AP Chemistry Quiz: Coupled Reactions

What this quiz covers

This quiz focuses on Coupled Reactions, 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 coupled process is favorable because coupling lowers the activation energy of the nonspontaneous step.
2. The coupled process is not favorable because coupling requires both individual reactions to have negative $\Delta G$ values.
3. The coupled process is not favorable because the net $\Delta G$ is positive when the two $\Delta G$ values are added. (correct answer)
4. The coupled process is favorable because the spontaneous step occurs first and then makes the second step spontaneous.
5. The coupled process is favorable because each reaction keeps its sign but the overall sign is determined by the larger magnitude.

Explanation: This question tests the understanding of coupled reactions in thermodynamics, specifically how the net Gibbs free energy change determines the favorability of the overall process. The overall reaction X → Z has a net ΔG of -25 + 30 = +5 kJ/mol, which is positive, indicating it is nonspontaneous and thermodynamically unfavorable. Even though the reactions are coupled through intermediate Y, the nonspontaneous step requires more energy than the spontaneous step provides, so the overall process cannot proceed favorably. Coupling does not alter the individual ΔG values but relies on their sum to assess overall spontaneity. A tempting distractor is choice A, which is incorrect due to the misconception that the order of reactions determines favorability, overlooking that net ΔG is the key factor regardless of sequence. To evaluate coupled reactions, always sum the individual ΔG values to find the net ΔG and determine if it is negative for spontaneity.

Question 2

1. The coupled process is not favorable because the nonspontaneous reaction prevents any net progress.
2. The coupled process is favorable because the net $\Delta G$ is negative. (correct answer)
3. The coupled process is favorable because coupling increases the magnitude of the negative $\Delta G$ step.
4. The coupled process is favorable only if Reaction B is run in a separate container.
5. The coupled process is favorable because a catalyst is equivalent to coupling for $\Delta G$.

Explanation: This question tests understanding of thermodynamic favorability in coupled reactions. When Reaction A (ΔG = -10 kJ/mol) is coupled to Reaction B (ΔG = +3 kJ/mol), the net ΔG = (-10 kJ/mol) + (+3 kJ/mol) = -7 kJ/mol. Since the overall ΔG is negative, the coupled process is thermodynamically favorable and consistent with thermodynamic principles. The misconception in choice A is that the nonspontaneous reaction prevents any net progress, but this ignores that coupling allows the favorable overall ΔG to drive both reactions forward. A transferable strategy is to focus on the net ΔG for coupled reactions—individual reaction spontaneity doesn't prevent the overall process if the sum is negative.

Question 3

1. Reaction 2 can proceed in the coupled process because the net $\Delta G$ is negative. (correct answer)
2. Reaction 2 can proceed only if its own $\Delta G$ becomes negative due to coupling.
3. Reaction 2 can proceed only if a catalyst makes its $\Delta G$ negative.
4. Reaction 2 cannot proceed unless Reaction 1 is slowed so the two rates match.
5. Reaction 2 cannot proceed because coupling cannot change thermodynamic favorability.

Explanation: This question tests the understanding of coupled reactions, specifically how a nonspontaneous reaction can be driven by a spontaneous one if the overall net change in Gibbs free energy ($\Delta G$) is negative. The correct answer is A because the net $\Delta G$ is $-41 + 6 = -35$ kJ/mol, which is negative, allowing Reaction 2 to proceed in the coupled process. The shared intermediate in the pathway enables the spontaneous Reaction 1 to provide sufficient energy to overcome the positive $\Delta G$ of Reaction 2. The principle relies on the net $\Delta G$ being negative for overall spontaneity. A tempting distractor is B, which states coupling cannot change favorability, reflecting the misconception that thermodynamics of individual reactions are unalterable by coupling, ignoring the net effect. To analyze similar problems, always calculate the net $\Delta G$ by summing the individual $\Delta G$ values of the coupled reactions to determine overall thermodynamic favorability.

Question 4

1. The overall coupled process is thermodynamically favorable because the intermediate is shared.
2. The overall coupled process is not favorable because coupling makes $\Delta G$ for Reaction P less negative.
3. The overall coupled process is thermodynamically neutral because the net $\Delta G$ is approximately zero. (correct answer)
4. The overall coupled process is thermodynamically favorable because Reaction P is spontaneous.
5. The overall coupled process is favorable only if a catalyst makes Reaction Q faster.

Explanation: This question tests the understanding of coupled reactions, specifically how the net Gibbs free energy change (ΔG) determines the thermodynamic favorability of the overall process. The correct answer is C because the net ΔG is -10 + 10 = 0 kJ/mol, making the overall coupled process thermodynamically neutral with no driving force. When perfectly coupled through a shared intermediate, the equal and opposite ΔG values result in equilibrium rather than spontaneous progression. The sum of ΔG values indicates no net energy change. A tempting distractor is B, which suggests favorability because Reaction P is spontaneous, reflecting the misconception that a spontaneous component dominates without evaluating the net ΔG. To analyze similar problems, always calculate the net ΔG by summing the individual ΔG values of the coupled reactions to determine overall thermodynamic favorability.

Question 5

1. The overall coupled process is thermodynamically favorable because the net $\Delta G$ is negative. (correct answer)
2. The overall coupled process is not favorable because one of the steps has positive $\Delta G$.
3. The overall coupled process is favorable only if the endergonic reaction is faster than the exergonic reaction.
4. The overall coupled process is not favorable because coupling increases the $\Delta G$ of the exergonic reaction.
5. The overall coupled process is favorable because coupling makes both reactions have the same $\Delta G$.

Explanation: This question tests understanding of coupled reactions where an exergonic reaction drives an endergonic one. When the reactions are coupled, the net ΔG = (-40 kJ/mol) + (+35 kJ/mol) = -5 kJ/mol. Since the overall ΔG is negative, the coupled process is thermodynamically favorable and will proceed spontaneously. The misconception in choice B is that having one step with positive ΔG prevents the overall process from being favorable, but this ignores the additive nature of ΔG values in coupled reactions. A transferable strategy is to always calculate the net ΔG for coupled reactions—the individual signs don't matter as long as the sum is negative for spontaneity.

Question 6

1. The coupled process must have a negative net $\Delta G$ when the two reactions are added. (correct answer)
2. Reaction 2 must be spontaneous on its own before coupling can occur.
3. Coupling must change $\Delta G_2$ to a negative value even if the net $\Delta G$ is positive.
4. A catalyst must be added so that $\Delta G_2$ becomes negative.
5. Reaction 1 must be slower than Reaction 2 so that Reaction 2 can proceed.

Explanation: This question tests understanding of the thermodynamic requirement for coupled reactions to proceed. For Reaction 2 to proceed as part of the coupled process, the overall ΔG must be negative: ΔG_total = ΔG₁ + ΔG₂ = -40 kJ/mol + 15 kJ/mol = -25 kJ/mol. Since the net ΔG is indeed negative, the coupled process is thermodynamically favorable, allowing the nonspontaneous Reaction 2 to proceed. The large negative ΔG from Reaction 1 provides sufficient energy to drive Reaction 2. Choice D incorrectly suggests that a catalyst can make ΔG₂ negative, but catalysts only affect reaction rates, not thermodynamic favorability—they cannot change the sign of ΔG. The sufficient condition for any nonspontaneous reaction to proceed when coupled is that the net ΔG of the coupled system must be negative.

Question 7

1. The coupled overall process is thermodynamically favorable because the net $\Delta G$ is negative. (correct answer)
2. The coupled overall process is thermodynamically unfavorable because one of the reactions has a positive $\Delta G$.
3. The coupled overall process is thermodynamically unfavorable because coupling changes only the rate, not the sign of $\Delta G$.
4. The coupled overall process is thermodynamically favorable only if Reaction 2 is catalyzed to lower its activation energy.
5. The coupled overall process is thermodynamically favorable because coupling makes each individual $\Delta G$ negative.

Explanation: This question tests understanding of coupled reactions and how to determine the thermodynamic favorability of the overall process. When reactions are coupled, the overall Gibbs free energy change is the sum of the individual ΔG values: ΔG_total = ΔG₁ + ΔG₂ = -18 kJ/mol + 11 kJ/mol = -7 kJ/mol. Since the net ΔG is negative, the coupled overall process is thermodynamically favorable and can proceed spontaneously. Choice C incorrectly suggests that coupling makes each individual ΔG negative, but coupling doesn't change the individual reaction energetics—it only allows the favorable reaction to drive the unfavorable one through their shared intermediate. The key strategy is to calculate the net ΔG by adding the individual values: if negative, the coupled process is favorable.

Question 8

1. The coupled process is favorable because equal and opposite $\Delta G$ values make the pathway strongly spontaneous.
2. The coupled process is favorable because the spontaneous step transfers spontaneity to the nonspontaneous step.
3. The coupled process is not favorable because the net $\Delta G$ is zero, so there is no thermodynamic drive. (correct answer)
4. The coupled process is favorable because coupling makes both steps have $\Delta G = 0$.
5. The coupled process is not favorable because a catalyst is required to make $\Delta G$ negative.

Explanation: This question tests understanding of coupled reactions when ΔG values are equal and opposite. For reactions coupled through intermediate I, the overall Gibbs free energy change is: ΔG_total = ΔG₁ + ΔG₂ = -12 kJ/mol + 12 kJ/mol = 0 kJ/mol. When ΔG_total = 0, the system is at equilibrium with no net thermodynamic drive in either direction—the process is neither favorable nor unfavorable. This means there's no tendency for the overall reaction S → T to proceed spontaneously, though the system can exist with all species present at equilibrium. Choice B incorrectly suggests that spontaneity can be "transferred" between reactions, which misunderstands that only energy (not spontaneity itself) is coupled. For coupled reactions, always calculate the net ΔG; when it equals zero, the system is at equilibrium with no directional preference.

Question 9

1. The coupled process is favorable because the spontaneous reaction makes the nonspontaneous reaction spontaneous.
2. The coupled process is not favorable because the net $\Delta G$ is positive when the two $\Delta G$ values are added. (correct answer)
3. The coupled process is favorable because the overall $\Delta G$ equals the more negative $\Delta G$ value.
4. The coupled process is favorable only because coupling lowers activation energy, which determines spontaneity.
5. The coupled process is not favorable because coupled reactions must each have negative $\Delta G$ individually.

Explanation: This question tests recognition of unfavorable coupling despite having a spontaneous step. For reactions coupled through intermediate K, the overall Gibbs free energy change is: ΔG_total = ΔG₁ + ΔG₂ = -10 kJ/mol + 13 kJ/mol = +3 kJ/mol. Since the net ΔG is positive, the coupled process is not thermodynamically favorable and will not proceed spontaneously to form W. The energy released from Reaction 1 is insufficient to overcome the energy requirement of Reaction 2. Choice A incorrectly assumes that any spontaneous reaction can make a nonspontaneous reaction spontaneous, which ignores the quantitative energy balance required for coupling. To evaluate coupled reactions, sum the ΔG values—the process is unfavorable when ΔG_total > 0, even with a spontaneous component.

Question 10

1. The nonspontaneous reaction will proceed as long as Reaction 1 is spontaneous, regardless of the net $\Delta G$.
2. The nonspontaneous reaction will proceed only if a catalyst is added, since coupling is a type of catalysis.
3. The nonspontaneous reaction will proceed only if coupling makes its own $\Delta G$ change to a negative value.
4. The nonspontaneous reaction will proceed only if both reactions have equal magnitudes of $\Delta G$ so they cancel.
5. The reactions must be coupled so that the combined $\Delta G$ for $L \rightarrow N$ is negative. (correct answer)

Explanation: This question tests the understanding of coupled reactions in thermodynamics, specifically how the net Gibbs free energy change determines the favorability of the overall process. The nonspontaneous Reaction 2 can proceed when coupled because the net ΔG for L → N is -40 + 12 = -28 kJ/mol, which is negative, making the overall process spontaneous. Coupling through intermediate M allows the large energy release from Reaction 1 to drive Reaction 2, as the thermodynamic favorability is based on the combined ΔG. Without coupling to ensure the net is negative, the nonspontaneous step would not occur. A tempting distractor is choice B, which is incorrect due to the misconception that a spontaneous reaction alone guarantees progress regardless of net ΔG, ignoring that the overall sum must be negative. To evaluate coupled reactions, always sum the individual ΔG values to find the net ΔG and determine if it is negative for spontaneity.

Question 11

1. The coupled process is favorable because coupling makes Reaction 2 spontaneous by changing its $\Delta G$ to a negative value.
2. The coupled process is not favorable because the net $\Delta G$ is positive when the two $\Delta G$ values are added. (correct answer)
3. The coupled process is favorable because any spontaneous reaction can drive any nonspontaneous reaction regardless of magnitudes.
4. The coupled process is favorable because coupling decreases the activation energy, which makes $\Delta G$ more negative.
5. The coupled process is not favorable because coupling requires the two $\Delta G$ values to have the same sign.

Explanation: This question tests the understanding of coupled reactions in thermodynamics, specifically how the net Gibbs free energy change determines the favorability of the overall process. The coupled process D → F has a net ΔG of -2 + 9 = +7 kJ/mol, which is positive, indicating it is not thermodynamically favorable. Even with coupling via intermediate E, the nonspontaneous Reaction 2 demands more energy than Reaction 1 supplies, preventing overall spontaneity. The key principle is that the sum of ΔG values must be negative for favorability. A tempting distractor is choice C, which is incorrect due to the misconception that a spontaneous reaction can drive any nonspontaneous one without regard to ΔG magnitudes, when net positivity blocks progress. To evaluate coupled reactions, always sum the individual ΔG values to find the net ΔG and determine if it is negative for spontaneity.

Question 12

1. The coupled process is not favorable because the net $\Delta G$ is positive when the two $\Delta G$ values are added. (correct answer)
2. The coupled process is favorable because coupling allows the negative $\Delta G$ to be transferred to the second step.
3. The coupled process is favorable because the overall reaction is spontaneous if at least one step is spontaneous.
4. The coupled process is not favorable because each individual step must have $\Delta G < 0$ for the overall reaction to proceed.
5. The coupled process is favorable because coupling acts like a catalyst and changes thermodynamic favorability.

Explanation: This question tests the understanding of coupled reactions in thermodynamics, specifically how the net Gibbs free energy change determines the favorability of the overall process. The coupled process S → T has a net ΔG of -14 + 20 = +6 kJ/mol, which is positive, making it thermodynamically unfavorable. Despite coupling through intermediate I, the energy required by the nonspontaneous step exceeds that released by the spontaneous step, preventing overall progress. Favorability depends on the net ΔG sum, not merely the presence of a spontaneous reaction. A tempting distractor is choice D, which is incorrect due to the misconception that one spontaneous step suffices for the whole process, disregarding the need for a negative net ΔG. To evaluate coupled reactions, always sum the individual ΔG values to find the net ΔG and determine if it is negative for spontaneity.

Question 13

1. The coupled process is favorable because the spontaneous step ensures the overall process proceeds in the forward direction.
2. The coupled process is not favorable because the net $\Delta G$ is zero, so there is no thermodynamic driving force. (correct answer)
3. The coupled process is favorable because coupling changes the sign of the nonspontaneous step without changing the net $\Delta G$.
4. The coupled process is not favorable because coupling only affects activation energy and cannot change $\Delta G$.
5. The coupled process is favorable because the net $\Delta G$ is determined by the step with the larger magnitude.

Explanation: This question tests the understanding of coupled reactions in thermodynamics, specifically how the net Gibbs free energy change determines the favorability of the overall process. The coupled process H → K has a net ΔG of -7 + 7 = 0 kJ/mol, meaning it is at equilibrium with no driving force for net progress in either direction. Coupling through intermediate J does not provide a thermodynamic bias since the energy release and requirement cancel exactly. Thus, the process is not favorable for proceeding to products. A tempting distractor is choice A, which is incorrect due to the misconception that a spontaneous step alone ensures overall favorability, ignoring that a zero net ΔG results in equilibrium. To evaluate coupled reactions, always sum the individual ΔG values to find the net ΔG and determine if it is negative for spontaneity.

Question 14

1. The coupled process is favorable because the reaction with the smaller magnitude of $\Delta G$ determines spontaneity.
2. The coupled process is favorable because the net $\Delta G$ is negative when the two $\Delta G$ values are added. (correct answer)
3. The coupled process is not favorable because a positive $\Delta G$ step cannot occur under any circumstances.
4. The coupled process is favorable because coupling changes Reaction 2 so its $\Delta G$ becomes negative even when considered alone.
5. The coupled process is not favorable because coupling affects kinetics rather than thermodynamics.

Explanation: This question tests the understanding of coupled reactions in thermodynamics, specifically how the net Gibbs free energy change determines the favorability of the overall process. The coupled reaction P → R has a net ΔG of -9 + 3 = -6 kJ/mol, which is negative, indicating thermodynamic favorability. Coupling via intermediate Q enables the spontaneous Reaction 1 to provide enough free energy to overcome the positive ΔG of Reaction 2. The overall spontaneity is determined by the sum of the ΔG values, not by individual steps in isolation. A tempting distractor is choice C, which is incorrect due to the misconception that coupling alters the intrinsic ΔG of a single reaction, when in fact it only affects the overall net through summation. To evaluate coupled reactions, always sum the individual ΔG values to find the net ΔG and determine if it is negative for spontaneity.

Question 15

1. The coupled process is favorable because coupling changes only Reaction 2 so that its $\Delta G$ becomes negative on its own.
2. The coupled process is not favorable because coupling is equivalent to adding a catalyst, which does not change $\Delta G$.
3. The coupled process is favorable because the net $\Delta G$ is negative when the two $\Delta G$ values are added. (correct answer)
4. The coupled process is not favorable because the nonspontaneous step cannot occur even if the net $\Delta G$ is negative.
5. The coupled process is favorable because the step with the smaller magnitude of $\Delta G$ controls spontaneity.

Explanation: This question tests the understanding of coupled reactions in thermodynamics, specifically how the net Gibbs free energy change determines the favorability of the overall process. The coupled process M → P has a net ΔG of -50 + 45 = -5 kJ/mol, which is negative, making it thermodynamically favorable. Coupling through intermediate O in the cellular compartment allows the highly spontaneous Reaction 1 to drive Reaction 2 by providing excess free energy. The overall outcome is spontaneous because the sum of ΔG values indicates a net release of energy. A tempting distractor is choice B, which is incorrect due to the misconception that a nonspontaneous step inherently prevents progress even with negative net ΔG, overlooking the compensatory effect of coupling. To evaluate coupled reactions, always sum the individual ΔG values to find the net ΔG and determine if it is negative for spontaneity.

Question 16

1. The coupled process is thermodynamically favorable because one step is spontaneous.
2. The coupled process is thermodynamically favorable because the net $\Delta G$ is negative.
3. The coupled process is thermodynamically neutral because the net $\Delta G$ is approximately zero. (correct answer)
4. The coupled process is not favorable because coupling requires a catalyst to change $\Delta G$.
5. The coupled process is favorable because coupling transfers spontaneity to Reaction N.

Explanation: This question tests understanding of coupled reactions with equal and opposite ΔG values. When Reaction M (ΔG = -16 kJ/mol) is coupled to Reaction N (ΔG = +16 kJ/mol) in a 1:1 ratio, the net ΔG = (-16 kJ/mol) + (+16 kJ/mol) = 0 kJ/mol. A process with ΔG = 0 is at equilibrium and is neither favorable nor unfavorable—it's thermodynamically neutral. The misconception in choice B is that having a negative net ΔG when the actual sum is zero, failing to recognize that equal and opposite ΔG values cancel out. A transferable strategy is to carefully calculate the net ΔG and recognize that ΔG = 0 indicates equilibrium, not spontaneity.

Question 17

1. The synthesis proceeds because the total $\Delta G$ of the coupled process is negative. (correct answer)
2. The synthesis proceeds because coupling makes the synthesis reaction's own $\Delta G$ negative.
3. The synthesis proceeds because coupling acts like a catalyst and changes thermodynamic favorability.
4. The synthesis proceeds only if the spontaneous reaction reaches equilibrium first.
5. The synthesis cannot proceed because any positive $\Delta G$ step prevents the overall process.

Explanation: This question tests understanding of how coupling enables nonspontaneous reactions to proceed. The synthesis reaction (ΔG = +22 kJ/mol) can proceed when coupled to the spontaneous reaction (ΔG = -50 kJ/mol) because the net ΔG = (-50 kJ/mol) + (+22 kJ/mol) = -28 kJ/mol is negative. The overall coupled process is thermodynamically favorable, which drives the nonspontaneous synthesis forward. The misconception in choice B is that coupling changes the individual ΔG of the synthesis reaction to make it negative, but coupling doesn't alter individual reaction thermodynamics—it only matters that the sum is negative. A transferable strategy is to recognize that coupling works by making the overall process favorable, not by changing individual reaction ΔG values.

Question 18

1. The combined $\Delta G$ for the coupled reactions must be negative. (correct answer)
2. Reaction Y must become spontaneous on its own, so its individual $\Delta G$ must become negative.
3. A catalyst must be present so that Reaction Y becomes thermodynamically favorable.
4. Reaction X must stop at equilibrium so that Reaction Y can proceed past equilibrium.
5. The magnitude of $\Delta G$ values is irrelevant because coupling guarantees completion.

Explanation: This question tests understanding of the thermodynamic requirement for coupled reactions to proceed. For Reaction Y (nonspontaneous, ΔG = +9 kJ/mol) to proceed when coupled to Reaction X (spontaneous, ΔG = -12 kJ/mol), the combined ΔG must be negative: (-12 kJ/mol) + (+9 kJ/mol) = -3 kJ/mol. Since the net ΔG is indeed negative, the coupled process is thermodynamically favorable. The misconception in choice B is that Reaction Y must become spontaneous on its own with negative ΔG, but coupling doesn't change individual reaction ΔG values—it only matters that the sum is negative. A transferable strategy is to recognize that coupling allows nonspontaneous reactions to proceed without changing their individual thermodynamics, as long as the net ΔG < 0.

Question 19

1. The coupled process is favorable only if the activation energy is lowered, since $\Delta G$ does not matter.
2. The coupled process is favorable because the more exothermic reaction always dominates regardless of $\Delta G$.
3. The coupled process is thermodynamically favorable because the spontaneous reaction forces the other to occur.
4. The coupled process is not thermodynamically favorable because the net $\Delta G$ is positive. (correct answer)
5. The coupled process is favorable because coupling changes the sign of Reaction B's $\Delta G$.

Explanation: This question tests understanding of coupled reactions and the calculation of net Gibbs free energy change. When two reactions are coupled in a 1:1 ratio, the overall ΔG equals the sum of individual ΔG values: (-18 kJ/mol) + (+25 kJ/mol) = +7 kJ/mol. Since the net ΔG is positive, the overall coupled process is not thermodynamically favorable and will not proceed spontaneously as written. The misconception in choice A is that the spontaneous reaction can somehow force the nonspontaneous one to occur regardless of the net ΔG, but thermodynamic favorability always depends on the overall free energy change being negative. A transferable strategy is to remember that for coupled reactions to be favorable, the magnitude of the negative ΔG must exceed the magnitude of the positive ΔG.

Question 20

1. The coupled process is not favorable because $\Delta G$ values cannot be added for coupled reactions.
2. The coupled process is thermodynamically favorable because the net $\Delta G$ is negative. (correct answer)
3. The coupled process is not favorable because Reaction 2 has a positive $\Delta G$.
4. The coupled process is favorable only if the activation energy for Reaction 2 is lowered.
5. The coupled process is favorable because coupling makes both reactions have $\Delta G<0$ individually.

Explanation: This question tests understanding of coupled reactions where an exergonic reaction drives an endergonic one. When Reaction 1 (ΔG = -6 kJ/mol) is coupled to Reaction 2 (ΔG = +4 kJ/mol), the net ΔG = (-6 kJ/mol) + (+4 kJ/mol) = -2 kJ/mol. Since the overall ΔG is negative, the coupled process is thermodynamically favorable and will proceed spontaneously. The misconception in choice B is that having any reaction with positive ΔG prevents favorability, ignoring that the net ΔG determines spontaneity in coupled systems. A transferable strategy is to always sum the individual ΔG values for coupled reactions—if negative, the process is favorable regardless of individual reaction spontaneity.