Δf G- vs T plot in the Ellingham diagram slopes downward for the reaction

Question

Δf G- vs T plot in the Ellingham diagram slopes downward for the reaction (A) M g+(1/2) O2 arrow M g O (B) 2 Ag +(1/2) O 2 arrow Ag 2 O (C) C +(1/2) O 2 arrow CO (D) CO +(1/2) O 2 arrow CO 2

Tardigrade Admin · Accepted Answer

C(s)+(1/2) O2(g) arrow C O(g) ; Δ S increases. Hence, as the temperature increases, T Δ S increases and hence Δ G(Δ H-T Δ S) decreases. In other words, the slope of the curve for formation of CO decreases. However, for all other oxides, it increases.

Q. $Δ_{f} G^{-}$ vs $T$ plot in the Ellingham diagram slopes downward for the reaction

$M g + \frac{1}{2} O_{2} \to M g O$

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

$C + \frac{1}{2} O_{2} \to CO$

$CO + \frac{1}{2} O_{2} \to C O_{2}$

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

Q. Δf​G− vs T plot in the Ellingham diagram slopes downward for the reaction

Mg+21​O2​→MgO

2Ag+21​O2​→Ag2​O

C+21​O2​→CO

CO+21​O2​→CO2​