AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/9/9-7)
Unit 9 · Topic 9.7
9.7 Coupled Reactions
A thermodynamically unfavored process can still be made to happen, either by supplying energy from outside (electricity in electrolysis, light in photosynthesis) or by coupling it to a favored reaction that shares an intermediate. When the coupled reactions are added, their ΔG° values add too, and if the total is negative the overall process is favored.
Key terms
- coupled reactions
- common intermediate
- ATP to ADP
- external energy source
- overall ΔG°
Using an outside energy source
A process with ΔG° > 0 won't go on its own in the direction you want, but you can drive it with energy from outside the system.
Electrical energy drives electrolysis, such as splitting water into H₂ and O₂ or producing aluminum from its ore, and it recharges a battery by forcing its reaction backward (topic 9.8).
Light energy drives photosynthesis. Overall, plants convert CO₂ and H₂O into glucose and O₂, which has a large positive ΔG°, and sunlight supplies that energy.
How big the gap can be
Photosynthesis shows how much energy an outside source can supply. For 6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(s) + 6O₂(g), ΔG° is about +2880 kJ per mole of glucose. No chemical reaction in the leaf could pay that cost alone; absorbed photons provide it, through a long series of coupled steps. Burning glucose (or using it in respiration) releases that free energy again.
Coupling reactions
The second strategy is to pair the unfavored reaction with a strongly favored one, so that the overall reaction has ΔG° < 0. The two reactions must share at least one common intermediate: a substance made in one reaction and used in the other, so that it cancels out when the reactions are added.
Because ΔG° behaves like ΔH° (topic 6.9), the overall ΔG° is the sum of the individual ΔG° values. If the sum is negative, the coupled process is favored.
ATP in living cells
Cells couple many unfavored reactions to the conversion of ATP to ADP. Under typical biochemical conditions, ATP + H₂O → ADP + phosphate releases about 30.5 kJ/mol of free energy (ΔG° ≈ −30.5 kJ/mol).
For example, attaching phosphate to glucose (glucose + phosphate → glucose-6-phosphate + H₂O) has ΔG° ≈ +13.8 kJ/mol, so it isn't favored alone. Coupled with ATP hydrolysis, the overall reaction, glucose + ATP → glucose-6-phosphate + ADP, has ΔG° ≈ −16.7 kJ/mol. In the cell this happens as one enzyme-controlled process, with the phosphate group passed directly from ATP to glucose.
Coupling in industry
Metal production often uses coupling. Heating a metal sulfide or oxide alone may not release the metal, but pairing the process with a very favored reaction, such as sulfur or carbon combining with oxygen, makes the overall process favored.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Getting copper from copper(I) sulfide
Cu₂S(s) → 2Cu(s) + S(s) has ΔG° = +86.2 kJ/mol. S(s) + O₂(g) → SO₂(g) has ΔG° = −300.1 kJ/mol. Write the coupled reaction, identify the common intermediate, and decide whether the overall process is favored.
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- Step 1: Add the two reactions: Cu₂S(s) + S(s) + O₂(g) → 2Cu(s) + S(s) + SO₂(g).
- Step 2: S(s) appears on both sides and cancels. It's the common intermediate.
- Step 3: Overall: Cu₂S(s) + O₂(g) → 2Cu(s) + SO₂(g).
- Step 4: ΔG° = +86.2 + (−300.1) = −213.9 kJ/mol. Negative, so the coupled process is favored. This overall reaction is how copper is obtained by heating its sulfide ore in air.
Answer: Cu₂S(s) + O₂(g) → 2Cu(s) + SO₂(g), ΔG° = −213.9 kJ/mol; S(s) is the common intermediate; favored.
- Example 2
Trap: coupling needs a shared species
A student claims that because ATP hydrolysis has ΔG° ≈ −30.5 kJ/mol, any reaction with ΔG° less than +30.5 kJ/mol can be driven simply by running ATP hydrolysis in the same container. What's missing from this claim?
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- Step 1: Running two reactions side by side doesn't transfer free energy from one to the other. The reactions must be linked through a common intermediate.
- Step 2: In cells, enzymes link them: for example, a phosphate group moves directly from ATP to glucose, so the phosphate is the shared species, and the overall reaction is glucose + ATP → glucose-6-phosphate + ADP.
- Step 3: Only for that combined reaction is ΔG° = +13.8 + (−30.5) = −16.7 kJ/mol.
Answer: The reactions must share a common intermediate (here, the transferred phosphate group) so that they form one overall reaction; then their ΔG° values add.
Common mistakes
- Thinking coupling means two reactions merely happen at the same time. They must share an intermediate and combine into one overall reaction.
- Multiplying ΔG° values when combining reactions. Like ΔH°, ΔG° values add.
- Forgetting that external energy sources (electricity, light) can also drive unfavored processes.
On the exam
- Coupling questions usually give two reactions with ΔG° values: add them, cancel the common intermediate, and state whether the total ΔG° is negative.
- Be ready to name examples of external energy driving an unfavored process: electrolysis, recharging a battery and photosynthesis.
Connected topics
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Check yourself
4 questions on 9.7 Coupled Reactions. Pick an answer to see if you got it, and why.
Reaction 1: glucose + phosphate → glucose-6-phosphate + H₂O ΔG° = +13.8 kJ/mol
Reaction 2: ATP + H₂O → ADP + phosphate ΔG° = −30.5 kJ/mol
Approximate values for a reaction in cells
In cells, reaction 1 is coupled to reaction 2. What is ΔG° for the overall reaction, glucose + ATP → glucose-6-phosphate + ADP, and is it favored?
Photosynthesis, 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂, has a large positive ΔG°. Which of the following explains how plants make it happen?
Reaction 1: 2Fe₂O₃(s) → 4Fe(s) + 3O₂(g) ΔG° = +1,484.4 kJ/mol
Reaction 2: 2CO(g) + O₂(g) → 2CO₂(g) ΔG° = −514.4 kJ/mol
Thermochemical data at 25 °C
What is ΔG° for the overall reaction, 2Fe₂O₃(s) + 6CO(g) → 4Fe(s) + 6CO₂(g)?
Which of the following best describes how the two reactions are coupled?
0 of 4 answered