121
EXAM 4
version
001-130 in GSB 2.124
131-260 in PAI 4.42
Supraj Prakash
sp35838
McCord 2pm / 52125
signature:
Thermodynamic Data at 25◦ C
∆Hf◦
S◦
kJ/mol
J/mol K
—
28
Al2 O3 (s)
CH4 (g)
C3 H8 (g)
-1676
-75
-104
51
186
270
C10 H8 (s)
C12 H22 O11 (s)
77
-2222
167
360
CO (g)
CO2 (g)
-111
-394
198
214
COCl2 (g)
Cl2 (g)
-220
—
284
223
H2 O (ℓ)
H2 O (g)
NH4 ClO4 (s)
-286
-242
-295
70
189
186
NH4 Cl (s)
NH4 NO3 (s)
-314
-366
95
151
NO2 (g)
N2 H4 (ℓ)
33
50
240
12
N2 O (g)
O2 (g)
82
—
220
205
PCl3 (g)
PCl5 (g)
-287
-375
312
365
Substance
Al (s)
(values are rounded to the nearest integer)
Single Bond Energies (kJ/mol)
H C N O
S
F Cl
Br
H 436
C 413 346
N 391 305 163
O 463 358 201 146
S 347 272 — — 226
F 565 485 283 190 284 155
Cl 432 339 192 218 255 253 242
Br 366 285 — 201 217 249 216 193
Multiple Bond Energies (kJ/mol)
C=C 602
C=N 615
C=O 799
C≡C 835
N=N 418
C≡N 887
O=O 498
C≡O 1072
N≡N 945
Version 121 – Exam 4 – mccord – (52125)
This print-out should have 28 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001 (part 1 of 3) 4.0 points
Consider the following chemical reaction for
this multi-part question:
NH4 NO3 (s) → N2 O(g) + 2H2 O(ℓ)
2
∆S = 220 + 2(70) − 151
= 360 − 151
= 209 J/K
This is 0.209 kJ/K, needed for part 3.
003 (part 3 of 3) 4.0 points
What is the change in the standard Gibb’s
free energy for this reaction?
1. -186 kJ correct
What is the change in the standard reaction
enthalpy for this reaction?
2. -62 kJ
1. -124 kJ correct
3. -257 kJ
2. -36 kJ
4. 100 kJ
3. 162 kJ
5. -224 kJ
4. 75 kJ
6. 13 kJ
5. -570 kJ
6. -176 kJ
Explanation:
∆H = 82 + 2(−286) − (−366)
= −490 + 366
= −124 kJ
002 (part 2 of 3) 4.0 points
What is the change in the standard reaction
entropy for this reaction?
Explanation:
∆G = ∆H − T ∆S
= −124 − 298.15(0.209)
= −124 − 62
= −186 kJ
004 4.0 points
The heat of combustion of butane (C4 H10 ) is
−2878 kJ/mol. Assume a typical cigarette
lighter holds about 1.75 g of butane. How
much heat energy would be released if ALL of
the butane in the lighter were combusted?
1. 1.05 kJ
1. 230 J/K
2. 95.59 kJ
2. 139 J/K
3. 86.63 kJ correct
3. 108 J/K
4. 40.61 kJ
4. 311 J/K
5. 209 J/K correct
6. 265 J/K
Explanation:
Explanation:
MWbutane = 58.14 g/mol
−2878 kJ
1 mol
· (1.75 g) ·
mol
58.14 g
= −86.6271 kJ
? kJ =
Version 121 – Exam 4 – mccord – (52125)
005 4.0 points
When sodium chloride is melted, the sign of
qsys and ∆Ssys are
and
, respectively.
1. +, + correct
2. +, −
3. −, +
4. −, −
Explanation:
The disorder is increased for the process.
Melting is an endothermic process, therefore
qsys will be positive.
006 2.0 points
Which term is the intensive measure of heat
flow relative to temperature change?
1. calorimetry
2. enthalpy
3. entropy
4. heat capacity
5. specific heat capacity correct
2. YES, I would like the points and the
higher score. correct
Explanation:
This should be a no-brainer. Most students
want higher scores. If you picked yes, you got
credit for the question and you got the extra
points you asked for (if they were granted
by your instructor). If you answered NO,
you also got what you wanted... no points
awarded.
008 4.0 points
A new snack food is analyzed for its energy
content (Calories). The snack is ground into a
powder, and then pressed into a 1.36 g sample
pellet that is suitable to use in a standard
bomb calorimeter.
The bomb calorimeter has 2050 g of water
in it and has a known hardware heat capacity
of 1750 J/◦ C. The combustion of the snack
food is carried out and the temperature climbs
from 26.25◦C to 30.12◦ C.
The “official” serving size of this snack is
1 ounce (28.4 g). Use the calorimetry data
above to calculate the number of food calories
per serving of this snack food. Note that
a food calorie (Cal) is actually a kilocalorie
where 1 kcal = 4.184 kJ.
1. 200 Cal/serving
Explanation:
Heat capacity is heat/temperature change
which is extensive. The intensive version has
per amount also, heat/(amount · temperature
change), or J/g ◦ C. This is known as the
specific heat capacity or molar heat capacity
if shown per mole.
2. 158 Cal/serving
007 0.0 points
This question starts out at zero points but
could very well increase after the grading.
Now, if more points are awarded (the curve)
on this assignment, would you like them
added to your score?
6. 226 Cal/serving
1. NO, leave my score alone, I prefer the
lower score
3
3. 310 Cal/serving
4. 285 Cal/serving
5. 182 Cal/serving correct
Explanation:
∆T = 30.12 − 26.25 = 3.87◦C
qcal = qw + qhard
qcal = mCs ∆T + Chard ∆T
qcal = 2050 · 4.184 · 3.87◦ + 1750 · 3.87◦
qcal = 33194 + 6773
qcal ≃ 40000 = 40.0kJ
Divide this total amount of heat released
Version 121 – Exam 4 – mccord – (52125)
by the amount used of 1.36 g.
energy per gram = 40kJ/1.36g = 29.4 kJ/g
scale up for a serving:
29.4 kJ/g (28.4 g/serving) = 835 kJ/serving
and finally change to Calories: 835/4.184 =
200 kcal/serving
009 4.0 points
During a particular exothermic reaction gases
are evolved, causing the volume of the system
to increase. What can you say about the value
for ∆U?
1. It will be positive.
3. The sign of ∆U cannot be determined
without more information.
Explanation:
∆U = q + w
For exothermic reactions, q is negative.
For reactions which evolve gases and cause the
system volume to increase, work is being done
by the system and the work has a negative
sign.
Because ∆U is the addition of two negative
values, it will also be negative.
010 4.0 points
Reaction of tertiary butyl alcohol with hydrobromic acid produces tertiary butyl bromide
by the following reaction. Use bond energies
(provided in preamble) to estimate the change
in enthalpy, ∆H, for this reaction.
Br
+ HBr
4. +24 kJ/mol
5. −186 kJ/mol
6. +105 kJ/mol
Explanation:
In this reaction, you break two bonds in the
reactants: a C-O bond and a H-Br bond. You
form new bonds in the products: a C-Br bond
and a H-O bond.
The change in enthalpy is the energy in for
breaking the bonds combined with the energy
out gained from forming the new bonds.
∆Hrxn = B.E.C−O + B.E.H−Br
−B.E.C−Br − B.E.H−O
2. It will be negative. correct
OH
4
+ H2 O
∆Hrxn = 358 + 366 − 285 − 463
∆Hrxn = −105 kJ mol−1
011 2.0 points
A process where the system does work and
simultaneously releases heat will always be a
non-spontaneous process.
1. True
2. False correct
Explanation:
False. Heat and work only directly affect
the internal energy and enthalpy. Spontaneity is governed by universal entropy, or the
system’s change in free energy. The signs
on ∆U or ∆H do not govern spontaneity by
themselves.
012 4.0 points
3560 joules of heat are added to 228 grams
of water originally at 28◦ C. What is the final
temperature of the water?
1. 35.4◦ C
1. −105 kJ/mol correct
2. 31.7◦ C correct
2. +186 kJ/mol
3. 30.2◦ C
3. −24 kJ/mol
4. 36.6◦ C
Version 121 – Exam 4 – mccord – (52125)
5. 33.5◦C
Explanation:
q = 3560 J
Ti = 28◦ C
m = 228 g
C = 4.184 J/g ◦ C
5
2 H2 (g) + O2 (g) → 2 H2 O(g)
∆H ◦ = −484 kJ · mol−1
N2 (g) + 3 H2(g) → 2 NH3(g)
∆H ◦ = −92.2 kJ · mol−1
1. −1119 kJ · mol−1
q = m C (Tf − Ti )
q
= Tf − Ti
mC
q
Tf = Ti +
mC
= 28
3560
+
(228) (4.184)
= 31.7◦ C
013 4.0 points
A particular protein folds spontaneously at
25 ◦ C and 1 atm. During this folding, the protein changes conformation from a higher entropy unfolded state to a lower entropy folded
state. For this process, ∆H is
1. No way to know
2. ∆H > 0
3. ∆H = 0
4. ∆H < 0 correct
Explanation:
If the reaction proceeds spontaneously, then
∆G must be negative. The reaction also transitions from a higher entropy state to a lower
entropy state, resulting in a negative ∆S. Using the Gibbs’ Free Energy Equation, ∆H
must be negative in order to have both a negative ∆G and a negative ∆S.
014 4.0 points
Calculate the standard reaction enthalpy
for the reaction
N2 H4 (ℓ) + H2 (g) → 2 NH3(g)
given
N2 H4 (ℓ) + O2 (g) → N2 (g) + 2H2 O(g)
∆H ◦ = −543 kJ · mol−1
2. −151 kJ · mol−1 correct
3. −935 kJ · mol−1
4. −243 kJ · mol−1
5. −59 kJ · mol−1
Explanation:
We need to reverse the second reaction and
add them:
N2 H4 (ℓ) + O2 (g) → N2 (g) + 2 H2O(g)
∆H = −543 kJ/mol
2 H2 O(g) → 2 H2 (g) + O2 (g)
∆H = +484 kJ/mol
N2(g) + 3 H2 (g) → 2 NH3(g)
∆H = −92.2 kJ/mol
N2 H4 (ℓ) + 3 H2 (g) → 2 H2 (g) + 2 NH3(g)
N2 H4 (ℓ) + H2 (g) → 2 NH3(g)
∆H = −151.2 kJ/mol
015 4.0 points
If the change in internal energy for a given
process is zero, which of the following must be
true?
1. q = 0
2. w = 0
3. q = -w correct
4. ∆E = ∆H
Explanation:
∆E = q + w = 0
q = -w
016 4.0 points
A 100 g cube of iron metal (Cs = 0.45 J/g
Version 121 – Exam 4 – mccord – (52125)
◦ C)
at 25◦ C is put in contact with 50 g of
an unknown metal at 125◦ C. After 1 minute
of contact, the two metals reach the same
temperature of 75◦ C. What is the specific
heat of the unknown metal?
6
Consider a chemical reaction where ∆S is
36.1 J/mol K, and ∆H is -2.88 kJ/mol. What
is the change in entropy for the universe
(∆Suniv ) for this reaction at 50◦ C?
1. +27.2 J/mol K
1. 0.90 J/g ◦ C correct
2. +45.0 J/mol K correct
2. 1.12 J/g ◦ C
3. -36.1 J/mol K
3. 0.23 J/g ◦ C
4. -47.9 J/mol K
4. 0.45 J/g ◦ C
5. +40.5 J/mol K
5. 0.68 J/g ◦ C
Explanation:
qFe = qx
m · Cs,Fe · ∆T = m · Cs,x · ∆T
100(.45)(50) = 50(Cs,x )(50)
0.90 = cs,x
017 4.0 points
Calculate the entropy of vaporization for compound X at its boiling point of 112◦ C . The
enthalpy of vaporization of compound X is
46.8 kJ/mol .
1. 88.8574
2. 123.527
3. 97.8427
4. 121.511
5. 117.38
6. 85.3346
7. 78.7667
8. 86.2764
9. 73.7224
10. 60.7528
Explanation:
The heat leaving the system enters the surroundings. Therefore
∆Ssurr = −∆H/T
∆Ssurr = −(−2880)/323.15
∆Ssurr = +8.91 J/mol K
Now ∆Suniv can be easily calculated:
∆Suniv = ∆Ssys + ∆Ssurr
∆Suniv = 36.1 + 8.91 = 45.0 J/mol K
019 4.0 points
Which of the following reactions corresponds
to the value of the heat of formation for the
product molecule as found in the tables?
1. Na(s) +
1
2
Cl2 (g) → NaCl(s) correct
2. 2 H2 (g) + O2 (g) → 2 H2 O(ℓ)
3. CO2 (s) → CO2 (g)
4. SO2 (g) +
1
2
O2 (g) → SO3 (g)
∆H
46800 J/mol
=
T
385.15 K
= 121.511 J/molK
Explanation:
All of the reactants in a formation reaction
must be ELEMENTS in their thermochemical
standard state. AND, to match table values,
only 1 MOLE of product can be formed because standard table values are in kJ/mol.
Only the NaCl fits this definition. The H2 O
reaction unfortunately is forming 2 moles of
water instead of one.
018
020 2.0 points
Under standard thermodynamic conditions
Correct answer: 121.511 J/molK.
Explanation:
T = 112◦ C + 273.15 = 385.15 K
∆H = 46.8 kJ/mol = 46800 J/mol
∆S =
4.0 points
Version 121 – Exam 4 – mccord – (52125)
which of the following choices best describes
the relationship of enthalpy (H) to internal
energy (U ) for a given system?
7
5. ∆Ssurr = −∆Ssys
1. H = U for all cases
Explanation:
The entropy change for the surroundings is
related to the heat for the process such that
2. H > U for all cases correct
∆Ssurr =
3. H < U for all cases
where q is from the perspective of the system
(out of the system into the surroundings). At
a constant temperature there is only one temperature T. At constant pressure the heat for
a process (from the perspective of the system)
is the change in enthalpy of the system. Thus
Explanation:
Definition of enthalpy is:
H = U + PV
Under standard conditions the P V term will
have to be non-zero and positive, thus making
H greater than U by that amount in all cases.
qsurr
Tsurr
=
−q
Tsurr
−∆Hsys
T
∆Ssurr =
1. work
023 4.0 points
A 15 g sample of steam at 110 ◦ C was placed
into a warehouse freezer at −40 ◦ C. In order to
properly calculate the total change in entropy
of this system, what equations would you use?
! "
f
I. ∆S = nC ln T
Ti
2. kinetic energy
II. ∆S =
021 2.0 points
Which of the following terms best matches
this definition:
energy dispersed relative to temperature
3. enthalpy
4. potential energy
5. entropy correct
6. heat
7. free energy
Explanation:
entropy matches this definition
∆H
T
III. ∆Suniv = ∆Ssys + ∆Ssurr
IV. ∆S = mC∆T
1. II, IV
2. I, II correct
3. IV
4. I, III
5. I, II, III, IV
022 4.0 points
At constant pressure and temperature, which
of the following is true about ∆Ssurr
1. ∆Ssurr = −∆Hsys /T correct
2. ∆Ssurr = −T∆Hsys
Explanation:
Equation I is used three times (cooling the
steam from 110 to 100 ◦ C, cooling the water
from 100 to 0 ◦ C, and cooling the ice from 0
to −40 ◦ C). Equation II would be used two
times (condensing the steam and freezing the
water).
3. ∆Ssurr = −T∆Ssys
4. ∆Ssurr = −∆Gsys /T
024 4.0 points
Consider the reaction
Version 121 – Exam 4 – mccord – (52125)
4Fe(s) + 3O2 (g) → 2Fe2 O3 (s)
∆H◦ = −1648 kJ
Calculate the work done on or by the system
for the reaction of 4 moles of Fe(s). Assume a
constant pressure of 1 atm and a temperature
of 300 K.
1. −12.5 kJ
2. −7.5 kJ
3. +7.5 kJ correct
4. +10 kJ
5. +12.5 kJ
6. 0 kJ
Explanation:
w = −P∆V = −∆nRT
This reaction has three moles of gases in the
reactants and zero moles of gases in the products, so the volume is decreasing as the reaction proceed, work is done on the system, and
work will be positive.
w = −∆nRT = −(0 − 3 moles gas)
(8.314 J mol−1K−1 ) (300K)
w = 7480 J = 7.5 kJ
025 4.0 points
Iron metal will react with oxygen gas to form a
variety of iron oxides. This oxidation reaction
is typically referred to as the iron “rusting”.
The fact that this reaction is spontaneous at
room temperature tells you that
1. iron oxides have a negative Gibbs energy
of formation correct
tropy compared to oxygen and iron
Explanation:
The fact that the iron and oxygen (both elements) spontaneously form a compound, iron
oxide, at room temperature tell us that the
free energy of formation of the iron oxide
must be negative. The iron oxide is lower in
free energy compared to the elements that it
is formed from.
026 4.0 points
◦
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. positive, either
2. either, positive
3. negative, either
4. either, negative
5. negative, positive correct
6. positive, negative
Explanation:
This comes from
∆G = ∆H − T ∆S .
In order for ∆G◦rxn to always be negative,
◦
◦
∆Hrxn
must always be negative and ∆Srxn
must always be positive.
027 4.0 points
For which of the following is ∆Ssys likely to be
greater than zero?
I. 2N2 O5 (g) → 4NO2 (g) + O2 (g)
2. the 2nd law of thermodynamics has been
violated
II. Br2 (l) → Br2 (g)
III. Al(25 ◦ C) → Al(80 ◦ C)
3. iron oxides have a positive enthalpy of
formation
4. iron oxides have a higher standard en-
8
1. III only
2. I only
Version 121 – Exam 4 – mccord – (52125)
3. I, III
4. II only
5. I, II, III correct
6. II, III
Explanation:
The entropy increases in I because two moles
of gaseous reactants are converted to five
moles of gaseous products. The process of vaporization always increases the entropy, thus
II is correct. III is also a correct answer because the increase in temperature will always
increase the entropy.
028 4.0 points
Older moth balls were made from naphthalene. Naphthalene happens to be very
flammable.
Calculate the standard enthalpy of combustion (∆c H) of naphthalene,
C10 H8 (s).
1. -6345 kJ/mol
2. -3940 kJ/mol
3. -2877 kJ/mol
4. -4400 kJ/mol
5. -2331 kJ/mol
6. -4985 kJ/mol
7. -5161 kJ/mol correct
Explanation:
The balanced equation is:
C10 H8 (s) + 12 O2 (g) → 10 CO2 (g) + 4H2 O(ℓ)
∆c H = 10(−394) + 4(−286) − 77
∆c H = −3940 − 1144 − 77
∆c H = −5161
9