Gen Chem 2Entropy and Free Energy

Coupled Reactions

An unfavored reaction can run when it is paired with a strongly favored one, as ATP does in cells.

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Living cells build proteins and sugars, and both have positive ΔG. They manage it by pairing each uphill reaction with a strongly downhill one. Free energies add the way enthalpies do in Hess’s law.

Key idea

When two reactions share an intermediate and run as one combined process, their free energies add: $\Delta G_\text{total} = \Delta G_1 + \Delta G_2$, and their constants multiply: $K_\text{total} = K_1K_2$. If the total is negative, the combined process is favored, even though one part alone is not.

Why it works

$G$ is a state function, so the overall change does not depend on the path; adding the equations adds the $\Delta G$ values. Coupling is not a bookkeeping trick: the two reactions must be physically linked, usually through a shared intermediate. In cells, ATP transfers a phosphate group to the reactant, making an activated intermediate whose next step is downhill.
An uphill step (+14.2 kJ/mol) paired with ATP hydrolysis (−30.5 kJ/mol). The combined process is downhill by 16.3 kJ/mol.

Worked examples

Worked example: Making glutamine

Glutamate + $\mathrm{NH_4^+}$ → glutamine + water has $\Delta G^{\circ\prime} = +14.2\ \text{kJ/mol}$. ATP hydrolysis has $\Delta G^{\circ\prime} = -30.5\ \text{kJ/mol}$. Find the total and $K$ at $310\ \text{K}$ (body temperature).
  1. $\Delta G_\text{total} = 14.2 + (-30.5) = -16.3\ \text{kJ/mol}$.
  2. $$K = e^{16\,300/(8.314 \times 310)} = e^{6.32} = 5.6 \times 10^{2}$$
  3. Alone, the first reaction has $K = e^{-14\,200/(8.314 \times 310)} = 4.0 \times 10^{-3}$: coupling raises $K$ by more than 100 000 times.

Answer: $-16.3\ \text{kJ/mol}$; $K \approx 560$.

Worked example: Adding constants

Reaction 1 has $K_1 = 2.0 \times 10^{-3}$ and reaction 2 has $K_2 = 5.0 \times 10^{5}$ at the same temperature. Coupled, they add to the overall process. Find $K$ and the sign of $\Delta G^\circ$.
  1. $K = K_1K_2 = (2.0 \times 10^{-3})(5.0 \times 10^{5}) = 1.0 \times 10^{3}$.
  2. $K > 1$, so $\Delta G^\circ < 0$: the coupled process is favored.

Answer: $K = 1.0 \times 10^{3}$; $\Delta G^\circ$ negative.

The $\circ\prime$ (degree-prime) marks the biochemists' standard state: pH 7 rather than $1\ \text{M}$ $\mathrm{H_3O^+}$, which suits reactions in cells. Other values you will meet: phosphoenolpyruvate hydrolysis, $-61.9\ \text{kJ/mol}$; glucose + phosphate → glucose 6-phosphate, $+13.8\ \text{kJ/mol}$. Industry couples reactions too: in smelting, carbon's reaction with oxygen pays for pulling oxygen out of a metal oxide.

A typical cell turns over its ATP many times a day, remaking it from ADP and phosphate with energy from food; the ATP is a carrier of free energy, not a store.

Common mistake

Thinking coupling changes the $\Delta G$ of the unfavored step itself. That step still has $\Delta G = +14.2\ \text{kJ/mol}$; what changes is that it no longer runs alone. Check: the total is the sum, and each part keeps its own value.

Practice

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Builds on

Leads to

This is one of the top rungs of General Chemistry 2.

Sources and further reading

  • OpenStax Chemistry 2e, Section 16.4 Free Energy
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