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

On the Ellingham diagram, the C+O$_2 \rightarrow$CO$_2$ / 2CO+O$_2 \rightarrow$2CO$_2$ and 2C+O$_2\rightarrow$2CO lines cross at roughly 983 K. Below this crossover, CO oxidising to CO$_2$ is thermodynamically more favourable (lower $\Delta G^\circ$) than carbon oxidising to CO, meaning CO is the stronger/more effective reducing agent b …

Step 1 – Recall the relevant Ellingham lines

Three related lines matter here: $C + O_2 \rightarrow CO_2$ (roughly flat, $\Delta S \approx 0$), $2C + O_2 \rightarrow 2CO$ (steep negative slope, since moles of gas increase, $\Delta S > 0$), and $2CO + O_2 \rightarrow 2CO_2$ (positive slope, moles of gas decrease, $\Delta S < 0$).

Step 2 – The crossover point

The C,CO line and the CO,CO$_2$ line intersect at approximately 983 K (about 710°C).

• Below ~983 K: the CO,CO$_2$ line lies below (more negative $\Delta G^\circ$) than the C,CO line, meaning oxidation of CO to CO$_2$ is thermodynamically more favourable than oxidation of C to CO — so CO is the better reducing agent in this range.
• Above ~983 K: the lines cross and the C,CO line drops lower, so carbon becomes the better reducing agent.

Step 3 – Apply to 673 K …

Recall the ~983 K crossover of the C,CO and CO,CO$_2$ Ellingham lines: below it CO is the better reducing agent, abo …

• Assuming carbon is always the stronger reducing agent regardless of temperature — the Ellingham diagram shows this reverses below ~983 K. …