= ATP synthase energy and proton budget
{c}
Define side $P$ as the higher <proton> <electrochemical potential> reservoir and side $N$ as the synthesis-side reservoir. For transfer $P\to N$ at common <temperature> $T$, define the positive downhill <proton-motive force> magnitude $\Delta p_{\rm drive}$ by
$$
\Delta\widetilde\mu_{P\to N}
=RT\ln\frac{a_N}{a_P}+F(\psi_N-\psi_P)
=-F\Delta p_{\rm drive}.
$$
Here $a$ is <thermodynamic activity>, $\psi$ is <electric potential>, $R$ is the <molar gas constant> and $F$ is the <Faraday constant>. Thus both a higher <proton> <thermodynamic activity> and a more positive <electric potential> on side $P$ contribute to the driving magnitude. This definition fixes the sign; conventions for a signed <proton-motive force> differ. The calculation under <chemiosmosis> separates these two contributions numerically.
Assume fixed reservoirs, tight mechanical <energy coupling>, one transported <proton> per $c$ subunit per revolution, and three <ATP> produced per revolution. Neglect leaks, slipping and <membrane transport> of the chemical reagents. Let $g_{\rm ATP}>0$ be the local molar <Gibbs free energy> cost of <ATP> formation at the stipulated composition, <pH> and <magnesium>-binding convention. It is not necessarily the <standard state> value. Adding changes in the <state function> gives, per mole of complete turns,
$$
\Delta G_{\rm turn}=3g_{\rm ATP}-nF\Delta p_{\rm drive}.
$$
For this model net synthesis is thermodynamically favoured when $\Delta G_{\rm turn}<0$. Equality is a reversible balance, not a promise of a useful <reaction rate>. A negative value establishes an available driving margin; it does not specify how fast the <enzyme> runs.
Choose $n=10$, $g_{\rm ATP}=50.0\,\mathrm{kJ\,mol^{-1}}$, $F=96485\,\mathrm{C\,mol^{-1}}$ and $\Delta p_{\rm drive}=0.180\,\mathrm V$. These are illustrative inputs, not measurements of the preparation in the experiment below. Then
$$
nF\Delta p_{\rm drive}=173.673\,\mathrm{kJ\,mol^{-1}},\qquad
3g_{\rm ATP}=150.0\,\mathrm{kJ\,mol^{-1}},
$$
and $\Delta G_{\rm turn}=-23.673\,\mathrm{kJ\,mol^{-1}}$. Per <ATP>, the ideal input is $57.891\,\mathrm{kJ\,mol^{-1}}$, leaving a driving margin of $7.891\,\mathrm{kJ\,mol^{-1}}$. The balance point is
$$
\Delta p_* =\frac{3g_{\rm ATP}}{nF}\approx0.15546\,\mathrm V.
$$
At $0.140\,\mathrm V$ the same synthesis cycle instead has $\Delta G_{\rm turn}=+14.921\,\mathrm{kJ\,mol^{-1}}$: the chosen reservoirs favour its reverse.
\Image[/atp-synthase-energy-coupling.png]
{title=Chosen ten-subunit model: the <free energy> balance determines the favoured direction}
{description=Original calculation, not experimental data. Fixed $g_{\rm ATP}=50$ kJ/mol, three <ATP> per turn, no leak or reagent transport. The second panel adds an explicitly hypothetical one-<proton>-equivalent transport cost per delivered <ATP>.}
The machine is reversible: sufficient <ATP> <hydrolysis> can drive opposite rotation and pump <protons> uphill. Actual direction also depends on inhibition and available <substrates>; a weak <proton-motive force> need not make every intact <enzyme> immediately run backward. <Respiratory uncoupling> supplies a different failure mode: a parallel <proton> leak dissipates the gradient without productive shaft rotation.<Rotary catalysis of FoF1-ATP synthase by Rikiya Watanabe>[Watanabe, 2013, p. 51 and Figure 2: reversal of the coupled motors]{ref}
The motor budget $10/3$ is not the whole cost of <ATP> delivered outside a <mitochondrion>. Import of <ADP> and <inorganic phosphate>, and export of <ATP>, also draw on the <electrochemical gradient>.<The Cell: A Molecular Approach, 2nd edition by Geoffrey M. Cooper/The Mechanism of Oxidative Phosphorylation>[Cooper, The Cell: A Molecular Approach, 2nd edition, Transport of Metabolites across the Inner Membrane, Figure 10.12]{ref} For a deliberately simplified system budget, assign these transports together one additional downhill-<proton> equivalent per delivered <ATP>, valued at the same $F\Delta p_{\rm drive}$. This is a chosen accounting convention, not a claim of one extra literal <proton> through $F_0$, nor a universal transport <stoichiometry>. The total becomes $10/3+1=13/3$ equivalents and $75.258\,\mathrm{kJ\,mol^{-1}}$ of gradient expenditure per delivered <ATP> at $0.180\,\mathrm V$. The ratio of stored chemical <free energy> to that expenditure is $50/75.258\approx66.4\%$, compared with $50/57.891\approx86.4\%$ for the motor alone. This system budget changes the resource expenditure, not the motor's balance point $\Delta p_*$. Real transport terms require their own <electrochemical potentials>, and leaks add further cost.
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