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\1 Version 229 — Exam 4 v This printout should have 26 questions.
Multiplechoice qumtions may continue on
the next column or page — ﬁnd all choices
before answering. 1 cal = 4.184 J
1 L atm = 101.325 .1
R = 8.314 J/mol K our 10.0 points
At the normal boiling point of water,
AHnP = 40 kJ/mole. What is the entropy
change for 1120(2) —r 1120(g) 7 1. 40 kJ/mol K
2. —40 lrJ/mol ~K
3. 400 J/mol K
4. 107 J/mol ‘ K
5.107 kJ/mol l( 6. 107 J/mol 1( correct
Explanation ‘ 002 ‘10. rats " _ The standard molar enthalpy of formation of
hydrazine, Ngimtj', is +5053 ch/mol. What
is the standard molar internal energy of for—
mation of hydrazine? 1. +5311 kJ/rnol
2. $43.19 kJ/mol ,. 3; +5559 kJ/mol 4“. +4815 kJ/mol 5. +£8.07 kJ/mol correct
. / ./ s. +671T55'ri7inoi 7. +5063 kJ/mol Version 229 — Exam 4 ~ The speciﬁc heat of water is 4.184 .l/g“C,
so we have to multiply by grams and temper
ature change in order to obtain Joules: AH of water = (SHmenm) (AT)
: (4.134 J/g “ c) (238 g)
x (155°C)
= 1543 J The heat capacity of the calorimeter is 92.3
.1/“C. This is not per gram, so we just have
to multiply by the temperature change to get
Joules: AH of calorimeter = (SH)(AT)
= (92.3 J/°C)(l.55°C)
= 143 .1 AH of Dm : AH of water
+ AH of calorimeter
=1543J+143J=1686J This is the amount of heat evolved. To
get the amount per gram of fuel burned, we
divide AH of the reaction by the amount of
fuel burned (3.4 grams): 1685 J ——:' .1
3.4g "195“ 007 10.0 points Calculate the standard reaction enthalpy
for the reaction N2H4(€) + P1205) —' 2 NHule‘) given
NzHAU) + Ozlgl —’ N2lg) + lezolg)
AH“ = 7543 kJ  moi“
2H2(g) + 02(3) p 211200;)
AH” = —434 k.)  moi“
Nzlg) +3Ii2(g)  ZNHAg)
AH” = 79221:.1 mol“ 1. V243 k.) moi“ 2. 4335 k] ~ mol" McCord — (53745) 1 s. +45.s7 kJ/mol Explanation:
AH = 50.63 kJ/mol
T = 298.15 K (standard temperature) N2(g) + 2 H2(g) a N2H4(1) n;=(1+2)mol=3mol n; = 0 mol
An. = (0 — 3) mol = —3 mol AE=q+w, q=AHand
—PAV = —AnRT
— (—3 mol)(8.314 J/mol A K) 1:1
x(298.15 K) (m)
= 7.43646 14.] Work is per 1 mole OlN2H4 formed, so AE = AH +111
= 50.63 kJ/mol + 7.43646 kJ/mol
= 58.0665 kJ/mol 003 10.0 points
Given the standard heats of reaction
Reaction AH 0 Ms) + 2 X204) —> Mug) —73.7 moi
X2(s) —* 2 X(g) +2823 kJ/mol
We) ’ M(5) —15.1 kJ/mol calculate the average bond energy for a single
M — X bond. /a
191.85 x
203.85 ‘r
200.85 175.85 184.35 163.35 178.85 206.35 . 194.35 10. 173.85 wwﬂpwewwr Correct answer: 163.35 kJ/mol.
Ebcplanation: McCord — (53745) 3 3. ~59 k1 marl
4. —1119 kJ ‘ moi1 5. —151 kJ  11ml“l correct Explanation:
We need to reverse the second reaction and
add them; N21{4(l)+92(g) —> w +enyeeg+
AH = —543 kJ/rnol
wet ~ 2 Hug) + 92%}
AH = +434 kJ/mol w + 3H2(g) —' 2 NHalg)
AH = —92.2 kl/mol
NZHilE) + 3 Halt) ’ 2 H2(g) + 2 Mine)
Nam“) + Hz(g) —’ 2 NH3(E)
AH = —151.2 kJ/mol 008 10.0 points
For the reaction 2 (3(3) + 2 H200 a Czﬂdg) on: = +523 kimol“ and AS,“ =
—53.07 J K“‘ mol‘1 at 298 K. This reac
tion will be spontaneous at 1. no temperature. correct
2. temperatura below 1015 K.
3. temperatures below 985 K.
4. temperatures above 985 K. 5. all temperatures. Explanation: AG : AH e TAS is used to predict sponv
taneiry. (AG is negative for a spontaneous
reaction.) T is always positive. We thus
have AG = (—1—) » T<~), For AG to be neg—
ative. T would have to be negative. which
is not physically possible, so the reaction is
nouspontaneous at any temperature. Version 229  Exam 4 — McCord  (53745) 2 The correct reaction will he i [mo —» Me) Mme] so you must ﬂip the ﬁrst and third reactions
and double the second reaction. AH for this reaction is —135.28 kcah 11811.7 kcal is released, how many molx of CD must have reacted? a: '1. 5.41 mol l
2. 16.3 mol
3. 40.5 mol
4. 6.00 mol 5. 270.5 moi
3: 6. 1.35 mol ‘7. 12.01130] correct
Explanation: ,
Ali‘: —l35.28 kczll 77cc 12.0 mol
(—135.28 heal) (1 mol rxn ,7; 1 moern 2 11101 CO
"*«~_x(12mol C0) = —811.68 kcal so 811.68 kcal were released. W1 3 005 j10.0 points
If you have endothermic proc in which
the diange in ﬁﬁopy is positive, how can you
y’make it spontaneous?
l
1. Reducing the entropy change
2. Decreasing the volume 3. Increasing the temperature correct 4. Increasing the pressure Version 229 — Exam 4 — 009 10.0 points
Calculate the average H —S bond enthalpy
in HgS(g) given the standard enthalpies of for mation for H»;S(g), H(g). and 5(3) as —20.1,
218, and 223 kJ  Incl—x, respectively. 1. 340 kJ ~ inol‘1 correct 2. 679 lol Imol 1 10.1 k1 mor‘ 4. 461 kJ  moi“ 5. 231 kJ ‘ moi“ Explanation:
Write an equation which represents the
bond energy (or a. multiple of it) H2$(g) H 2Hng + 3(a) AH. = 2 (313113)
Reactants: . AH; 325m) = —20.15 kJ/mnl
Products: AH, ME, = 218 kJ/mol
AHPS z 223 kJ/mol 1mg“ = Z n AHSFrad — Z n AHPm
= [2 (218 kJ/rnol) + 223 kJ/mol]
7 (201 kJ/mol)
z 579.] kJ/mol rxn, so Hgm = 2 x BE
679.1 M .
BE = ~—=“°'m‘ : 339.55 1‘7 .
2 mo] rxn 010 10.0 points
Which of the following is not a deﬁnition of
internal energy or change in internal energy? 1. The total energy content of a system. 2. A measure of the change in heat of a
system at constant volume. 3. A measure 01. the spontaneity of a reac—
tion. correct 5. Decreasing the temperature Explanation: AG AH—TAS AH > 0 for endothermic processes.
AG < 0 for spontaneous processes.
T is always positive, so AG AH—TAS
=(+)—TAS AG is negative if T is very large, so in
creasing the temperature malces the process
endothermic. 006 10.0 points 3.4 g of a hydrocarbon Fuel is burned in a
calorimeter that contains 238 grams of water
initially at 25.00°C. After the combustion,
the temperature is 26.55°C. How much heat
is evolved per gram of fuel burned? The heat
capacity of the calorimeter (hardware only) is
92.3 J/°C. 1. 1686 J/g 2. 7363 J/g 3. 25037 J/g 4. 143 .1 /g 5. 1543 J/g 6. 495 .l/g correct
7. 453 J/g s. 7818 J/g Explanation:
mmd = 3.4 g mm“ = 238 g
AT = 26.55°C — 26.00°C = 155°C The amount of heat evolved by the reaction
is equal to the amount of heat gained by the
water plus the amount of heat gained by the
calorimeter. McCord ' (53745) 4 4. The sum of heat (q) and work (w). Explanation: Gibbs free energy is a measure of the spon
taneity of a reaction. All of the other state
ments are mathematical identities describing
internal energy or prose restatements of the
same. .1 '\ 011 '_ 10.0 points Phosphine (thé'hommon name for P113, :1
highly mic gas used for fumigation), has
a. AHJM = 14.6kJ~m01 1 and a 53",, =
78.83Jmol“ l(“l. What is the normal
boiling point of phosphine expressed in centi—
grade? 1. —0.2 DC 2. 273 QC
3. —87.8 “C correct
4. 185.2 “C Explanation: Because boiling is an equilibrium process,
A05", n AHﬁu, — TA55u1,. And so TASS“ = AHﬁup
and T = AHSBI, A53”,
= 14,600.1 marl/78.33 J .mor‘ K“
:2 185.2 K
= 437.8 “C 012 10.0 points
Calculate AG for the process
He(g. 1 atm, 298 K) a He(g, 10 Mm. 298 K) 1. 2.10 kJ/rnol 2. 4.80 kJ/mOl 3. 3.501(rl/rnol 41. ~2.1O kl/mol 5. —19.1k.l/mol l?
:7 Version 229 — Exam 4 — McCord — (53745) 5 6. 5.70 k.l/ruol correct 7. seem/mar .. ~ " / s. —4.80 k'J/mol kylanation:
The temperature is the same before and
aﬂer the change so the change is isothermal (AH=0)a.nd
' AG=0—TAS=—TAS. Foii an isothermal change
AS = 1112 ln and All of the answers are intensive so 11 drops out
of the equation: AG = —(8.314 .l/mol ~K) (293 K) ln = «5704.82 .l/mol = —5.70482 kJ/mol 013 10.0 points
The deﬁnition of internal energy is 1 AU = q + w .
Which of these three values are state func
tions?
1. q and w only 2. q and 111, but only at constant volume
3. AU , but only at constant volume
4. AU, q, and w 5. AU only correct
Explanation:
014 10.0 points Which of the statements below concerning
thermodynamic sign convention is NOT true: 1. Work is done on the system when AV is
negative. 2. AS is positive whai there is increasing
disorder. 3. w is positive When work is done by the
i, system. correct
, 4. AH is negative when heat is released to
5 the surroundings. 5. AG is negative when a reaction is spon
tmieous. Explanation: w is positive only when work is done ON the
system, not BY the system. When the system
does work, volume increasa or the number of
moles increasm (AV > 0, An > O). u: = —P AV = ~‘AnlZT and will therefore be negative when AV or
A11 is positive.
#—
«, 015 2.10.0 points The temperature of a. container of BF; (g)
goa from 29W 373 K at a constant pra—
surc of 100 atm when 6.24 k] is added as heat.
How many moles of BF3 does the container
hold? You should treat BF3 ideally. 1. 2.86 mol
2. 1.50 mol
3. 3.33 mol
4. 6.67 mol
5. 2.50 mol correct 6. 4.00 mol Explanation:
Cp for a non~linear ideal gas is 4R. The
temperature increase is 75 K. q=nCpAT Version 229 — Exam 4 — McCord — (53745) 7 5. +2339 .l/K correct Explanation: T1 =10"C = 283 K T2 = 150 C = 423 K n=2mol R=8.314m
For a linear ideal gas at constant pressure CW" = g R. so T2 AS = nCmm ln more =,23.3912 J/K 020 10.0 points Calculate the change in entropy when 2.00
moles of ‘metbane are compremed isother—
mally to one ﬁfth ofiLs original volume. Treat
methane as an ideal gas. 1. +131 JK“ 2. — 13.4 .lK“ 3. —1,39 JK“ 4. +208 JK" 5. — 26.8 .I~l(_1 correct Explanation:
If compressed to ou‘e/ ﬁfth the original vol—
. 1 V 1
urne this means V2 2 m i . X.
115:7”? ln = (2.00 mol)(8.3ltl.l~mol‘1‘K‘ll = —25.m17.1 x“ lVe expect a negative :msn'er since the .volume
decreased. 10.0 points
f the following statements is the
best ﬁatemétimﬁthe Second Law of Thermo
dynamics? 1. For any spontaneous process. the change
in entropy of the universe must be negative. 2. The absolute entropy of a. perfect crystal
at 0 K would be 0. 3. For any spontaneous process, the total
entropy of the universe must increase. cor
reel: 4. For any reversible procms, the entropy
change of the universe must be positive. 5. For any spontaneous process, the entropy
of the system must increase. Explanation; The Second Law of Thermo dynamics states
that in spontaneous changes, the universe
tends toward a state of greater disorder. 022 10.0 points
For the methanol combustion reaction 2 ongonuiisﬁug) e
2 C02(gl + 4 H»;O(g) estimate the amount of PAV work done and
tell whether the work was done on or by the
system. ASSILTDE a temperature of 27°C. 1. 7.5 kl, work done by the system correct 2. 2.5 k1. work done on the system 3. 2.5 in]. work done by the system 4. 7.5 kl. work done on the system 5. No work is done in this reaction. Explanation:
T:27~C+273= 300K
Considering only moles ofg'us, i1.=n._n,:(2+1)‘3= “\l Version 229 — Exam 4 ~ 6240 = n(4  s.314)(75)
n = 2.5018 mol 016 10.0 points
3.5 g of a hydrocarbon fuel is burned in n. ves—
sel that contains 250. grams of water initially at 25.00 C. After the combustion, the temper«
nture of the water is 26.55 G. How much heat
is evolved per gram of fuel burned? 1. 26555 J/g 2.1764 .l/g 3. 1105 J/g 4. 143 .l/g 5. 1263 J/g
6. 511 .l/g
7. 73.5 J/g 8. 463 1/; correct Explanation:
(1 m Cp AT
(1 = 250 (4.184) 1.55 = 1621.3 .1
pergram will he
1621.3 .l / 3.5 g = 463 J/g 017 10.0 points
The internal energy of a. substance can be
Jﬁrought to zero by cooling the substance all
j! the way down to absolute zero ( T = 0 K).
i 1. False correct 2. True Explanation: The very [act that you have any matter
at all means you have some internal energy 
even at absolute zero. 018 10.0 points
When a. sugar cube dissolves in a. cup of notice
(an endothermic process), entropy changes of
the sugar plus water. the surroundings. and Version 229 ‘ Exam 4 — «— AnRT
— (3 mol) (8.314.1/mol K) (300 K)
= —7500J = —7.5kJ The system expands because An is positive.
so the system does the work on the surround—
ings. Also, when w is negative, work is done
by the system ,. . 023
For which reactions
1) 020:) +‘Hz<g) —> H2020)
n) C(s. diamond) + 02(g) —» C02(g)
m) 112(4) + 3 F2(g) _. 2 NF3(L7)
3
IV) C(s. graphite) + 5 Mg) +H2(gl —»
001(g)+ H20(g]
V) 2 Fe(s) + g 02(g) —. F9203(s)
would AH," = AHA“‘2 1. 10.0 points 1. IV and V only 2. [and 11 only
3.1.11] and V only 4. I. III. IV and V only
5. ll, Ill and TV only
5. H and 1V only
7.1.11. Ill IV and V 8. land V only correct Explanation: 024 10.0 poinm
How is ASHngv related to AG5\.5'? 1. AG”. T A5...» , AS .
2. no... : 3. [AGsys = —AS..,.2\ McCord — (53745) 6 the universe are, respectively,
1. None of these is correct. 2. positive, negative. positive. correct
3. negative, positive, positive.
4. positive, positive, positive. 5. negative. negative, negative. Explanation: Let’s consider each of the entropy changes
individually. A55”: The entropy of any substance in—
creases as the substance goes from solid to
liquid to gas. Solutions are inherently more
disordered than pure substances. so in our
system (the process inside the coffee cup) eu
tropy increasa when we dissolve the sugar
cube in water; i.e., ASS},s is positive. A591": Here we know that the disolution
procem is endothermic. The system draws
heat from the surroundings7 causing molec
ular motion immediately around the coli‘ee
cup to slow down. in turn, as molecula
slow down, less random motion occurs, and
entropy is decreased, causing AS...“ to be
negative. Asmﬁvz The process of dissolving sugar in
Water is spoutiuieuus, so ASWv must be pos
itive by the Second Law of Thermodyainics.
(In spontaneous changes, the universe tends
toward a. state of greater disorder.) 019 10.0 points
The temperature of 2.00 mol 002(5) is in
creased ﬁrom 10°C to 150”C at constant pres
sure. Calculate the change in the entropy of
carbon dioxide. Assume ideal behavior. 1. 1170 J/K
2. +1711 J/K
3. +1170 J /K 4. —23.39 J/K McCord — (53745) 8 5. AG... = A5... 6. AGsys = ‘TASuniv correct
Explanation: 025 10.0 points
The entropy of fusion of asubstance is always
smaller than its entropy of vaporization. 1. False 2. True correct Explanation: The entropy effusion is the entropy change
upon melting a solid into a liquid. The en
tropy change of vaporization is the entropy
change when a liquid becomes a gas. The en
tropy of a. gas is signiﬁcantly greater than that
of a liquid, which is a little more than that of
a solid. This is because in a gas the particles
are able to move around much more. 026 10.0 points
A friend says that as a plant grows and takes
an unordered set of molecules and puts them
in a highly organized system the second law
of thermodynamics is violated. Your best
response is that 1. while the entropy involved in the plant
growth decreased, the entropy in the universe
increased as a result of this process. correct 2. the plant is not at equilibrium and hence
the argument is invalid. 3. you must be thinking of the third law. 4. the second law applies only to chemical
systems and not biological systems. Explanation: The plant became more organized. but the
universe became less organized. Chemical
I’euvinns Within who plan: are ezrprgnnir (—AG)
and drive the plant growth forward. ...
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This note was uploaded on 12/07/2009 for the course CH 53130 taught by Professor Mccord during the Fall '09 term at University of Texas.
 Fall '09
 McCord

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