Question

Calculate the standard Gibbs free energy change ($\Delta G^{\circ}$) for the following reaction at 298 K:2Fe(s) + 3/2O2(g) → Fe2O3(s) Given: Standard enthalpy of formation of Fe2O3(s) = -822.2 kJ/molStandard enthalpy of formation of Fe(s) = 0 kJ/molStandard entropy change for the reaction = +87.4 J/mol K

Answer 1

To calculate the standard Gibbs free energy change  G  for the reaction, we can use the following equation:G = H - TSwhere H is the standard enthalpy change, T is the temperature in Kelvin, and S is the standard entropy change.First, we need to find the standard enthalpy change  H  for the reaction. We can do this using the given standard enthalpies of formation:H reaction  = H products  - H reactants For this reaction, we have:H reaction  = [1  H Fe2O3 ] - [2  H Fe  + 3/2  H O2 ]Since the standard enthalpy of formation of Fe s  and O2 g  are both 0 kJ/mol, the equation simplifies to:H reaction  = [1   -822.2 kJ/mol ] - [0 + 0]H reaction  = -822.2 kJ/molNow we can calculate the standard Gibbs free energy change  G  using the equation:G = H - TSG =  -822.2 kJ/mol  -  298 K  87.4 J/mol K First, we need to convert the entropy change from J/mol K to kJ/mol K:87.4 J/mol K   1 kJ / 1000 J  = 0.0874 kJ/mol KNow we can plug this value into the equation:G =  -822.2 kJ/mol  -  298 K  0.0874 kJ/mol K G = -822.2 kJ/mol - 26.0 kJ/molG = -848.2 kJ/molThe standard Gibbs free energy change  G  for the reaction is -848.2 kJ/mol.