Δ G° versus T plot in the Ellingham's diagram slopes downward for the reaction

Question

Δ G° versus T plot in the Ellingham's diagram slopes downward for the reaction (A) C +(1/2) O 2 longrightarrow CO (B) CO +(1/2) O 2 longrightarrow CO 2 (C) 2 Ag +(1/2) O 2 longrightarrow Ag 2 O (D) Mg +(1/2) O 2 longrightarrow MgO

Tardigrade Admin · Accepted Answer

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

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

$C + \frac{1}{2} O_{2} ⟶ CO$

$CO + \frac{1}{2} O_{2} ⟶ C O_{2}$

$2 A g + \frac{1}{2} O_{2} ⟶ A g_{2} O$

$M g + \frac{1}{2} O_{2} ⟶ M g O$

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

Q. ΔG∘ versus T plot in the Ellingham's diagram slopes downward for the reaction

C+21​O2​⟶CO

CO+21​O2​⟶CO2​

2Ag+21​O2​⟶Ag2​O

Mg+21​O2​⟶MgO