Question

Calculate the standard change in entropy for the reaction below using the given standard enthalpy and temperature information:              3Fe(s) + 4H2O(g) → Fe3O4(s) + 4H2(g)      ΔH° = -770.2 kJ     T = 298 K

Answer 1

To calculate the standard change in entropy  S  for the reaction, we can use the Gibbs free energy equation:G = H - TSFirst, we need to find the standard Gibbs free energy change  G  for the reaction. We can do this using the standard free energy of formation  Gf  values for each compound involved in the reaction. The Gf values are:Fe s : 0 kJ/mol  since it's an element in its standard state H2O g : -228.6 kJ/molFe3O4 s : -1015.1 kJ/molH2 g : 0 kJ/mol  since it's an element in its standard state Now, we can calculate G for the reaction:G = [Gf Fe3O4  + 4Gf H2 ] - [3Gf Fe  + 4Gf H2O ]G = [ -1015.1  + 4 0 ] - [3 0  + 4 -228.6 ]G = -1015.1 -  -914.4 G = -100.7 kJNow that we have G, we can use the Gibbs free energy equation to find S:G = H - TSRearrange the equation to solve for S:S =  H - G  / TPlug in the values:S =  -770.2 kJ -  -100.7 kJ   / 298 KS =  -669.5 kJ  / 298 KConvert kJ to J:S =  -669.5 * 1000 J  / 298 KS = -669500 J / 298 KCalculate S:S = -2246.98 J/KThe standard change in entropy  S  for the reaction is -2246.98 J/K.