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How can a thermodynamically unfavorable reaction be driven by coupling it to a favorable one?

Topic 9.7 Coupled Reactions: explain how an unfavorable reaction can be driven by coupling it to a favorable reaction so that the combined free energy change is negative.

A focused answer to AP Chemistry Topic 9.7, covering how coupling an unfavorable reaction to a more favorable one gives a net negative free energy change, the role of a shared intermediate, and biological and industrial examples, with full worked examples.

Generated by Claude Opus 4.89 min answer

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  1. What this topic is asking
  2. How coupling works
  3. The free energies add
  4. Why coupling matters
  5. Try this

What this topic is asking

The College Board (Topic 9.7) wants you to explain how an unfavorable reaction can be driven by coupling it to a favorable reaction so that the combined free energy change is negative. This applies the additivity of free energy to a powerful idea used throughout biology and industry.

How coupling works

The shared intermediate is the mechanical link: the product of one reaction is the reactant of the other, so they cannot run independently. When summed, the intermediate cancels (as in Hess's law), and the overall reaction has the combined free energy change. If that combined value is negative, the whole process proceeds.

The free energies add

So coupling is not magic: it cannot make an arbitrarily unfavorable reaction go. The favorable partner must release more free energy than the unfavorable reaction requires. This is a direct consequence of free energy being additive, the same property that underlies Hess's law for enthalpy.

Why coupling matters

Coupling is the principle behind much of biochemistry. Cells drive unfavorable reactions (building large molecules, pumping ions against a gradient) by coupling them to the highly favorable hydrolysis of ATP, which releases free energy. Industrially, an unfavorable extraction or synthesis can be coupled to a favorable reaction to make it proceed. In every case, the test is the same: the sum of the free energy changes must be negative.

Try this

Q1. An unfavorable reaction (ΔG=+18 kJ mol−1\Delta G = +18\ \text{kJ mol}^{-1}) is coupled to a favorable one (ΔG=−30 kJ mol−1\Delta G = -30\ \text{kJ mol}^{-1}). State whether the coupled process proceeds. [2 points]

  • Cue. Sum =18+(−30)=−12 kJ mol−1= 18 + (-30) = -12\ \text{kJ mol}^{-1}, negative, so the coupled process is favorable and proceeds.

Q2. Explain why a catalyst cannot be used instead of coupling to make an unfavorable reaction proceed. [2 points]

  • Cue. A catalyst changes only the rate (kinetics), not ΔG\Delta G; an unfavorable reaction stays unfavorable, so it needs coupling to a favorable reaction to make the overall ΔG\Delta G negative.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (style)4 marksSection II (long FRQ, part). Reaction 1 is unfavorable with ΔG1=+30. kJ mol−1\Delta G_1 = +30.\ \text{kJ mol}^{-1}, and reaction 2 is favorable with ΔG2=−50. kJ mol−1\Delta G_2 = -50.\ \text{kJ mol}^{-1}; they share a common intermediate. (a) Determine the overall ΔG\Delta G when the two reactions are coupled. (b) State whether the coupled process is favorable, and justify. (c) Explain the role of the shared intermediate in coupling. (d) Explain why coupling cannot drive reaction 1 if reaction 2 has ΔG2=−20. kJ mol−1\Delta G_2 = -20.\ \text{kJ mol}^{-1}.
Show worked answer →

A 4-point conceptual FRQ on coupled reactions.

(a) Overall (1 point): coupling adds the free energies: ΔG=ΔG1+ΔG2=30.+(−50.)=−20. kJ mol−1\Delta G = \Delta G_1 + \Delta G_2 = 30. + (-50.) = -20.\ \text{kJ mol}^{-1}.
(b) Favorable (1 point): the overall ΔG<0\Delta G < 0, so the coupled process is thermodynamically favorable even though reaction 1 alone is not.
(c) Shared intermediate (1 point): the product of one reaction is a reactant of the other (a common intermediate), so the two reactions proceed together as one pathway, and their free energies add.
(d) Insufficient coupling (1 point): with ΔG2=−20.\Delta G_2 = -20., the sum would be 30.+(−20.)=+10. kJ mol−130. + (-20.) = +10.\ \text{kJ mol}^{-1}, still positive, so the favorable reaction is not favorable enough to drive the unfavorable one.

Markers reward the summed free energy, the favorable conclusion, the shared-intermediate reasoning, and the insufficient-coupling case.

AP 2021 (style)1 marksSection I (multiple choice). An unfavorable reaction can be driven by coupling it to a favorable reaction provided that (A) the favorable reaction is faster (B) the sum of the two free energy changes is negative (C) both reactions are exothermic (D) a catalyst is added. Justify your choice.
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A 1-point conceptual MCQ. The answer is (B).

Coupling drives an unfavorable reaction only if the combined free energy change is negative, that is, the favorable reaction releases more free energy than the unfavorable one requires. The trap is (D): a catalyst changes rate, not favourability, so it cannot make a coupled process favorable.

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