Question

Using the standard values of enthalpy changes (ΔH°) and entropy change (ΔS°) to calculate the Gibbs free energy change for the production of following metallic elements from their ore sources: (a) 2ZnO(s) 2Zn(s) + O2(g) (b) 2CaO(s) 2Ca(s) + O2(g) (c) 2Al2O3(s) 4Al(s) + 3O2(g) (d) 2MgO(s) 2Mg(s) + O2(g)

Answer 1

(a) 2ZnO(s) ------------>2Zn(s) + O2(g)

ΔH°rxn   = ΔH°f products - ΔH°f reactants

              = (2*0 + 0)- (2*-348.28)

               = 696.56KJ/mole

ΔS°rxn   = S°f products - S°f reactants

               = (2*41.63+ 205) - (2*43.64)

                = 200.98J/mole-K    = 0.20098KJ/mole-K

G⁰      = H⁰ - TS

                = 696.56-298*0.20098    = 636.67KJ/mole

(b) 2CaO(s) ------->2Ca(s) + O2(g)

ΔH°rxn   = ΔH°f products - ΔH°f reactants

              = (2*0 + 0)- (2*-635.09)

              = 1270.18KJ/mole

ΔS°rxn   = S°f products - S°f reactants

               = (2*41.6+ 205) - (2*39.75)

                = 208.7J/mole-K    = 0.2087KJ/mole-K

G⁰      = H⁰ - TS

                = 1270.18-298*0.2087    = 1207.98KJ/mole

(c) 2Al2O3(s) ------->4Al(s) + 3O2(g)

ΔH°rxn   = ΔH°f products - ΔH°f reactants

              = (3*0 + 4*0)- (2*-1675.7)

              = 3351.4KJ/mole

ΔS°rxn   = S°f products - S°f reactants

               = (3*205+4*28.33) - (2*50.92)

                = 626.5J/mole-K    = 0.6265KJ/mole-K

G⁰      = H⁰ - TS

                = 3351.4-298*0.6265     = 3164.7KJ/mole

(d) 2MgO(s) ------>2Mg(s) + O2(g)

ΔH°rxn   = ΔH°f products - ΔH°f reactants

              = (2*0 + 0)- (2*-601.7)

              = 1203.4KJ/mole

ΔS°rxn   = S°f products - S°f reactants

               = (2*32.68+205) - (2*26.94)

                = 216.5J/mole-K    = 0.2165KJ/mole-K

G⁰      = H⁰ - TS

                = 1203.5-298*0.2165     = 1138.98KJ/mole

(b) 2CaO(s) --------->2Ca(s) + O2(g)

(c) 2Al2O3(s)------> 4Al(s) + 3O2(g)

(d) 2MgO(s)-------> 2Mg(s) + O2(g)