Coupled Reactions
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AP Chemistry › Coupled Reactions
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 a catalyst is added, since coupling is a type of catalysis.
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 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 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.
In a synthesis scheme, two reactions are coupled by sharing intermediate $E$. Reaction 1 is spontaneous: $D \rightarrow E$, $\Delta G = -2\ \text{kJ/mol}$. Reaction 2 is nonspontaneous: $E \rightarrow F$, $\Delta G = +9\ \text{kJ/mol}$. They are carried out together so the intended overall conversion is $D \rightarrow F$. Which statement is correct?
The coupled process is favorable because any spontaneous reaction can drive any nonspontaneous reaction regardless of magnitudes.
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 coupling requires the two $\Delta G$ values to have the same sign.
The coupled process is favorable because coupling decreases the activation energy, which makes $\Delta G$ more negative.
The coupled process is favorable because coupling makes Reaction 2 spontaneous by changing its $\Delta G$ to a negative value.
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.
In an engineered reaction sequence, two steps are linked by a shared intermediate $I$. Step 1 is spontaneous: $S \rightarrow I$ with $\Delta G = -14\ \text{kJ/mol}$. Step 2 is nonspontaneous: $I \rightarrow T$ with $\Delta G = +20\ \text{kJ/mol}$. The steps are coupled to attempt the overall conversion $S \rightarrow T$. Which statement is correct about the coupled process?
The coupled process is not favorable because each individual step must have $\Delta G < 0$ for the overall reaction to proceed.
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 coupling acts like a catalyst and changes thermodynamic favorability.
The coupled process is favorable because the overall reaction is spontaneous if at least one step is spontaneous.
The coupled process is favorable because coupling allows the negative $\Delta G$ to be transferred to the second step.
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.
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?
The coupled process is favorable because the spontaneous step ensures the overall process proceeds in the forward direction.
The coupled process is not favorable because the net $\Delta G$ is zero, so there is no thermodynamic driving force.
The coupled process is favorable because the net $\Delta G$ is determined by the step with the larger magnitude.
The coupled process is favorable because coupling changes the sign of the nonspontaneous step without changing the net $\Delta G$.
The coupled process is not favorable because coupling only affects activation energy and cannot change $\Delta G$.
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 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.
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?
The coupled process is not favorable because coupling affects kinetics rather than thermodynamics.
The coupled process is favorable because coupling changes Reaction 2 so its $\Delta G$ becomes negative even when considered alone.
The coupled process is not favorable because a positive $\Delta G$ step cannot occur under any circumstances.
The coupled process is favorable because the net $\Delta G$ is negative when the two $\Delta G$ values are added.
The coupled process is favorable because the reaction with the smaller magnitude of $\Delta G$ determines 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 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.
Two reactions are coupled in a cellular compartment by sharing intermediate $O$. Reaction 1 is spontaneous: $M \rightarrow O$, $\Delta G = -50\ \text{kJ/mol}$. Reaction 2 is nonspontaneous: $O \rightarrow P$, $\Delta G = +45\ \text{kJ/mol}$. When coupled as $M \rightarrow P$, which statement best describes the overall thermodynamic outcome?
The coupled process is favorable because the net $\Delta G$ is negative when the two $\Delta G$ values are added.
The coupled process is not favorable because the nonspontaneous step cannot occur even if the net $\Delta G$ is negative.
The coupled process is favorable because coupling changes only Reaction 2 so that its $\Delta G$ becomes negative on its own.
The coupled process is not favorable because coupling is equivalent to adding a catalyst, which does not change $\Delta G$.
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.
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 each reaction keeps its sign but the overall sign is determined by the larger magnitude.
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 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.
In a cell, a nonspontaneous synthesis reaction ($\Delta G = +22\ \text{kJ/mol}$) is coupled to a spontaneous reaction ($\Delta G = -50\ \text{kJ/mol}$) by sharing an intermediate so that they occur together. Which statement best describes what allows the synthesis reaction to proceed?
The synthesis proceeds because coupling acts like a catalyst and changes thermodynamic favorability.
The synthesis cannot proceed because any positive $\Delta G$ step prevents the overall process.
The synthesis proceeds because the total $\Delta G$ of the coupled process is negative.
The synthesis proceeds only if the spontaneous reaction reaches equilibrium first.
The synthesis proceeds because coupling makes the synthesis reaction’s own $\Delta G$ negative.
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.
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?
A catalyst must be present so that Reaction Y becomes thermodynamically favorable.
Reaction X must stop at equilibrium so that Reaction Y can proceed past equilibrium.
The combined $\Delta G$ for the coupled reactions must be negative.
Reaction Y must become spontaneous on its own, so its individual $\Delta G$ must become negative.
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.
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 only if Reaction B is run in a separate container.
The coupled process is not favorable because the nonspontaneous reaction prevents any net progress.
The coupled process is favorable because a catalyst is equivalent to coupling for $\Delta G$.
The coupled process is favorable because the net $\Delta G$ is negative.
The coupled process is favorable because coupling increases the magnitude of the negative $\Delta G$ step.
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.