Question

Calculate the value of ΔG°, ΔH°, & ΔS° for the oxidation of solid elemental sulfur to gaseous sulfur trioxide at at 25°C.​2S (s, rhombic) + 3O2( g) → 2SO3 (g)

Answer 1

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