$\text{ Free energies of formation }\left({\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{-}\right)\text{of }\mathrm{MgO}\left(\mathrm{s}\right)\text{ and }{\mathrm{CO}}_{\left(\mathrm{g}\right)}\text{ at 1273 K and 2273 K are given below:}$

$\begin{array}{l}{\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{\circ }\left({\mathrm{MgO}}_{\left(\mathrm{s}\right)}\right)=-941\mathrm{kJ}/\mathrm{mol}\text{ at }1273\mathrm{K}\\ {\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{\odot }\left({\mathrm{MgO}}_{\left(\mathrm{s}\right)}\right)=-314\mathrm{kJ}/\mathrm{mol}\text{ at }2273\mathrm{K}\\ {\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{\odot }\left({\mathrm{CO}}_{\left(\mathrm{g}\right)}\right)=-439\mathrm{kJ}/\mathrm{mol}\text{ at }1273\mathrm{K}\\ {\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{\odot }\left({\mathrm{CO}}_{\left(\mathrm{g}\right)}\right)=-628\mathrm{kJ}/\mathrm{mol}\text{ at }2273\mathrm{K}\end{array}$

On the basis of above data, The temperature at which carbon be used as a reducing agent for MgO_(s) is _______K

$\begin{array}{r}{\mathrm{Mg}}_{\left(\mathrm{s}\right)}+\frac{1}{2}{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to {\mathrm{MgO}}_{\left(\mathrm{s}\right)};{\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{-}=-941\mathrm{kJ}/{\mathrm{mol}}^{-1} . . . \left(1\right)\\ {\mathrm{C}}_{\left(\mathrm{s}\right)}+\frac{1}{2}{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to 4{\mathrm{CO}}_{\left(\mathrm{g}\right)};{\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{-}=-439\mathrm{kJ}/{\mathrm{mol}}^{-1} . . . \left(2\right)\end{array}$

The redox equation for reduction of MgO to Mg using C can
be obtained by subtracting Eq. (i) from Eq. (ii).

$\text{ Thus, }{\mathrm{MgO}}_{\left(\mathrm{s}\right)}+{\mathrm{C}}_{\left(\mathrm{s}\right)}\to {\mathrm{Mg}}_{\left(\mathrm{s}\right)}+{\mathrm{CO}}_{\left(\mathrm{g}\right)}$

Since ${\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{-}$of the above reduction ls *ve, reduction of MgO by C is not feasible at 1273 K.

b. At 2273 K

$\begin{array}{l}{\mathrm{Mg}}_{\left(\mathrm{s}\right)}+\frac{1}{2}{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to {\mathrm{MgO}}_{\left(\mathrm{s}\right)};{\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{-}=-314\mathrm{kJ}/{\mathrm{mol}}^{-1} . . . \left(3\right)\\ {\mathrm{C}}_{\left(\mathrm{s}\right)}+\frac{1}{2}{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to {\mathrm{CO}}_{\left(\mathrm{g}\right)};{\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{-}=-628\mathrm{kJ}/{\mathrm{mol}}^{-1} . . . .\left(4\right)\end{array}$

Substracting Eq. (iii) from Eq. (iv), the redox equation is

$\begin{array}{l}{\mathrm{MgO}}_{\left(\mathrm{s}\right)}+{\mathrm{C}}_{\left(\mathrm{s}\right)}\to {\mathrm{Mg}}_{\left(\mathrm{s}\right)}+{\mathrm{CO}}_{\left(\mathrm{g}\right)}\\ \text{ and }{\mathrm{\Delta }}_{\mathrm{r}}{\mathrm{G}}^{-}={\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{-}\text{(products) }-{\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{-}\text{(reactants) }\\ ={\mathrm{\Delta HG}}^{-}{\mathrm{CO}}_{\left(\mathrm{g}\right)}-{\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{-}{\mathrm{MgO}}_{\left(\mathrm{s}\right)}\\ =(-628)-(-314)=-314\mathrm{kJ}{\mathrm{mol}}^{-1}\end{array}$

Since ${\mathrm{\Delta }}_{\mathrm{f}}{\mathrm{G}}^{-}$for the above reduction is -ve, reduction of MgO
by carbon at 2273 K is feasible.