(njt298) triesault H12: Thermo - 1st Law McCord (53740)
This print-out should have 29 questions.
Multiple-choice questions may continue on
the next column or page nd all choices
before answering.
Remember that internal energy is E in our
book, but is U in many other books. Quest
(and this assignment) has questions with both
symbols.
001 10.0 points
On January 1, 2001, a bookstore had 266,478
books in stock. On February 1, 2001, the same
bookstore had 257,814 books in stock. What
is the value of (books) for the bookstore
during that period?
Correct answer: 8664 books.
Explanation:
Bf = 257, 814 books
Bi = 266, 478 books
B = B f B i
= 257, 814 books 266, 478 books
= 8664 books
Thus the bookstore had 8664 fewer books
on February 1.
002 10.0 points
Work is
1. organized molecular motion. correct
2. chaotic molecular motion.
3. heat.
4. kinetic energy.
5. potential energy.
1
(Sign does matter.)
Correct answer: 145 J.
Explanation:
004 10.0 points
A 100 W electric heater (1 W = 1 J/s) operates for 12.5 min to heat the gas in a cylinder.
At the same time, the gas expands from 3 L to
15 L against a constant atmospheric pressure
of 3.742 atm. What is the change in internal
energy of the gas?
Correct answer: 70.4501 kJ.
Explanation:
Pext = 3.742 atm
Vini = 3 L
1 L atm = 101.325 J
Vnal = 15 L
If the heater operates as rated, then the total amount of heat transferred to the cylinder
will be
q = (100 J/s) (12.5 min) (60 s/min)
= 75000 J = 75 kJ
Work will be given by
w = Pext V
in this case because it is an expansion against
a constant opposing pressure:
w = (3.742 atm) (15 L 3 L)
= 44.904 L atm
Convert to kilojoules (kJ)
w = (44.904 L atm)(101.325J/L atm)
= 4549.9 J = 4.5499 kJ
The internal energy change is
Explanation:
Work is organized molecular motion since
it is displacement of a solid against a force.
U = q + w
= 75 kJ + (4.5499 kJ)
= 70.4501 kJ
003 10.0 points
A chemical reaction takes place in a container
of cross-sectional area 100 cm2 . As a result of
the reaction, a piston is pushed out through 21
cm against an external pressure of 519 torr.
What is the value for w for this reaction?
005 10.0 points
A system had 150 kJ of work done on it and
its internal energy increased by 60 kJ. How
much energy did the system gain or lose as
heat?
triesault (njt298) H12: Thermo - 1st Law McCord (53740)
2
U = 2 kJ + (0.60795 kJ) = +1.39205 kJ.
1. The system gained 90 kJ of energy as
heat.
2. The system lost 90 kJ of energy as heat.
correct
3. The system gained 210 kJ of energy as
heat.
4. The system gained 60 kJ of energy as
heat.
5. The system lost 210 kJ of energy as heat.
Explanation:
007 10.0 points
A 1.00 g sample of n-hexane (C6 H14 ) undergoes complete combustion with excess O2 in
a bomb calorimeter. The temperature of the
1502 g of water surrounding the bomb rises
from 22.64 C to 29.30 C. The heat capacity
of the calorimeter is 4042 J/ C. What is U
for the combustion of n-C6 H14 ? One mole of
n-C6 H14 is 86.1 g. The specic heat of water
is 4.184 J/g C.
1. 9.96 103 kJ/mol
2. 1.15 104 kJ/mol
U = q + w
+60 kJ = q + 150 kJ
q = 90 kJ
(Negative means the system lost energy as
heat.)
006 10.0 points
When 2.00 kJ of energy is transferred as heat
to nitrogen in a cylinder tted with a piston
at an external pressure of 2.00 atm, the nitrogen gas expands from 2.00 to 5.00 L against
this constant pressure. What is U for the
process?
1. +2.61 kJ
2. +1.39 kJ correct
3. 0
4. 2.61 kJ
3. 7.40 104 kJ/mol
4. 4.52 103 kJ/mol
5. 5.92 103 kJ/mol correct
Explanation:
mwater = 1502 g
mC6 H8 = 1.00 g
SH = 4.184 J/g C
HC = 4042 J/ C
T = 29.30 C 22.64 C = 6.66 C
The increase in the water temperature is
29.30C 22.64C = 6.66 C. The amount of
heat responsible for this increase in temperature for 1502 g of water is
J
g C
= 41854 J = 41.85 kJ
q = (6.66 C) 4.184
The amount of heat responsible for the warming of the calorimeter is
q = (6.66 C)(4042 J/ C)
= 26920 J = 26.92 kJ
5. 0.608 kJ
Explanation:
U = q + w We know q , we need w:
For expansion against a constant external
pressure w = Pext V
w = (2 atm)(5 L 2 L)
(101.325 J L1 atm1)
= 607.95 J = 0.60795 kJ .
(1502 g)
The amount of heat released on the reaction
is thus 41.85 kJ + 26.92 kJ = 68.77 kJ per g
of n-hexane.
Per mol of n-hexane, this becomes
68.77
kJ
g
86. 1
g
= 5921 kJ/mol
mol
triesault (njt298) H12: Thermo - 1st Law McCord (53740)
However, since heat is released, the sign is
negative.
008 10.0 points
For a reaction in which gases are neither produced nor consumed, H is
3
010 10.0 points
A property that determines the direction of
heat ow is
1. entropy.
1. greater than E .
2. combustion.
2. less than E .
3. calorie.
3. unrelated to E .
4. temperature. correct
4. the same as E . correct
Explanation:
For a reaction at constant temperature and
constant pressure, E = H (n) R T .
If gases are neither produced nor consumed,
n will be zero, so E must equal H .
009 10.0 points
An ideal gas is allowed to expand isothermally
from 2.00 liters at 5.00 atm to 5.00 liters at
2.00 atm against a vacuum (i.e., no outside
pressure). Which of the quantities q , w, E ,
and H are zero for this process?
1. w, q , E , and H correct
2. w
3. w and E ,
4. Some other combination
011 10.0 points
In the manufacture of nitric acid by the oxidation of ammonia, the rst product is nitric
oxide. The nitric oxide is then oxidized to
nitrogen dioxide:
2 NO(g) + O2 (g) 2 NO2 (g)
Calculate the standard reaction enthalpy
for the reaction above (as written) using the
following data:
N2 (g) + O2 (g) 2 NO(g)
H = 180.5 kJ
N2 (g) + 2 O2 (g) 2 NO2 (g)
H = 66.4 kJ
1. 520.2 kJ/mol rxn
5. w, q , and E
Explanation:
P = 0 so w = P V = 0.
For isothermal expansion, T = const.
There is no change in molecular kinetic energy or potential energy due to intermolecular attraction (which is zero for ideal gases).
Therefore E = 0.
E = q + w
0 = q + 0;
H = q = 0
Explanation:
Heat ows spontaneously from a hotter
body to a colder body. In other words, heat
ows from a body of high temperature to a
body of lower temperature.
q=0
Therefore w, q , E , and H all equal zero.
2. 690.72 kJ/mol rxn
3. 128.2 kJ/mol rxn
4. 114.1 kJ/mol rxn correct
5. 252.4 kJ/mol rxn
6. 975.0 kJ/mol rxn
7. 100.3 kJ/mol rxn
Explanation:
triesault (njt298) H12: Thermo - 1st Law McCord (53740)
Using Hess Law and the given standard reaction enthalpies, the rst reaction is reversed
and added to the second:
2 NO(g) N2 (g) + O2 (g)
H = 180.5 kJ
N2 (g) + 2 O2 (g) 2 NO2 (g)
H =
66.4 kJ
2 NO(g) + O2 (g) 2 NO2 (g)
H = 114.1 kJ
012 10.0 points
A piece of metal of mass 29 g at 106 C is
placed in a calorimeter containing 47.4 g of
water at 23 C. The nal temperature of the
mixture is 37.3C. What is the specic heat
capacity of the metal? Assume that there is
no energy lost to the surroundings.
J
Correct answer: 1.42348 .
g C
Explanation:
mmetal = 29 g
mH2 O = 47.4 g
TH2 O = 23 C
Tnal = 37.3 C
J
Tmetal = 106 C
CH2 O = 4.184
g C
qlost metal = qgained water
mm Cm Tm = mH2 O CH2 O TH2 O
mH2 O CH2 O (Tnal TH2 O )
Cmetal =
mmetal (Tnal Tmetal )
J
g C
g) (37.3 C 106 C)
(47.4 g) 4.184
=
(29
(37.3 C 23 C)
J
= 1.42348 .
g C
013 10.0 points
When 0.483 g of compound X is burned completely in a bomb calorimeter containing 3000
g of water, a temperature rise of 0.354 C is
observed. What is Urxn for the combustion
of compound X? The hardware component of
the calorimeter has a heat capacity of 3.95
kJ/ C. The specic heat of water is 4.184
4
J/g C, and the MW of X is 56.0 g/mol.
Correct answer: 677.299 kJ/mol.
Explanation:
mX = 0.483 g
mwater = 3000 g
Cs = 4.184 J/g C
MWX = 56.0 g/mol
T = 0.354 C
HC = 3.95 kJ/ C
qcal = qwater + qhardware
qcal = Cs mwater T + Chardware T
Then divide the heat by the number of
moles, which is mass/MW. Also remember
that because the temperature increased in the
calorimeter, the reaction must be exothermic
and the value of q (and therefore E ) is negative.
qrxn = qcal
014 10.0 points
What mass of ethanol (C2 H5 OH()) must be
burned to supply 500 kJ of heat? The standard enthalpy of combustion of ethanol at 298
K is 1368 kJ mol1 .
1. 2.74 g
2. 29.7 g
3. 16.12 g correct
4. 10.9 g
5. 126 g
Explanation:
500 kJ
= 0.365497 mol of propane
1368 kJ/mol
(0.365497 mol)(44.1 g/mol)
= 16.1184 g propane
015 10.0 points
Carbon monoxide reacts with oxygen to form
carbon dioxide by the following reaction:
2 CO(g) + O2 (g) 2 CO2 (g)
H for this reaction is 135.28 kcal.
triesault (njt298) H12: Thermo - 1st Law McCord (53740)
How much heat would be released if 12.0
moles of carbon monoxide reacted with sucient oxygen to produce carbon dioxide? Use
only the information provided in this question.
1. 541.12 kcal
2. 405.84 kcal
3. 1623.36 kcal
4. 135.28 kcal
017
5
10.0 points
Calculate the standard reaction enthalpy
for the reaction.
CH4 (g) H2 + O(g) CO(g) + 3 H2(g)
given
2 H2 (g) + CO(g) CH3 OH()
H = 128.3 kJ mol1
2 CH4 (g) + O2 (g) 2 CH3 OH()
H = 328.1 kJ mol1
2 H2 (g) + O2 (g) 2 H2 O(g)
H = 483.6 kJ mol1
5. 270.56 kcal
1. +206.1 kJ mol1 correct
6. 811.68 kcal correct
2. +216 kJ mol1
Explanation:
H = 135.28 kcal
nCO = 12.0 mol
135.28 kcal 1 mol rxn
1 mol rxn
2 mol CO
(12 mol CO) = 811.68 kcal
so 811.68 kcal were released.
016 10.0 points
Burning 1 mol of methane in oxygen to form
CO2 (g) and H2 O(g) produces 803 kJ of energy. How much energy is produced when 3
mol of methane is burned?
1. 268 kJ
2. 803 kJ
3. 2,409 kJ correct
4. 1,606 kJ
Explanation:
n1 = 1 mol
q1 = 803 kJ
n2 = 3 mol
If we know how much heat is evolved when 1
mole of methane is combusted, then we know
how much heat would be evolved if 3 mol of
methane were combusted:
803 kJ
(3 mol methane)
1 mol methane
= 2409 kJ
3. +42.0 kJ mol1
4. +412.1 kJ mol1
5. +155.5 kJ mol1
Explanation:
We need to reverse the rst reaction, halve
the second, halve and reverse the third and
add the results:
CH3 OH() 2 H2(g) + CO(g)
H = 128.3 kJ/mol
CH4 (g) + 0.5 O2 (g) CH3OH()
H = 164.05 kJ/mol
H2 O(g) H2 (g) + 0.5 O2(g)
H = +241.8 kJ/mol
CH4 (g) + H2 O(g) CO(g) + 3 H2 (g)
H = 206.05 kJ/mol
018 10.0 points
The value of H for the reaction
C3 H8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O()
is 2220 kJ/mol rxn. How much heat is
given o when 11.0 g of propane gas (C3 H8 )
is burned at constant pressure?
1. 6660.0 kJ
triesault (njt298) H12: Thermo - 1st Law McCord (53740)
2. 25.96 kJ
4. H 0 = 1116.6 kJ/mol rxn
3. 22420.0 kJ
5. H 0 = +516.6 kJ/mol rxn
4. 555.0 kJ correct
6. H 0 = +1015.4 kJ/mol rxn
5. 1665.0 kJ
7. H 0 = +1587.2 kJ/mol rxn
6. 2220.0 kJ
8. H 0 = 516.6 kJ/mol rxn
7. 50.5 kJ
6
9. H 0 = 1088.2 kJ/mol rxn
Explanation:
H = 2220 kJ/mol
mC3 H8 = 11 g
2220 kJ
1 mol rxn
mol rxn
1 mol C3 H8
1 mol C3 H8
(11 g C3 H8 )
44 g C3 H8
= 555 kJ or 555 kJ released
q=
019 10.0 points
Calculate the standard enthalpy change for
the reaction
2 HCl(g) + F2 (g) 2 HF() + Cl2 (g)
given
4 HCl(g) + O2 (g) 2 H2 O() + 2 Cl2 (g)
10. H 0 = +1116.6 kJ/mol rxn
Explanation:
The rst equation needs to be multiplied by
1
in order to get the equation were interested
2
1
in. Thus its H 0 is multiplied by as well.
2
The second equation needs to be multiplied
by two in order to get the equation were
interested in. We also multiply its H 0 by
two.
The third equation needs to be reversed, so
the sign of its H 0 should change.
Then we add the equations to get the equation were interested in. We also add the adjusted H 0 values to get the answer, 1015.4
kJ/mol rxn.
020 10.0 points
Consider the reaction
H 0 = 202.4 kJ/mol rxn
1
1
H2 (g) + F2 (g) HF()
2
2
H 0 = 600.0 kJ/mol rxn
H2 (g) +
1
O2 (g) H2 O()
2
H 0 = 285.8 kJ/mol rxn
1. H 0 = 1015.4 kJ/mol rxn correct
4 FeO(s) + O2 (g) 2 Fe2 O3 (s)
and heat-of-formation data
H = 274 kJ/mol
FeO Fe + 1 O2 (g)
2
3
2 Fe + 2 O2 Fe2 O3
H = 823 kJ/mol
Find the change in enthalpy.
Correct answer: 550.
Explanation:
2. H 0 = 1587.2 kJ/mol rxn
3. H 0 = +1088.2 kJ/mol rxn
4 FeO(s) + O2 (g) 2 Fe2 O3 (s)
triesault (njt298) H12: Thermo - 1st Law McCord (53740)
Hf0
(kJ/mol)
Reaction
4 [FeO Fe + 1 O2 ]
2
4 (274) =
1096
3
2 [2 Fe + 2 O2 Fe2 O3 ]
4 (823) = 1646
4 FeO + O2 2 Fe2 O3
550
021 10.0 points
Calculate the standard enthalpy of formation
of bicyclo[1.1.0]butane
7
1 Hf C4 H6 (g)
= [4 (717 kJ/mol) + 6 (218 kJ/mol)]
Hf C4 H6 (g)
= 4176 kJ/mol Hf C4 H6 (g)
Now we use the comment on which bonds
were broken:
Hrxn = 6 BECC + 6 BECH
= 5 (348 kJ/mol) + 6 (412 kJ/mol)
= 4212 kJ/mol
H
We can set the two sides of the equation
equal since they represent the same reaction:
C
4212 kJ/mol = 4176 kJ/mol Hf C4 H6 (g)
H2 C
CH2
C
H
given the standard enthalpies of formation of
717 kJ mol1 for C(g) and 218 kJ mol1 for
H(g) and the average bond enthalpies of 412
kJ mol1 for C H and 348 kJ mol1 for
C C.
1. +312 kJ mol1
2. 124 kJ mol1
3. 472 kJ mol1
4. +175 kJ mol1
5. 36 kJ mol1 correct
Explanation:
We can write an equation in which we completely decompose bicyclo [1,1,0] butane:
C4 H6 (bicyclobutane, g) 4 C(g) + 6 H(g)
Hrxn = 5 (BECC ) + 6 (BECH )
(The comment on the right comes from
dissecting the structure given in the question
and noting how many of each kind of bond is
present). To nd H for this reaction we can
use Hess Law with formation enthalpies:
Hrxn = 4 Hf C(g) + 6 Hf H(g)
Hf C4 H6 (g) = 36 kJ/mol .
022 10.0 points
Calculate the enthalpy change for the reaction
2 SO2 (g) + O2 (g) 2 SO3 (g)
Hf for SO2 (g) = 16.9 kJ/mol;
Hf for SO3 (g) = 21.9 kJ/mol.
1. 10.0 kJ/mol rxn correct
2. +5.0 kJ/mol rxn
3. +10.0 kJ/mol rxn
4. 77.6 kJ/mol rxn
5. 5.0 kJ/mol rxn
Explanation:
Reactants:
Hf SO2 (g) = 16.9 kJ/mol
Hf O2 (g) = 0 kJ/mol
Products:
Hf SO3 (g) = 21.9 kJ/mol
Hrxn =
n Hf products
n H f
reactants
= (2 mol)(21.9 kJ/mol)
(2 mol)(16.9 kJ/mol)
(1 mol)(0 kJ/mol)
= 10.0 kJ/mol rxn
triesault (njt298) H12: Thermo - 1st Law McCord (53740)
023 10.0 points
Consider the following substances:
8
2. 1209 kJ/ mol bonds
3. 416 kJ/mol bonds
HCl(g) F2 (g) HCl(aq) Na(s)
4. 289.0 kJ/mol bonds
Which response includes ALL of the substances listed that have Hf0 = 0?
5. 582 kJ/mol bonds
1. HCl(g), Na(s) and F2 (g)
6. 196 kJ/mol bonds
2. Na(s)
7. 1962 kJ/mol bonds
3. Na(s) and F2 (g) correct
8. 327 kJ/mol bonds correct
4. HCl(g), Na(s), HCl(aq) and F2 (g)
9. 981 kJ/mol bonds
5. HCl(g) and Na(s)
Explanation:
Hf0 = 0 for elements in their standard
states. This would be true for both Na(s) and
F2 (g).
024 10.0 points
If the products of a reaction have a higher
energy than the reactants, then the reaction
1. is exothermic.
10. 1565 kJ/mol bonds
Explanation:
The S F bond energy (in SF6 ) is dened
as the H for the reaction in which ONE
MOLE of S F bonds in SF6 are broken to
give gaseous atoms. Each molecule of SF6
has SIX S F bonds, so each mole of SF6 has
SIX moles of S F bonds. Thus we need to
1
break the S F bonds in only mol of SF6 .
6
So we need the H for the reaction
1
1
SF6 (g) F(g) + S(g)
6
6
2. must be spontaneous.
3. is not spontaneous.
Use Hesss Law and the given values of Hf
to calculate H for this reaction.
4. is endothermic. correct
Explanation:
Energy must ow into the system for the
products to have a higher energy than the reactants. H will be positive and the reaction
will be endothermic.
025 10.0 points
Calculate the average S F bond energy in
SF6 , using the following Hf values:
SF6 (g) = 1209 kJ/mol
S(g) = 279 kJ/mol
F(g) = 79 kJ/mol
1. 202 kJ/mol bonds
026
10.0 points
The bond energy of H Cl is 432 kJ/mol,
the bond energy of Cl Cl is 242 kJ/mol,
and the bond energy of H H is 436 kJ/mol.
The gas-phase reaction
H2 + Cl2 2 HCl
is
1. exothermic by 246 kJ/mol rxn.
2. impossible.
3. endothermic by 246 kJ/mol rxn.
triesault (njt298) H12: Thermo - 1st Law McCord (53740)
5. exothermic by 186 kJ/mol rxn. correct
Explanation:
H Cl = 432 kJ/mol
Cl Cl = 242 kJ/mol
H H = 436 kJ/mol
H + Cl
0
Hrxn =
E = q + w,
Cl 2 H
BErct
= (H
H) + (Cl
2 (H
13
mol = 2 mol
+
22
nf = 1 mol
n = (1 2) mol = 1 mol
ni =
4. endothermic by 186 kJ/mol rxn.
H
9
Cl
BEprod
Cl)
Cl)
= (436 kJ/mol + 242 kJ/mol)
2 (432 kJ/mol)
= 186 kJ/mol
The negative sign means the reaction is
exothermic by 186 kJ/mol.
027 10.0 points
The standard molar enthalpy of formation
of NH3 (g) is 46.11 kJ/mol. What is the
standard molar internal energy of formation
of NH3 (g)?
1. 38.67 kJ/mol
2. 48.59 kJ/mol
3. 43.63 kJ/mol correct
q = H and
w = P V = n R T
= (1 mol)(8.31451 J/mol K)
kJ
(298.15 K)
1000 J
= 2.47897 kJ
Work is per 1 mole of NH3 formed, so
E = H + w
= 46.11 kJ/mol + 2.47897 kJ/mol
= 43.631 kJ/mol
028 10.0 points
The internal energy change is 5.67 kJ when
an ideal gas expands from 3.50 to 22.7 liters
at a constant external pressure of 2.38 atm.
What is the heat absorbed by the gas from its
surroundings?
Correct answer: 10.3001 kJ.
Explanation:
E = 5.67 kJ
Vi = 3.5 L
P = 2.38 atm
Vf = 22.7 L
V = 22.7 L 3.5 L = 19.2 L
E = q + w = H P V
4. 39.91 kJ/mol
5. 41.15 kJ/mol
H = E + P V
= (5.67 kJ) + 2.38 atm (19.2 L)
6. 46.11 kJ/mol
7. 51.07 kJ/mol
Explanation:
H = 46.11 kJ/mol
T = 298.15 K (standard temperature)
3
1
N2 (g) + H2 (g) NH3 (g)
2
2
kJ
101.33 J
1 atm L 1000 J
= 10.3001 kJ
029 10.0 points
What is the total motional contribution to the
molar internal energy of gaseous BF3 ?
triesault (njt298) H12: Thermo - 1st Law McCord (53740)
1. 1.5 R T
2. R T
3. 3 R T correct
4. 3.5 R T
5. 2.5 R T
Explanation:
The contribution of each mode of motion to
1
the total molar internal energy is R T .
2
BF3 is a nonlinear molecule so it has
three modes of translational motion and three
modes of rotational motion (assuming no contribution from vibration).
1
Therefore, Um = 6
RT = 3 RT
2
10