In an Ellingham diagram the line for the formation of carbon monoxide ($2C + O_2 ightarrow 2CO$) has a negative slope. This indicates that

1. Slope of an Ellingham line is $-\Delta S^\circ$. 2. Forming 2 mol of CO gas from 1 mol $O_2$ increases the number of gas moles, so $\Delta S^\circ$ is positive and the slope $-\Delta S^\circ$ is negative. 3. A downward line means $\Delta G^\circ$ keeps decreasing with T, so **CO grows more stable at higher temperature** — this is why carbon is an excellent reducing agent at high T. _Source: Samacheer Kalvi Class 12 Chemistry Vol.1, Unit 1 'Metallurgy', §1.4.1 (observations) / Evaluation Q.23_