Answer:
ΔG° = -750324.96 J/mol = -750.324 Kj/mol
ΔH° = - 790.4 kJ/mol = -790400 j/mol
ΔS° = = - 134.48 J K-1mol-1
Explanation:
(S, rhombic) + 3O2( g) → 2SO3 (g)
Temperature = 25°C + 273 = 298K (Converting to kelvin temperature.)
The formulae relating all four paramenters (ΔG°, T, ΔH°, & ΔS°) is given as;
ΔG° = ΔH° - TΔS°
All H and S values used are measurements at 25°C
Calculating ΔH
You started with 1 mole of Sulphur and 3 moles of oxygen.
Total starting enthalpy = 0+ 3(0) = 0 kJ/mol
You ended up with 2 moles of SO3.
Total enthalpy at the end = 2(-395.2) = - 790.4 kJ/mol
enthalpy change = what you end up with - what you started with.
ΔH = enthalpy change = - 790.4 kJ/mol - 0 kJ/mol = - 790.4 kJ/mol = -790400 j/mol
Calculating ΔS
You started with 1 mole of Sulphur and 3 moles of oxygen.
Total starting entropy =31.88 + 3(205) = 646.88 J K-1mol-1
You ended up with 2 moles of SO3.
Total entropy at the end = 2(256.2) = 512.4 J K-1mol-1
entropy change = what you end up with - what you started with.
ΔS = entropy change = 512.4 J K-1mol-1 - 646.88 J K-1mol-1
= - 134.48 J K-1mol-1
Inputing the values into the formular, we have;
ΔG° = ΔH° - TΔS°
ΔG° = - 790400 - 298 (- 134.48)
ΔG° = - 790400 + 40075.04
ΔG° = -750324.96 J/mol = -750.324 Kj/mol
