This can be explained thermodynamically, taking entropy and free energy changes into account.
(a) \(\mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(g)\)
(b) \(2 \mathrm{C}(s)+\mathrm{O}_2(g) \longrightarrow 2 \mathrm{CO}(\mathrm{g})\)
Case ( \(i\) : Volume of \(\mathrm{CO}_2\) produced \(=\) Volume of \(\mathrm{O}_2\) used. \(\therefore \Delta \mathrm{S}\) is very small and \(\Delta \mathrm{G}\) does not change with temperature.
\(\therefore\) Plot of \(\triangle \mathrm{G} \mathrm{Vs} \mathrm{T}\) is almost horizontal.
Case (ii): Volume of \(\mathrm{CO}\) produced \(=2 \times\) Volume of \(_2\) used. \(\therefore \Delta \mathrm{S}\) is positive and hence \(\Delta \mathrm{G}\) becomes increasingly negative as the temperature increases.
\(\therefore\) Plot of \(\Delta^{\circ} \mathrm{G} \mathrm{Vs} \mathrm{T}\) slopes downwards.


As can be seen from \(\Delta \mathrm{G}^{\circ} \mathrm{Vs}\) T plot (Ellingham diagram), lines for the reactions, \(\mathrm{C} \longrightarrow \mathrm{CO}_2\) and \(\mathrm{C} \longrightarrow \mathrm{CO}\) cross at \(983 \mathrm{~K}\). Below \(983 \mathrm{~K}\), the reaction \((a)\) is energetically more favourable but above \(673 \mathrm{~K}\), reaction \((b)\) is favourable and preferred. Thus, below \(673 \mathrm{~K}\) both \(\mathrm{C}\) and \(\mathrm{CO}\) can act as a reducing agent but since \(\mathrm{CO}\) can be more easily oxidised to \(\mathrm{CO}_2\) than \(\mathrm{C}\) to \(\mathrm{CO}_2\), therefore, below \(673 \mathrm{~K}, \mathrm{CO}\) is more effective reducing agent than carbon.