Calculate the change in entropy and enthalpy for the reaction of iron (III) oxide with carbon monoxide to form iron and carbon dioxide, given the following information:Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)ΔH°f (Fe2O3) = -824.2 kJ/molΔH°f (CO) = -110.5 kJ/molΔH°f (Fe) = 0 kJ/molΔH°f (CO2) = -393.5 kJ/molΔS° (Fe2O3) = 87.4 J/K molΔS° (CO) = 197.9 J/K molΔS° (Fe) = 27.3 J/K molΔS° (CO2) = 213.6 J/K mol

Question

Calculate the change in entropy and enthalpy for the reaction of iron (III) oxide with carbon monoxide to form iron and carbon dioxide, given the following information:Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)ΔH°f (Fe2O3) = -824.2 kJ/molΔH°f (CO) = -110.5 kJ/molΔH°f (Fe) = 0 kJ/molΔH°f (CO2) = -

Answer 1

To calculate the change in enthalpy  H  for the reaction, we can use the following equation:H =  Hf  products  -  Hf  reactants H = [2Hf  Fe  + 3Hf  CO2 ] - [Hf  Fe2O3  + 3Hf  CO ]H = [2 0  + 3 -393.5 ] - [ -824.2  + 3 -110.5 ]H = [-1180.5] - [-1155.7]H = -24.8 kJ/molNow, to calculate the change in entropy  S  for the reaction, we can use the following equation:S =  S  products  -  S  reactants S = [2S  Fe  + 3S  CO2 ] - [S  Fe2O3  + 3S  CO ]S = [2 27.3  + 3 213.6 ] - [ 87.4  + 3 197.9 ]S = [54.6 + 640.8] - [87.4 + 593.7]S = 695.4 - 681.1S = 14.3 J/K molSo, the change in enthalpy  H  for the reaction is -24.8 kJ/mol, and the change in entropy  S  is 14.3 J/K mol.