Determine the equilibrium constant for the following reaction at 498 K. 2 Hg(g)+ O₂(g)→ 2 HgO(s)ΔH° = -304.2 kJ; ΔS° = -414.2 J/K

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The equilibrium constant (K) can be calculated using the Gibbs free energy change (ΔG°) formula: ΔG° = ΔH° - TΔS°. First, convert ΔH° to J (since ΔS° is given in J/K): ΔH° = -304.2 kJ = -304200 J. Then, calculate ΔG°: ΔG° = -304200 J - (498 K)(-414.2 J/K) = -304200 J + 206370 J = -97830 J. Now, use the relationship ΔG° = -RTlnK, where R = 8.314 J/(mol·K), to find K: -97830 J = -(8.314 J/(mol·K))(498 K)lnK. Solving for K gives lnK = 23.6, so K = e^23.6 ≈ 1.87 × 10^10.

Determine the Equilibrium Constant for the Following Reaction at 498