Chemistry - Metallurgy Question with Solution | TestHub

Question Explanation:

Interpret the plot of $-{\mathrm{lnK}}_{\mathrm{p}}$ vs T (equivalent to Ellingham diagram) for metal reduction and identify the correct statement.

Concept:

Ellingham Diagram ( $\Delta {\mathrm{G}}^{\circ }$ and ${\mathrm{K}}_{\mathrm{p}}$ ).

Solution:

Since $\Delta {\mathrm{G}}^{\circ }=-\mathrm{RT}\ln {\mathrm{K}}_{\mathrm{p}}$, plotting $-\ln {\mathrm{K}}_{\mathrm{p}}$ vs T is equivalent to plotting $\Delta {\mathrm{G}}^{\circ }$ vs T . The metal M is reducible by Carbon (C) when the C → CO line is below the $\mathrm{M}\to \mathrm{MO}$ line. This is equivalent to $-\ln {\mathrm{K}}_{\mathrm{p}}$ for the $\mathrm{C}\to \mathrm{CO}$ reaction being above the $\mathrm{M}\to \mathrm{MO}$ curve.

The curves intersect at 1200 K :

Above $1200\mathrm{K}(\mathrm{T}>1200\mathrm{K})$: The $\mathrm{C}\to \mathrm{CO}$ curve is higher ($-\ln {\mathrm{K}}_{\mathrm{p}}$ is more positive) than the M → MO curve. This means $\Delta {\mathrm{G}}^{\circ }{}_{\mathrm{C}\to \mathrm{CO}}$ is more negative than $\Delta {\mathrm{G}}^{\circ }{}_{\mathrm{M}\to \mathrm{MO}}$. The net reaction $\mathrm{MO}+\mathrm{C}\to \mathrm{M}+\mathrm{CO}$ has $\Delta {\mathrm{G}}^{\circ }<0$ ($\ln \mathrm{K}>0$ or $\mathrm{K}>1$). Hence, C reduces MO. (Correct)

Below $1200\mathrm{K}(\mathrm{T}<1200\mathrm{K})$: MO is more stable, and the reduction by C is unfavorable ($\Delta {\mathrm{G}}^{\circ }>0$).