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Coupled reactions

An unfavorable reaction is not doomed — it can hitch a ride on a favorable one. But the ride only works if they share a chemical connection, the way ATP hydrolysis powers reactions across biology.

§1

Driving the unfavorable with the favorable.

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An unfavorable reaction (ΔG > 0) can be driven by a favorable one (ΔG < 0) when the two are coupled through a shared intermediate — a species produced by one step and consumed by the other.

When coupled, the free energies add: ΔG_overall = ΔG₁ + ΔG₂. The combined process is favored when the favorable step is negative enough to make the sum negative.

The connection is essential. A favorable reaction elsewhere in the beaker, with no shared species, cannot push an unfavorable one — and coupling does not change the target reaction's own ΔG, it just adds a second, favorable process.

UNIT 9 TOPIC 9.7 • COUPLED REACTIONS COUPLED REACTIONS Coupling a favorable reaction to an unfavorable one through a common intermediate makes the overall process spontaneous. STEP 1 · FAVORABLE phosphoryl transfer to A A + ATP → A–P + ADP ΔG₁ < 0 ↓ releases free energy SHARED INTERMEDIATE A–P high-energy intermediate STEP 2 · UNFAVORABLE A–P drives formation of AB A–P + B → AB + Pᵢ ΔG₂ > 0 ↑ requires free energy OVERALL (NET) COUPLED REACTION A + B + ATP → AB + ADP + Pᵢ ΔGₙₑₜ = ΔG₁ + ΔG₂ < 0 → spontaneous ✓ OTHER ENERGY SOURCES ⚡ Electrical — active ion transport ☀ Light — photosynthesis also drive ΔG > 0 processes. A ATP A–P B AB + + ADP + Pᵢ shared intermediate CED ANCHOR Coupling links an unfavorable process to a favorable one through a common intermediate, so the summed ΔGₙₑₜ = ΔG₁ + ΔG₂ < 0 — the overall process is spontaneous. Enthalpy/entropy sum to drive the whole path. AP Chemistry · Unit 9 · Applications of Thermodynamics
Fig. 9.7.1 Coupling lets a favorable reaction drive an unfavorable one, but only through a shared intermediate. The overall free energy is the sum of the steps' ΔG values, which must be negative for the coupled process to be favored.
§2

Making coupling work.

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Share a species and add the ΔG values.

  1. Identify the unfavorable target. The reaction with ΔG > 0 you want to drive.
  2. Find a favorable partner. A reaction with ΔG < 0 that shares an intermediate with the target.
  3. Check the shared species. One step must produce what the other consumes — no shared species, no coupling.
  4. Add the free energies. ΔG_overall = sum; the coupled process is favored only if the total is negative.
§3

The pieces you'll meet.

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What coupling requires.

couple
Coupling
Linking reactions via a shared intermediate.
shared
Shared intermediate
Produced by one step, used by the other.
sum
ΔG adds
ΔG_overall = ΔG₁ + ΔG₂.
driver
Favorable driver
Must be negative enough to flip the sum.
own ΔG
Target ΔG
Unchanged — coupling adds a second process.
no link
No shared species
Then no coupling is possible.
§4

Worked example: coupling two steps.

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Target. Step 1 is unfavorable: ΔG₁ = +20 kJ.

Driver. Step 2 is favorable and shares an intermediate: ΔG₂ = −35 kJ.

Sum. ΔG_overall = +20 + (−35) = −15 kJ.

Verdict. The coupled process is favored (ΔG < 0). The favorable step's −35 kJ more than pays for the unfavorable +20 kJ — but only because they share an intermediate.

§5

Mistakes that cost real points.

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Pitfall · 01

"Any favorable reaction in the mixture can drive any unfavorable one."

Coupling requires a shared intermediate linking the two reactions; a favorable reaction with no chemical connection cannot drive an unfavorable one. Without a shared species, their free energies do not combine into one driven process.

Fix. Require a shared intermediate before treating one reaction as driving another.

Pitfall · 02

"Coupling changes the unfavorable reaction's own ΔG."

The target reaction's ΔG is unchanged; coupling adds a second, favorable reaction so that the overall ΔG (the sum) becomes negative. The unfavorable step is still unfavorable on its own — it is just carried by the pair.

Fix. Add the two ΔG values for the overall process; the target's individual ΔG stays the same.

Pitfall · 03

"Any favorable partner is enough, no matter how small its ΔG."

The driver must be negative enough that the sum ΔG₁ + ΔG₂ is negative. A weakly favorable step may not outweigh the unfavorable one, leaving the overall process unfavored. Check the magnitude, not just the sign.

Fix. Confirm the summed ΔG is negative — the driver must outweigh the unfavorable step.

§6

Skill Check.

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