Iron(iii) oxide can be reduced by carbon monoxide: fe2o3 (s) + 3 co (g) ⇌ 2 fe (s) + 3 co2 (g) use the following thermodynamic data at 298 k to determine the equilibrium constant at this temperature. substancefe2o3 (s)co (g)fe (s)co2 (g) ∆hfº(kj /mol)–824.2–110.50 –393.5∆gfº(kj/mol)–742.2 –137.20 –394.4sº (j/mol•k)87.4 197.727.78213.7

k = 1,423x10⁵

Explanation:

For the reaction:

Fe₂O₃(s) + 3CO (g) ⇌ 2Fe(s) + 3CO₂(g)

You can obtain equilibrium constant, k, using:

ΔG° = -RT lnK (1)

ΔG° of the reaction is:

ΔGf° products - ΔGf° reactants, that means:

ΔG° = 2∆Gfº Fe(s) + 3∆Gfº CO₂(g) - (∆Gfº Fe₂O₃(s) + 3∆Gfº CO (g)

ΔG° = 2×0 + 3×-394,4kJ/mol - (-742,2kJ/mol + 3×-137,2kJ/mol)

ΔG° = -29,4 kJ/mol

Replacing in (1) knowing R = 8,314472x10⁻³ kJ/molK; T = 298K

-29,4 kJ/mol = -8,314472x10⁻³ kJ/molK 298K lnK

11,866 = lnK

1,423x10⁵ = k

I hope it helps!