AP Chemistry Quiz: Free Energy And Equilibrium

What this quiz covers

This quiz focuses on Free Energy And Equilibrium, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The system is at equilibrium because $\Delta G<0$ indicates the minimum free energy state is reached.
2. The system cannot reach equilibrium because $\Delta G<0$ means the reverse reaction cannot occur.
3. The system must contain only $\mathrm{H_2F_2}$ because negative $\Delta G$ implies completion.
4. The system will shift toward products until $\Delta G \approx 0$. (correct answer)
5. The system will shift toward reactants until $\Delta G \approx 0$.

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.

Question 2

1. Only products are present because $\Delta G \approx 0$ indicates completion.
2. The reverse reaction has stopped completely because $\Delta G \approx 0$ means products are maximum.
3. Only reactants are present because $\Delta G \approx 0$ indicates no reaction can proceed.
4. Both the forward and reverse reactions are occurring, but there is no net change in composition. (correct answer)
5. The forward reaction has stopped completely because $\Delta G \approx 0$ means no collisions occur.

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.

Question 3

1. The system will shift toward reactants until $\Delta G$ approaches zero. (correct answer)
2. The system will shift toward products until $\Delta G$ becomes more positive.
3. The system is already at equilibrium because $\Delta G$ can be positive at equilibrium.
4. The system will produce only products because combustion reactions always go to completion.
5. No net change will occur because a positive $\Delta G$ means the reaction is too slow.

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.

Question 4

1. There is no net driving force for dissolution or precipitation under the current conditions. (correct answer)
2. Dissolution is spontaneous and will proceed until all liquid bromine is gone.
3. Precipitation is spontaneous and will proceed until all aqueous bromine is gone.
4. The process is impossible because $\Delta G \approx 0$ means the reaction cannot occur.
5. The equilibrium mixture must contain equal amounts of $\mathrm{Br_2(l)}$ and $\mathrm{Br_2(aq)}$.

Explanation: This question examines interpreting ΔG ≈ 0 in phase or solubility equilibria. For Br₂(l) ⇌ Br₂(aq), ΔG ≈ 0 at equilibrium means no net driving force for dissolution or precipitation, as the system is balanced. This indicates saturation where rates of dissolving and precipitating are equal. Choice A best describes this. Choice E tempts but is incorrect, assuming equal amounts, from the misconception that equilibrium requires equal quantities rather than equal rates. For solubility problems, use ΔG = 0 to identify saturation points without assuming equal concentrations.

Question 5

1. Because $\Delta G < 0$, the system is at equilibrium and will not change composition.
2. Because $\Delta G < 0$, the system will shift toward $\mathrm{N_2O_4}$ until $\Delta G = 0$. (correct answer)
3. Because $\Delta G < 0$, the equilibrium constant must be less than 1 at this temperature.
4. Because $\Delta G < 0$, the reverse reaction is spontaneous until $\Delta G$ becomes positive.
5. Because $\Delta G < 0$, the reaction must go to completion with only $\mathrm{N_2O_4}$ present.

Explanation: This question tests understanding of spontaneous direction when ΔG < 0. When ΔG is negative for the reaction as written (2NO₂ → N₂O₄), the forward reaction is spontaneous, meaning the system will shift toward products (N₂O₄) until equilibrium is reached. The negative ΔG indicates the system can lower its free energy by forming more N₂O₄, and this process continues until ΔG = 0 at equilibrium. The system is not currently at equilibrium because ΔG ≠ 0. The misconception in choice A is that negative ΔG indicates equilibrium, when it actually indicates the forward reaction is favored. When ΔG < 0, the system spontaneously proceeds in the forward direction until reaching equilibrium.

Question 6

1. The system is at equilibrium, so the forward and reverse reaction rates are equal. (correct answer)
2. The system is product-favored, so the reaction will proceed forward until completion.
3. The system is reactant-favored, so the reverse reaction must be faster than the forward reaction.
4. The system is not at equilibrium, because $\Delta G$ must be negative at equilibrium.
5. The system is at equilibrium only if the enthalpy change $\Delta H$ is also approximately zero.

Explanation: This question tests understanding of the relationship between Gibbs free energy and equilibrium. When ΔG ≈ 0 for a reaction mixture, the system is at equilibrium, meaning the forward and reverse reaction rates are equal and there is no net change in concentrations. At equilibrium, the system has reached its minimum free energy state, and neither the forward nor reverse reaction is thermodynamically favored. The misconception in choice D is that ΔG must be negative at equilibrium, when in fact ΔG = 0 defines the equilibrium condition. To determine if a system is at equilibrium, check if ΔG = 0 for the current mixture composition.

Question 7

1. $\Delta G \approx 0$ means the mixture is at equilibrium for its current composition. (correct answer)
2. $\Delta G \approx 0$ means the equilibrium constant $K$ is approximately 0.
3. $\Delta G \approx 0$ means products and reactants must be present in equal amounts.
4. $\Delta G \approx 0$ means the reaction is strongly product-favored and will proceed forward rapidly.
5. $\Delta G \approx 0$ means the reaction cannot proceed in either direction under any conditions.

Explanation: This question tests understanding of the meaning of ΔG ≈ 0 for a reaction mixture. When ΔG ≈ 0, the system is at equilibrium for its current composition, meaning there is no driving force for net change in either direction. At this point, the forward and reverse reactions occur at equal rates, and the reaction quotient Q equals the equilibrium constant K. The value of ΔG depends on the current mixture composition, not just the balanced equation. The misconception in choice B is confusing ΔG = 0 with K = 0, when actually ΔG = 0 occurs when Q = K regardless of K's value. To identify equilibrium, check if ΔG = 0 for the specific mixture composition.

Question 8

1. Correct, because negative $\Delta G$ means the forward and reverse rates are equal.
2. Incorrect, because $\Delta G<0$ indicates a net tendency to form products until $\Delta G \approx 0$. (correct answer)
3. Incorrect, because $\Delta G<0$ indicates the reverse reaction is spontaneous until $\Delta G \approx 0$.
4. Incorrect, because $\Delta G<0$ means the reaction cannot be reversible.
5. Correct, because equilibrium occurs whenever $\Delta G$ is negative.

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.

Question 9

1. $\Delta G \approx 0$ means the reaction cannot proceed in either direction.
2. $\Delta G \approx 0$ means the forward reaction is spontaneous and the reverse is not.
3. $\Delta G \approx 0$ means products are present in a greater amount than reactants.
4. $\Delta G \approx 0$ means the equilibrium constant $K$ must equal 1 for this reaction.
5. $\Delta G \approx 0$ means the forward and reverse reaction rates are equal. (correct answer)

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.

Question 10

1. The system is at equilibrium, so there is no net change in the amounts of $\mathrm{A}$ and $\mathrm{B}$. (correct answer)
2. The system is product-favored, so $\mathrm{B}$ must be present at a higher concentration than $\mathrm{A}$.
3. The system is reactant-favored, so $\mathrm{A}$ must be present at a higher concentration than $\mathrm{B}$.
4. The forward reaction is spontaneous, so $\mathrm{A}$ will be completely converted to $\mathrm{B}$.
5. The reverse reaction is spontaneous, so $\mathrm{B}$ will be completely converted to $\mathrm{A}$.

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.

Question 11

1. Net formation of ions will occur because $\Delta G>0$ means the forward reaction is favored.
2. Net formation of undissociated acid will occur because the reverse direction is spontaneous. (correct answer)
3. No net change will occur because $\Delta G>0$ indicates equilibrium.
4. The reaction will go to completion toward reactants so that no ions remain.
5. The reaction will stop permanently because positive $\Delta G$ prevents any molecular collisions.

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.

Question 12

1. The system is at equilibrium, so the forward and reverse reaction rates are equal. (correct answer)
2. The system is product-favored, so essentially all reactants have been converted to products.
3. The forward reaction is spontaneous, so the reaction will proceed to completion.
4. The reverse reaction is spontaneous, so the reaction mixture must contain only reactants.
5. The reaction is nonspontaneous in both directions, so no reaction is occurring.

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.

Question 13

1. The reverse reaction is spontaneous, so the system will shift toward reactants until equilibrium is reached.
2. The forward reaction is spontaneous, so the system will shift toward products until equilibrium is reached. (correct answer)
3. The system is at equilibrium because $\Delta G<0$ indicates no net change.
4. The system must already be product-only because $\Delta G<0$ means products are maximum.
5. The system cannot reach equilibrium because a negative $\Delta G$ prevents reversibility.

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.

Question 14

1. The system is at equilibrium because solids make $\Delta G$ always negative.
2. The system will shift toward products until equilibrium is reached. (correct answer)
3. The system will shift toward reactants until equilibrium is reached.
4. The system must already contain only $\mathrm{CO_2(g)}$ because $\Delta G<0$ implies completion.
5. The system cannot be reversible because $\Delta G<0$ prevents the reverse reaction.

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.

Question 15

1. $\Delta G$ will decrease toward zero as the system shifts toward reactants. (correct answer)
2. $\Delta G$ will increase away from zero as the system shifts toward reactants.
3. $\Delta G$ will become more positive as the system shifts toward products.
4. $\Delta G$ will remain positive at equilibrium because equilibrium favors reactants.
5. $\Delta G$ will become negative at equilibrium because equilibrium requires spontaneity.

Explanation: This question tests predicting changes in ΔG as a system approaches equilibrium from a positive value. For 2A(g) ⇌ B(g), ΔG > 0 for forward means reverse is spontaneous, shifting toward reactants, which decreases ΔG (makes it less positive) toward zero. As Q decreases toward K, the positive ΔG reduces in magnitude per ΔG = ΔG° + RT ln Q. Choice A best describes this. Choice C is a tempting distractor but incorrect, suggesting shift toward products, from the misconception that positive ΔG drives forward. Monitor how Q's change alters ΔG's value to understand equilibrium dynamics.

Question 16

1. The system is at equilibrium because $\Delta G<0$ means the reaction quotient equals the equilibrium constant.
2. The system will shift toward products until $\Delta G$ becomes approximately zero. (correct answer)
3. The system will shift toward reactants until $\Delta G$ becomes more negative.
4. The system must form only complex ion because negative $\Delta G$ implies complete conversion.
5. The system cannot reach equilibrium because negative $\Delta G$ eliminates the reverse process.

Explanation: This question tests linking ΔG's sign to the direction of spontaneous change in complex ion formation. For Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq), ΔG < 0 means the forward reaction is spontaneous, so the system shifts toward the complex ion (products) until ΔG ≈ 0 at equilibrium. This follows from ΔG < 0 implying Q < K, driving product formation. Choice B correctly connects this to equilibrium approach. Choice C is a tempting but wrong distractor, confusing negative ΔG with reverse spontaneity, due to the misconception of sign interpretation. Always verify the written reaction's direction when using ΔG to predict shifts in coordination chemistry equilibria.

Question 17

1. The system will shift toward gases because $\Delta G>0$ indicates spontaneous decomposition.
2. The system will shift toward liquid water because the reverse direction is spontaneous. (correct answer)
3. The system is at equilibrium because $\Delta G>0$ indicates equal forward and reverse rates.
4. The system cannot establish equilibrium because $\Delta G>0$ prevents reversibility.
5. The system will contain only water at equilibrium because equilibrium means complete reactant formation.

Explanation: This question assesses using ΔG > 0 to determine spontaneity in decomposition reactions. For 2H₂O(l) ⇌ 2H₂(g) + O₂(g), ΔG > 0 means the forward reaction is nonspontaneous, so the reverse is favored, shifting the system toward liquid water. This indicates the current gas concentrations make Q > K, driving condensation. Choice B best describes this tendency. Choice A is a tempting distractor but wrong, as it suggests shifting toward gases, from the misconception that positive ΔG favors the forward process. For electrolysis or decomposition contexts, use ΔG's sign to identify the nonspontaneous direction and predict stability.

Question 18

1. The system is at equilibrium, so there is no net change in amounts of reactants and products. (correct answer)
2. The system is at equilibrium only if equal moles of $\mathrm{H_2}$ and $\mathrm{O_2}$ are present.
3. The system must contain only gases at equilibrium because $\Delta G \approx 0$.
4. The system will spontaneously produce more gases because $\Delta G \approx 0$ implies $\Delta G < 0$.
5. The system is not at equilibrium because equilibrium requires $\Delta G$ to be at a maximum.

Explanation: This question tests understanding of equilibrium in phase change reactions. When ΔG ≈ 0 for the water electrolysis reaction, the system is at equilibrium with no net change in amounts of reactants and products. At equilibrium, water molecules continue to decompose into hydrogen and oxygen gases while these gases simultaneously recombine to form water, but these opposing processes occur at equal rates. The equilibrium state can include both liquid water and gaseous products in various proportions. The misconception in choice D is interpreting ΔG ≈ 0 as slightly negative, when it actually means the system is at equilibrium. For any reaction at equilibrium, ΔG = 0 indicates balanced forward and reverse rates.

Question 19

1. It will shift toward reactants until $\Delta G$ becomes more negative.
2. It will shift toward products until $\Delta G$ reaches 0 at equilibrium. (correct answer)
3. It is already at equilibrium because $\Delta G$ is less than 0.
4. It cannot shift because equilibrium requires $\Delta G$ to stay constant.
5. It will shift toward products until all reactants are consumed because $\Delta G < 0$.

Explanation: This question tests understanding of how the sign of ΔG predicts the direction of spontaneous change. When ΔG < 0 for a reaction mixture, the forward reaction is spontaneous, meaning the system will shift toward products to reach equilibrium. The negative ΔG indicates that forming products will lower the system's free energy until ΔG = 0 at equilibrium. The system is not yet at equilibrium because ΔG ≠ 0, and it will continue shifting toward products until equilibrium is established. The misconception in choice E is that a negative ΔG means the reaction goes to completion, when actually it only proceeds until ΔG = 0. To predict reaction direction, remember that systems spontaneously move toward lower free energy (ΔG < 0 forward, ΔG > 0 reverse).

Question 20

1. It will shift toward $\mathrm{CaCO_3(s)}$ until $\Delta G$ becomes 0 at equilibrium. (correct answer)
2. It will shift toward $\mathrm{CaO(s)}$ and $\mathrm{CO_2(g)}$ until $\Delta G$ becomes more positive.
3. It is at equilibrium because solids do not affect free energy, so $\Delta G$ must be 0.
4. It will shift toward products because $\Delta G > 0$ indicates product-favored conditions.
5. No shift is possible because equilibrium requires both solids to be completely consumed.

Explanation: This question tests understanding of heterogeneous equilibria and spontaneous direction. When ΔG > 0 for the decomposition of CaCO₃ as written, the forward reaction is nonspontaneous, meaning the reverse reaction (formation of CaCO₃) is spontaneous. The system will shift toward reactants, consuming CO₂ gas and forming more CaCO₃ solid until ΔG = 0 at equilibrium. The presence of solids doesn't eliminate their participation in equilibrium; rather, their activities are constant (equal to 1). The misconception in choice D is that positive ΔG indicates product-favored conditions, when it actually means reactant-favored. For heterogeneous equilibria, ΔG > 0 still means the reverse reaction is spontaneous.