M a + O 2 2 5 ∘ C 2 M a O Δ G ∘ = − 800 M b + O 2 2 5 ∘ C 2 M b O Δ G ∘ = − 600 M a + O 2 100 0 ∘ C 2 M a O Δ G ∘ = − 400 M b + O 2 100 0 ∘ C 2 M b O Δ G ∘ = − 500 C +

Question

M a ​ + O 2 ​ 2 5 ∘ C ​ 2 M a ​ O Δ G ∘ = − 800 M b ​ + O 2 ​ 2 5 ∘ C ​ 2 M b ​ O Δ G ∘ = − 600 M a ​ + O 2 ​ 100 0 ∘ C ​ 2 M a ​ O Δ G ∘ = − 400 M b ​ + O 2 ​ 100 0 ∘ C ​ 2 M b ​ O Δ G ∘ = − 500 C + O 2 ​ 71 0 ∘ C ​ C O 2 ​ Δ G ∘ = − 400 All data in kJ/mole Select correct position of metal M a and metal M b in ellingham diagram

TestHub · Accepted Answer

Based on the given data, we can determine the correct positions on the Ellingham diagram ( $Δ G^{\circ}$ vs $T$ ):

At low temperature ( $25℃$ ): The Gibbs free energy for $M_{a}$ ( $Δ G^{\circ} = - 800 k J / m o l$ ) is more negative than that for $M_{b}$ ( $Δ G^{\circ} = - 600 k J / m o l$ ), meaning the line for $M_{a}$ must start below $M_{b}$ .

At high temperature ( $1000℃$ ): $Δ G^{\circ}$ for $M_{a}$ becomes $- 400 k J / m o l$ while $M_{b}$ is $- 500 k J / m o l$ , meaning the line for $M_{b}$ becomes lower than $M_{a}$ .

Slope analysis: The line for $M_{a}$ increases more rapidly (steeper slope), crossing both $M_{b}$ and the horizontal carbon line ( $C \to C O_{2}$ , which stays at $- 400 k J / m o l$ ).

This intersection where $M_{a}$ starts below $M_{b}$ and crosses it to go above is correctly represented in Option (B).

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