An Ellingham diagram is a plot of
Aentropy change $\Delta S^\circ$ against temperature $T$
B$\Delta G^\circ$ against reciprocal temperature $1/T$
C$\Delta G^\circ$ of oxide formation against temperature $T$
Dequilibrium constant $K_p$ against temperature $T$
Answer & Solution
Correct answer: C. $\Delta G^\circ$ of oxide formation against temperature $T$
1. Ellingham used $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$ for oxide formation.
2. Plotting $\Delta G^\circ$ (y-axis) against **temperature T** (x-axis) gives a straight line of slope $-\Delta S^\circ$ and intercept $\Delta H^\circ$.
3. Hence an Ellingham diagram is $\Delta G^\circ$ vs T.
_Source: Samacheer Kalvi Class 12 Chemistry Vol.1, Unit 1 'Metallurgy', §1.4.1 / Evaluation Q.22_
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