Determine ΔG for the following reaction: CO(g) + 1/2O2(g) → CO2(g); ΔH° = -82.84 kJ [Given: S°CO2 = 213.8 J K⁻¹mol⁻¹, S° CO(g) = 197.9 J K⁻¹mol⁻¹, S° O2 = 205.8 J K⁻¹mol⁻¹]

The reaction is: CO(g) + 1/2O2(g) → CO2(g)

We are given ΔH° = -82.84 kJ
S°CO2 = 213.8 J K⁻¹mol⁻¹
S°CO(g) = 197.9 J K⁻¹mol⁻¹
S°O2 = 205.8 J K⁻¹mol⁻¹

ΔS° = S°(products) - S°(reactants)
ΔS° = S°(CO2) - [S°(CO) + 1/2S°(O2)]
ΔS° = 213.8 J K⁻¹mol⁻¹ - [197.9 J K⁻¹mol⁻¹ + (1/2)(205.8 J K⁻¹mol⁻¹)]
ΔS° = 213.8 J K⁻¹mol⁻¹ - [197.9 J K⁻¹mol⁻¹ + 102.9 J K⁻¹mol⁻¹]
ΔS° = 213.8 J K⁻¹mol⁻¹ - 300.8 J K⁻¹mol⁻¹
ΔS° = -87 J K⁻¹mol⁻¹ = -0.087 kJ K⁻¹mol⁻¹

We can use the Gibbs Free Energy equation:
ΔG° = ΔH° - TΔS°
Assuming the reaction occurs at standard temperature (298 K):
ΔG° = -82.84 kJ - (298 K)(-0.087 kJ K⁻¹mol⁻¹)
ΔG° = -82.84 kJ + 25.966 kJ
ΔG° = -56.874 kJ ≈ -56.91 kJ