Princeton 2012/Barron 4th ed.
AP Practice Problems
Unit 10 – Thermochemistry
Multiple Choice (no calculator)
For questions 1-3, one or more of the
following responses will apply; each
response may be used more than once or not
at all in these questions.
A.
B.
C.
D.
E.
6. Which of the following is true of the
reaction shown in the diagram
below? (P8.6)
Free energy change (∆G)
Entropy change (∆S)
Heat of vaporization
Heat of fusion
Heat capacity
1. If this has a negative value for a
process, then the process occurs
spontaneously. (P8.1)
2. This is a measure of how the
disorder of a system is changing.
(P8.2)
3. This is the energy given off when a
substance condenses. (P8.3)
4. This is the energy taken in by a
substance when it melts. (P8.4)
5. The reaction below is not
spontaneous under standard
conditions, but becomes spontaneous
as the temperature decreases towards
absolute zero. Which of the
following is true at standard
conditions? (P8.5)
2 Al(s) + 3 Cl2(g)  2 AlCl3(s)
a.
b.
c.
d.
e.
∆S and ∆H are both negative
∆S and ∆H are both positive
∆S is negative, and ∆H is positive
∆S is positive, and ∆H is negative
∆S and ∆H are both equal to zero
a. The reaction is endothermic
because the reactants are at a
higher energy level than the
products.
b. The reaction is endothermic
because the reactants are at a
lower energy level than the
products.
c. The reaction is exothermic
because the reactants are at a
higher energy level than the
products.
d. The reaction is exothermic
because the reactants are at a
lower energy level than the
products.
e. The reaction is endothermic
because the reactants are at
the same energy level as the
products.
Princeton 2012/Barron 4th ed.
7. Based on the information given in
the table below, what is ∆H° for the
above reaction? (P8.7)
10. Which point on the graph shown
below corresponds to activate
complex or transition state? (P8.10)
2 H2(g) + O2(g)  2 H2O(g)
Bond
(kJ/mol)
H-H
O=O
O-H
a.
b.
c.
d.
e.
Average bond energy
500
500
500
-2,000 kJ
-1,500 kJ
-500 kJ
+1,000 kJ
+2,000 kJ
8. Which of the following is true of a
reaction that becomes spontaneous at
298 K but becomes nonspontaneous
at a higher temperature? (P8.8)
a. ∆S° and ∆H° are both
negative
b. ∆S° and ∆H° are both
positive
c. ∆S° is negative, and ∆H° is
positive
d. ∆S° is positive, and ∆H° is
negative
e. ∆S° and ∆H° are both equal
to zero
9. Which of the following will be true
when a pure substance in liquid
phase freezes spontaneously? (P8.9)
a. ∆G, ∆H, and ∆S are all positive
b. ∆G, ∆H, and ∆S are all negative
c. ∆G and ∆H are negative, but ∆S
is positive
d. ∆G and ∆S are negative, but ∆H
is positive
e. ∆S and ∆H are negative, but ∆G
is positive
a.
b.
c.
d.
e.
1
2
3
4
5
11.
C(s) + O2(g)  CO2(g)
∆H° = -390 kJ/mol
H2(g) + ½ O2(g)  H2O (l) ∆H° = -290 kJ/mol
2 C(s) + H2(g)  C2H2(g) ∆H° = +230 kJ/mol
Based on the information given
above, what is ∆H for the following
reaction? (P8.11)
C2H2(g) + 5/2 O2(g)  2 CO2(g) + H2O(l)
a.
b.
c.
d.
e.
-1,300 kJ
-1,070 kJ
-8,40 kJ
-780 kJ
-680 kJ
Princeton 2012/Barron 4th ed.
12. If an endothermic reaction is
spontaneous at 298 K, which of the
following must be true for the
reaction? (P8.12)
I. ∆G is greater than zero
II. ∆H is greater than zero
III.
∆S is greater than zero
a.
b.
c.
d.
e.
I only
II only
I and II only
II and III only
I, II, and III
13. C(s) + 2H2(g)  CH4(g) ∆H = x
C(s) + O2(g)  CO2(g)
∆H° = y
H2(g) + ½ O2(g)  H2O(l) ∆H° = z
Based on the information given
above, what is ∆H for the following
reaction? (P8.14)
15. In which of the following reactions is
entropy increasing? (P8.16)
a. 2 SO2(g) + O2(g)  2 SO3(g)
b. CO(g) + H2O (g)  H2(g) +
CO2(g)
c. H2(g) + Cl2(g)  2 HCl(g)
d. 2 NO2(g)  2 NO(g) + O2(g)
e. 2 H2S(g) + 3 O2(g)  2
H2O(g) + 2 SO2(g)
16. When pure sodium is placed in an
atmosphere of chlorine gas, the
following spontaneous reaction
occurs. (P8.17)
2 Na(s) + Cl2(g)  2 NaCl(s)
Which of the following statements is
true about the reaction?
I.
II.
III.
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(l)
a.
b.
c.
d.
e.
x+y+z
x+y-z
z + y- 2x
2z + y - x
2z + y - 2x
14. For which of the following processes
will ∆S be positive? (P8.15)
I. NaCl(s)  Na+(aq) + Cl-(aq)
II. 2 H2(g) + O2(g)  2 H2O(g)
III. CaCO3(s)  CaO(s) + CO2(g)
a.
b.
c.
d.
e.
I only
II only
I and II only
I and III only
I, II, and III
a.
b.
c.
d.
e.
∆S > 0
∆H < 0
∆G > 0
I only
II only
I and II only
II and III only
I, II, and III
17. Gaseous hydrogen and fluorine
combine in the reaction above to
form hydrogen fluoride with an
enthalpy change of -540 kJ. What is
the value of the heat of formation of
HF(g)? (P8.18)
H2(g) + F2(g)  2 HF(g)
a.
b.
c.
d.
e.
-1,080 kJ/mol
-540 kJ/mol
-270 kJ/mol
270 kJ/mol
540 kJ/mol
Princeton 2012/Barron 4th ed.
18. Which of the following is true of the
reaction shown below at room
temperature? (P8.19)
H2O(s)  H2O(l)
I.
II.
III.
a.
b.
c.
d.
e.
∆G is greater than
zero
∆H is greater than
zero
∆S is greater than
zero
II only
III only
I and II only
I and III only
II and III only
19. 2 S(s) + 3 O2(g)  2 SO3(g)
∆H = +800 kJ/mol
2 SO3(g)  2 SO2(g) + O2(g)
∆H = -200 kJ/mol
Based on the information given
above, what is ∆H for the following
reaction? (P8.20)
S(s) + O2(g)  SO2(g)
a.
b.
c.
d.
e.
300 kJ
500 kJ
600 kJ
1,000 kJ
1,200 kJ
Princeton 2012/Barron 4th ed.
Essays
1. The reaction above proceeds spontaneously from standard conditions at 298K. (P8.4)
2 H2(g) + O2(g)  2 H2O(l)
a. Predict the sign of the entropy change, ∆S°, for the reaction. Explain.
b. How would the value of ∆S° for the reaction change if the product of the reaction
was H2O(g)?
c. What is the sign of ∆G° at 298 K? Explain.
d. What is the sign of ∆H° at 298 K? Explain.
2. The reaction above is spontaneous at 298 K, and the heat of reaction, ∆H°, is -178 kJ.
(P8.5a, b, c)
a. Predict the sign of the entropy change, ∆S°, for the reaction. Explain.
b. What is the sign of ∆G° at 298 K? Explain.
c. What change, if any, occurs to the value of ∆G° as the temperature is increased
from 298 K?
3.
Substance
C6H12O6(s)
O2(g)
CO2(g)
H2O(l)
Absolute Entropy, S
(J/mol-K)
212.13
205
213.6
69.9
Molecular Weight
180
32
44
18
Energy is released when glucose is oxidized in the following reaction, which is a
metabolism reaction that takes place in the body. (P8.1a, b, d)
C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(l)
The standard enthalpy change, ∆H°, for the reaction is -2,801 kJ at 298 K.
a. Calculate the standard entropy change, ∆S°, for the oxidation of glucose.
b. Calculate the standard free energy change, ∆G°, for the reaction at 298K.
c. How much energy is given off by the oxidation of 1.00 gram of glucose?
Princeton 2012/Barron 4th ed.
4.
Bond
C-H
O=O
C=O
O-H
Average Bond Dissociation Energy
(kJ/mol)
415
495
799
463
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O (g)
The standard free energy change, ∆G°, for above is -801 kJ at 298 K. (P8.1a, c, d)
a. Use the table of bond dissociation energies to find ∆H° for the reaction above.
b. What is the value of ∆S° for the reaction at 298 K?
c. Give an explanation for the size of the entropy change found in (b).
5. The heat of formation, ∆H°f, of NH3(g) is -46.2 kJ/mol. The free energy of formation,
∆G°f, of NH3(g) is -16.7 kJ/mol. (P8.3)
N2(g) + 3 H2 (g) ↔ 2 NH3(g)
a. What are the values of ∆H° and ∆G° for the reaction?
b. What is the value of the entropy change, ∆S°, for the reaction above at 298 K?
c. As the temperature is increased, what is the effect on ∆G for the reaction? How
does this affect the spontaneity of the reaction?
d. At what temperature can N2, H2, and NH3 gases be maintained together in
equilibrium, each with a partial pressure of 1 atm?
Princeton 2012/Barron 4th ed.
Answer Key – Thermochemistry
Multiple Choice
1. A
2. B
3. C
4. D
5. A
6. C
7. C
8. A
9. B
10. C
11. A
12. D
13. D
14. D
15. D
16. B
17. C
18. E
19. A
Essays
1. A. Negative; the products are less random than the reactants.
B. ∆S° would increase because H2O(g) is more random than water, but remaining
negative because the entropy would still decrease from reactants to products.
C. ∆G° is negative because the reaction proceeds spontaneously.
D. ∆H° must be negative at 298 K. For a reaction to occur spontaneously from standard
conditions, either ∆S° must be positive or ∆H° must be negative. This reaction is
spontaneous although ∆S° is negative, so ∆H° must be negative.
2. A. ∆S° is negative because the products are less random than the reactants. That’s
because two moles of reactants are converted to one mole of products and gas is
converted into solid in the reaction.
B. ∆G° is negative because the reaction proceeds spontaneously.
C. Use ∆G° = ∆H° - T∆S°
∆G° will become less negative because as temperature is increased, the entropy change of
a reaction becomes more important in determining its spontaneity. The entropy change
for this reaction is negative, which discourages spontaneity, so increasing temperature
will make the reaction less spontaneous, thus making ∆G° less negative.
3. A. 259 J/K (Use ∆S° = ∑∆S°products - ∑∆S°reactants)
B. -2,880 kJ (Use ∆G° = ∆H°-T∆S°)
C. 15.6 kJ (Use stoichiometry)
4. A. -800 kJ (Use ∆H° = ∑Energies of bonds broken - ∑Energies of bonds formed)
B. 3 J/K (Use ∆G° = ∆H°-T∆S°)
Princeton 2012/Barron 4th ed.
C. ∆S° is very small, which means that the entropy change for the process is very small.
This makes sense because the number of moles remains constant, the number of moles of
gas remains constant, and the complexity of the molecules remains about the same.
5. A. -33.4 kJ (Use ∆G° = ∆H°-T∆S°)
B. -198 J/K (Use ∆G° = ∆H°-T∆S°)
C. From (B), ∆S° is negative, so increasing the temperature increases the value of ∆G°,
making the reaction less spontaneous.
D. 467 K (Use ∆G° = ∆H°-T∆S°)