Calculate Δ_rG at 298 K under the conditions shown below for the following reaction: 2Hg(g) + O₂(g) → 2HgO(s) ΔG° = -180.8 kJ mol^-1
P(Hg) = 0.025 atm, P(O₂) = 0.037 atm

Question

Calculate Δ_rG at 298 K under the conditions shown below for the following reaction: 2Hg(g) + O₂(g) → 2HgO(s) ΔG° = -180.8 kJ mol^-1 P(Hg) = 0.025 atm, P(O₂) = 0.037 atm

Quizplus · Accepted Answer

To calculate Δ_rG, we first need to calculate Δ_rH using the standard enthalpy of formation values: 2Hg(g) + O₂(g) → 2HgO(s) ΔH° = -181.6 kJ mol^-1 Δ_rH can be calculated using the following equation: Δ_rH = ΣnSjS°(products) - ΣnSjS°(reactants) = 2(70.1) + (205.0) - 0 = 345.2 J K-1 mol-1 Next, we need to calculate Δ_rS using the following equation: Δ_rS = ΣnSjS°(reactants) ln(Q) Q = P(O₂)²/P(Hg)² = (0.037 atm)²/(0.025 atm)² = 2.76 Δ_rS = -181.6 kJ mol^-1/298 K - (345.2 J K-1 mol-1) ln(2.76) = -181.6 kJ mol^-1 + 122.3 J K-1 mol-1 = -181.5 kJ mol^-1 Finally, Δ_rG can be calculated using the following equation: Δ_rG = Δ_rH - TΔ_rS = (-181.6 kJ mol^-1 * 1000 J/kJ - 298 K * 122.3 J K-1 mol-1) / 1000 = -154.4 kJ mol^-1 The best choice is B because it matches our calculated value.

Calculate ΔrG at 298 K Under the Conditions Shown Below

Calculate Δ_rG at 298 K Under the Conditions Shown Below