For the reaction ATP \( +\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{ADP}+\mathrm{P}_{\mathrm{i}} \), the \( \Delta G^{\circ}=-30.5 \mathrm{~kJ} / \mathrm{mol} \). What would be the associated equilibrium constant for this reaction? Choose one: \( \begin{array}{l}\text { A. } 35,000 \\ \text { B. } 135,000 \\ \text { C. } 235,000 \\ \text { D. } 3500\end{array} \)

To find the equilibrium constant \( K \) for the reaction, we can use the relationship between \( \Delta G^{\circ} \) and \( K \): \[ \Delta G^{\circ} = -RT \ln K \] Given \( \Delta G^{\circ} = -30.5 \, \text{kJ/mol} \) (which is \( -30,500 \, \text{J/mol} \)) and using \( R = 8.314 \, \text{J/(mol·K)} \) at a temperature of \( 298 \, \text{K} \) (standard conditions), we can rearrange and solve for \( K \). Calculating: \[ -30,500 = -8.314 \times 298 \ln K \] This simplifies to: \[ \ln K = \frac{30,500}{8.314 \times 298} \approx 12.24 \] Exponentiating both sides gives us: \[ K \approx e^{12.24} \approx 35,000 \] So, the associated equilibrium constant for this reaction is: **A. 35,000**