Calculate the standard enthalpy change (∆H°) and standard entropy change (∆S°) for the reaction between solid aluminum and gaseous oxygen to form solid aluminum oxide, given the following balanced chemical equation:4Al(s) + 3O2(g) → 2Al2O3(s) The standard enthalpies of formation of Al(s), O2(g), and Al2O3(s) are -394.4 kJ/mol, 0 kJ/mol, and -1675.7 kJ/mol, respectively. The standard entropy change for the reaction of O2(g) to form Al2O3(s) is -159.6 J/K·mol.

Question

Calculate the standard enthalpy change (∆H°) and standard entropy change (∆S°) for the reaction between solid aluminum and gaseous oxygen to form solid aluminum oxide, given the following balanced chemical equation:4Al(s) + 3O2(g) → 2Al2O3(s) The standard enthalpies of formation of Al(s), O2(g), and Al2O3(s) are -394.4 kJ/mol, 0 kJ/mol, and -1675.7 kJ/mol, respectively. The standard entropy change for th

Answer 1

To calculate the standard enthalpy change  H  for the reaction, we can use the following formula:H =  [Hf products ] -  [Hf reactants ]For the given reaction:4Al s  + 3O2 g   2Al2O3 s H = [2   -1675.7 kJ/mol ] - [4   -394.4 kJ/mol  + 3   0 kJ/mol ]H =  -3351.4 kJ  -  -1577.6 kJ H = -1773.8 kJThe standard enthalpy change for the reaction is -1773.8 kJ.Now, let's calculate the standard entropy change  S  for the reaction. The given information states that the standard entropy change for the reaction of O2 g  to form Al2O3 s  is -159.6 J/Kmol. Since there are 3 moles of O2 g  reacting to form 2 moles of Al2O3 s , we can calculate the total standard entropy change for the reaction as follows:S =  2/3    -159.6 J/Kmol S = -106.4 J/KmolThe standard entropy change for the reaction is -106.4 J/Kmol.