Question

Which of the following is incorrect on the basis of the Ellingham diagram below for Carbon?

• A. Up to ${710}^{\circ }C$, the reaction of formation of $C{O}_2$ is energectically more favourable, 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. $\Delta S(C_{\left(s\right)}+\frac{1}{2}{O}_{\left(g\right)}\to C{O}_{\left(g\right)})<\Delta S(C_{\left(s\right)}+O_{2\left(g\right)}\to C{O}_{2\left(g\right)})$
• D. Carbon reduces many oxides at elevated temperature because $\Delta G^{\circ }$ vs temperature has a negative slope.

Answer 1

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

$\Delta G=\Delta H-T\Delta S$
In a plot of $\Delta G$ vs T, slope = - $\Delta S$
Higher the negative slope higher the entropy change.
Formation of $C{O}_2$ line is nearly horizontal. But the line of $CO$ formation has more negative slope, thus, it has more entropy.
In the Ellingham diaghram, below ${710}^{\circ }C$, CO formation has more negative $\Delta G$ value than $C{O}_2$ formation. Hence, $CO$ formation is preferred.

At elevated temperature, free energy formation of $CO$ is high so it will reduce many metal oxide.
Thus, option (c) is incorrect.