Summary AP Classroom for Chemical Thermodynamics
Test | 2023.
The reaction between SO2 and O2 is represented by the chemical equation above.
The table provides the approximate absolute entropies, S°, for O2(g) and SO3(g).
Which of the following mathematical expressions can be used to correctly
calculate S° for SO2(g) ?
C. S°=12[187+(2×257)−205]J/(mol⋅K)
Since ΔS°reaction=Σ(moles of substance×S°)products−Σ(moles of
substance×S°)reactants, if x=S° for SO2(g), then −187=(2×257)−[2x+205]J/K. Solving
for x, S°=12[187+(2×257)−205]J/(mol⋅K) for SO2(g).
The synthesis of NH3 is represented by the equation above. Based on the
equilibrium constant, K, and ΔH°rxn given above, which of the following can best
be used to justify that the reaction is thermodynamically favorable at 298K and
constant pressure?
B. ΔG°=−RTlnK<0 because K>>1
Correct. A process is thermodynamically favorable if ΔG°<0 at constant T and P. When
K>>1, the term lnK is positive and ΔG° is negative (ΔG°<0).
The combustion of C2H5OH is represented by the equation above and the
standard entropy and enthalpy changes for the reaction are provided. When the
reactants are combined at 25°C, essentially no CO2(g) or H2O(g) is produced after
a few hours. Which of the diagrams above could best help explain the low yield of
the reaction under these conditions, and why?
C. Diagram 2, because it represents a reaction with a high activation energy barrier for
molecules to overcome and a very slow reaction rate, even if it is thermodynamically
favorable with ΔG°<0.
Based on the values of ΔH° and ΔS° provided, the reaction is thermodynamically
favored (ΔG°<0) at all temperatures. Diagram 2 represents a reaction with a higher
activation energy than the reaction represented in diagram 1, and this can help explain
why very little product formed. At 25°C, very few molecules possess enough energy to
overcome the high activation energy barrier, resulting in a very slow reaction rate.
The chemical equation above represents the exothermic reaction of CH4(g) with
O2(g). Which of the following best helps to explain why the reaction is
thermodynamically favored (ΔG<0) at 2000K and 1atm?
D. The amount of energy released when the product bonds form is much greater than
the amount of energy needed to break the reactant bonds.
ΔG for a process that is thermodynamically favored, and ΔG=ΔH−TΔS. The value of ΔS
for the reaction is small because the number of moles of gas is the same on both sides
of the equation. ΔHfor the process is negative and large because of the formation of
much stronger bonds in the products than existed in the reactants. When the relatively
, large negative term ΔH is added to the relatively small term −TΔS, ΔG<0 and the
process is thermodynamically favored.
The table above lists the equilibrium constants and changes in thermodynamic
properties for the dissolution of FeCO3 and MnCO3 at 25°C. The two particle
diagrams below represent saturated solutions of each compound at
equilibrium.Which of the following explains which of the properties listed in the
table is best represented by the particle diagram?
D. The particle diagrams best represent that the molar solubility is greater for FeCO3
compared to MnCO3.
The particle diagrams for these saturated solutions represent the relative magnitudes of
their molar solubility by showing that the dissolution of FeCO3 favors, if only slightly, the
release of ions in solution.
2Fe2O3(s)+3C(s)→4Fe(s)+3CO2(g)
In a blast furnace, the reaction represented above occurs, producing Fe(s) from
its ore, Fe2O3(s). The reaction is thermodynamically favorable and based on
coupling the two reactions represented below.
2Fe2O3→4Fe+3O2
C+O2→CO2
Which of the following identifies a limitation in how the representations above
describe a system of coupled reactions?
C. The values of ΔG° for each reaction are not shown.
In a coupled pair of reactions, the thermodynamic favorability of each component
reaction and the overall reaction is central to the understanding of how coupled
reactions work. In this case the blast furnace reaction is favorable because it is a
combination of the unfavorable decomposition of iron oxide into its elements (ΔG°>0)
and the favorable combustion of carbon to form carbon dioxide (ΔG°<0).
2POCl3(g)⇄2PCl3(g)+O2(g)ΔG°rxn=+490kJ/mol
A sample of POCl3(g) is placed in a closed, rigid container at 298K and allowed to
reach equilibrium according to the equation above. Based on the value for
ΔG°rxn=+490kJ/mol, which of the following is true?
A. K=e−490,0008.314×298<<1 and at equilibrium PPOCl3>>PPCl3.
The large and positive value of ΔG°rxn for the reaction means that at equilibrium the
partial pressure of POCl3 is much greater than the partial pressures of PCl3 and O2.
ΔG°rxn=−RTlnK; hence, rearranging for K gives K=e−ΔG°rxnRT, and because ΔG°rxn
has units of kJ/mol, it must be converted to Jto use R=8.314J/mol⋅K. Substitution of the
values given yields K=e−490,0008.314×298, which is much smaller than 1(K<<1)
because e has a negative exponent. As a result, the reaction reaches equilibrium after a
very small amount of products have been formed, meaning that PPOCl3>>PPCl3.
The diagram above represents the gas-phase reaction of NO2(g) to form N2O4(g)
at a certain temperature. Based on the diagram, which of the following best
predicts and explains the sign of the entropy change for the reaction, ΔS°rxn ?
Test | 2023.
The reaction between SO2 and O2 is represented by the chemical equation above.
The table provides the approximate absolute entropies, S°, for O2(g) and SO3(g).
Which of the following mathematical expressions can be used to correctly
calculate S° for SO2(g) ?
C. S°=12[187+(2×257)−205]J/(mol⋅K)
Since ΔS°reaction=Σ(moles of substance×S°)products−Σ(moles of
substance×S°)reactants, if x=S° for SO2(g), then −187=(2×257)−[2x+205]J/K. Solving
for x, S°=12[187+(2×257)−205]J/(mol⋅K) for SO2(g).
The synthesis of NH3 is represented by the equation above. Based on the
equilibrium constant, K, and ΔH°rxn given above, which of the following can best
be used to justify that the reaction is thermodynamically favorable at 298K and
constant pressure?
B. ΔG°=−RTlnK<0 because K>>1
Correct. A process is thermodynamically favorable if ΔG°<0 at constant T and P. When
K>>1, the term lnK is positive and ΔG° is negative (ΔG°<0).
The combustion of C2H5OH is represented by the equation above and the
standard entropy and enthalpy changes for the reaction are provided. When the
reactants are combined at 25°C, essentially no CO2(g) or H2O(g) is produced after
a few hours. Which of the diagrams above could best help explain the low yield of
the reaction under these conditions, and why?
C. Diagram 2, because it represents a reaction with a high activation energy barrier for
molecules to overcome and a very slow reaction rate, even if it is thermodynamically
favorable with ΔG°<0.
Based on the values of ΔH° and ΔS° provided, the reaction is thermodynamically
favored (ΔG°<0) at all temperatures. Diagram 2 represents a reaction with a higher
activation energy than the reaction represented in diagram 1, and this can help explain
why very little product formed. At 25°C, very few molecules possess enough energy to
overcome the high activation energy barrier, resulting in a very slow reaction rate.
The chemical equation above represents the exothermic reaction of CH4(g) with
O2(g). Which of the following best helps to explain why the reaction is
thermodynamically favored (ΔG<0) at 2000K and 1atm?
D. The amount of energy released when the product bonds form is much greater than
the amount of energy needed to break the reactant bonds.
ΔG for a process that is thermodynamically favored, and ΔG=ΔH−TΔS. The value of ΔS
for the reaction is small because the number of moles of gas is the same on both sides
of the equation. ΔHfor the process is negative and large because of the formation of
much stronger bonds in the products than existed in the reactants. When the relatively
, large negative term ΔH is added to the relatively small term −TΔS, ΔG<0 and the
process is thermodynamically favored.
The table above lists the equilibrium constants and changes in thermodynamic
properties for the dissolution of FeCO3 and MnCO3 at 25°C. The two particle
diagrams below represent saturated solutions of each compound at
equilibrium.Which of the following explains which of the properties listed in the
table is best represented by the particle diagram?
D. The particle diagrams best represent that the molar solubility is greater for FeCO3
compared to MnCO3.
The particle diagrams for these saturated solutions represent the relative magnitudes of
their molar solubility by showing that the dissolution of FeCO3 favors, if only slightly, the
release of ions in solution.
2Fe2O3(s)+3C(s)→4Fe(s)+3CO2(g)
In a blast furnace, the reaction represented above occurs, producing Fe(s) from
its ore, Fe2O3(s). The reaction is thermodynamically favorable and based on
coupling the two reactions represented below.
2Fe2O3→4Fe+3O2
C+O2→CO2
Which of the following identifies a limitation in how the representations above
describe a system of coupled reactions?
C. The values of ΔG° for each reaction are not shown.
In a coupled pair of reactions, the thermodynamic favorability of each component
reaction and the overall reaction is central to the understanding of how coupled
reactions work. In this case the blast furnace reaction is favorable because it is a
combination of the unfavorable decomposition of iron oxide into its elements (ΔG°>0)
and the favorable combustion of carbon to form carbon dioxide (ΔG°<0).
2POCl3(g)⇄2PCl3(g)+O2(g)ΔG°rxn=+490kJ/mol
A sample of POCl3(g) is placed in a closed, rigid container at 298K and allowed to
reach equilibrium according to the equation above. Based on the value for
ΔG°rxn=+490kJ/mol, which of the following is true?
A. K=e−490,0008.314×298<<1 and at equilibrium PPOCl3>>PPCl3.
The large and positive value of ΔG°rxn for the reaction means that at equilibrium the
partial pressure of POCl3 is much greater than the partial pressures of PCl3 and O2.
ΔG°rxn=−RTlnK; hence, rearranging for K gives K=e−ΔG°rxnRT, and because ΔG°rxn
has units of kJ/mol, it must be converted to Jto use R=8.314J/mol⋅K. Substitution of the
values given yields K=e−490,0008.314×298, which is much smaller than 1(K<<1)
because e has a negative exponent. As a result, the reaction reaches equilibrium after a
very small amount of products have been formed, meaning that PPOCl3>>PPCl3.
The diagram above represents the gas-phase reaction of NO2(g) to form N2O4(g)
at a certain temperature. Based on the diagram, which of the following best
predicts and explains the sign of the entropy change for the reaction, ΔS°rxn ?
