Question

Consider the following unbalanced equation. What is the standard free energy for the reaction of 7.2 moles of Al2O3(s) at 298K?Al2O3 (s) + CO (g) → Al (s) + CO2 (g)Substance ΔH°f (kJ/mol) S° (J/mol·K)Al2O3 (s) -1676.0 50.92CO (g) -110.5 197.6Al (s) 0.0 28.3CO2 (g) -393.5 213.65800 kJ810 kJ-15,000 kJ-1.1 x 105 kJ

Answer 1

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