Consider the following reaction at $1000°\mathrm{C}$

1) ${\mathrm{Zn}}_{\left(\mathrm{s}\right)}+\frac{1}{2}{\mathrm{O}}_{2( \mathrm{g})}\to {\mathrm{ZnO}}_{\left(\mathrm{s}\right)}$;${\mathrm{\Delta G}}^0=-360 \mathrm{kJ} {\mathrm{mol}}^{-1}$

2) ${\mathrm{C}}_{\left(\mathrm{s}\right)}+\frac{1}{2}{\mathrm{O}}_{2( \mathrm{g})}\to {\mathrm{CO}}_{\left(\mathrm{g}\right)}$;${\mathrm{\Delta G}}^0=-460 \mathrm{kJ} {\mathrm{mol}}^{-1}$

Choose the correct statement

Formation of CO is more thermodynamically favored at $1000°\mathrm{C}$.

The zinc oxide can be reduced by $\mathrm{C}$ at this temperature.

In Ellingham diagram, the Δf_G of C lies below that of ZnO. If the reduction is carried out at very high temperatures, the C will act as a reductant and can reduce ZnO.

ZnO(s)→ Zn(s)+12O₂(g) ΔfG_o=−360

C(s)+12O₂(g)→ CO(s) ΔfG_o=−460
Net reaction
ZnO(s)+ C(s)→ Zn(s)+ CO ΔfGo=−100 

As ΔfG_o=−100kJ/mol , the reaction is spontaneous.