for the following unbalanced equation.5800 kj is the standard free energy for the reaction of 7. 2 moles of al2o3(s) at 298k
First, let's balance the equation. Both sides must have the same amount of each element, so: Al₂O(s) + 3CO(g) 2Al(s) + 3CO2(g)
The free energy can be calculated by: AG = AH-TAS Where AH is the enthalpy of the reaction, and AS is the entropy of the reaction.
Al2O3(s): Hf=-1676.0 kJ/mol; S = 50.92 J/mol.K Al(s): Hf = 0.00; S° = 28.3 J/mol.K
CO(g): Hf=-110.5 kJ/mol; S = 197.6 J/mol.K
CO2(g): Hf=-393.5 kJ/mol; S = 213.6 J/mol.K
AH = Σn*Hf products - En*Hf reactants (n is the coefficient of the compound).
AH = (3*(-393.5) + 2*0) - (3*(-110.5) + (-1676)) = 827KJ
AS = {n*S* products - Σn*S° reactants
AS = (3*213.6 +2*28.3) - (3*197.6 + 50.92) = 53.68J/K = 0.05368 kJ/K
AG 827-298*0.05368 AG = 811 KJ
Which is the free energy for 1 mol of Al2O3
1 mol of Al2O3=811 KJ X 7.2 moles of Al2O3
By a simple direct three rule: standard free energy is x = 5839.2 kJ = 5800 kJ
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