ΔG from enthalpies of formation and entropies

Worked example

For 2SO₂(g) + O₂(g) → 2SO₃(g), use ΔfH(SO₂) = −296.8 kJ mol⁻¹, ΔfH(SO₃) = −395.7 kJ mol⁻¹, S(SO₂) = 248.2 J K⁻¹ mol⁻¹, S(O₂) = 205.2 J K⁻¹ mol⁻¹ and S(SO₃) = 256.8 J K⁻¹ mol⁻¹ to calculate ΔG at 881 K.

1. 1. Weighted sum of 2·ΔfH(SO₃) − 2·ΔfH(SO₂) → ΔH

   = −197.8 kJ mol⁻¹

2. 2. Weighted sum of 2·S(SO₃) − 2·S(SO₂) − 1·S(O₂) → ΔS

   = −188.0 J K⁻¹ mol⁻¹

3. 3. Gibbs energy from ΔH and ΔS: solve for ΔG

   $\Delta G = \Delta H - T\Delta S$

   = −32.17 kJ mol⁻¹

Answer: ΔG = −32.2 kJ mol⁻¹

Method

ΔH = 2ΔfH(SO₃) − 2ΔfH(SO₂). ΔS = 2S(SO₃) − 2S(SO₂) − S(O₂). Then ΔG = ΔH − TΔS, with ΔS converted to kJ K⁻¹ mol⁻¹. ΔS is negative, so ΔG rises with temperature and becomes positive above about 1050 K.

Common mistakes

• O₂ has ΔfH = 0 because it is an element, but its entropy is not zero: include S(O₂) in ΔS.
• The equation has 2SO₂ and 2SO₃: multiply their values by 2.

Formulas used

• Gibbs energy from ΔH and ΔS

  $\Delta G = \Delta H - T\Delta S$

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