Question

Which of the following is incorrect on the basis of the above Ellingham diagram for carbon?

• A. Up to ${710}^{\circ }C$, the reaction of formation of $C{O}_2$ is energetically more favorable, but above ${710}^{\circ }C$, the formation of $CO$ is preferred.
• B. In principle, carbon can be used to reduce any metal oxide at a sufficiently high temperature.
• C. $\Delta S(C_{\left(s\right)}+\frac{1}{2}{O}_{2\left(g\right)}\to C{O}_{\left(g\right)})<\Delta S(C_{\left(S\right)}+O_{2\left(g\right)}\to C{O}_{2\left(g\right)})$
• D. Carbon reduces many oxides at elevated temperature because $\Delta G^{\circ }$ vs temperature line has a negative slope.

Answer 1

The correct option is C $\Delta S(C_{\left(s\right)}+\frac{1}{2}{O}_{2\left(g\right)}\to C{O}_{\left(g\right)})<\Delta S(C_{\left(S\right)}+O_{2\left(g\right)}\to C{O}_{2\left(g\right)})$

The line corresponding to the formation of $C{O}_2$ is nearly horizontal.
Till ${710}^{\circ }C$, $\Delta G_f^{\circ }$ for $C{O}_2$ is more negative.
But after ${710}^{\circ }C$, $\Delta G_f^{\circ }$ for $CO$ is more negative
and thus the formation of $CO$ is thermodynamically favoured.

Don't forget - the reactions are all for the formation of one mole of the oxide.
In the formation of the monoxide, 0.5 mol of a $O_2$ yields 1 mol of $CO$.
Contrast this with the formation of $C{O}_2$:
One mole of $O_2$ yields one mol of $C{O}_2$. Hence the answer.