A reaction has K = 2.5 × 10⁻³ at 500 K. Find Δ G° at this temperature…

At equilibrium $\Delta G = 0$ and $\Delta G^\circ = -RT\ln K = -2.303\,RT\log K$, with $R = 8.314\ \text{J/(K mol)}$ and $\Delta G^\circ$ in J/mol. A negative $\Delta G^\circ$ gives $K > 1$, a positive one $K < 1$. At another temperature, find $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ first.