Thermodynamics

 

 50) (a) ΔH° = -844 kJ          ΔS° = -165 J/K

 

             2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g)

 

             ΔG° = ΔH° - TΔS°

             ΔG° = -844 kJ - 298 K x -165 J/K x 1 kJ/103 J = -795 kJ

             The reaction will be spontaneous at this temperature because the reactants and products are in their standard states and ΔG°rxn < 0.

 

       (b) ΔH° = 572 kJ          ΔS° = 179 J/K

 

             2POCl3(g) 2PCl3(g) + O2(g)

 

            ΔG° = ΔH° - TΔS°

            ΔG° = 572 kJ - 298 K x 179 J/K x 1 kJ/103 J = +519 kJ

            The reaction will be nonspontaneous at this temperature because ΔG°rxn < 0.

 

            For a reaction to be spontaneous, ΔG° < 0.

            ΔH° - TΔS° < 0

            T > ΔH°/ΔS° > 572 kJ/(179 J/K x 1 kJ/103J) > 3.20 x 103 K