Question

$\Delta G^{\circ}$ versus $T$ plot in the Ellingham's diagram slopes downward for the reaction

• A. $C +\frac{1}{2} O _{2} \longrightarrow CO$
• B. $CO +\frac{1}{2} O _{2} \longrightarrow CO _{2}$
• C. $2 Ag +\frac{1}{2} O _{2} \longrightarrow Ag _{2} O$
• D. $Mg +\frac{1}{2} O _{2} \longrightarrow MgO$

Answer 1

Answer: (A)

$C +\frac{1}{2} O _{2} \longrightarrow CO$

Answer 2

In the reaction, $C +\frac{1}{2} O _{2} \longrightarrow CO$, the $\Delta S$ increases. Therefore as the temperature increases, $T \Delta S$ increases and thus $\Delta G(\Delta H-T \Delta S)$ decreases. In other words, the slope of the curve for the formation of $CO$ decreases, whereas for all other oxides, it increases.