Coupled Reactions
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AP Chemistry › Coupled Reactions
A lab group studies two linked reactions that share an intermediate in a closed system. Reaction 1 (spontaneous): $X \rightarrow Y$, $\Delta G = -25\ \text{kJ/mol}$. Reaction 2 (nonspontaneous): $Y \rightarrow Z$, $\Delta G = +30\ \text{kJ/mol}$. The reactions are coupled through intermediate $Y$ so they proceed together as $X \rightarrow Z$. Which statement best describes the thermodynamic favorability of the coupled process?
The coupled process is favorable because the spontaneous step occurs first and then makes the second step spontaneous.
The coupled process is favorable because coupling lowers the activation energy of the nonspontaneous step.
The coupled process is not favorable because the net $\Delta G$ is positive when the two $\Delta G$ values are added.
The coupled process is favorable because each reaction keeps its sign but the overall sign is determined by the larger magnitude.
The coupled process is not favorable because coupling requires both individual reactions to have negative $\Delta G$ values.
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.
A student designs a coupled process in which Reaction A (spontaneous) has $\Delta G = -10\ \text{kJ/mol}$ and Reaction B (nonspontaneous) has $\Delta G = +3\ \text{kJ/mol}$. The reactions are coupled through a shared intermediate so they occur together. Which conclusion is most consistent with thermodynamics?
The coupled process is favorable because coupling increases the magnitude of the negative $\Delta G$ step.
The coupled process is not favorable because the nonspontaneous reaction prevents any net progress.
The coupled process is favorable only if Reaction B is run in a separate container.
The coupled process is favorable because the net $\Delta G$ is negative.
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.
Two reactions are coupled in a pathway by a shared intermediate. Reaction 1: $\Delta G = -41\ \text{kJ/mol}$ (spontaneous). Reaction 2: $\Delta G = +6\ \text{kJ/mol}$ (nonspontaneous). Which statement correctly describes the feasibility of Reaction 2 when coupled to Reaction 1?
Reaction 2 cannot proceed unless Reaction 1 is slowed so the two rates match.
Reaction 2 can proceed only if a catalyst makes its $\Delta G$ negative.
Reaction 2 can proceed in the coupled process because the net $\Delta G$ is negative.
Reaction 2 cannot proceed because coupling cannot change thermodynamic favorability.
Reaction 2 can proceed only if its own $\Delta G$ becomes negative due to coupling.
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.
A student considers coupling two reactions that share an intermediate. Reaction P is spontaneous with $\Delta G = -10\ \text{kJ/mol}$. Reaction Q is nonspontaneous with $\Delta G = +10\ \text{kJ/mol}$. If they are perfectly coupled to occur together, which statement best describes the overall thermodynamic driving force?
The overall coupled process is thermodynamically favorable because Reaction P is spontaneous.
The overall coupled process is favorable only if a catalyst makes Reaction Q faster.
The overall coupled process is not favorable because coupling makes $\Delta G$ for Reaction P less negative.
The overall coupled process is thermodynamically favorable because the intermediate is shared.
The overall coupled process is thermodynamically neutral because the net $\Delta G$ is approximately zero.
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.
In an engineered system, an exergonic reaction with $\Delta G = -40\ \text{kJ/mol}$ is coupled to an endergonic reaction with $\Delta G = +35\ \text{kJ/mol}$ by directly transferring an intermediate from one reaction to the other. Assuming the coupling makes the reactions occur together, which statement best describes the overall thermodynamics?
The overall coupled process is thermodynamically favorable because the net $\Delta G$ is negative.
The overall coupled process is favorable because coupling makes both reactions have the same $\Delta G$.
The overall coupled process is not favorable because one of the steps has positive $\Delta G$.
The overall coupled process is favorable only if the endergonic reaction is faster than the exergonic reaction.
The overall coupled process is not favorable because coupling increases the $\Delta G$ of the exergonic reaction.
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.
In an engineered process, Reaction 1 is spontaneous with $\Delta G_1 = -40\ \text{kJ/mol}$. Reaction 2 is nonspontaneous with $\Delta G_2 = +15\ \text{kJ/mol}$. The two reactions are coupled by sharing an intermediate so that they occur together in a single step sequence. Which condition is sufficient for Reaction 2 to proceed as part of the coupled process?
Coupling must change $\Delta G_2$ to a negative value even if the net $\Delta G$ is positive.
The coupled process must have a negative net $\Delta G$ when the two reactions are added.
Reaction 1 must be slower than Reaction 2 so that Reaction 2 can proceed.
Reaction 2 must be spontaneous on its own before coupling can occur.
A catalyst must be added so that $\Delta G_2$ becomes negative.
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.
In a biochemical pathway, Reaction 1 is known to be spontaneous with $\Delta G_1 = -18\ \text{kJ/mol}$. Reaction 2 is nonspontaneous with $\Delta G_2 = +11\ \text{kJ/mol}$. The cell couples the reactions by using a shared intermediate so that they occur together as a single overall process. Which statement best describes the thermodynamic favorability of the coupled overall process under these conditions?
The coupled overall process is thermodynamically favorable only if Reaction 2 is catalyzed to lower its activation energy.
The coupled overall process is thermodynamically favorable because the net $\Delta G$ is negative.
The coupled overall process is thermodynamically unfavorable because one of the reactions has a positive $\Delta G$.
The coupled overall process is thermodynamically unfavorable because coupling changes only the rate, not the sign of $\Delta G$.
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.
In a metabolic pathway, two reactions are coupled by sharing intermediate $\mathrm{I}$ and occurring together. Reaction 1 is spontaneous: $\mathrm{S \rightarrow I}$ with $\Delta G = -12\ \mathrm{kJ,mol^{-1}}$. Reaction 2 is nonspontaneous: $\mathrm{I \rightarrow T}$ with $\Delta G = +12\ \mathrm{kJ,mol^{-1}}$. Based on thermodynamics, which statement best describes the overall coupled process?
The coupled process is favorable because the spontaneous step transfers spontaneity to the nonspontaneous step.
The coupled process is not favorable because a catalyst is required to make $\Delta G$ negative.
The coupled process is favorable because coupling makes both steps have $\Delta G = 0$.
The coupled process is favorable because equal and opposite $\Delta G$ values make the pathway strongly spontaneous.
The coupled process is not favorable because the net $\Delta G$ is zero, so there is no thermodynamic drive.
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.
A researcher couples two reactions by sharing intermediate $\mathrm{K}$ so they occur together in one device. Reaction 1 is spontaneous: $\mathrm{V \rightarrow K}$ with $\Delta G = -10\ \mathrm{kJ,mol^{-1}}$. Reaction 2 is nonspontaneous: $\mathrm{K \rightarrow W}$ with $\Delta G = +13\ \mathrm{kJ,mol^{-1}}$. Which statement best describes the overall coupled process?
The coupled process is not favorable because the net $\Delta G$ is positive when the two $\Delta G$ values are added.
The coupled process is not favorable because coupled reactions must each have negative $\Delta G$ individually.
The coupled process is favorable because the spontaneous reaction makes the nonspontaneous reaction spontaneous.
The coupled process is favorable because the overall $\Delta G$ equals the more negative $\Delta G$ value.
The coupled process is favorable only because coupling lowers activation energy, which determines spontaneity.
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.
Two reactions are coupled in a metabolic sequence by sharing intermediate $M$ (Reaction 1 produces $M$, Reaction 2 consumes $M$). Reaction 1 is spontaneous: $L \rightarrow M$, $\Delta G = -40\ \text{kJ/mol}$. Reaction 2 is nonspontaneous: $M \rightarrow N$, $\Delta G = +12\ \text{kJ/mol}$. When coupled as $L \rightarrow N$, which condition allows the nonspontaneous reaction to proceed thermodynamically?
The nonspontaneous reaction will proceed only if both reactions have equal magnitudes of $\Delta G$ so they cancel.
The reactions must be coupled so that the combined $\Delta G$ for $L \rightarrow N$ is negative.
The nonspontaneous reaction will proceed only if coupling makes its own $\Delta G$ change to a negative value.
The nonspontaneous reaction will proceed only if a catalyst is added, since coupling is a type of catalysis.
The nonspontaneous reaction will proceed as long as Reaction 1 is spontaneous, regardless of the net $\Delta G$.
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.