001 Version Exam 4 sparks (50990)
This print-out should have 27 questions.
Multiple-choice questions may continue on
the next column or page nd all choices
before answering.
1
3. None of these is dominant.
4. covalent.
5. ionic.
Dispersion Forces
001 1.0 points
In which of these compounds would you nd
ONLY dispersion forces existing between the
molecules?
I. NH3 ;
II. CH2 Cl2 ;
III. CO2 ;
IV. CCl4 .
1. IV only
2. II only
3. I and III only
4. I only
5. I and II only
6. I and IV only
7. III only
Explanation:
HCl would also possess van der Waals intermolecular forces but, due to the small size,
these would not be the dominant force.
ChemPrin3e T05 15
003 1.0 points
Which of the following can form intermolecular hydrogen bonds?
1. (CH3 )2 NH correct
2. H2 CO
3. CH3 COCH3
4. PH3
Explanation:
Only molecules with H attached to the
electronegative atoms N, O, and F can hydrogen bond. Of the molecules given only
(CH3 )2 NH2 has the H directly bonded to N,
so it can undergo hydrogen bonding.
8. III and IV only correct
9. II and IV only
10. II and III only
Explanation:
A nonpolar covalent molecule would have
only dispersion forces with another nonpolar
covalent molecule.
Campion 05 Ex 01 10
002 1.0 points
The DOMINANT intermolecular force that
causes gaseous HCl molecules to attract one
another is
Sparks interionic 001
004 1.0 points
Which would be expected to have the
strongest interionic forces?
1. MgO correct
2. NaF
3. SF2
4. C2 H4
5. Na2 O
Explanation:
1. dipole-dipole. correct
2. van der Waals.
SPARKS 04 0003
005 1.0 points
Consider the substances
Version 001 Exam 4 sparks (50990)
CaCl2 , H2 O, N2 , and Cl2 .
The order of increasing boiling points of
these substances is
1. N2 , H2 O, Cl2 , and CaCl2
2
is minimal. For a given volume, a sphere has
the least possible surface area.
Sparks cap 001
007 1.0 points
Cohesive forces are forces
2. CaCl2 , N2 , Cl2 , and H2 O
3. N2 , Cl2 , H2 O, and CaCl2 correct
4. CaCl2 , H2 O, N2 , and Cl2
5. Cl2 , N2 , H2 O, and CaCl2
Explanation:
The greater the intermolecular forces/
bonds, the more energy is required to break
those bonds and thus the greater the melting
points or boiling points. The intermolecular
forces, from strongest to weakest, are ion-ion,
hydrogen bonding, dipole-dipole, and dispersion forces.
CaCl2 ion-ion
H2 O hydrogen bonding
Cl2 dispersion
(larger molecule)
N2 dispersion
(smaller molecule)
N2 experiences the weakest forces and is the
lightest of the nonpolar molecules.
1. between the molecules of the liquid and
another surface.
2. Neither of these is correct.
3. between the molecules of a liquid. correct
Explanation:
Sparks viscosity 010
008 1.0 points
Which would you expect to be most viscous?
1. C8 H18 at 50 C
2. C8 H18 at 30 C correct
3. C4 H8 at 50 C
4. C4 H8 at 30 C
Explanation:
Mlib 04 2001
006 1.0 points
A drop of liquid tends to have a spherical
shape due to the property of
1. viscosity.
2. close packing.
Sparks Sp Heat 001
009 1.0 points
A 100 g sample of an unknown liquid absorbs
2000 J of heat energy, raising the liquids
temperature from 50 C to 70 C .
What is the specic heat capacity of this
liquid? No phase change took place.
3. vapor pressure.
1. 4000 kJ/g C
4. capillary action.
2. 0.3 J/g C
5. surface tension. correct
3. 1 J/g C correct
Explanation:
Molecules on the surface of a liquid are
inuenced by intermolecular attractions towards the interior; these attractions pull the
surface layer toward the center. The most stable situation is one in which the surface area
4. 2.5 J/g C
5. 4.8 J/g C
Explanation:
Version 001 Exam 4 sparks (50990)
Sparks w 001
010 1.0 points
Calculate the value for the work done at 25 C
when 2 moles of K(s) reacts according to the
equation shown.
2 K(s) + 2 H2 O( ) 2 KOH(aq) + H2 (g)
1. 2478 J correct
2. 24.5 J
3
U = q + w
= 2 kJ + (0.60795 kJ) = +1.39205 kJ .
Sparks ex3 8
012 1.0 points
The heat of combustion of butane (C4 H10 ),
is 2623 kJ/mol. How much heat would
be absorbed or evolved when 90 g butane is
burned?
3. +2478 J
1. 4070 kJ evolved correct
4. 208 J
2. 2623 kJ absorbed
5. +1020 J
3. 2623 kJ evolved
6. 0 J
Explanation:
ChemPrin3e T06 17
011 1.0 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. 0
2. +1.39 kJ correct
3. 2.61 kJ
4. 4070 kJ absorbed
5. 2.36 105 kJ evolved
6. 2.36 105 kJ absorbed
Explanation:
Sparks calor 002
013 1.0 points
A bomb calorimeter with a heat capacity of
25 J/C contains 2000 g of water with an initial temperature of 30 C . A 0.2 g sample of a
gummy bear is placed in a bomb calorimeter
and ignited, resulting in a new water temperature of 31 C . What is E for this reaction?
4. +2.61 kJ
1. +42 kJ/g
5. 0.608 kJ
2. 42 kJ/g correct
Explanation:
q = 2 kJ
P = 2 atm
V0 = 2 L
Vf = 5 L
For expansion against a constant external
pressure,
w = Pext V
= (2 atm)(5 L 2 L)
(101.325 J L1 atm1)
= 607.95 J = 0.60795 kJ .
3. 0 kJ/g
4. +8.4 kJ/g
5. +125 kJ/g
6. 125 kJ/g
7. 8.4 kJ/g
Version 001 Exam 4 sparks (50990)
4
Explanation:
Msci 15 0804
014 1.0 points
Calculate the enthalpy change at 25 C for the
reaction
N2 O(g) + H2 (g) N2 (g) + H2 O(g)
Useful heat of formation data
Hf N2 O(g) = 82.05 kJ/mol
Hf H2 O(g) = 241.8 kJ/mol
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 = +1088.2 kJ/mol rxn
1. 323.85 kJ/mol correct
2. H 0 = +1015.4 kJ/mol rxn
2. 158.9 kJ/mol
3. H 0 = 516.6 kJ/mol rxn
3. 241.8 kJ/mol
4. H 0 = +1116.6 kJ/mol rxn
4. 89.1 kJ/mol
5. H 0 = 1015.4 kJ/mol rxn correct
5. 159.8 kJ/mol
6. H 0 = +516.6 kJ/mol rxn
Explanation:
Reactants:
Hf N2 O(g) = 82.05 kJ/mol
Hf H2 (g) = 0 kJ/mol
Products:
Hf H2 O(g) = 241.8 kJ/mol
Hf N2 (g) = 0 kJ/mol
7. H 0 = 1088.2 kJ/mol rxn
8. H 0 = +1587.2 kJ/mol rxn
9. H 0 = 1587.2 kJ/mol rxn
10. H 0 = 1116.6 kJ/mol rxn
0
Hrxn =
n Hf0prod
kJ
= 241.8
mol
kJ
= 323.85
mol
n Hf0rct
kJ
82.05
mol
Hess add 1
015 rxns 1.0 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)
0
H = 202.4 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.
DAL 15 001
016 1.0 points
Version 001 Exam 4 sparks (50990)
Which of these statements about thermodynamics is NOT TRUE?
1. The entropy of the system is always increasing. correct
2. Energy is conserved in chemical reactions.
3. V , S , and H are examples of
changes in thermodynamic state functions.
4. Work done on the system is positive in
sign.
5. Heat given o to the surroundings is negative in sign.
Explanation:
The entropy of a system or its surroundings
may increase or decrease. The Second Law of
Thermodynamics states that in spontaneous
changes, the universe tends toward a state
of greater disorder; that is, entropy increases
(S > 0).
ChemPrin3e T07 68
017 1.0 points
WITHDRAWN
G = H T S is used to predict spontaneity. (G is negative for a spontaneous
reaction.) T is always positive. We thus
have G = (+) T (). For G to be negative, T would have to be negative, which
is not physically possible, so the reaction is
nonspontaneous at any temperature.
ChemPrin3e T07 61
019 1.0 points
For the reaction
2 SO3 (g) 2 SO2 (g) + O2 (g)
Hr = +198 kJ mol1 and Sr =
1
1
190 J K mol
at 298 K. The forward
reaction will be spontaneous at
1. temperatures below 1042 K.
2. temperatures above 1315 K.
3. all temperatures.
4. no temperature.
5. temperatures above 1042 K. correct
Explanation:
Hr = +198 kJ/mol
ChemPrin3e T07 58
018 1.0 points
For the reaction
2 C(s) + 2 H2 (g) C2 H4 (g)
Hr = +52.3 kJ mol1 and Sr =
1
1
53.07 J K mol at 298 K. This reaction will be spontaneous at
1. no temperature. correct
2. temperatures below 1015 K.
3. temperatures above 985 K.
4. all temperatures.
5. temperatures below 985 K.
Explanation:
5
Sr = 0.019
kJ
mol K
G0 = H 0 T S
G0 < 0 for a spontaneous reaction, so
0 > H 0 T S
H 0
T>
S 0
198 kJ/mol
= 1042.11 K
=
kJ
0.019 molK
Thus the temperature would need to be >
1042.11 K.
Mlib 05 3005
020 1.0 points
A negative value of S corresponds to
1. a decrease in randomness in the system.
correct
Version 001 Exam 4 sparks (50990)
2. a release of heat by the system.
6
2. The entropy of the universe must increase
as a result of this reaction occuring.
3. an increase in disorder in the system.
4. an absorption of heat by the system.
3. The reaction should occur very rapidly
because G is very negative. correct
Explanation:
Entropy increases when the disorder or randomness of a system increases. S is positive when randomness increases and negative
when randomness decreases.
4. If S is positive, then the reaction will
probably be more spontaneous at higher temperatures.
Msci 15 1416
021 1.0 points
The G0 for Cu2 S(s) and SO2 (g) are
f
86.2 kJ/mol and 300.1 kJ/mol, respectively.
What is G0 for the balanced reaction
Explanation:
Thermodynamics determines if a reaction
will happen; kinetics determines how fast a
reaction will occur.
G is a thermodynamic term, not a kinetics term.
Cu2 S(s) + O2 (g) SO2 (g) + 2 Cu(s)
at 298 K?
1. G0 = +386.3 kJ/mol
2. G0 = 213.9 kJ/mol correct
2. G.
4. G0 = +231.9 kJ/mol
5. Information is needed on G0 for Cu(s)
f
and O2 (g)
G0 =
rxn
G0 react = 86.2 kJ/mol
f
G0 prod = 300.1 kJ/mol
f
n G0 prod
f
Msci 15 1430
023 1.0 points
In a system at equilibrium at constant temperature and pressure, H is necessarily
equal to
1. U.
3. G0 = 386.3 kJ/mol
Explanation:
G0 O2 = 0
f
G0 Cu = 0
f
5. The reaction is termed spontaneous.
n G0rct
f
= 300.1 kJ/mol (86.2 kJ/mol)
= 213.9 kJ/mol .
3. P V.
4. T S. correct
5. 0.
Explanation:
G = 0 for a system at equilibrium, so
G = H T S
0 = H T S
H = T S
Mlib 05 3037
022 1.0 points
For a given reaction G is very negative.
Which of the following is FALSE?
Mlib 05 3007
024 1.0 points
A process CANNOT be spontaneous if
1. S for the reaction may be positive or
negative.
1. it is endothermic and there is an increase
in disorder.
Version 001 Exam 4 sparks (50990)
2. it is endothermic and there is a decrease
in disorder. correct
7
2. 85 g
3. 1 g
3. it is exothermic and there is an increase
in disorder.
4. it is exothermic and there a decrease in
disorder.
Explanation:
G = H T S
For a process to be spontaneous, G must
be negative. Given G = H T S , if H
is positive (endothermic) and S is negative
(decrease in disorder), G has to be positive
(nonspontaneous). T is in Kelvin so it is
always positive.
Solid Naphthalene Sparks rev 001
025 1.0 points
Naphthalene (C10 H8 ), a component of mothballs, has a melting point of 80.2C and a
standard molar enthalpy of fusion of 18.923
kJ/mol. If you have 5 kg of solid naphthalene
at its melting point and you add 190 kJ of
energy as heat, how much liquid naphthalene
will you have?
1. 10 g
2. 4990 g
3. 142 g
4. 2500 g
5. 3720 g
6. 1280 g correct
Explanation:
Sparks Therm Eq 001
026 1.0 points
How many grams of water at 50 C must be
added to 10.0 grams of ice at 10 C to result
in ONLY liquid water at 0 C?
1. 3540 g
4. 50 g
5. 17 g correct
Explanation:
Sparks enthalpy 002
027 1.0 points
For the following problem, all reactants and
products are in the gas phase. Remember to
balance the equation!
0
Calculate Hrxn for the reaction
CH3 Cl + F2 CH3 F + Cl2 .
Is the reaction endothermic or exothermic?
1. 300 kJ, exothermic
2. 242 kJ, endothermic
3. 379 kJ, exothermic correct
4. 379 kJ, exothermic
5. 379 kJ, endothermic
6. 242 kJ, exothermic
7. 300 kJ, endothermic
8. 379 kJ, endothermic
9. 242 kJ, exothermic
10. 242 kJ, endothermic
Explanation: