Out of C and CO, which is a better reducing agent at 673K?

Explanation

When carbon reacts with dioxygen two reactions are possible.
The CO thus produced can also react with `O_2` to form `CO_2`
`2CO_(g)+O_2to2CO_2` …(iii)
In the first reaction the volume of `CO_2` produced is the same as that of `O_2` used hence `triangleS` is very small and `triangle_(f)G^(ɵ)` does not change with temperature. Thus the graph of `triangle_(f)G^(ɵ)` against T is almost horizontal.
The second reaction produces two volumes of CO for every one volume of `O_2` used. Thus `triangleS` is positive and hence `triangle_(f)G^(ɵ)` becomes increasingly negative as the temperature increases. Consequently, C,CO line on the ellingham diagram slopes downwards . The third reaction produces two volumes of `CO_2` for every three volumes of reactants used. Thus `triangleS` is -ve and hence `triangle_(f)G^(ɵ)` becomes incresingly `+ve` as the temperature increases consequently, `CO,CO_2` line on the ellingham diagram slopes upwards The three lines cross at 673K. Below this temperature formation of CO from C is energetically more favourable. In ther words, below 673K, both C and CO can act as reducing agents but since CO can be more easily oxidised to `CO_2` than C to `CO_2`, therefore, below 673K. CO is a more effective reducing agent than C. However, above 673K, CO is more stable and hence its oxidation to `CO_2` is less rapid than that of C to `CO_2` therefore, above 673K C is better reducing agent than CO.