Equilibrium constant: \[ K = 10^{10} \]
Gas constant: \[ R = 8.314\ \text{J K}^{-1}\text{mol}^{-1} \]
Temperature: \[ T = 300\ \text{K} \]
Relation between ΔG° and K: \[ \Delta G^\circ = -RT \ln K \]
First, convert \(\log_{10}\) to natural log:
\[ \ln K = \ln(10^{10}) = 10 \ln 10 \approx 10 \times 2.303 = 23.03 \]
Then: \[ \Delta G^\circ = - (8.314)(300)(23.03) \] \[ 8.314 \times 300 = 2494.2 \] \[ \Delta G^\circ \approx - 2494.2 \times 23.03 \approx -5.75 \times 10^{4}\ \text{J mol}^{-1} \] \[ \Delta G^\circ \approx -57.5\ \text{kJ mol}^{-1} \]
\[ \boxed{\Delta G^\circ \approx -5.8 \times 10^{4}\ \text{J mol}^{-1} \;\;(\text{or } -58\ \text{kJ mol}^{-1})} \]