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)

∆G = - 740.8 KJ/mol

∆H = - 790.4 KJ/mol

∆S = - 166.36 KJ/mol

Explanation:

I will use the following standard values for this question:

O2 (g): ( ∆G = 0 KJ/mol, ∆H = 0 KJ/mol, ∆S = + 205.0 KJ/mol )

S (s): ( ∆G = 0 KJ/mol, ∆H = 0 KJ/mol, ∆S = + 31.88 KJ/mol )

S03 (g): ( ∆G = - 370.4 KJ/mol, ∆H = - 375.2 KJ/mol, ∆S = + 256.2 KJ/mol )

For the balanced equation:

∆G reaction = 2 x ∆GSO3 (g) – 2 x ∆G S (s) + 3 x ∆G O2 (g)

∆G reaction = 2 x ( - 370.4) – 0 + 0

∆G reaction = - 740.8 – 0

∆G reaction = - 740.8 KJ/mol

∆H reaction = 2 x ∆HSO3 (g) – 2 x ∆H (S(s) + 3 x ∆H O2 (g)

∆H reaction = 2 x ( - 395.2 ) – 0 + 0

∆H reaction = - 790.4 – 0

∆H reaction = - 790.4 KJ/mol

∆S reaction = 2 x ∆S SO3 (g) – 2 x ∆S S (s) + 3 x ∆S O2 (g)

∆S reaction = 2 x ( + 256.2 ) – 2 x ( + 31.88 ) + 3 x ( + 205 )

∆S reaction = ( 512.4 ) – ( 63.76 + 615 )

∆S reaction = ( 512.4 ) – ( 678.76 )

∆S reaction = 512.4 – 678.76

∆S reaction = - 166.36 KJ/mol

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