UNIT 11 Practice Test Advanced Thermodynamics Page 1 of 11 1.
All of the following symbols represent state functions except
A. w.
B. H.
C. U.
D. P.
E. T.
2.
The heat of vaporization of freon, CCl2F2, is 17.2 kJ/mol at 25˚C. What is the change of entropy for
one mole of liquid freon when it vaporizes at 25˚?
A. 57.7 J(mol·K)
B. 0.688 J/(mol·K)
C. 5.13 x 10 kJ/(mol·K)
D. 3.16 J/(mol· K)
E. 2.39 x 10J/(mol·K)
3.
The heat of vaporization of methanol, CH3OH, is 36.8 kJ/mol. Its boiling point is 64.7˚C. The change
in entropy for the vaporization of methanol in J/mol·K is
A. –17.8.
B. 3.41.
C. 17.8.
D. 109.
E. 569.
4.
Which of the following would be expected to have the largest entropy value per mole?
A. SO2Cl2(s)
B. SO2Cl2(l)
C. SO2Cl2(g)
D. SO2(g)
E. Cl2(g)
5.
Arrange the following in order of increasing entropy, S˚:
Hg(l), Hg(s), C6H6(l), CH3OH(l)
A. Hg(s), CH3OH(l), C6H6(l), Hg(l)
B. CH3OH(l), Hg(s), Hg(l), C6H6(l)
C. Hg(l), Hg(s), C6H6(l), CH3OH(l)
D. Hg(s), Hg(l), C6H6(l), CH3OH(l)
E. Hg(s), Hg(l), CH3OH(l), C6H6(l)
6.
In which of the following reactions is ∆S˚ positive?
A. I2(g) → I2(s)
B. H2O(l) → H2O(s)
C. CO2(g) + CaO(s) → CaCO3(s)
D. O2(g) + 2SO2(g) → 2SO3(g)
E. none of these
7.
Which of the following processes would be expected to have a ∆S value very close to zero?
A. H2O(s) → H2O(l)
UNIT 11 Practice Test Advanced Thermodynamics B.
C.
D.
E.
Page 2 of 11 2 H2(g) + O2(g) → 2 H2O(g)
H2O(s) → H2O(g)
N2(g) + O2(g) → 2 NO(g)
OF2(g) + H2O(g) → O2(g) + 2 HF(g)
8.
The Gibbs free energy is given by the equation
A. ∆G = ∆S – T∆H.
B. ∆G = Ginitial – Gfinal.
C. ∆G = ∆H – T∆S.
D. ∆G = ∆S – H∆T.
E. ∆G = ∆H – S∆T.
9.
The free energy change of a reaction is a measure of
A. the excess entropy given off to the surroundings.
B. the excess heat given off to the surroundings.
C. the direction in which a net reaction occurs.
D. the increased molecular disorder that occurs in the system.
E. how rapidly the reaction occurs.
10.
If a reaction is exothermic and nonspontaneous at 25˚C and 1 atmosphere pressure, it may be
A. spontaneous at higher temperatures.
B. spontaneous at lower temperatures.
C. endothermic at lower temperatures.
D. endothermic at higher temperatures.
E. nonspontaneous at all temperatures.
11.
All of the following have free energy of formation values of zero except
A. Br2(l).
B. Fe(s).
C. N(g).
D. Na(s).
E. Ne(g).
12.
For the reaction PCl5(g) → PCl3(g) + Cl2(g) at 1 atmosphere pressure, the value of ∆H is positive and
the process is spontaneous at 250˚C. Which of the following statements concerning this reaction is
true?
A. The reverse reaction is endothermic.
B. The reverse reaction is nonspontaneous at room temperature.
C. The change in entropy is the driving force for the decomposition of PCl5.
D. The process is exothermic at 250˚C.
E. The process is endothermic at 250˚C and exothermic at room temperature.
UNIT 11 Practice Test Advanced Thermodynamics 13.
Page 3 of 11 For the following reaction,
2 Cl(g) → Cl2(g)
which is spontaneous at room temperature, we find that
A. ∆H is + and ∆S is + for the reaction.
B. ∆H is – and ∆S is – for the reaction.
C. ∆H is + and ∆S is – for the reaction.
D. ∆H is – and ∆S is + for the reaction.
E. ∆G is – for all temperatures.
14. The reaction C(s) + CO2(g) → 2 CO(g), is spontaneous only at temperatures in excess of 1100 K. We
can conclude that
A. ∆H˚ is positive and ∆S˚ is negative.
B. ∆H˚ is negative and ∆S˚ is negative.
C. ∆H˚ is positive and ∆S˚ is positive.
D. ∆H˚ is negative and ∆S˚ is positive.
E. ∆G˚ is negative for all temperatures.
15.
When ammonium nitrate dissolves in water, the solution becomes cold. We can conclude the
following:
A. ∆H˚ is positive and ∆S˚ is positive.
B. ∆H˚ is positive and ∆S˚ is negative.
C. ∆H˚ is negative and ∆S˚ is negative.
D. ∆H˚ is negative and ∆S˚ is positive.
E. ∆H˚ is positive and ∆S˚ is zero.
16.
Sublimation is an example of a process for which
A. ∆H, ∆S, and ∆G are negative at all temperatures.
B. ∆H, ∆S, and ∆G are positive at all temperatures.
C. ∆H and ∆S are positive at all temperatures.
D. ∆H is positive and ∆S is negative at all temperatures.
E. ∆H is negative and ∆S is positive at all temperatures.
17.
For the spontaneous reaction Zn(s) + ½ S2(g) → ZnS(s), one would predict that
A. ∆H is + and ∆S is + for the reaction.
B. ∆H is – and ∆S is – for the reaction.
C. ∆H is + and ∆S is – for the reaction.
D. ∆H is – and ∆S is + for the reaction.
E. ∆G is + at all temperatures.
18.
When one mole of ice melts to liquid at 0˚C,
A. the entropy decreases.
B. the entropy stays the same.
C. the entropy increases.
D. the order increases.
E. none of these happens.
19.
For a spontaneous process, which one of the following is always positive?
UNIT 11 Practice Test Advanced Thermodynamics A.
B.
C.
D.
E.
20.
Page 4 of 11 ∆G
∆G˚
∆Suniv
∆Ssurr
∆Ssys
Calculate G˚ for the reaction
Cu(s) + H2O(g) → CuO(s) + H2(g)
at 500 K.
Cu(s)
H2O(g)
CuO(s)
H2(g)
A.
B.
C.
D.
E.
∆Hf˚(kJ/mol)
0
–241.8
–155.2
0
S˚(J/K)
33.3
188.7
43.5
130.6
+110.6 kJ
–86.6 kJ
+23.9 kJ
–62.6 kJ
+301 kJ
21.
The best criterion for the spontaneity of a chemical reaction is the sign of
A. ∆H.
B. ∆H/T.
C. ∆S.
D. ∆S/T.
E. ∆G.
22.
If the temperature of the reaction system
CaCO3(s) → CaO(s) + CO2(g)
is increased from 25˚C to 1000˚C, the quantity that changes in value the most is
A. ∆G˚.
B. ∆H˚.
C. ∆H.
D. ∆S˚.
E. ∆S.
UNIT 11 Practice Test Advanced Thermodynamics Page 5 of 11 23.
For the reaction CaCO3(g) → CaO(s) + O2(g) at 1 atm pressure, the values of ∆H and ∆S are both
positive, and the process is spontaneous at 900˚C. Which of the following statements about this
reaction is true?
A. ∆G at room temperature is negative.
B. The process is exothermic at 900˚C and endothermic at room temperature.
C. The change in entropy is the driving force for the reaction.
D. The reverse reaction is nonspontaneous at room temperature.
E. The reverse process is endothermic.
24.
For a reaction, if ∆G˚ = 0 then
A. ∆S˚ = 0.
B. ∆H˚ = 0.
C. ∆G˚ = 0.
D. K = 1.
E. K = 0.
UNIT 11 Practice Test Advanced Thermodynamics Answer Key:
1.
A
2.
A
3.
D
4.
C
5.
E
6.
E
7.
D
8.
C
9.
C
10. B
11. C
12. C
13. B
14. C
15. A
16. C
17. B
18. C
19. C
20. A
21. E
22. A
23. C
24. D
Page 6 of 11 UNIT 11 Practice Test Advanced Thermodynamics Page 7 of 11 Answer Justifications:
1.
A
State functions depend only on the current state of the system, not on the way the system got to that
state. P for pressure and T for temperature are both state functions. Notice that when you take a
pressure or a temperature measurement, you don’t need to take two measurements or watch the
measurements change. The same is true of energy, such as enthalpy (H) or internal energy (U). Work
(w) on the other hand is created according to the process that occurs to get to the state. Picking up a
100 pound box takes more work when you try to lift it up directly than when you lift it with pulleys.
2.
A
Since freon is vaporizing, it is undergoing a phase change. This means that the system can be viewed
as if it were at equilibrium, meaning that ∆G = 0. The values of enthalpy, entropy, and temperature can
be entered into the equation, ∆G = ∆H – T∆S:
0  17200 J mol   x  298K 
x
3.
17200
 57.7 J mol  K
298
D
Since methanol is vaporizing, it is undergoing a phase change. This means that the system can be
viewed as if it were at equilibrium, meaning that ∆G = 0. The values of enthalpy, entropy, and
temperature can be entered into the equation, ∆G = ∆H – T∆S:
0  36800 J mol   x  337.7 K 
x
36800
 109 J mol  K
337.7
4.
C
The value of entropy can be altered by the phase, the size of the molecule, etc. Out of the choices, the
chemicals that are solid and liquid, responses A and B, respectively, will automatically have a lower
entropy than the gaseous chemicals. Then, notice that the that the chemicals listed in responses D and
E are smaller than the chemical listed in response C. Since larger molecules have more entropy,
response C will have the largest value per mole.
5.
E
Absolute entropy, or chaos of a particular species, is determined by phase and size. The particles with
the least amount of chaos are those that are very ordered, such as solids; this would include Hg(s). The
remaining particles are all liquids, so a distinction determined by phase is no longer effective.
Therefore, the next factor to look at would be size. Hg(l) is the smallest of the three remaining species,
so it would have the least amount of disorder or arrangements. Finally, there are only six atoms in
CH3OH(l) while there are twelve atoms in C6H6(l). The implication here is that there would be more
arrangements in a molecule that has more atoms. Therefore the C6H6(l) would have more chaos than
CH3OH(l).
6.
E
UNIT 11 Practice Test Advanced Thermodynamics Page 8 of 11 When the ∆S˚ is positive, it means that the change is becoming more disordered, more chaotic. This
could be a change in phase, sizes, quantity, etc. In Response A, a gas goes to a solid. Gases move
randomly and quickly, while solids are locked in fixed positions. This means that gases are
significantly more chaotic than solids, with liquids in between. Notice Responses B and C also create a
solid, converting either a gas or a liquid to make a more ordered system. Response D is made up of all
gases. However, there are three gases on the reactants side and only two gases on the products side.
The mixing of three gases provides a more random arrangement than the mixing of two gases.
Therefore, this response is also becoming more ordered. None of the responses become more chaotic
so the correct response must be E.
7.
D
In order to determine whether a ∆S value would be close to zero, the amount of chaos within the
reactants and products would need to be almost the same. This would not occur if there were a change
in phase (like Responses A and C) or a change in number of molecules (like Responses B or E). In
Response D there are two molecules on both sides of the reaction AND all of the molecules involved
are gases. This means that very little entropy will change.
8.
C
Response C is the best choice. Look on your equation sheet.
9.
C
For a question such as this, it is beneficial to review each of the responses. Responses A and D refers
to entropy (S). Response B refers to enthalpy (H). Response C refers to whether a reaction should
occur in the forward direction or the reverse reaction, otherwise known as spontaneity or Gibb’s free
energy. Response E refers to rates of the reaction. So Response C is the correct choice.
10. B
When comparing a reaction’s enthalpy, spontaneity, and/or disorder, use the equation:
∆G = ∆H – T ∆S. This allows you to visualize the effects of the positive and negative signs of these
variables. A reaction that is exothermic has a negative ∆H and a nonspontaneous reaction has a
positive ∆G. Plugging these into the equation shows: (+) = (–) – T∆S. In order to make the ∆G
positive, the value of ∆S must be negative so that the quantity of (–T∆S) becomes positive and
overcomes the negative value of ∆H. If the quantity of (–T∆S) will not overcome the negative ∆H if
the quantity is small enough; this means that lower temperatures will cause the negative ∆H to force the
reaction to be spontaneous.
Now looking at the responses, Response B says that the reaction will be spontaneous at lower
temperatures, which matches what was just described.
By the way, the endothermic or exothermic nature of a chemical process can never change, so neither
Response C nor D will ever be true.
11. C
When the free energy of formation value, ∆Gf˚, is zero, the substance is an element at its natural state
and no energy is needed to form it. Looking over the responses, the one that stands out is Response C,
N(g). Nitrogen is diatomic, N2(g), as an element at its natural state, so a monoatomic form must have a
value for ∆G.
12. C
UNIT 11 Practice Test Advanced Thermodynamics Page 9 of 11 According to the information given, the ∆H is (+), so endothermic. This means that Response D is
false. The reverse reaction must be exothermic, making Response A false. Response E is also false
since the energy of a reaction can not change with a change in temperature.
According to the information given, the ∆G is spontaneous, so (–). If the forward process is
spontaneous at 250˚C, then the reverse process is nonspontaneous. This seems to make Response B
true, but use the Gibb’s free energy equation to make a decision. The equation becomes (–) = (+) –
T∆S. This means that the only way for this reaction to have a negative ∆G is if ∆S is positive. This
also means that the sign of ∆G can only be negative when there is a high temperature to make the
quantity –T∆S exceed the value of ∆H. This makes Response C true, since the positive ∆S is the
reason that the reaction can be spontaneous. Going back to Response B, however, the chance of the
forward reaction being spontaneous at room temperature is unlikely since the value of room
temperature is so much lower than 250˚C.
13. B
Before considering the spontaneity of this reaction, take a moment to really look at the reaction. In
terms of enthalpy, this reaction is forming a bond. Bond formation is exothermic and negative. In
terms of entropy, this reaction goes from two gaseous reactants to one gaseous product. The main
difference is the quantity of two to one, causing a reduction in chaos and a negative ∆S. With this
information, the correct response is Response B.
14. C
Based on the equation ∆G = ∆H – T∆S, in order for the reaction to be only spontaneous (∆G –) at such
a high temperature the value of ∆H must be positive, while the value of ∆S must be positive. This
causes the equation to appear as (–) = (+) – T(+). Notice how the negative T∆S quantity takes over the
∆H quantity.
15. A
The fact that the solution becomes cold means that the energy of the surroundings (water, test tube,
etc.) is being absorbed by the dissolving process. This means that the dissolving process is
endothermic and positive. In addition, the dissolving process is a chaotic process. This is more easily
seen with the balanced equation:
NH4NO3(s) → NH4+(aq) + NO3–(aq)
Notice how the reaction has one solid turning into two aqueous ions, showing an increase in disorder
and a positive value for ∆S.
16. C
As a phase change, sublimation absorbs energy to change the chemical from a solid to a gas. This
means that it is an endothermic process with a positive value for enthalpy (∆H). The fact that the
phases are changing from one solid to one gas also means that the chaos is increasing, forming a
positive entropy (∆S) value. This narrows down the options to either Response B or Response C. If
the value of ∆G is positive, the reaction is NOT spontaneous. Substances like dry ice, however,
sublime automatically. Therefore, sublimation can be spontaneous and have a negative free energy
(∆G). In addition, there is no way for a reaction to have all positive values for all temperatures. Look
at the equation ∆G = ∆H – T∆S: (+) = (+) – T(+). The value for ∆G would have to become negative
when the temperature became large enough because the quantity of –T∆S would eventually take over
the value of ∆H.
UNIT 11 Practice Test Advanced Thermodynamics Page 10 of 11 17. B
The first thing that you notice is one gas and one solid turns into one solid. This means that the entropy
is decreasing and the value of ∆S is negative. Now you need to determine whether the reaction is
endothermic or exothermic. But no information has been given about energy. Now, use the Gibb’s
free energy equation ∆G = ∆H – T∆S. Knowing that the reaction is spontaneous, you know that the
value for ∆G is negative. Now plug in: (–) = ∆H – T(–). Notice that the –T∆S quantity will be
positive. Therefore the driving force of the reaction must be a negative enthalpy value, so that the ∆G
value becomes negative.
18. C
Skimming the responses, almost all of them have to do with entropy. This, of course, doesn’t include
Response D, which relates to orders and rates… rates have nothing to do with the question. Looking at
the reaction, one mole of solid turns into one mole of liquid. This means that the entropy is increasing.
19. C
Based on the Laws of Thermodynamics, every spontaneous process must result in an increase in
entropy within the entire universe. In other words, the ∆Suniv must always be positive.
20. A
Clearly, this is the sort of question that would require a calculator. You have been given heats of
formations and absolute entropies. Use the two equations to solve for the heat of reaction, ∆Hrxn, and
the entropy change for the reaction, ∆Srxn.
H rxn   H f  products    H f  reactants 
H rxn   H f  CuO   H f  H 2     H f  Cu   H f  H 2O  
H rxn   155.2  0  0   241.8    86.6kJ
Srxn   S   products    S   reactants 
S rxn   S   CuO   S   H 2     S   Cu   S   H 2O  
Srxn   43.5  130.6  33.3  188.7   47.9 J
K
Now, plug the values into the Gibb’s free energy equation, remembering to convert the energies to the
same unit.
∆G = 86.6 – (500K)(–0.0479) = +110.6 kJ
21. E
The only indication whether a reaction will be spontaneous or not is the sign of Gibb’s free energy.
22. A
Although all three of the variables are dependent on temperature to some extent, the one that will be the
most influenced is the value of ∆G. Notice that this can be seen in the Gibb’s free energy equation: ∆G
= ∆H – T∆S.
23. C
Response B is not possible; no reaction changes its ability to absorb or release energy. Response E is
also untrue since the forward process is endothermic (+). This means that the reverse process must be
exothermic.
UNIT 11 Practice Test Advanced Thermodynamics Page 11 of 11 Looking at the Gibb’s free energy equation, the values would appear as (–) = (+) – (900)(+). Notice
that the reaction is spontaneous with a very high temperature (900˚C). This means that Response A is
probably untrue, since room temperature (25˚C) is so much lower than 900˚C so the ∆G will probably
be positive at room temperature. “Probably”… there must be more…
If the forward reaction is spontaneous at 900˚C, the reverse reaction must be nonspontaneous at the
same time. This means that Response D can only be true if Response A is also true. Since there can’t
be two answers that are true, both responses are false.
So Response C must be the true response, which it is. Again, look at the signs in the Gibb’s free
energy equation. Notice that the ∆G is negative because of the –T∆S quantity, since the positive ∆S
makes this quantity negative while the ∆H is positive. Since this is the only way to get a spontaneous
reaction, ∆S is the driving force.
24. D
The value of ∆G does not influence the values of ∆H or ∆S. So neither Response A nor B can be true.
Response C just seems silly.
In order to review Responses D and E, use the equation ∆G = – R T ln K. If ∆G = 0, then the equation
becomes 0 = –R T ln K, or K  e
0
RT
 e0  1 . So Response D is true.