Given the Data:
PbO(s) $\Delta$ H°_f = -217 $\Delta$ H°_f = -277 $\Delta$

Given the data:
PbO(s) , $\Delta$ H°_f = -217.3 kJ mol^-1, S° = +68.7 J mol^-1 K^?1
PbO₂(s) , $\Delta$ H°_f = -277.0 kJ mol^-1, S° = +68.6 J mol^-1K^-1
O₂(g) , $\Delta$ H°_f = 0.00 kJ mol^-1, S° = +205 J mol^-1 K^-1
Calculate the standard free energy change, ?G°, for the reaction:PbO(s) + ½ O₂(g) PbO₂(s)

A) +1.45 kJ mol^-1
B) -29.1 kJ mol^-1
C) -68.3 kJ mol^-1
D) -90.3 kJ mol^-1
E) +29.1 kJ mol^-1