 Solution

Let $r=a_{\mathrm{ADP}}/a_{\mathrm{ATP}}$ and use a one-molar standard state, so the inorganic phosphate activity is approximately $a_P=0.006$. With the source's stated standard hydrolysis value,The downhill proton-transfer Gibbs free energy is $F\Delta p$ for the signed driving force $\Delta p=-0.150$ V. Ideal reversible balance of one complete turn requiresSolving yieldsAt 298.15 K the estimate is $3.39\times 10^2$. Temperature is not specified in this paragraph; the first value explicitly adopts physiological temperature rather than pretending it is supplied. The unexpectedly large ADP/ATP ratio follows from using the printed standard value $-50$ kJ/mol with the stated dilute inorganic phosphate. One must not silently replace that value by another convention or treat it as the already concentration-corrected cellular hydrolysis Gibbs free energy.

This estimate assumes ideal thermodynamic activities, consistent pH and magnesium conventions, tight coupling of ten protons to three ATP, no leak or transport overhead, and operation close to reversible balance. A working cell is not in global equilibrium. Net synthesis at this fixed driving force requires $\Delta G_{\mathrm{turn}}<0$, hence $r$ above this reversible-balance threshold; finite chemical kinetics and dissipation prevent the stoichiometry alone from determining an exact physiological ratio.