Q.Which of the following is not true with respect to Ellingham diagram?

Selecting a reducing agent for a metal oxide MxOy → xM + (y/2)O2 is a thermodynamic problem: this reduction is generally not spontaneous by itself, so it must be COUPLED to the oxidation of a chosen reducing agent (carbon to CO2 or CO, or CO to CO2) such that the combined (coupled) reaction has an overall NEGATIVE ΔG -- the reducing agent is chosen specifically to provide a sufficiently large negative ΔG for its own oxidation.

The Ellingham diagram is the tool built to compare these possibilities visually. Using ΔG = ΔH - TΔS and ΔG0 = -RT ln Kp, Ellingham calculated ΔG0 for the FORMATION of many metal oxides across a range of temperatures and plotted temperature (x-axis) against ΔG0 of formation (y-axis); each metal-oxide system gives a near-straight line with ΔS as slope and ΔH as intercept.

Four key readings: (1) most oxide-formation lines have a POSITIVE slope, because forming the oxide consumes gaseous O2, decreasing randomness (ΔS negative, so -TΔS grows positive with T); (2) the CO-formation line, 2C+O2=2CO, uniquely has a NEGATIVE slope, since 2 mol of gaseous CO forms from 1 mol of gaseous O2 (ΔS positive), so CO grows MORE stable at higher T; (3) as T rises, a typical oxide's ΔG becomes less negative and eventually crosses zero -- above that crossover the oxide is unstable and decomposes; (4) a few oxides (MgO, HgO) show a sudden slope change at a phase-transition temperature. …