2 E ( s ) + O 2 ( g ) → 2 EO ( g ) Δ H o = − ive Select a possible graph for the change in Δ G ∘ against temperature in an Ellingham diagram. ( Δ G ∘ = − v e )

Question

2 E ( s ) ​ + O 2 ( g ) ​ → 2 EO ( g ) ​ Δ H o = − ive Select a possible graph for the change in Δ G ∘ against temperature in an Ellingham diagram. ( Δ G ∘ = − v e )

TestHub · Accepted Answer

Thermodynamic Relation: The change in Gibbs free energy is given by $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$ . In an Ellingham diagram ( $Δ G^{\circ}$ vs $T$ ), the slope of the line represents $- Δ S^{\circ}$ .

Entropy Change ( $Δ S^{\circ}$ ): In the reaction $2 E_{(s)} + O_{2 (g)} \to 2 E O_{(g)}$ , there is $1$ mole of gas on the reactant side and $2$ moles of gas on the product side. This results in an increase in entropy ( $Δ S^{\circ} = p os i t i v e$ ).

Graph Slope: Since $Δ S^{\circ}$ is positive, the slope ( $- Δ S^{\circ}$ ) must be negative, meaning the line should slope downwards as temperature increases.

Intercept: Given $Δ H^{\circ}$ is negative, the $y$ -intercept ( $Δ G^{\circ}$ at $T = 0$ ) starts below zero, consistent with graph (B).

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