text: Ellingham diagrams are a particular graphical form of the principle that the thermodynamic feasibility of a reaction depends on the sign of ΔG, the Gibbs free energy change, which is equal to ΔH − TΔS, where ΔH is the enthalpy change and ΔS is the entropy change.
  The Ellingham diagram plots the Gibbs free energy change (ΔG) for each oxidation reaction as a function of temperature. For comparison of different reactions, all values of ΔG refer to the reaction of the same quantity of oxygen, chosen as one mole O ( mol ) by some authors and one mole  by others. The diagram shown refers to 1 mole , so that e.g. the line for the oxidation of chromium shows ΔG for the reaction  Cr(s) + (g) →  (s), which is  of the molar Gibbs energy of formation ΔG°(, s).
  In the temperature ranges commonly used, the metal and the oxide are in a condensed state (solid or liquid), and oxygen is a gas with a much larger molar entropy. For the oxidation of each metal, the dominant contribution to the entropy change (ΔS) is the removal of  mol , so that ΔS is negative and roughly equal for all metals. The slope of the plots  is therefore positive for all metals, with ΔG always becoming more negative with lower temperature, and the lines for all the metal oxides are approximately parallel. Since these reactions are exothermic, they always become feasible at lower temperatures. At a sufficiently high temperature, the sign of ΔG may invert (becoming positive) and the oxide can spontaneously reduce to the metal, as shown for Ag and Cu.
  For oxidation of carbon, the red line is for the formation of CO: C(s) +  (g) → CO(g) with an increase in the number of moles of gas, leading to a positive ΔS and a negative slope. The blue line for the formation of  is approximately horizontal, since the reaction C(s) + (g) → (g) leaves the number of moles of gas unchanged so that ΔS is small.
  As with any chemical reaction prediction based on purely thermodynamic grounds, a spontaneous reaction may be very slow if one or more stages in the reaction pathway have very high activation energies E.
  If two metals are present, two equilibria have to be considered. The oxide with the more negative ΔG will be formed and the other oxide will be reduced.