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

Questions 

\[
\mathrm{C} \text { (diamond) } \rightarrow \mathrm{C} \text { (graphite) } \quad \Delta G^{\circ}=-2.9 \mathrm{~kJ} / \mathrm{mol}_{r x n}
\]

Which of the following best explains why the reaction represented above is not observed to occur at room temperature?

(A) The rate of the reaction is extremely slow because of the relatively small value of \(\Delta G^{\circ}\) for the reaction.
(B) The entropy of the system decreases because the carbon atoms in graphite are less ordered than those in diamond.
(C) The reaction has an extremely large activation energy due to strong three-dimensional bonding among carbon atoms in diamond.
(D) The reaction does not occur because it is not thermodynamically favorable.

▶️Answer/Explanation

Ans: C

The given reaction, \(C(diamond) \rightarrow C(graphite)\), represents the transformation of diamond into graphite, with a standard Gibbs free energy change, \(Delta G^{\circ}\), of-2.9 kJ/mol for the
reaction.

Let’s analyze each option:

(A) This option suggests that the reaction rate is slow due to the relatively small value of \(\Delta G^{\circ}\). However, the magnitude of \(Delta G^{\circ}\) does not directly dictate the reaction rate. It is the
activation energy that primarily influences the reaction rate.

(B) This option refers to the entropy change, stating that the carbon atoms in graphite are less ordered than those in diamond, leading to a decrease in entropy. However, the entropy change is not the primary
factor influencing whether the reaction occurs at room temperature.

(C) This option proposes that the reaction has a large activation energy due to strong three-dimensional bonding among carbon atoms in diamond. This is a plausible explanation because the activation energy
represents the energy barrier that the reactant molecules must overcome to form products. Strong bonding in diamond could indeed result in a large activation energy barrier, making the reaction kinetically
unfavorable at room temperature.

(D) This option states that the reaction does not occur because it is not thermodynamically favorable. However, the given value of \(\Delta G^{\circ}\)(-2.9 kJ/mol) indicates that the reaction is
thermodynamically favorable. Therefore, thermodynamic favorability is not the reason why the reaction is not observed at room temperature.

Based on the analysis, option (C) is the best explanation. The reaction likely has an extremely large activation energy due to strong three-dimensional bonding among carbon atoms in diamond, making the
reaction kinetically unfavorable at room temperature despite being thermodynamically favorable.

Questions 34

The saturated\( CuSO_4\)(aq) shown above is left uncovered on a lab bench at a constant temperature. As the solution evaporates, 1.0 mL samples of the solution are removed every three days and the \([SO_{4} ^{2-}]\) in the samples is measured. It is observed that the \([SO_{4}^{2-}]\) in the solution did not change over time. Which of the following best helps to explain the observation?
(A) As the solution evaporates, \(Cu^{2+}\)(aq) and \(SO_{4} ^{2-}\)(aq) leave the beaker along with the water molecules.
(B) As the solution evaporates, the dissolution of \(CuSO_{4}\)(s) in the beaker decreases.
(C) The evaporation of water is endothermic, so more \(CuSO_{4}\)(s) dissolves exothermically in the solution, which increases the\( [SO_{4}^{2-}]\).
(D) As water evaporates, more\( CuSO_{4}\)(s) precipitates out of the solution in the beaker.

▶️Answer/Explanation

Ans: B

The observation that the concentration of \( SO_{4}^{2-} \) in the solution remains constant despite the evaporation of water suggests that the dissolution of \( CuSO_{4}(s) \) in the beaker is a key factor. Let’s analyze each option:

(A) This statement suggests that both \( Cu^{2+} \) and \( SO_{4}^{2-} \) ions leave the solution along with water molecules as it evaporates. However, this contradicts the premise that the concentration of \( SO_{4}^{2-} \) remains constant.

(B) This option aligns with the observation. If the dissolution of \( CuSO_{4}(s) \) in the beaker decreases as the solution evaporates, the concentration of \( SO_{4}^{2-} \) would remain constant because there’s no additional source of sulfate ions in the system.

(C) This option suggests that the evaporation of water causes more \( CuSO_{4}(s) \) to dissolve in the solution, which increases the \( SO_{4}^{2-} \) concentration. However, this is unlikely as the dissolution of \( CuSO_{4}(s) \) generally decreases with decreasing solution volume due to saturation.

(D) This option proposes that as water evaporates, more \( CuSO_{4}(s) \) precipitates out of the solution. However, if this were the case, the concentration of \( SO_{4}^{2-} \) would decrease over time, which contradicts the observation.

Therefore, option (B) is the best explanation for the observation, as it suggests that the decrease in the dissolution of \( CuSO_{4}(s) \) in the beaker is responsible for the constant concentration of \( SO_{4}^{2-} \) in the solution.

Questions 

As a sample of \(KNO_3\)(s) is stirred into water at 25°C, the compound dissolves endothermically. Which of the following best helps to explain why the process is thermodynamically favorable at 25°C?
(A) All endothermic processes are thermodynamically favorable.
(B) Stirring the solution during dissolution adds the energy needed to drive an endothermic process.
(C) Dissolving the salt decreases the enthalpy of the system.
(D) Dissolving the salt increases the entropy of the system.

▶️Answer/Explanation

>Ans: D

The correct answer is (D) Dissolving the salt increases the entropy of the system.

For a process to be thermodynamically favorable at a given temperature, the total entropy change (AStotal) must be positive. The total entropy change is the sum of the entropy change due to the process itself
(ASsystem) and the entropy change due to the transfer of heat between the system and the surroundings (ASsurroundings).

In the case of the endothermic dissolution of KNO3(s) in water at 25C, the following considerations apply:

(A) All endothermic processes are thermodynamically favorable: This statement is incorrect. Both exothermic and endothermic processes can be thermodynamically favorable or unfavorable, depending on the
entropy change.

(B) Stirring the solution during dissolution adds the energy needed to drive an endothermic process: This statement is incorrect. Stirring may facilitate the dissolution process by increasing the rate of mixing,
but it does not provide the energy required for the endothermic process.

(C) Dissolving the salt decreases the enthalpy of the system: This statement is incorrect for an endothermic process. Endothermic processes involve an increase in the enthalpy of the system.

(D) Dissolving the salt increases the entropy of the system: This statement is correct. When a solid salt dissolves in water, the ions become more disordered and dispersed, resulting in an increase in entropy
(ASsystem > o). Additionally, the absorption of heat from the surroundings during the endothermic process also contributes to an increase in the entropy of the surroundings (ASsurroundings >0).

The increase in total entropy (AStotal = ASsystem + ASsurroundings) is what makes the endothermic dissolution process thermodynamically favorable at 25C, despite the increase in enthalpy.

Question

Reaction 1 : 4\( NH _3\)(g) + 5 \(O_2\)(g) → 4 NO(g) + \(6 H_2O\)(l) ΔG° = −1010 \(kJ/mol_{rxn}\)
Reaction 2 : 2\( NO _2\)(g) → 2 NO(g) +\( O_2\)(g) ΔG° = 70 \(kJ/mol_rxn\)
Reaction 3 : 4 \(NO_ 2\)(g) + \(O_2\)(g) + 2 \(H_2\)O(l) → 4 \(HNO _3\)(aq) ΔG° = −170 \(kJ/mol_{rxn}\)

Based on the values of ΔG° for the three reactions represented above, what is the value of ΔG° for the reaction represented below?

\(4 NH_3\)(g) +\( 8 O_2\)(g) →\( 4 HNO _3\)(aq) + 4 \(H_2\)O(l)

(A) −1040\( kJ/mol_rxn\)
(B) −1110 \(kJ/mol_rxn\)
(C) −1250\( kJ/mol_rxn\)
(D) −1320\( kJ/mol_rxn\)

▶️Answer/Explanation

Ans:D

The value of ΔG° for the reaction can be calculated by adding the ΔG° values of the reactions that sum up to the given reaction.

The given reaction, \(4 NH_3(g) + 8 O_2(g) → 4 HNO_3(aq) + 4 H_2O(l)\), can be obtained by adding Reaction 1, the reverse of Reaction 2 (multiplied by 2), and Reaction 3.

The ΔG° for the reverse of a reaction is the negative of the ΔG° for the forward reaction, and the ΔG° for a reaction multiplied by a factor is that factor times the ΔG° for the original reaction.

So, the ΔG° for the given reaction is:

$ΔG° = ΔG°(Reaction 1) + 2 × ΔG°(Reverse of Reaction 2) + ΔG°(Reaction 3) = -1010 kJ/mol_rxn – 2 × 70 kJ/mol_rxn – 170 kJ/mol_rxn = -1320 kJ/mol_rxn$

So, the correct answer is (D) -1320 kJ/mol_rxn.

Question

    \(2 NO_2(g) → N_2O_4\)  (g)

dark                                              brown colorless

The dimerization of NO2(g), an exothermic process, is represented by the equation above.

The forward reaction is thermodynamically favored at which of the following temperatures?
(A) All temperatures
(B) Low temperatures only
(C) High temperatures only
(D) No temperature

▶️Answer/Explanation

Ans:B

The dimerization of \(NO_2(g)\) to form \(N_2O_4(g)\) is an exothermic process, meaning it releases heat. According to Le Chatelier’s principle, for an exothermic reaction, increasing the temperature will shift the equilibrium position towards the reactants, favoring the reverse reaction. Conversely, decreasing the temperature will favor the forward reaction, leading to the formation of more \(N_2O_4(g)\).

Therefore, the forward reaction is thermodynamically favored at low temperatures. Thus, the correct answer is:

(B) Low temperatures only

Questions

Which of the following statements identifies the greatest single reason that the value of \(K_{p}\) for the overall reaction at 298 K has such a large magnitude?
(A) The activation energy for step 1 of the mechanism is large and positive.
(B) The activation energy for step 2 of the mechanism is small and positive.
(C) The value of \(\bigtriangleup S^{\circ }\) for the overall reaction is small and positive.
(D) The value of \(\bigtriangleup H^{\circ }\) for the overall reaction is large and negative.

▶️Answer/Explanation

Ans: D

To determine the greatest single reason for the large magnitude of the equilibrium constant (Kp) for the overall reaction at \(298 \mathrm{~K}\), we need to consider the relationship between the equilibrium constant and the standard Gibbs free energy change \(\left(\Delta G^{\circ}\right)\) for the reaction.

The equilibrium constant ( \(\mathrm{Kp}\) ) is related to the standard Gibbs free energy change \(\left(\Delta G^{\circ}\right)\) by the following equation:
\[
\Delta G^{\circ}=-R T \ln (\mathrm{Kp})
\]

Where:
\(\mathrm{R}\) is the universal gas constant \((8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K})\)
 \(\mathrm{T}\) is the absolute temperature (in Kelvin)

A large negative value of \(\Delta G^{\circ}\) corresponds to a large positive value of \(K p\), and vice versa.
From the given information, we can calculate the overall standard enthalpy change \(\left(\Delta H^{\circ}\right)\) for the reaction:
\[
\begin{aligned}
& \Delta \mathrm{H}^{\circ}=\Delta \mathrm{H}^{\circ}(\text { Step } 1)+\Delta \mathrm{H}^{\circ}(\text { Step } 2)+\Delta \mathrm{H}^{\circ}(\text { Step 3) } \\
& \Delta \mathrm{H}^{\circ}=242 \mathrm{~kJ} / \mathrm{mol}+4 \mathrm{~kJ} / \mathrm{mol}+(-432 \mathrm{~kJ} / \mathrm{mol}) \\
& \Delta \mathrm{H}^{\circ}=-186 \mathrm{~kJ} / \mathrm{mol}
\end{aligned}
\]

The large negative value of \(\Delta H^{\circ}\) indicates that the overall reaction is highly exothermic, which means that the products are more stable than the reactants.

Therefore, the greatest single reason for the large magnitude of \(\mathrm{Kp}\) for the overall reaction at \(298 \mathrm{~K}\) is (D) The value of \(\Delta H^{\circ}\) for the overall reaction is large and negative.

A large negative value of \(\Delta \mathrm{H}^{\circ}\) contributes to a large negative value of \(\Delta \mathrm{G}^{\circ}\), which in turn results in a large positive value of \(\mathrm{Kp}\) according to the equation \(\Delta \mathrm{G}^{\circ}=-\mathrm{RT} \ln (\mathrm{Kp})\).

Questions

                     

Based on the thermodynamic data, which of the following is true at 298 K?
(A) \(K_{eq}\)=0
(B) 0<\(K_{eq}\)<1
(C) \(K_{eq}\)=1
(D) \(K_{eq}\)>1

▶️Answer/Explanation

Ans: B

\[ \Delta G^{\circ} = -RT \ln(K_{eq}) \]

Where:
 \( \Delta G^{\circ} \) is the standard Gibbs free energy change
 \( R \) is the gas constant (\( 8.314 \, \text{J/K} \cdot \text{mol} \))
\( T \) is the absolute temperature (\( 298 \, \text{K} \))
 \( K_{eq} \) is the equilibrium constant

Given:
\[ \Delta G^{\circ}_{298} = +15 \, \text{kJ/mol}_{\text{rxn}} \]

We substitute the values into the equation:

\[ +15 \, \text{kJ/mol}_{\text{rxn}} = -RT \ln(K_{eq}) \]
\[ +15000 \, \text{J/mol}_{\text{rxn}} = -(8.314 \, \text{J/K} \cdot \text{mol}) \times (298 \, \text{K}) \times \ln(K_{eq}) \]
\[ \ln(K_{eq}) = \frac{-15000 \, \text{J/mol}_{\text{rxn}}}{(8.314 \, \text{J/K} \cdot \text{mol}) \times (298 \, \text{K})} \]
\[ \ln(K_{eq}) = -6.06 \]

Taking the exponential of both sides:
\[ K_{eq} = e^{-6.06} = 0.0023 \]

Since \( 0 < K_{eq} < 1 \), the correct answer is \( (B) \) \( 0 < K_{eq} < 1 \).

The positive value of \( \Delta G^{\circ}_{298} \) indicates that the reaction is non-spontaneous at \( 298 \, \text{K} \), and the equilibrium favors the reactants over the products, resulting in an equilibrium constant value less than \( 1 \).

Questions

Two molecules of the amino acid glycine join through the formation of a peptide bond, as shown above. The thermodynamic data for the reaction are listed in the following table.

Under which of the following temperature conditions is the reaction thermodynamically favored?
(A) It is only favored at high temperatures.
(B) It is only favored at low temperatures.
(C) It is favored at all temperatures.
(D) It is not favored at any temperature.

▶️Answer/Explanation

Ans: D

Given:

$ΔH°298 = +12 kJ/mol_{rxn} ~~~~~ (ΔH° > 0)$

$ΔS°298 = -10 J/(K · mol_rxn)~~~ (ΔS° < 0)$

The temperature conditions under which a process is thermodynamically favored (ΔG° < 0) can be predicted from the signs of ΔH° and ΔS°. When ΔH° < 0 and ΔS° > 0, the process is favored at all temperatures.

When ΔH° < 0 (exothermic) and ΔS° > 0 (increase in entropy), the process is favored at all temperatures.

In this case, since ΔH° > 0 (endothermic) and ΔS° < 0 (decrease in entropy), the correct answer is (D): The reaction is not favored at any temperature.

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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