The standard Gibbs free energies for the reaction at `1773K` are given below:
`C(s) +O_(2)(g) rarr CO_(2)(g), DeltaG^(Theta) =- 380 kJ mol^(-1)`
`2C(s) +O_(2)(g) hArr 2CO(g),DeltaG^(Theta) =- 500 kJ mol^(-1)`
Discuss the possibility of reducing `AI_(2)O_(3)` and `PbO` with carbon at this temperature,
`4AI +3O_(2)(g) rarr 2AI_(2)O_(3)(s),DeltaG^(Theta) =- 22500 kJ mol^(-1)`
`2Pb +O_(2)(g) rarr 2PbO(s),DeltaG^(Theta) =- 120 kJ mol^(-1)`

Explanation

`4Al+3O_(2)to2Al_(2)O_(3),..(1)`
`Delta_(r)G^(@)=-22500kJ mol^(-1)`
`3C+3O_(2)to3CO_(2),......(2)`
`Delta_(r)G^(@)=-380xx3 kJ mol^(-1)`
By eq. `(2)-(1)`
`3C+2Al_(2)O_(3)to3CO_(2)+4Al,`
`Delta_(r)G^(@)=-1140-(22500)`
`=+21360 kJ mol^(-1)`
Since, `Delta_(r)G^(@)` is positive, the reaction will not take place or `Al_(2)O_(3)` will not be reduced by carbon.
`2Pb+O_(2)to2PbO, Delta_(r)G^(@)=-120 kJ mol^(-1)......(3)`
`C+O_(2)toCO_(2), Delta_(r)G^(@)=-380 kJ mol^(-1)....(4)`
By eq. (4)-(3)
`C+2PbOtoCO_(2)+2Pb`
`Delta_(r)G^(@)=-380-(-120)=-260 kJ`
Thus, reaction will be spontaneous and `PbO` will be reduced by carbon.