Question

$\Delta_{f} G^{-}$ vs $T$ plot in the Ellingham diagram slopes downward for the reaction

• A. $M g+\frac{1}{2} O_{2} \rightarrow M g O$
• B. $2 Ag +\frac{1}{2} O _{2} \rightarrow Ag _{2} O$
• C. $C +\frac{1}{2} O _{2} \rightarrow CO$
• D. $CO +\frac{1}{2} O _{2} \rightarrow CO _{2}$

Answer 1

Answer: (C)

$C(s)+\frac{1}{2} O_{2}(g) \rightarrow C O(g) ; \Delta S$ increases.
Hence, as the temperature increases, $T \Delta S$ increases and hence $\Delta G(\Delta H-T \Delta S)$ decreases. In other words, the slope of the curve for formation of $CO$ decreases. However, for all other oxides, it increases.