Question

Calculate the ΔG°rxn using the following information. 2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g) ΔH°f(kJ/mol) -20.6 296.8 -241.8 S°(J/mol∙K) 205.8 205. 248.2 188.8 ΔG°rxn = ?

	+676.2 kJ
	+108.2 kJ
	-466.1 kJ
	+196.8 kJ
	-147.1 kJ

Answer 1

Calculate the ΔG°rxn using the following information. 2 H2S(g) + 3 O2(g) → 2 SO2(g) + 2 H2O(g) ΔH°f(kJ/mol) -20.6 296.8 -241.8 S°(J/mol∙K) 205.8 205. 248.2 188.8 ΔG°rxn = ?

Given chemical transformation is,

2 H₂S (g) + 3 O₂(g) → 2 SO₂(g) + 2 H₂O(g)

Gibbs equation,

ΔG° = ΔH° - TΔS°

1) Let us calculate ΔH° for given reaction using ΔH°f(kJ/mol) values of species involved.

Formula,

ΔH° = ∑ ΔH°f(Product) - ∑ ΔH°f(Reactant)

ΔH° = [2ΔH°f(H₂O)+ 2ΔH°f(SO₂)] – [3ΔH°f(O₂)+ 2ΔH°f(H₂S)]

ΔH° = [2(-241.8)+ 2(296.8)] – [3(0)+ 2(-20.6)]

ΔH° = (110) – (-41.2)

ΔH° = 151.2 kJ/mol

ΔH° = 151200 J/mol.

2) Let us calculate ΔS° for given reaction using S°f(J.K^-1.mol^-1) values of species involved.

Formula:

ΔS° = ∑ S°f(Product) - ∑ S°f(Reactant)

ΔS° = [2S°f(H₂O)+ 2S°f(SO₂)] – [3S°f(O₂)+ 2S°f(H₂S)]

ΔS° = [2(188.8)+ 2(248.2)] – [3(205)+ 2(205.8)]

ΔS° = (874) – (1026.6)

ΔS° = 152.6 J.K^-1.mol^-1

ΔS° = 152.6 J.K^-1.mol^-1

With the values ΔH° = 151200 J/mol, ΔS° = 152.6 J.K^-1.mol^-1 and at 298.15 K

Let us calculate ΔG° = ?

Using Gibb’s equation,

ΔG° = ΔH° - TΔS°

ΔG° = 151200 - (298.15x152.6)

ΔG° = 105702.31 J/mol

ΔG° = 105.7 kJ/mol.

Temperature was not specified hence I used T=298.15k.

Note at 273.15 K, ΔG° = 109.6 kJ/mol.

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