Free Energy and Equilibrium
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AP Chemistry › Free Energy and Equilibrium
For $\mathrm{X(aq) + Y(aq) \rightleftharpoons Z(aq)}$ at constant temperature, a student says: “If $\Delta G$ is negative, the reaction must already be at equilibrium.” Which statement best addresses this idea?
Correct, because equilibrium occurs whenever $\Delta G$ is negative.
Incorrect, because $\Delta G<0$ indicates a net tendency to form products until $\Delta G \approx 0$.
Incorrect, because $\Delta G<0$ indicates the reverse reaction is spontaneous until $\Delta G \approx 0$.
Correct, because negative $\Delta G$ means the forward and reverse rates are equal.
Incorrect, because $\Delta G<0$ means the reaction cannot be reversible.
Explanation
This question probes evaluating claims about ΔG and equilibrium states. For X(aq) + Y(aq) ⇌ Z(aq), the student's idea that negative ΔG means the reaction is at equilibrium is incorrect because ΔG < 0 indicates spontaneity toward products until ΔG ≈ 0 at equilibrium. Negative ΔG shows the system is not yet at minimum free energy. Choice C best addresses this. Choice D tempts but is wrong, as it flips to reverse spontaneity, due to the misconception of sign reversal. To assess such claims, contrast ΔG's role in non-equilibrium (spontaneity) versus equilibrium (zero) conditions.
A student observes the reaction $\mathrm{2HF(aq) \rightleftharpoons H_2F_2(aq)}$ in solution at constant temperature. At one moment, $\Delta G<0$ for the reaction as written. Which statement best describes the system’s position relative to equilibrium?
The system will shift toward reactants until $\Delta G \approx 0$.
The system is at equilibrium because $\Delta G<0$ indicates the minimum free energy state is reached.
The system will shift toward products until $\Delta G \approx 0$.
The system must contain only $\mathrm{H_2F_2}$ because negative $\Delta G$ implies completion.
The system cannot reach equilibrium because $\Delta G<0$ means the reverse reaction cannot occur.
Explanation
This question tests determining a system's equilibrium position from ΔG's sign. For 2HF(aq) ⇌ H₂F₂(aq), ΔG < 0 means the system is not at equilibrium and will shift toward products (H₂F₂) until ΔG ≈ 0. This indicates current concentrations make Q < K, favoring dimer formation. Choice B best describes this. Choice C tempts but is incorrect, suggesting shift to reactants, due to the misconception of negative ΔG favoring reverse. In aqueous equilibria, use ΔG's sign to predict net shifts toward the spontaneous direction.
For $\mathrm{H_2(g) + Cl_2(g) \rightleftharpoons 2HCl(g)}$ at a given temperature, the system is at equilibrium so $\Delta G \approx 0$. Which statement correctly connects $\Delta G$ to the equilibrium state?
$\Delta G \approx 0$ means the forward and reverse reaction rates are equal.
$\Delta G \approx 0$ means the reaction cannot proceed in either direction.
$\Delta G \approx 0$ means the equilibrium constant $K$ must equal 1 for this reaction.
$\Delta G \approx 0$ means products are present in a greater amount than reactants.
$\Delta G \approx 0$ means the forward reaction is spontaneous and the reverse is not.
Explanation
This question assesses connecting ΔG ≈ 0 to microscopic aspects of equilibrium. For H₂(g) + Cl₂(g) ⇌ 2HCl(g), ΔG ≈ 0 at equilibrium means forward and reverse rates are equal, maintaining constant concentrations. This dynamic state persists despite ongoing reactions. Choice C correctly connects this. Choice A is a tempting distractor but wrong, as it assumes K=1, from the misconception that ΔG=0 implies Q=1 specifically. Relate ΔG=0 to rate equality, not K's value, for accurate equilibrium descriptions.
For $\mathrm{C(s) + CO_2(g) \rightleftharpoons 2CO(g)}$ at constant temperature, the system is at equilibrium and thus $\Delta G \approx 0$. Which statement is true about the reaction at this point?
The forward reaction has stopped completely because $\Delta G \approx 0$ means no collisions occur.
Only products are present because $\Delta G \approx 0$ indicates completion.
The reverse reaction has stopped completely because $\Delta G \approx 0$ means products are maximum.
Only reactants are present because $\Delta G \approx 0$ indicates no reaction can proceed.
Both the forward and reverse reactions are occurring, but there is no net change in composition.
Explanation
This question tests understanding that at equilibrium (ΔG ≈ 0), reactions continue microscopically without net change. For C(s) + CO₂(g) ⇌ 2CO(g), ΔG ≈ 0 means both forward and reverse reactions occur, but rates are equal, resulting in no net composition change. This dynamic equilibrium maintains constant concentrations despite ongoing reactions. Choice A is true about the system. Choice D tempts but is incorrect, assuming ΔG ≈ 0 means only products, due to the misconception that equilibrium implies completion. To distinguish, remember equilibrium involves equal rates, not cessation, and verify with ΔG = 0.
For the reaction $\mathrm{A(g) \rightleftharpoons B(g)}$ at constant temperature, a student measures $\Delta G \approx 0$ for the forward direction. Which statement is consistent with this measurement?
The system is product-favored, so $\mathrm{B}$ must be present at a higher concentration than $\mathrm{A}$.
The system is at equilibrium, so there is no net change in the amounts of $\mathrm{A}$ and $\mathrm{B}$.
The reverse reaction is spontaneous, so $\mathrm{B}$ will be completely converted to $\mathrm{A}$.
The forward reaction is spontaneous, so $\mathrm{A}$ will be completely converted to $\mathrm{B}$.
The system is reactant-favored, so $\mathrm{A}$ must be present at a higher concentration than $\mathrm{B}$.
Explanation
This question examines how ΔG ≈ 0 indicates a system at equilibrium with no net composition change. For A(g) ⇌ B(g), ΔG ≈ 0 for the forward direction means the system is at equilibrium, with no net change in amounts of A and B as rates are equal. At this point, the free energy is at a minimum, preventing spontaneous conversion. Choice A is consistent with this measurement. Choice B tempts by assuming product-favorability implies higher B concentration, but this is incorrect as ΔG = 0 doesn't specify favorability, stemming from mixing ΔG with ΔG°. To interpret such data, recall that ΔG = 0 solely denotes equilibrium, independent of K's value.
For $\mathrm{CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)}$ at a certain temperature, the solution is adjusted so that $\Delta G>0$ for the dissociation as written. Which statement correctly describes the net change needed to reach equilibrium?
The reaction will stop permanently because positive $\Delta G$ prevents any molecular collisions.
Net formation of ions will occur because $\Delta G>0$ means the forward reaction is favored.
Net formation of undissociated acid will occur because the reverse direction is spontaneous.
The reaction will go to completion toward reactants so that no ions remain.
No net change will occur because $\Delta G>0$ indicates equilibrium.
Explanation
This question probes understanding of ΔG's sign in relation to spontaneity in acid dissociation equilibria. For CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO⁻(aq), ΔG > 0 for dissociation indicates the reverse reaction is spontaneous, leading to net formation of undissociated acid until equilibrium where ΔG = 0. This occurs because positive ΔG means the current ion concentrations make Q > K, driving recombination. Choice B accurately describes this net change. Choice A is a tempting distractor but incorrect, as it reverses the spontaneity direction, stemming from the misconception that positive ΔG favors forward ionization. Remember to associate positive ΔG with reverse spontaneity to correctly predict equilibrium adjustments in weak acid systems.
For the reaction $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$ at a certain temperature, the system is observed to have $\Delta G \approx 0$. Which statement best describes the state of the system?
The system is product-favored, so essentially all reactants have been converted to products.
The reverse reaction is spontaneous, so the reaction mixture must contain only reactants.
The reaction is nonspontaneous in both directions, so no reaction is occurring.
The system is at equilibrium, so the forward and reverse reaction rates are equal.
The forward reaction is spontaneous, so the reaction will proceed to completion.
Explanation
This question tests the understanding of how the Gibbs free energy change (ΔG) relates to the equilibrium state of a chemical reaction. For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), when ΔG ≈ 0, the system is at equilibrium, meaning the forward and reverse reaction rates are equal and there is no net change in concentrations. This occurs because at equilibrium, the reaction quotient Q equals the equilibrium constant K, resulting in ΔG = 0 according to the equation ΔG = ΔG° + RT ln(Q/K). Thus, choice A correctly describes the system's state. A tempting distractor is choice B, which is incorrect because it confuses ΔG ≈ 0 with a highly product-favored reaction, stemming from the misconception that equilibrium implies complete conversion rather than a dynamic balance. To analyze similar problems, remember that ΔG = 0 indicates equilibrium regardless of the reaction's favorability, and use the sign of ΔG to predict shifts away from equilibrium.
For $\mathrm{2CO(g) + O_2(g) \rightleftharpoons 2CO_2(g)}$ at constant temperature, the system is prepared so that $\Delta G$ for the forward reaction is positive. Which statement best describes the equilibrium tendency from that moment?
No net change will occur because a positive $\Delta G$ means the reaction is too slow.
The system will shift toward products until $\Delta G$ becomes more positive.
The system will shift toward reactants until $\Delta G$ approaches zero.
The system is already at equilibrium because $\Delta G$ can be positive at equilibrium.
The system will produce only products because combustion reactions always go to completion.
Explanation
This question evaluates the interpretation of positive ΔG in predicting the path to equilibrium. For 2CO(g) + O₂(g) ⇌ 2CO₂(g), ΔG > 0 for the forward reaction means the reverse is spontaneous, so the system shifts toward reactants until ΔG approaches zero at equilibrium. This is because positive ΔG indicates Q > K, favoring reactant formation. Choice A best describes this tendency. Choice B is a tempting distractor but incorrect, as it suggests shifting toward products, arising from the misconception that positive ΔG favors the forward direction. To solve these, use the rule that the spontaneous direction opposes the sign of ΔG for the written reaction.
Consider $\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$ at a fixed temperature. A chemist reports that under the current conditions $\Delta G$ for the reaction is negative. What does this indicate about the system relative to equilibrium?
The forward reaction is spontaneous, so the system will shift toward products until equilibrium is reached.
The reverse reaction is spontaneous, so the system will shift toward reactants until equilibrium is reached.
The system must already be product-only because $\Delta G<0$ means products are maximum.
The system cannot reach equilibrium because a negative $\Delta G$ prevents reversibility.
The system is at equilibrium because $\Delta G<0$ indicates no net change.
Explanation
This question tests the skill of interpreting the sign of ΔG to determine a system's position relative to equilibrium and the direction of spontaneous change. For the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), a negative ΔG indicates that the forward reaction is spontaneous under the current conditions, so the system will shift toward products to reduce the free energy until equilibrium is reached where ΔG = 0. This is because ΔG < 0 means Q < K, driving the net formation of products. Therefore, choice B accurately reflects the system's behavior. Choice C is a tempting distractor but incorrect as it assumes the reverse is spontaneous, arising from the misconception that negative ΔG favors the reverse direction instead of the forward. A useful strategy for such questions is to recall that ΔG < 0 predicts a spontaneous forward reaction, helping predict shifts in non-equilibrium systems.
A mixture undergoing $\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}$ at constant temperature has $\Delta G<0$ for the reaction as written. What does this say about the system’s position relative to equilibrium?
The system will shift toward reactants until equilibrium is reached.
The system must already contain only $\mathrm{CO_2(g)}$ because $\Delta G<0$ implies completion.
The system will shift toward products until equilibrium is reached.
The system cannot be reversible because $\Delta G<0$ prevents the reverse reaction.
The system is at equilibrium because solids make $\Delta G$ always negative.
Explanation
This question examines how the sign of ΔG indicates the direction of shift in a heterogeneous equilibrium system. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), ΔG < 0 means the forward reaction is spontaneous, so the system will shift toward products (decomposition) until equilibrium is reached when ΔG = 0. This is because a negative ΔG implies the current partial pressure of CO₂ makes Q < K, favoring product formation. Choice B correctly identifies this behavior. Choice D is a tempting but incorrect distractor, based on the misconception that negative ΔG means the reaction goes to completion without equilibrium, ignoring that equilibria exist even for favored reactions. To approach similar questions, calculate or infer if Q < K from ΔG's sign to predict the net direction.