AP Chemistry: 9.3 Gibbs Free Energy and Thermodynamic Favorability – Exam Style questions with Answer- MCQ

Question

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?
A Melting ice releases energy, ΔH°<0 , and ΔS°<0 because the motion of the \(H_2O\) molecules decreases as it transitions from solid to liquid. At a temperature higher than 0°C (273K), the term TΔS° is smaller than ΔH°, resulting in a thermodynamically favorable process with ΔG°<0.

B Melting ice releases energy, ΔH°<0 , and ΔS°>0 because the motion of the \(H_2O\) 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 .

C Melting ice requires energy, ΔH°>0 , and ΔS°<0 because the motion of the \(H_2O\) molecules decreases as it transitions from solid to liquid. At a temperature higher than 0°C (273K), the term TΔS° is smaller than ΔH°, resulting in a thermodynamically favorable process with ΔG°>0.

D Melting ice requires energy, ΔH°>0 , and ΔS°>0 because the motion of the \(H_2O\) 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.

▶️Answer/Explanation

Ans:D

For the melting of an ice cube at 22°C , ΔH°>0 because the process requires energy and ΔS°>0 as a result of the increase in the motion of the \(H_2O\) molecules in the liquid compared to the solid. At constant temperature and pressure, ΔG°<0 for a process that is thermodynamically favored and ΔG°=ΔH°−TΔS°. The increase in entropy is large enough that when ΔH° is added to the term −TΔS° at 295K, ΔG° is negative (ΔG°<0).

Question

A 5.00g-sample of \(KOH(s)\) at 25.0°C was added to 100.0g of \(H_2O(l)\) 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?

A The forces between the ions and the water molecules are stronger than the forces between water molecules, thus ΔH<0 . Also, the ions become less dispersed as \(KOH(s)\) dissolves, thus ΔS>0. Therefore, ΔG<0.

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 .

C The average kinetic energy of the particles increases, resulting in ΔH>0 . Also, the ions become more widely dispersed as \(KOH(s)\) dissolves, thus ΔS>0. Therefore, ΔG>0.

D The average kinetic energy of the particles increases, resulting in ΔH>0 . Also, the ions become more widely dispersed as \(KOH(s)\) dissolves, thus ΔS<0. Therefore, ΔG>0.

▶️Answer/Explanation

Ans:B

A process is thermodynamically favored if ΔG<0. When \(KOH(s)\)  dissolves in water, the temperature increases as a result of the net release of energy (ΔH<0). The dissolution also results in the dispersion of the \(K^+\) and \(OH^−\) ions throughout the solution, making ΔS>0. Because ΔG=ΔH−TΔS, the negative values of both ΔH and the term −TΔS result in ΔG<0.

Question

Based on the information, which of the following statements best helps to explain whether or not the reaction is thermodynamically favored at 298K?
A ΔG°\(_{rxn}\)<<0 and the reaction is thermodynamically favored because the product molecules have more complex structures and greater absolute entropies than the reactant molecules do.

B ΔG°\(_{rxn}\)>>0 and the reaction is not thermodynamically favored because the total number of moles of gas-phase products is not greater than the total number of moles of gas-phase reactants.

C ΔG°\(_{rxn}\)<<0 and the reaction is thermodynamically favored because the energy released when the bonds in the products are formed is greater than the energy absorbed to break the bonds in the reactants.

D ΔG°\(_{rxn}\)>>0 and the reaction is not thermodynamically favored because the energy released when the bonds in the products are formed is less than the energy absorbed to break the bonds in reactants.

▶️Answer/Explanation

Ans:C

Based on the ΔG°\(_f\) for the reactants and products, ΔG°\(_{rxn}\)≈−915kJ/mol and the reaction is thermodynamically favored. Because the total number of moles of gas-phase products is equal to the total number of moles of gas-phase reactants, the contribution of ΔS°rxn to ΔG°\(_{rxn}\) is negligible compared to the contribution of ΔH°rxn. Hence, ΔG°\(_{rxn}\)≈ΔH°\(_{rxn}\)<<0 because the energy released when the bonds in the products are formed is greater than the energy absorbed to break the bonds in the reactants.

Question

\(FeF_{2}(s) \rightleftharpoons Fe2^{+}(aq) + 2 F ^{-}(aq\))           \(K_{1} = 2 × 10^{-6} \)

\(F ̄(aq) + H^{+}(aq) \rightleftharpoons HF(aq)  \)                                         \(K_{2} = 1 × 10^{3} \)

\(FeF_{2}(s) + 2 H^{+}(aq) \rightleftharpoons Fe^{2+}(aq) + 2 HF(aq)\)           \( K_{3} \)= ?

On the basis of the information above, the dissolution of\( FeF_2\)(s) in acidic solution is

(A) thermodynamically favorable, because \(K_2 > 1\)

(B) thermodynamically favorable, because\( K_3 > 1\)

(C) not thermodynamically favorable, because \(K_1 < 1\)

(D) not thermodynamically favorable, because \(K_3 < 1\)

▶️Answer/Explanation

Ans:B

Question

\(XY_{2} → X + Y_{2}\) The equation above represents the decomposition of a compound \(XY_{2}\). The diagram below shows two reaction profiles (path one and path two) for the decomposition of\( XY_{2}\).

The reaction is thermodynamically favorable under standard conditions at 298 K. Therefore, the value of AS° for the reaction must be

(A) equal to zero

(B) equal to \(\Delta H^{\circ}/298 K\)

(C) greater than \(\Delta H^{\circ}/298 K\)

(D) less than \(\Delta H^{\circ}/298 K\)

▶️Answer/Explanation

Ans:C

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