45.6 kJ mol1 .
1. 122 J mol1 K1
1. I, II
2. 124 J mol1 K1
2. I only
3. 122 J mol1 K1
3. II, III
4. 2 J mol1 K1 correct
4. I, II, III
5. 2 J mol1 K1
5. III only
6. 124 J mol1 K1
6. II only
7. I, III correct
Explanation:
Sign conventions dictate that an endothermic process has a positive change in gibbs free
energy, that exothermic processes have a negative value for change in enthalpy, and that
an increase entropy is positive.
026 6.0 points
Which of the following is a not state function?
1. internal energy
2. temperature
3. enthalpy
4. Gibbs free energy
5. work correct
Explanation:
Since the system in this case is condensing
heavy water,
D2 O(g) D2 O()
Ssystem = Sm,products Sm,reactants
= 76 J mol1 K1 198 J mol1 K1
= 122 J mol1 K1
Hsystem = T Ssystem
= 374 K 122 J mol1 K1
= 45, 628 J mol1
Since the only process during a phase
change is heat exchange,
Ssurroundings = Hsystem/Tsurroundings
= 45, 628 J mol1 /368 K
= 123.98 J mol1 K1
Suniverse = Ssystem + Ssurroundings
= 122 J mol1 K1 + 124 J mol1 K1
= 2 J mol1 K1
Version 095 Make up Test 3 laude (50995)
9
(2)(743 kJ mol1 ) + (2)(2)(463 kJ mol1 )
028
6.0 points
What is the maximum positional entropy
of 3 mol of CHCl3 at absolute zero?
030 6.0 points
Which of the following would require the most
energy to raise its temperature by 1 K?
1. 11.53 J K1
2. 5.74 1023 J K1
1. 10 grams of water correct
3. 34.58 J K1 correct
2. 1 gram of water
4. 1.91 1023 J K1
Explanation:
CHCl3 has 4 possible orientations.
k ln(4n ) = 3R ln 4 = 34.58 J K1
029
3. 1 0 grams of iron metal
S=
6.0 points
Using bond enthalpy data, calculate the
change in enthalpy for the following reaction.
CH4 (g) + 2O2 (g) CO2 (g) + 2H2 O(g)
bond
C
O
C
O
Bond enthalpy
(kJ mol1)
H
O
O
H
= 698 kJ mol1
412
496
743
463
1. 698 kJ mol1 correct
2. 203 kJ mol1
3. 1420 kJ mol1
4. 203 kJ mol1
5. 1420 kJ mol1
6. 698 kJ mol1
Explanation:
Hrxn = BEreactants BEproducts
=
(4)(412 kJ mol1 ) + (2)(496 kJ mol1 )
4. 1 gram of iron metal
Explanation:
Heat capacity is an extensive property, that
reects the amount of energy required to raise
the temperature of an object. Because water
has a much higher heat capacity than metals
(to some extent a consequence of stronger
IMF), and because 10 grams is more than
1 gram, the 10 gram smaple of water would
require the greatest amount of enegy to raise
its temperature by 1 K.