Answer to Question #336882 in General Chemistry for Meg

Question

Calculate the △G° at 298 K for the following reaction. Show your work.

Fe2O3 (s) + 13CO (g) = 2Fe(CO)5(g) + 3CO2(g)

-824.2 -110.5 -733.8 -393.5

△H° (kJ/mol)

87.4 197.6 445.2 213.6

△S° (J/mol x K)

Answer 1

Solution:

Balanced chemical equation:

Fe₂O₃(s) + 13CO(g) → 2Fe(CO)₅(g) + 3CO₂(g)

The standard enthalpy of reaction is then given by:

ΔH^∘_rxn = ∑νΔH^∘_f(products) − ∑νΔH^∘_f(reactants)

Therefore,

ΔH^∘_rxn = [3×ΔH^∘_f(CO₂, g) + 2×ΔH^∘_f(Fe(CO)₅, g)] − [13×ΔH^∘_f(CO, g) + ΔH^∘_f(Fe₂O₃, s)]

ΔH^∘_rxn = [3×(−393.5) + 2×(−733.8)] − [13×(−110.5) + (−824.2)] = −387.4

ΔH^∘_rxn = −387.4 kJ mol⁻¹

The standard entropy of reaction is then given by:

ΔS^∘_rxn = ∑νS^∘_f(products) − ∑νS^∘_f(reactants)

Therefore,

ΔS^∘_rxn = [3×S^∘_f(CO₂, g) + 2×S^∘_f(Fe(CO)₅, g)] − [13×S^∘_f(CO, g) + S^∘_f(Fe₂O₃, s)]

ΔS^∘_rxn = [3×213.6 + 2×445.2] − [13×197.6 + 87.4] = −1125

ΔS^∘_rxn = −1125 J mol⁻¹ K⁻¹

ΔG = ΔH − TΔS

Therefore,

ΔG^∘_rxn = (−387400 J mol⁻¹) − (298 K) × ( −1125 J mol⁻¹ K⁻¹) = −52150 J mol⁻¹ = −52.15 kJ mol⁻¹

ΔG^∘_rxn = −52.15 kJ mol⁻¹

Answer: ΔG^∘_rxn = −52.15 kJ mol⁻¹