Explain the relationship between external sources of energy or coupled reactions and their ability to drive thermodynamically unfavorable processes.
An external source of energy can be used to make a thermodynamically unfavorable process occur. The CED gives two examples:
Alternatively, a desired product can be formed by coupling a thermodynamically unfavorable reaction that produces it to a favorable reaction. The CED's example is the conversion of ATP to ADP in biological systems.
How coupling works. In the coupled system, the individual reactions share one or more common intermediates. The sum of the individual reactions produces an overall reaction that achieves the desired outcome and has ΔG° < 0.
The arithmetic is Hess's law applied to free energy: ΔG° values add when reactions are added, exactly as ΔH° values do (6.9). So coupling an unfavorable step (ΔG° = +30 kJ/mol) to a favorable one (ΔG° = −50 kJ/mol) gives an overall ΔG° of −20 kJ/mol.
The shared intermediate is essential. Two reactions that merely happen in the same beaker do not couple — they must be chemically linked so that the product of one is the reactant of the other. In biology, enzymes provide that linkage.
Industrial example: metal extraction. Decomposing an ore to the metal is usually unfavorable. Coupling it to the highly favorable oxidation of carbon to CO or CO₂ makes the overall process favorable — the basis of smelting.
The connection to K: since ΔG° values add, the corresponding K values multiply (7.6). A coupled reaction has K = K₁ × K₂.
The phosphorylation of glucose, glucose + Pᵢ → glucose-6-phosphate + H₂O, has ΔG° = +13.8 kJ/mol. ATP hydrolysis, ATP + H₂O → ADP + Pᵢ, has ΔG° = −30.5 kJ/mol. (a) Show how coupling makes phosphorylation favorable. (b) Identify the shared intermediates. (c) Estimate K for the coupled reaction at 310 K.
(a) Add the two reactions.
glucose + Pᵢ → glucose-6-phosphate + H₂O ΔG° = +13.8 kJ/mol
ATP + H₂O → ADP + Pᵢ ΔG° = −30.5 kJ/mol
Summing, Pᵢ and H₂O appear on both sides and cancel:
glucose + ATP → glucose-6-phosphate + ADP ΔG° = 13.8 + (−30.5) = −16.7 kJ/mol
The overall ΔG° is negative, so the coupled process is thermodynamically favored.
(b) Shared intermediates: inorganic phosphate (Pᵢ) and water. Pᵢ is consumed by the first reaction and produced by the second; water is produced by the first and consumed by the second. These shared species are what chemically link the two processes — without them the reactions would simply be two independent processes in the same solution.
(c) K for the coupled reaction at 310 K.
ΔG° = −RT ln K → ln K = −ΔG°/(RT)
ln K = 16 700 / [(8.314)(310)] = 16 700 / 2577 = 6.48
K = e6.48 ≈ 6.5 × 10²
K well above 1, confirming that products are favored — which is why cells use ATP to drive exactly this kind of reaction.