AP Classroom for Chemical Thermodynamics

Two students prepared aqueous solutions of LiCl and measured the properties, as shown in the table above. Both students observed that the solid LiCl readily dissolved in H2O. The students drew particle diagrams to explain the changes in the enthalpy and entropy of dissolution for LiCl based on their results and observations. Based on this information, the better particle diagram was drawn by which student, and why is that diagram more accurate?

a. The better particle diagram was drawn by Student 1 because when LiCl dissolves in water, it dissociates into Li+ and Cl− ions causing an increase in entropy. LiCl readily dissolves in water because the dissolution process is exothermic and there is an increase in entropy for this process resulting in a favorable free energy change (ΔG°<0). The particle diagram can only show the increase in entropy resulting from the dissociation of the solid, the dispersion of the ions in solution, and the rearrangement of the water molecules around the ions. The particle diagram does not provide a good representation for the changes in enthalpy (changes in energy) associated with the dissolution process.

H(g)+Cl(g)→HCl(g) The formation of HCl(g) from its atoms is represented by the equation above. Which of the following best explains why the reaction is thermodynamically favored?

d. ΔG<0 because although entropy decreases because the number of gaseous product particles is less than the number of gaseous reactant particles, energy is released as the bond between the Hand Cl atoms forms. Although entropy decreases, not increases, as the reaction proceeds (ΔS<0), the amount of energy that is released (ΔH<0) as the bonds between the H and Cl atoms form is large enough to make the reaction thermodynamically favored (ΔG<0).

HA(aq)+H2O(l)⇄A−(aq)+H3O+(aq)ΔG°=+35kJ/molrxn Based on the chemical equation and ΔG° given above, which of the following justifies the claim that HA(aq) is a weak acid?

c. Because ΔG°>>0, Ka<<1, and HA only partially dissociates. For a reaction with ΔG°>0, equilibrium is reached at a point where the ratio of products to reactants is very small and K<<1, indicating that HA only partially dissociates. For an acid ionization equilibrium, K=Ka; thus, HA(aq) is a weak acid and its dissociation is not thermodynamically favored.

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 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 ?

B. ΔS°rxn is negative because the number of molecules in the gas phase decreases as the reaction proceeds. The number of gas-phase reactants is larger than the number of gas-phase products, resulting in a decrease in entropy (ΔS°rxn<0).

The reaction in which H2O(l) is decomposed into H2(g) and O2(g) is thermodynamically unfavorable (ΔG°>0). However, an electrolytic cell, such as the one represented above, can be used to make the reaction occur. Which of the following identifies a flaw in the representation?

b. The equation for the reaction is not correctly balanced. It is true that the equation is not balanced. The correct equation is 2H2O(l)→2H2(g)+O2(g) .

Is the value of the equilibrium constant, K, for the reaction greater than 1, or less than 1 ? Justify your answer.

The response indicates that the equilibrium constant is greater than 1 because ΔG∘<0

(b) Is the reaction thermodynamically favorable at 298K? Justify your answer based on the calculation of the value and sign of ΔG°.

The response includes the following calculation. ΔG∘=ΔH∘−TΔS∘ ΔG∘=−112 kJ/molrxn−(298 K)(−0.147 kJ/(K molrxn))=−68 kJ/molrxn The response includes the following conclusion: The reaction is thermodynamically favorable because ΔG∘ is negative.

2NO(g)+O2(g)→2NO2(g)ΔH°=−112kJ/molrxn Answer the following questions about the reaction represented above at 298K. a) Using the information in the table above, calculate the value of ΔS° for the reaction.

ΔS∘=ΣS∘ products−ΣS∘ reactants ΔS∘=2(240)−[2(211)+205]=−147 J/(K molrxn)

Which of the following particle diagrams represents a process during which the entropy of the system increases?

A. Goes from white circles stacked all together in two rows, second pic is white circles in groups of 2 floating around. The particle diagram represents a phase change from solid to gas. Entropy increases because the sample of matter becomes more dispersed. The particles become freer to move around and move farther away from each other.

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.

4Fe(s)+3O2(g)→2Fe2O3(s)ΔG°=−1500kJ/molrxn The reaction of iron with oxygen to form rust is represented by the equation shown above. A student cleans two iron nails and places each nail in a capped test tube. The following table gives the experimental conditions and the student's observations after one week at room temperature. The student claims that the formation of rust in test tube 2 shows that the reaction is thermodynamically favored. Which of the following justifications should the student use to explain why rust did not form in test tube 1 ?

a. The reaction does not occur at an observable rate when water is not present because it proceeds through a mechanism with a high activation energy. The reaction between iron and oxygen proceeds through different mechanisms in the absence and presence of water. The activation energy for the reaction in the absence of water is much higher and the reactants do not have sufficient energy at room temperature to overcome the barrier and form rust at an observable rate.

2PCl3(g)+O2(g)→2POCl3(g) The oxidation of PCl3(g) is represented by the equation above, and the table provides the approximate values of the absolute molar entropies, S°, for these substances. Based on the information given, what is the approximate ΔS° for the reaction?

b. −170J/(molrxn⋅K) ΔS°rxn=Σ(moles of substance×S°)products−Σ(moles of substance×S°)reactants; therefore, ΔS°rxn=(2×330)−[(2×310)+210]=−170 J/(molrxn⋅K).

At 298K, NH4NO3 readily dissolves in water, suggesting that the change in free energy (ΔG) favors the dissolution process. However, when NH4NO3 dissolves in water, the temperature of the water decreases. The particulate diagram above attempts to provide a microscopic view of the dissolution of NH4NO3(s) considering both the change in enthalpy (ΔH) and the change in entropy (ΔS). Which of the following explains what the particle diagram is able to illustrate and why?

c. The particle diagram is able to illustrate that entropy increases when NH4NO3(s) dissolves in water because the ions disperse in solution. As a result of the dissociation of NH4NO3 and the dispersion of NH4+ and NO3− ions in solution, entropy increases.

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.

CO(g)+2H2(g)⇄CH3OH(g)K=2.2×104at298K A stoichiometric mixture of CO(g) and H2(g) was allowed to react in two different 2.0L rigid containers at a constant temperature of 298K. The reaction is represented by the equation above. Diagram 1 represents the uncatalyzed reaction and diagram 2 represents the catalyzed reaction one hour after the reactants were mixed. Which of the following correctly explains the experimental results represented in the particle diagrams?

Although the reaction is thermodynamically favorable because ΔG°<0 based on the value of K, only the catalyzed reaction could proceed in one hour because it has a lower activation-energy reaction pathway. The large value of K indicates that the reaction is thermodynamically favored and ΔG°<0. In the absence of a catalyst, the reaction proceeds very slowly. When a catalyst is added to the reaction mixture, the reaction proceeds through a different pathway with a lower activation energy, resulting in an increase in the rate of the reaction.

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 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).

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).

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.

4Fe(s)+3O2(g)⇄2Fe2O3(s)ΔH=−1,650kJ/molrxn The oxidation of Fe(s) is represented by the chemical equation above. Which of the following correctly explains whether or not the reaction is thermodynamically favorable?

a. There are more particles (including particles in the gas state) in the reactants than in the product, thus ΔSrxn<0. Because ΔH is large and negative, the reaction will be thermodynamically favorable at low temperatures. ΔS<0. Because ΔG=ΔH−TΔS<0 is the criterion for a process to be thermodynamically favorable, for ΔG to be negative, the term −TΔS should have a much smaller magnitude than ΔH. This will be true at low temperatures.

2AgNO3(aq)+CaCl2(aq)→2AgCl(s)+Ca(NO3)2(aq) The reaction between AgNO3 and CaCl2 is represented by the equation above, and the table provides the approximate S°values for the reactants and products. Which of the following is the approximate ΔS° for the reaction?

a. −68J/(molrxn⋅K) ΔS° for a reaction is calculated using the following mathematical relationship: ΔS°=Σ(moles of substance×S°)products−Σ(moles of substance×S°)reactants. Using the values provided in the table and the stoichiometric coefficients for each substance, ΔS°=[(2×96)+240]−[(2×220)+60]=−68J/(molrxn⋅K).

The particle diagrams above represent a change in physical state that occurred after heating two separate solid samples of a diatomic element. Which of the following best compares the relative magnitude of ΔS° and gives the sign for the entropy change undergone by each sample, and why?

c. The entropy values for both samples are positive and ΔS°sample2>ΔS°sample1, because the change in the spatial distribution of the molecules was greater for sample 2 than for sample 1. Based on the particle diagrams, sample 1 underwent melting and sample 2 underwent sublimation. The relative magnitudes of the change in entropy for these processes is ΔS°sublimation>ΔS°melting because there is a greater increase in the spatial distribution of the particles in a solid-to-gas transition compared to a solid-to-liquid transition. Both ΔS°sublimation>0 and ΔS°melting>0as a result of the increase in the spatial distribution of the particles.