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
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