Free energies of formation `(Delta_(f) G^(Ө))` of `MgO(s)` and `CO_((g))` at `1273 K` and `2273 K` are given below :
`Delta_(f) G^(Ө) (MgO_((s) )) = -941 kJ//mol at 1273 K`
`Delta_(f) G^(Ө) (MeO_((s))) = -314 kJ//mol at 2273 K`
`Delta_(f) G^(Ө) (CO_((g))) = - 439 kJ//mol at at 1273 K`
`Delta_(f) G^(Ө) (CO_((g))) = -628 kJ//mol at 2273 K`
On the basis of above data, predict the temperature at which carbon can be used as a reducing for agent `MgO_((s))`.

Explanation

(a) At `1273 K`
`Mg_((s)) + (1)/(2) O_(2(g)) rarr MgO_((s)) , Delta _(f) G^(Ө) = -941 kJ//mol^-1` ....(i)
`C_((s)) + (1)/(2) O_(2(g)) rarr 4CO_((g)) , Delta _(f) G^(Ө) = -439 kJ//mol^-1` ....(ii)
The redox equation for equation of `MgO` to `Mg` by `C` can be obtained by subtracting Eq. (i) from Eq. (ii).
Thus, `MgO_((s)) + C_((s)) rarr Mg_((s)) + CO_((g))`
Since `Delta_(f) G^(Ө)` of the above reduction reaction is `+ ve`, reduction of `MgO` by `C` is not feasible at `1273 K`.
(b) At `2273 K`
`Mg_((s)) +(1)/(2) O_(2(g)) rarr MgO_((s)) , Delta_(f) G^(Ө) = -314 kJ//mol^(-1)` ...(iii)
`C_((s)) + (1)/(2) O_(2(g)) rarr CO_((g)) , Delta_(f) G^(Ө) = -628 kJ//mol^-1` ...(iv)
Substracting Eq. (iii) from Eq. (iv), the redox equation is
`MgO_((s)) + C_((s)) rarr Mg_((s)) + CO_((g))`
and `Delta_(r) G^(Ө) CO_((g)) - Delta _(f) G^(Ө) MgO_((s))`
=`(-628) - (-314) = -314 kJ mol^(-1)`
Since `Delta_(r) G^(Ө)` for the above reduction reaction is `- ve`, reduction of `MgO` by carbon at `2273 K` is feasible.