Coupled Reactions - AP Chemistry

In an engineered system, Reaction 1 releases energy with $\Delta G = -60\ \text{kJ/mol}$ (spontaneous). Reaction 2 requires energy with $\Delta G = +55\ \text{kJ/mol}$ (nonspontaneous). The two are coupled so that Reaction 1 directly drives Reaction 2 through a shared intermediate. Which statement is correct about the coupled overall process?
Answer: The coupled overall process is thermodynamically favorable because the net $\Delta G$ is negative. 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 (ΔG) is negative. The correct answer is A because the net ΔG is -60 + 55 = -5 kJ/mol, which is negative, indicating the coupled overall process is thermodynamically favorable. In this engineered system, the shared intermediate allows Reaction 1 to directly drive Reaction 2, with the net exergonic change enabling spontaneity. The principle is that the sum of ΔG values determines the overall thermodynamics. A tempting distractor is B, which states the process is not favorable because Reaction 2 remains nonspontaneous, reflecting the misconception that individual spontaneity is required rather than 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.
A chemist couples two reactions by physically linking them so they share an intermediate on a surface. Reaction P is spontaneous with $\Delta G = -27\ \text{kJ/mol}$ . Reaction Q is nonspontaneous with $\Delta G = +31\ \text{kJ/mol}$ . Which statement correctly describes the feasibility of running Reaction Q forward via coupling to Reaction P in a 1:1 ratio?
Answer: Reaction Q cannot proceed because the net $\Delta G$ of the coupled process is positive. This question tests understanding of the thermodynamic constraints on coupled reactions. When Reaction P (ΔG = -27 kJ/mol) is coupled to Reaction Q (ΔG = +31 kJ/mol) in a 1:1 ratio, the net ΔG = (-27 kJ/mol) + (+31 kJ/mol) = +4 kJ/mol. Since the overall ΔG is positive, the coupled process is not thermodynamically favorable, and Reaction Q cannot proceed forward through this coupling. The misconception in choice E is that the more negative ΔG reaction always determines direction, but the net ΔG of the coupled system is what matters. A transferable strategy is to remember that successful coupling requires the magnitude of the negative ΔG to exceed the magnitude of the positive ΔG.
Two reactions are linked in a process so that the product of the first is an intermediate consumed immediately by the second (coupling). Reaction M has $\Delta G = -16\ \text{kJ/mol}$ and Reaction N has $\Delta G = +16\ \text{kJ/mol}$ . If they occur together in a 1:1 ratio, which statement best describes the overall coupled process?
Answer: The coupled process is thermodynamically neutral because the net $\Delta G$ is approximately zero. 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.
A reaction sequence is designed so that an exergonic reaction ( $\Delta G = -16\ \text{kJ/mol}$ ) produces an intermediate that is immediately consumed by an endergonic reaction ( $\Delta G = +18\ \text{kJ/mol}$ ). The steps are coupled and proceed together. Which statement best describes the overall thermodynamic outcome?
Answer: The overall coupled process is not thermodynamically favorable because the net $\Delta G$ is positive. 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 B because the net ΔG is -16 + 18 = +2 kJ/mol, which is positive, making the overall coupled process not thermodynamically favorable. The intermediate is produced and consumed, but the endergonic reaction results in a net nonspontaneous sequence. Coupling does not inherently change ΔG signs but sums them. A tempting distractor is C, which claims the exergonic step forces the endergonic one, reflecting the misconception that coupling always drives the process forward without verifying 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.
Two reactions are coupled in a metabolic sequence by a shared enzyme-bound intermediate. Reaction X has $\Delta G = -12\ \text{kJ/mol}$ (spontaneous). Reaction Y has $\Delta G = +9\ \text{kJ/mol}$ (nonspontaneous). Which condition must be true for Reaction Y to proceed as written when coupled to Reaction X?
Answer: The combined $\Delta G$ for the coupled reactions must be negative. 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.
Two reactions are coupled in solution by sharing intermediate $\mathrm{N}$ so that they proceed together. Reaction 1 is spontaneous: $\mathrm{F \rightarrow N}$ with $\Delta G = -7\ \mathrm{kJ\,mol^{-1}}$ . Reaction 2 is nonspontaneous: $\mathrm{N \rightarrow P}$ with $\Delta G = +2\ \mathrm{kJ\,mol^{-1}}$ . Which statement correctly describes whether the overall coupled process is thermodynamically favorable?
Answer: The coupled process is favorable because the net $\Delta G$ is negative when the two $\Delta G$ values are added. This question tests application of coupling principles to determine product formation feasibility. For reactions coupled through intermediate N, the overall Gibbs free energy change equals: ΔG_total = ΔG₁ + ΔG₂ = -7 kJ/mol + 2 kJ/mol = -5 kJ/mol. Since the net ΔG is negative, the coupled process is thermodynamically favorable, allowing formation of product P. The spontaneous Reaction 1 releases sufficient energy to drive the mildly nonspontaneous Reaction 2. Choice D incorrectly claims that any nonspontaneous step prevents the pathway from being favorable, which is a common misconception that ignores the principle of energy coupling. To assess coupled reactions, always calculate the net ΔG by summing individual values—a negative total confirms thermodynamic favorability.
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?
Answer: The coupled process is favorable because the net $\Delta G$ is negative. 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.
In a cell, a nonspontaneous synthesis reaction (Reaction S) with $\Delta G_S = +8\ \text{kJ/mol}$ is coupled to a spontaneous reaction (Reaction H) with $\Delta G_H = -12\ \text{kJ/mol}$ by sharing intermediates so they proceed together. Which statement best describes whether the coupled process can proceed thermodynamically as written?
Answer: The coupled process can proceed because the net $\Delta G$ is negative. This question tests understanding of coupled reactions and thermodynamic favorability based on net Gibbs free energy. When reactions are coupled through shared intermediates, the overall ΔG equals the sum of individual values: ΔG_total = Δ $G_S$ + Δ $G_H$ = +8 kJ/mol + (-12 kJ/mol) = -4 kJ/mol. Since the net ΔG is negative, the coupled process can proceed thermodynamically as written. The spontaneous Reaction H releases enough energy to drive the nonspontaneous Reaction S. Choice B incorrectly suggests that catalysis can change ΔG to make it negative, but catalysts only affect reaction rates by lowering activation energy—they cannot change the thermodynamic favorability (ΔG) of a reaction. Remember that for coupled reactions, calculate the net ΔG: if negative, the process is thermodynamically favorable.
A student couples a spontaneous reaction (Reaction M, $\Delta G_M = -22\ \text{kJ/mol}$ ) to a nonspontaneous reaction (Reaction N, $\Delta G_N = +5\ \text{kJ/mol}$ ) by linking them through shared intermediates so they proceed together. Which statement correctly describes the coupled overall process?
Answer: The coupled overall process is thermodynamically favorable because the net $\Delta G$ is negative. This question tests understanding of coupled reactions and net Gibbs free energy calculations. When reactions are coupled through shared intermediates, ΔG_total = Δ $G_M$ + Δ $G_N$ = -22 kJ/mol + 5 kJ/mol = -17 kJ/mol. Since the net ΔG is negative, the coupled overall process is thermodynamically favorable and can proceed spontaneously. The large negative ΔG from Reaction M provides more than enough energy to drive the nonspontaneous Reaction N. Choice C incorrectly suggests that coupling automatically makes any nonspontaneous reaction spontaneous, but this only occurs when the net ΔG is negative—if the nonspontaneous reaction required more energy than the spontaneous one released, coupling wouldn't work. The key strategy is to calculate net ΔG for coupled reactions: negative values indicate favorable processes.
A coupled pair of reactions in a cell share intermediate $\mathrm{Y}$ and occur together. Reaction 1 is spontaneous: $\mathrm{D \rightarrow Y}$ with $\Delta G = -6\ \mathrm{kJ\,mol^{-1}}$ . Reaction 2 is nonspontaneous: $\mathrm{Y \rightarrow E}$ with $\Delta G = +14\ \mathrm{kJ\,mol^{-1}}$ . Which statement best describes whether the coupled process is thermodynamically favorable?
Answer: The coupled process is not favorable because the net $\Delta G$ is positive when the two $\Delta G$ values are added. This question tests the ability to determine thermodynamic favorability when coupling isn't sufficient. For reactions coupled through intermediate Y, the overall Gibbs free energy change is: ΔG_total = ΔG₁ + ΔG₂ = -6 kJ/mol + 14 kJ/mol = +8 kJ/mol. Since the net ΔG is positive, the coupled process is not thermodynamically favorable and will not proceed spontaneously. The energy released from Reaction 1 is insufficient to drive the highly nonspontaneous Reaction 2. Choice B incorrectly states that the negative ΔG step dominates simply because it's spontaneous, which misunderstands that the actual energy values must be considered, not just their signs. When analyzing coupled reactions, calculate the net ΔG by addition—the process is unfavorable if ΔG_total > 0, regardless of individual reaction spontaneity.
A student couples two reactions in a process that shares intermediate $Q$ . Reaction 1 is spontaneous: $P \rightarrow Q$ with $\Delta G = -9\ \text{kJ/mol}$ . Reaction 2 is nonspontaneous: $Q \rightarrow R$ with $\Delta G = +3\ \text{kJ/mol}$ . The coupled overall reaction is $P \rightarrow R$ . Which statement best describes the thermodynamic favorability of the coupled reaction?
Answer: The coupled process is favorable because the net $\Delta G$ is negative when the two $\Delta G$ values are added. 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.
A coupled pair of reactions share intermediate $Y$ in a reaction network. Reaction 1 is spontaneous: $S \rightarrow Y$ with $\Delta G = -3\ \text{kJ/mol}$ . Reaction 2 is nonspontaneous: $Y \rightarrow P$ with $\Delta G = +1\ \text{kJ/mol}$ . When these reactions are coupled to occur together, what is the correct conclusion about the overall process?
Answer: The coupled process is thermodynamically favorable because the net $\Delta G$ is negative. This question tests understanding of coupled reactions with small ΔG values. For the coupled process S → Y → P, we calculate net ΔG: ΔG_net = (-3 kJ/mol) + (+1 kJ/mol) = -2 kJ/mol. Since the net ΔG is negative, the overall coupled process is thermodynamically favorable and can proceed spontaneously. Choice B incorrectly assumes that any positive ΔG step prevents the overall process, but coupling allows the energy from favorable steps to drive unfavorable ones. When analyzing coupled reactions, sum all individual ΔG values—even a small negative net ΔG indicates 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?
Answer: The overall coupled process is thermodynamically favorable because the net $\Delta G$ is negative. 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 a coupled set of reactions, intermediate $\mathrm{L}$ produced in the first step is immediately consumed in the second step. Reaction 1 is spontaneous: $\mathrm{X \rightarrow L}$ with $\Delta G = -35\ \mathrm{kJ\,mol^{-1}}$ . Reaction 2 is nonspontaneous: $\mathrm{L \rightarrow Y}$ with $\Delta G = +34\ \mathrm{kJ\,mol^{-1}}$ . Which statement best describes the thermodynamic favorability of the overall coupled process?
Answer: The coupled process is favorable because the net $\Delta G$ is negative when the two $\Delta G$ values are added. This question tests understanding of marginally favorable coupling. When reactions are coupled through intermediate L, the overall Gibbs free energy change is: ΔG_total = ΔG₁ + ΔG₂ = -35 kJ/mol + 34 kJ/mol = -1 kJ/mol. Since the net ΔG is negative (though small), the coupled process is thermodynamically favorable and will proceed spontaneously. The energy released from Reaction 1 just barely exceeds the energy required for Reaction 2, making the overall process feasible. Choice A incorrectly states that the net ΔG is positive, which would result from a calculation error or misunderstanding of how to add signed numbers. When analyzing coupled reactions, carefully add the ΔG values with their correct signs—even a slightly negative ΔG_total indicates a favorable process.
In a metabolic pathway, an exergonic step (Reaction X) with $\Delta G = -12\ \text{kJ/mol}$ is coupled to an endergonic step (Reaction Y) with $\Delta G = +9\ \text{kJ/mol}$ by forming and consuming a common intermediate. Which statement is correct about the overall coupled process under these conditions?
Answer: The overall coupled process is thermodynamically favorable because the net $\Delta G$ is negative. 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 (ΔG) is negative. The correct answer is A because the net ΔG is -12 + 9 = -3 kJ/mol, which is negative, making the overall coupled process thermodynamically favorable. In this metabolic pathway, the common intermediate allows the exergonic Reaction X to provide energy for the endergonic Reaction Y, resulting in a net spontaneous process. Coupling ensures the reactions occur together, with spontaneity determined by the sum of ΔG values. A tempting distractor is B, which states the process is not favorable because Reaction Y has ΔG > 0, reflecting the misconception that each reaction must be spontaneous individually rather than 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 a biochemical pathway, Reaction 1 is the hydrolysis of ATP: $\text{ATP} + \text{H}_2\text{O} \rightarrow \text{ADP} + \text{P}_i$ with $\Delta G = -30\ \text{kJ/mol}$ (spontaneous). Reaction 2 is the phosphorylation of glucose: $\text{glucose} + \text{P}_i \rightarrow \text{glucose-6-phosphate} + \text{H}_2\text{O}$ with $\Delta G = +14\ \text{kJ/mol}$ (nonspontaneous). In a cell, these reactions are coupled by sharing a phosphorylated intermediate so they occur together. Which statement best describes the thermodynamic favorability of the overall coupled process?
Answer: The overall coupled process is thermodynamically favorable because the net $\Delta G$ is negative. 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 (ΔG) is negative. The correct answer is A because the net ΔG for the coupled process is -30 + 14 = -16 kJ/mol, which is negative, indicating the overall process is thermodynamically favorable. In this biochemical pathway, the reactions share a phosphorylated intermediate, allowing the energy released from the spontaneous ATP hydrolysis (ΔG = -30 kJ/mol) to drive the nonspontaneous phosphorylation of glucose (ΔG = +14 kJ/mol). Thus, while Reaction 2 is endergonic individually, the coupling results in a net exergonic process. A tempting distractor is B, which incorrectly states the process is not favorable because Reaction 2 has a positive ΔG, reflecting the misconception that individual ΔG values determine favorability rather than 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.
A student analyzes a coupled process in which two reactions share intermediate $J$ . Reaction 1 is spontaneous: $H \rightarrow J$ with $\Delta G = -7\ \text{kJ/mol}$ . Reaction 2 is nonspontaneous: $J \rightarrow K$ with $\Delta G = +7\ \text{kJ/mol}$ . The reactions are coupled to give overall $H \rightarrow K$ . Which statement best describes the thermodynamic favorability of the coupled process?
Answer: The coupled process is not favorable because the net $\Delta G$ is zero, so there is no thermodynamic driving force. 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.
Two reactions in a lab are forced to occur together by coupling them through a common intermediate in the same reaction vessel. Reaction X is spontaneous with $\Delta G_X = -25\ \text{kJ/mol}$ , and Reaction Y is nonspontaneous with $\Delta G_Y = +30\ \text{kJ/mol}$ . Based on thermodynamics, which statement is correct about the overall coupled process?
Answer: The coupled overall process is thermodynamically unfavorable because the net $\Delta G$ is positive. This question tests the ability to determine thermodynamic favorability of coupled reactions by calculating the net Gibbs free energy change. For coupled reactions, ΔG_total = Δ $G_X$ + Δ $G_Y$ = -25 kJ/mol + 30 kJ/mol = +5 kJ/mol. Since the net ΔG is positive, the coupled overall process is thermodynamically unfavorable and cannot proceed spontaneously as written. The positive ΔG indicates that even though Reaction X releases energy, it doesn't release enough to drive Reaction Y. Choice B incorrectly suggests that coupling "transfers spontaneity," but coupling doesn't transfer properties—it simply allows reactions to occur together with their energies adding algebraically. To determine if coupled reactions can proceed, always calculate the net ΔG: only negative values indicate favorable processes.
Two reactions are coupled in a metabolic pathway via a shared intermediate. Reaction P is spontaneous with $\Delta G_P = -9\ \text{kJ/mol}$ , and Reaction Q is nonspontaneous with $\Delta G_Q = +14\ \text{kJ/mol}$ . Which statement best describes the overall thermodynamic favorability of the coupled process as written?
Answer: The coupled overall process is thermodynamically unfavorable because the net $\Delta G$ is positive. This question tests the ability to determine thermodynamic favorability of coupled reactions through calculation of net Gibbs free energy. For the coupled process, ΔG_total = Δ $G_P$ + Δ $G_Q$ = -9 kJ/mol + 14 kJ/mol = +5 kJ/mol. Since the net ΔG is positive, the coupled overall process is thermodynamically unfavorable and cannot proceed spontaneously as written. Even though Reaction P is spontaneous, it doesn't release enough energy to drive the more endergonic Reaction Q. Choice A incorrectly assumes that coupling automatically makes nonspontaneous reactions spontaneous, but coupling only works if the net ΔG is negative. To assess coupled reactions, always calculate the sum of individual ΔG values—only a negative sum indicates a favorable process.
In a biochemical pathway, Reaction 1 is spontaneous: $\mathrm{A \rightarrow B}$ with $\Delta G = -18\ \mathrm{kJ\,mol^{-1}}$ . Reaction 2 is nonspontaneous: $\mathrm{B \rightarrow C}$ with $\Delta G = +11\ \mathrm{kJ\,mol^{-1}}$ . The reactions are coupled by sharing the intermediate $\mathrm{B}$ so that they occur together in the same process. Which statement best describes whether the overall coupled process is thermodynamically favorable under these conditions?
Answer: The coupled process is favorable because the net $\Delta G$ is negative when the two $\Delta G$ values are added. This question tests understanding of coupled reactions and how to determine overall thermodynamic favorability. When reactions are coupled through a shared intermediate, 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 overall coupled process is thermodynamically favorable and will proceed spontaneously. The energy released from the spontaneous reaction (Reaction 1) drives the nonspontaneous reaction (Reaction 2) forward. Choice B incorrectly assumes that any positive ΔG step prevents the overall process, which is a common misconception—what matters is the net ΔG of the coupled system. When analyzing coupled reactions, always add the individual ΔG values to determine if the overall process is favorable (negative ΔG_total).

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