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CHM 2046
1)For which process is DS negative?
A
Chemistry
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University of South Florida
CHM 2046
1)For which process is DS negative?
A. evaporation of 1 mol of CCl4(l)
mixing 5 mL ethanol with 25 mL water
compressing 1 mol Ne at constant temperature from 1.5 L to 0.5 L
raising the temperature of 100 g Cu from 275 K to 295 K
grinding a large crystal of KCl to powder
Ten identical coins are shaken vigorously in a cup and then poured out onto a table top. Which of the following distributions has the highest probability of occurrence? (T = Tails, H = Heads)
T10H0
T8H2
T7H3
T5H5
T4H6
If two pyramid-shaped dice (with numbers 1 through 4 on the sides) were tossed, which outcome has the highest entropy?
The sum of the dice is 3.
The sum of the dice is 4.
The sum of the dice is 5.
The sum of the dice is 6.
The sum of the dice is 7.
A two-bulbed flask contains 5 particles. What is the probability of finding all 5 particles on the left side? A) 2.50%
B) 2.24%
C) 3.13%
D) 0.20%
E) 6.25%
Which of the following shows a decrease in entropy?
precipitation
gaseous reactants forming a liquid
a burning piece of wood
melting ice
two of these
Which of the following result(s) in an increase in the entropy of the system?
I.
Br2(g) ® Br2(l)
NaBr(s) ® Na+(aq) + Br–(aq)
IV. O2(298 K) ® O2(373 K)
V. NH3(1 atm, 298 K) ® NH3(3 atm, 298 K)
I
II, V
I, III, IV
I, II, III, IV
I, II, III, V
Consider the following processes:
I. condensation of a liquid
II increasing the volume of 1.0 mol of an ideal gas at constant temperature
dissolving sugar in water
heating 1.0 mol of an ideal gas at constant volume For how many of these is DS positive?
0
1
2
3
4
The second law of thermodynamics states that:
The entropy of a perfect crystal is zero at 0 K.
The entropy of the universe is constant.
The energy of the universe is increasing.
The entropy of the universe is increasing.
The energy of the universe is constant.
Which of the following statements is true?
The total energy and entropy of the universe are both increasing.
The total energy of the universe is increasing, but the entropy is constant.
The total energy of the universe increases, while the entropy decreases.
The total energy of the universe is constant, but the entropy is increasing.
None of these.
A 100-mL sample of water is placed in a coffee cup calorimeter. When 1.0 g of an ionic solid is added, the temperature decreases from 21.5°C to 20.8°C as the solid dissolves. For the dissolving of the solid
DH < 0
DSuniv > 0
DSsys< 0
DSsurr > 0
none of these
A chemical reaction is most likely to be spontaneous if it is accompanied by
increasing energy and increasing entropy
lowering energy and increasing entropy
increasing energy and decreasing entropy
lowering energy and decreasing entropy
none of these (A-D)
Assume that the enthalpy of fusion of ice is 6020 J/mol and does not vary appreciably over the temperature range 270-290 K. If one mole of ice at 0°C is melted by heat supplied from surroundings at 276 K, what is the entropy change in the surroundings, in J/K?
A) 22.1
B) 21.8
C) 0.0
D) –21.8
E) –22.1
If the change in entropy of the surroundings for a process at 431 K and constant pressure is –326 J/K, what is the heat flow absorbed by for the system?
326 kJ
1.32 kJ
–141 kJ
105 kJ
141 kJ
The heat of vaporization for 1.0 mole of water at 100.°C and 1.0 atm is 40.56 kJ/mol. Calculate DS for the process H2O(l) ® H2O(g) at 100.°C.
109 J/K mol
–109 J/K mol
406 J/K mol
–406 J/K mol
none of these
A change of state that occurs in a system is accompanied by 52.3 kJ of heat, which is transferred to the surroundings at a constant pressure and a constant temperature of 300. K. For this process DSsurr is:
52.3 kJ/K
B) –52.3 kJ/K
C) –174 J/K
174 J/K
248 kJ/K
The enthalpy of vaporization of ammonia is 23.35 kJ/mol at its boiling point (–33.4°C). Calculate the value of DSsurr when 1.00 mole of ammonia is vaporized at –33.4°C and 1.00 atm.
Which of the following is true for exothermic processes?
DSsurr < 0
DSsurr = –DH/T
DSsurr = 0
DSsurr > 0
two of these
At 1 atm, liquid water is heated above 100°C.
DSsurr for this process is
greater than zero
less than zero
equal to zero
more information needed to answer this question
none of these (A-D)
DSsys for this process is
greater than zero
less than zero
equal to zero
more information needed to answer this question
none of these (A-D)
DSuniv for this process is
greater than zero
less than zero
equal to zero
more information needed to answer this question
none of these (A-D)
Given that DHvap is 58.2 kJ/mol, and the boiling point is 83.4°C, 1 atm, if one mole of this substance is vaporized at 1 atm, calculate DSsurr.
–163 J/K mol
163 J/K mol
698 J/K mol
–698 J/K mol
0
As long as the disorder of the surroundings is increasing, a process will be spontaneous.
ANS: F
For any given process, DSsurr and DSsys have opposite signs.
ANS: F
If DSsurr = –DSsys, the process is at equilibrium.
ANS: T
Which of the following is true?
By spontaneous we mean that the reaction or process will always proceed to the right (as written) even if very slowly. Increasing the temperature may speed up the reaction, but it does not affect the spontaneity of the reaction.
By spontaneous we mean that the reaction or process will always proceed to the left (as written) even if very slowly. Increasing the temperature may speed up the reaction, but it does not affect the spontaneity of the reaction.
By spontaneous we mean that the reaction or process will always proceed to the left (as written) even if very slowly. Increasing the temperature may speed up the reaction and it generally affects the spontaneity of the reaction.
By spontaneous we mean that the reaction or process will always proceed to the right (as written) even if very slowly. Increasing the temperature may speed up the reaction, and it generally affects
the spontaneity of the reaction.
None of the above is true.
Substance X has a heat of vaporization of 55.9 kJ/mol at its normal boiling point (423°C). For the process X(l) ® X(g) at 1 atm and 423°C calculate the value of DSuniv.
0
80.3 J/K mol
132 J/K mol
–80.3 J/K mol
–132 J/K mol
Substance X has a heat of vaporization of 58.4 kJ/mol at its normal boiling point (423°C). For the process X(l) ® X(g) at 1 atm and 423°C calculate the value of DSsurr.
0
83.9 J/K mol
138 J/K mol
–83.9 J/K mol
–138 J/K mol
Substance X has a heat of vaporization of 47.1 kJ/mol at its normal boiling point (423°C). For the process X(l) ® X(g) at 1 atm and 423°C calculate the value of DS.
0
67.7 J/K mol
111 J/K mol
–67.7 J/K mol
–111 J/K mol
Substance X has a heat of vaporization of 64.4 kJ/mol at its normal boiling point (423°C). For the process X(l) ® X(g) at 1 atm and 423°C calculate the value of DG.
0 J
B) 92.5 J
152 J
D) –92.5 J
E) –152 J
The enthalpy of vaporization of ammonia is 23.35 kJ/mol at its boiling point (–33.4°C). Calculate the value of DS when 1.00 mole of ammonia is vaporized at –33.4°C and 1.00 atm.
For the process CHCl3(s) ® CHCl3(l), DH° = 9.19 kJ/mol and DS° = 43.9 J/mol/K. What is the melting point of chloroform?
A) –64 °C
B) 209 °C
C) 130 °C
64 °C
E) –130 °C
Given that DHvap is 60.3 kJ/mol, and the boiling point is 83.4°C, 1 atm, if one mole of this substance is vaporized at 1 atm, calculate DS.
–169 J/K mol
169 J/K mol
723 J/K mol
–723 J/K mol
0
DH° is zero for a chemical reaction at constant temperature.
ANS: F
Consider two perfectly insulated vessels. Vessel #1 initially contains an ice cube at 0°C and water at 0°C. Vessel #2 initially contains an ice cube at 0°C and a saltwater solution at 0°C. In each vessel, consider the "system" to be the ice, and the "surroundings" to be the liquid.
Determine the sign of DSsys, DSsurr, and DSuniv for the contents of Vessel #1.
DSsys
A) 0
DSsurr
0
0
DSuniv
B) +
C) +
D) +
E) +
–
+
– 0
0
+
+
+
ANS:
B
Determine the sign of DSsys, DSsurr, and DSuniv for the system (ice/saltwater) in Vessel #2.
DSsys
A) 0
DSsurr
0
0
DSuniv
B) +
C) +
D) +
E) +
–
+
– 0
0
+
+
+
ANS:
D
The melting point of water is 0°C at 1 atm pressure because under these conditions:
DS for the process H2O(s) ® H2O(l) is positive.
DS and DSsurr for the process H2O(s) ® H2O(l) are both positive.
DS and DSsurr for the process H2O(s) ® H2O(l) are equal in magnitude and opposite in sign.
DG is positive for the process H2O(s) ® H2O(l).
None of these is correct.
| second law of thermodynamics | entropy change for a phase transition
For the vaporization of a liquid at a given pressure:
DG is positive at all temperatures.
DG is negative at all temperatures.
DG is positive at low temperatures, but negative at high temperatures (and zero at some temperature).
DG is negative at low temperatures, but positive at high temperatures (and zero at some temperature).
None of these (A-D).
| second law of thermodynamics | entropy change for a phase transition
Which of the following statements is always true for a spontaneous process?
DSsys > 0
DSsurr > 0
DSuniv > 0
DGsys > 0
I
III
IV
I and III
III and IV
For a spontaneous exothermic process, which of the following must be true?
DG must be positive.
DS must be positive.
DS must be negative.
Two of the above must be true.
None of the above (A-C) must be true.
A mixture of hydrogen and chlorine remains unreacted until it is exposed to ultraviolet light from a burning magnesium strip. Then the following reaction occurs very rapidly:
H2(g) + Cl2(g) ® 2HCl(g) DG = –45.54 kJ
DH = –44.12 kJ DS = –4.76 J/K
Which of the following is consistent with this information?
The reactants are thermodynamically more stable than the products.
The reaction has a small equilibrium constant.
The ultraviolet light raises the temperature of the system and makes the reaction more favorable.
The negative value for DS slows down the reaction.
The reaction is spontaneous, but the reactants are kinetically stable.
For a particular chemical reaction DH = 4.3 kJ and DS = –19 J/K. Under what temperature condition is the reaction spontaneous?
When T < –226 K.
When T < 226 K.
The reaction is spontaneous at all temperatures.
The reaction is not spontaneous at any temperature.
When T > 226 K.
For a certain process at 355 K, DG = –11.8 kJ and DH = –9.2 kJ. Therefore, DS for the process is
0 J/K mol
7.3 J/K mol
–7.3 J/K mol
–25.9 J/K mol
25.9 J/K mol
Consider the freezing of liquid water at –10°C. For this process what are the signs for DH, DS, and DG?
DH
DS
DG
A) +
B) +
–
–
–
–
–
+
+
–
0
– 0
–
–
For the process of a certain liquid vaporizing at 1 atm, DH°vap = 54.2 kJ/mol and DS°vap= 74.1 J/mol K. Assuming these values are independent of T, what is the normal boiling point of this liquid?
A) 731 °C
B) 1004 °C
C) 458 °C
D) 0.731 °C
E) none of these
For the reaction A + B ® C + D, DH° = +40 kJ and DS° = +50 J/K. Therefore, the reaction under standard conditions is
spontaneous at temperatures less than 10 K
spontaneous at temperatures greater than 800 K
spontaneous only at temperatures between 10 K and 800 K
spontaneous at all temperatures
nonspontaneous at all temperatures
In which case must a reaction be spontaneous at all temperatures?
DH is positive, DS is positive.
DH = 0, DS is negative.
DS = 0, DH is positive.
DH is negative, DS is positive.
None of these.
For the process S8 (rhombic) ® S8 (monoclinic) at 110°C, DH = 3.21 kJ/mol and DS = 8.70 J/K × mol (at 110°C). Which of the following is correct?
This reaction is spontaneous at 110°C (S8 (monoclinic) is stable).
This reaction is spontaneous at 110°C (S8 (rhombic) is stable).
This reaction is nonspontaneous at 110°C (S8 (rhombic) is stable).
This reaction is nonspontaneous at 110°C (S8 (monoclinic) is stable).
Need more data.
As O2(l) is cooled at 1 atm, it freezes at 54.5 K to form Solid I. At a lower temperature, Solid I rearranges to Solid II, which has a different crystal structure. Thermal measurements show that DH for the I ®II phase transition is –
743.07 J/mol, and DS for the same transition is -17.0 J/K mol. At what temperature are Solids I and II in equilibrium?
A) 13.6 K
B) 43.7 K
C) 19.8 K
D) 98.2 K
E) They can never be in equilibrium because they are both solids.
At constant pressure, the following reaction 2NO2(g) ® N2O4(g) is exothermic. The reaction (as written) is
always spontaneous
spontaneous at low temperatures, but not high temperatures
spontaneous at high temperatures, but not low temperatures
never spontaneous
cannot tell
Given that DHvap is 53.3 kJ/mol, and the boiling point is 83.4°C, 1 atm, if one mole of this substance is vaporized at 1 atm, calculate DG.
A) –150 J
150 J
639 J
D) –639 J
E) 0 J
The following questions refer to the following reaction at constant 25°C and 1 atm.
Determine DSsurr for the reaction (in kJ/mol K) A) 3.14
B) 0.937
C) 0.378
D) 1.31
E) 2.65
Determine DSuniv for the reaction (in kJ/mol K) A) 0.23
B) 2.3
C) 0.36
D) 2.8
E) 3.6
What must be true about DG for this reaction?
DG = DH
DG = 0
DG > 0
DG < 0
DG = DSuniv
Given the following data, calculate the normal boiling point for formic acid (HCOOH).
DHf°(kJ/mol)
S°(J/mol K)
HCOOH(l)
-410.
130.0
HCOOH(g)
-363
251.1
A) 0.39 °C
B) 388 °C
C) 661 °C
D) 279 °C
E) 115 °C
The following reaction takes place at 120°C: H2O(l) ® H2O(g) DH = 44.0 kJ/mol DS = 0.119 kJ/mol K Which of the following must be true?
The reaction is not spontaneous.
The reaction is spontaneous.
DG = 0
DG < 0
Two of these.
When ignited, solid ammonium dichromate decomposes in a fiery display. This is the reaction for a "volcano" demonstration. The decomposition produces nitrogen gas, water vapor, and chromium(III) oxide. The temperature is constant at 25°C.
Substance
DHf° (kJ/mol)
S° (kJ/mol K)
Cr2O3(g)
–1147
0.08118
H2O(l)
–242
0.1187
N2(g)
0
0.1915
(NH4)2Cr2O7(s)
–22.5
0.1137
Determine DSuniv° (in kJ/mol K). A) 7.66
B) 6.39
C) 84.3
D) 5.22
E) 6.03
Determine DS° reaction (in kJ/mol K). A) 0.2777
B) 0.8612
C) 0.7475
D) 0.6338
E) 0.1590
Determine DG° (in kJ/mol). A) –191.4
B) –2281.4
C) –38.9
D) 1903.6
E) –1555.4
The third law of thermodynamics states:
The entropy of the universe is increasing.
The entropy of the universe is constant.
The entropy is zero at 0 K for a perfect crystal.
The absolute entropy of a substance decreases with increasing temperature.
The entropy of the universe equals the sum of the entropy of system and surroundings.
For which of the following processes would DS° be expected to be most positive? A) O2(g) + 2H2(g) ® 2H2O(g)
H2O(l) ® H2O(s)
NH3(g) + HCl(g) ® NH4Cl(g)
D) 2NH4NO3(s) ® 2N2(g) + O2(g) + 4H2O(g)
E) N2O4(g) ® 2NO2(g)
Which of the following statements is (are) always true?
In order for a process to be spontaneous, the entropy of the universe must increase.
A system cannot have both energy disorder and positional disorder. III.
DSuniv =
IV. S° is zero for elements in their standard states.
I
I, IV
I, III, IV
II, IV
II
In which process is DS expected to be positive?
a reaction that forms a solid precipitant from aqueous solutions
an ideal gas being compressed at a constant temperature and against a constant pressure
water freezing below its normal freezing point
a spontaneous endothermic process at a constant temperature and pressure
none of these
In which reaction is DS° expected to be positive?
A) I2(g) ® I2(s)
B) H2O(l) ® H2O(s) C)
CH3OH(g) + O2(g) ® CO2(g) + 2H2O(l)
D) 2O2(g) + 2SO(g) ® 2SO3(g)
E) none of these
For the dissociation reaction of the acid HF: HF(aq) H+(aq) + F–(aq)
DS is observed to be negative. The best explanation is:
This is the expected result since each HF molecule produces two ions when it dissociates.
Hydration of the ions produces the negative value of DS.
The reaction is expected to be exothermic and thus DS should be negative.
The reaction is expected to be endothermic and thus DS should be negative.
None of these can explain the negative value of DS.
Consider the dissociation of hydrogen:
H2(g) 2H(g)
One would expect that this reaction:
will be spontaneous at any temperature
will be spontaneous at high temperatures
will be spontaneous at low temperatures
will not be spontaneous at any temperature
will never happen
When a stable diatomic molecule spontaneously forms from its atoms, what are the signs of DH°, DS°, and DG°?
DH DS DG
A) +
+
+
B) +
–
–
C) –
+
+
D) –
–
+
E) –
–
–
Consider the reaction
2N2O5(g) 4NO2(g) + O2(g)
at 25°C for which the following data are relevant:
DHf°
S°
N2O5
11.289 kJ/mol
355.32 J/K mol
NO2
33.150 kJ/mol
239.90 J/K mol
O2
0 kJ/mol
204.80 J/K mol
Calculate DS° for the reaction. A) 809.08 J/K
B) 89.38 J/K
C) 453.76 J/K
D) –265.94 J/K
E) 1164.40 J/K
Calculate DH° for the reaction.
A) 110.022 kJ
B) 10.572 kJ
C) 121.311 kJ
D) 21.861 kJ
E) 155.178 kJ
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