GATE 2026 MT – Question 56
With reference to the Ellingham diagram, the standard Gibbs free-energy changes for oxidation of solid and liquid metal M are:
Reaction I: $\mathrm{M(s)+O_2(g)\to MO_2(s)}$, with $\Delta G^\circ=-338900-15.2T\ln T+247T$ J, from 300 K to the melting point.
Reaction II: $\mathrm{M(l)+O_2(g)\to MO_2(s)}$, with $\Delta G^\circ=-390800-15.2T\ln T+285.3T$ J, from the melting point to 1800 K.
The melting point of M is ________ K. (Rounded off to one decimal place.)
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Correct answer: 1353.9 to 1356.3
Explanation
At melting, solid and liquid metal have equal chemical potentials, so the two oxidation free energies coincide. The common $-15.2T\ln T$ terms cancel, leaving $-338900+247T=-390800+285.3T$.
Thus $38.3T=51900$ and $T=51900/38.3=1355.091\ldots\ \mathrm K$. Enter **1355.1 K**.