AP CHEMISTRY UNIT 9 PROGRESS CHECK: MCQ EXAM Questions with
Correct Answers
Question 1:
Based on the information given, which of the following is a major difference between the
zinc-mercury cell and the lithium-iodine cell?
Answer:
B. During the initial cell operation, each reaction is thermodynamically favorable, but the larger
operating potential of the lithium-iodine cell indicates that its cell reaction is more
thermodynamically favorable.
Question 2:
The use of zinc-mercury cells in hearing aids has been replaced by zinc-air cells that operate
using the oxidation of Zn by O2 from the air, generating a potential of +1.60V. Table 2 provides
the standard reduction potentials for the half-reactions used in zinc-mercury and zinc-air cells.
Which of the following best explains the modification to the cell design that is mostly
responsible for the difference in standard cell potentials for zinc-mercury and zinc-air cells?
A. The greater standard cell potential of the Zn-air cell compared to that of the zinc-mercury
cell most likely results from the thermodynamically more favorable reduction of O2 compared
to HgO.
The diagrams above represent physical changes for potassium. Which of the following correctly
identifies the physical process and provides the correct ΔS° for the process in the indicated
diagram?
Answer:
A. In diagram 1, the process is freezing and ΔS°freezing is negative.
Two samples containing an equal number of moles of N2(g) are kept inside separate 1-liter rigid
containers. The particle diagrams above show the distribution of molecular speeds in each
Question 3:
sample. Based on the information given, which of the following identifies the sample with the
greater S°, and why?
Answer:
B. Sample 1, because the N2 molecules have a larger distribution of energies compared to
sample 2.
, Question 4:
A sample of an ideal gas at Pi is initially confined to one chamber of the apparatus represented
above, and the other chamber is initially evacuated. The valve connecting the chambers is
opened, and the gas expands at constant temperature to fill both chambers. Which of the
following best describes ΔS for the
Answer:
D. ΔS>0 because the gas particles become dispersed in a larger volume
Question 5:
The chemical equation above represents the formation of SeF6(g) from its elements and the
table provides the approximate values of S° for Se(s) and SeF6(g). Based on the data, which of
the following mathematical expressions can be used to correctly calculate S° for F2(g)?
D. S°=−1/3[−337−314+42]J/(mol.K)
The table above provides approximate S° values for several substances. Based on the
information, which of the following reactions has the largest increase in entropy, ΔS°?
Answer:
A. 2C(s)+O2(g)→2CO(g)
Question 6:
The chemical equation above represents the combustion of glucose, and the table provides the
approximate standard absolute entropies, S°, for some substances. Based on the information
given, which of these equations can be used to calculate an approximation of S° for H2O(g)?
B. S°=1/6[900+209+(6×205)−(6×214)]J/(mol.K)
An ice cube at 0°C melts when placed inside a room at 22°C (295K). Based on the concepts of
enthalpy, entropy, and Gibbs free energy, which of the following best explains why the process
is thermodynamically favorable?
Answer:
D. Melting ice requires energy, ΔH°>0, and ΔS°>0 because the motion of the H2O molecules
increases as it transitions from solid to liquid. At a temperature higher than 0°C (273K), the
term TΔS° is greater than ΔH°, resulting in a thermodynamically favorable process with ΔG°<0.
Question 7:
A 5.00g-sample of KOH(s) at 25.0°C was added to 100.0g of H2O(l) at room temperature inside
an insulated cup calorimeter, and the contents were stirred. After all the KOH(s) dissolved, the
temperature of the solution had increased. Based on the information given, which of the
following best justifies the claim that the dissolution of KOH(s) is a thermodynamically
favorable process?
Answer:
B. The energy required to break the bonds between the ions in the solid is less than that
released as the ion-dipole attractions form during solvation, thus ΔH<0. Also, the ions become
more widely dispersed as KOH(s) dissolves, thus ΔS>0. Therefore, ΔG<0.
Correct Answers
Question 1:
Based on the information given, which of the following is a major difference between the
zinc-mercury cell and the lithium-iodine cell?
Answer:
B. During the initial cell operation, each reaction is thermodynamically favorable, but the larger
operating potential of the lithium-iodine cell indicates that its cell reaction is more
thermodynamically favorable.
Question 2:
The use of zinc-mercury cells in hearing aids has been replaced by zinc-air cells that operate
using the oxidation of Zn by O2 from the air, generating a potential of +1.60V. Table 2 provides
the standard reduction potentials for the half-reactions used in zinc-mercury and zinc-air cells.
Which of the following best explains the modification to the cell design that is mostly
responsible for the difference in standard cell potentials for zinc-mercury and zinc-air cells?
A. The greater standard cell potential of the Zn-air cell compared to that of the zinc-mercury
cell most likely results from the thermodynamically more favorable reduction of O2 compared
to HgO.
The diagrams above represent physical changes for potassium. Which of the following correctly
identifies the physical process and provides the correct ΔS° for the process in the indicated
diagram?
Answer:
A. In diagram 1, the process is freezing and ΔS°freezing is negative.
Two samples containing an equal number of moles of N2(g) are kept inside separate 1-liter rigid
containers. The particle diagrams above show the distribution of molecular speeds in each
Question 3:
sample. Based on the information given, which of the following identifies the sample with the
greater S°, and why?
Answer:
B. Sample 1, because the N2 molecules have a larger distribution of energies compared to
sample 2.
, Question 4:
A sample of an ideal gas at Pi is initially confined to one chamber of the apparatus represented
above, and the other chamber is initially evacuated. The valve connecting the chambers is
opened, and the gas expands at constant temperature to fill both chambers. Which of the
following best describes ΔS for the
Answer:
D. ΔS>0 because the gas particles become dispersed in a larger volume
Question 5:
The chemical equation above represents the formation of SeF6(g) from its elements and the
table provides the approximate values of S° for Se(s) and SeF6(g). Based on the data, which of
the following mathematical expressions can be used to correctly calculate S° for F2(g)?
D. S°=−1/3[−337−314+42]J/(mol.K)
The table above provides approximate S° values for several substances. Based on the
information, which of the following reactions has the largest increase in entropy, ΔS°?
Answer:
A. 2C(s)+O2(g)→2CO(g)
Question 6:
The chemical equation above represents the combustion of glucose, and the table provides the
approximate standard absolute entropies, S°, for some substances. Based on the information
given, which of these equations can be used to calculate an approximation of S° for H2O(g)?
B. S°=1/6[900+209+(6×205)−(6×214)]J/(mol.K)
An ice cube at 0°C melts when placed inside a room at 22°C (295K). Based on the concepts of
enthalpy, entropy, and Gibbs free energy, which of the following best explains why the process
is thermodynamically favorable?
Answer:
D. Melting ice requires energy, ΔH°>0, and ΔS°>0 because the motion of the H2O molecules
increases as it transitions from solid to liquid. At a temperature higher than 0°C (273K), the
term TΔS° is greater than ΔH°, resulting in a thermodynamically favorable process with ΔG°<0.
Question 7:
A 5.00g-sample of KOH(s) at 25.0°C was added to 100.0g of H2O(l) at room temperature inside
an insulated cup calorimeter, and the contents were stirred. After all the KOH(s) dissolved, the
temperature of the solution had increased. Based on the information given, which of the
following best justifies the claim that the dissolution of KOH(s) is a thermodynamically
favorable process?
Answer:
B. The energy required to break the bonds between the ions in the solid is less than that
released as the ion-dipole attractions form during solvation, thus ΔH<0. Also, the ions become
more widely dispersed as KOH(s) dissolves, thus ΔS>0. Therefore, ΔG<0.
