Calculate the standard enthalpy change of the reaction below at 298K.Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)Given the following standard enthalpies of formation (ΔHf°):ΔHf°[Fe2O3(s)] = -824.2 kJ/molΔHf°[Fe(s)] = 0 kJ/molΔHf°[CO(g)] = -110.5 kJ/molΔHf°[CO2(g)] = -393.5 kJ/mol. Also, calculate the change in standard entropy of the system (ΔS°) and determine whether the reaction is spontaneous or not.

Question

Calculate the standard enthalpy change of the reaction below at 298K.Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)Given the following standard enthalpies of formation (ΔHf°):ΔHf°[Fe2O3(s)] = -824.2 kJ/molΔHf°[Fe(s)] = 0 kJ/molΔHf°[CO(g)] = -110.5 kJ/molΔHf°[CO2(g)] = -393.5 kJ/mol. Also, calculate the change in standard entropy of the system (ΔS°) and determine wh

Answer 1

To calculate the standard enthalpy change  H  of 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] - [-824.2 - 331.5]H = -1180.5 -  -1155.7 H = -24.8 kJ/molThe standard enthalpy change of the reaction is -24.8 kJ/mol.To determine whether the reaction is spontaneous or not, we need to calculate the change in standard Gibbs free energy  G  using the equation:G = H - TSHowever, we do not have the values for the change in standard entropy  S  of the system. If you can provide the entropy values for the reactants and products, we can calculate S and then determine whether the reaction is spontaneous or not.