Answer:
5800 kJ
Explanation:
First, let's balance the equation. Both sides must have the same amount of each element, so:
Al₂O₃(s) + 3CO(g) → 2Al(s) + 3CO₂(g)
The free energy can be calculated by:
ΔG = ΔH - TΔS
Where ΔH is the enthalpy of the reaction, and ΔS is the entropy of the reaction.
Al₂O₃(s): Hf = -1676.0 kJ/mol; S° = 50.92 J/mol.K
CO(g): Hf = -110.5 kJ/mol; S° = 197.6 J/mol.K
Al(s): Hf = 0.00 ; S° = 28.3 J/mol.K
CO₂(g): Hf = -393.5 kJ/mol; S° = 213.6 J/mol.K
ΔH = ∑n*Hf products - ∑n*Hf reactants (n is the coefficient of the compound).
ΔH = (3*(-393.5) + 2*0) - (3*(-110.5) + (-1676)) = 827 kJ
ΔS = ∑n*S° products - ∑n*S° reactants
ΔS = (3*213.6 + 2*28.3) - (3*197.6 + 50.92) = 53.68 J/K = 0.05368 kJ/K
ΔG = 827 - 298*0.05368
ΔG = 811 kJ
Which is the free energy for 1 mol of Al₂O₃
1 mol of Al₂O₃ ----------------------- 811 kJ
7.2 moles of Al₂O₃ ---------------------- x
By a simple direct three rule:
x = 5839.2 kJ ≅ 5800 kJ
