Can u plz explain me ellingham diagram?Nd how we get to know abt the feasibility??

Ellingham diagrams:

  • Ellingham plotted the experimentally determined standard  free energy of

formation, ∆G0, of various oxides using one mole of oxygen with temperature.

(2x/y) M(s) + O2(g) ------> (2/y)  MxOy(s) ;   ∆G0 =∆H0-T∆S0

  • Ellingham pointed out that, the standard enthalpy and entropy of formation of a compound do not change significantly with the temperature as long as there is no change of state of product or reactant.   ∆G0T=∆H0289-T∆S0298
  • Thus, the general forms of ∆G0–T ∆S0relationships could be approximated to straight lines over temperature ranges

  ∆G0 = c + mx  ; c =∆H0289  m=-∆S0298

The slope of free energy vs temperature plot would thus be (-∆S0) entropy change of the reaction and the intercept is the enthalpy value.

  • It provides a sound idea about selecting a reducing agent in the reduction of oxides. Such diagrams helps in predicting the feasibility of a thermal reduction of an ore. ∆G must be negative at a given temperature for a reaction to be feasible.


Characteristics of Ellingham diagrams:

  1. 2xM(s) + O2(g) ------> 2MxO(s)

Metal  Oxygen  Metal oxide

  In this reaction ∆S=-ve because solid and gaseous reactants are changing to metal oxide

  which is solid, hence entropy decreases. ∆G will be +ve  if temperature is increased     and  T∆S>∆H. This results in +ve slope in the curve for the most of reactions shown in the diagram for the  formation of metallic oxides.

  1. Each plot is a straight line when solid changes to liquid or liquid changes to gas i.e., change in phase occurs. The temperature at which change occurs is indicated by increase in slope on +ve side. The abrupt increase in Zn to ZnO plot shows the melting point of Zn metal.
  2. There is a point in the curve below which ∆G is –ve .All metals with negative value of ΔG may be oxidized spontaneously so metal oxide is stable. Above this point the oxides with positive ΔG are not stable and are easily decomposed to elemental metals.The lower the position of a metal in the Ellingham diagram more is the stability of its oxide. For example, the Ellingham diagram for Al is found to be below Fe2O3.
  3. The value of fGo for the formation of oxides at different temperatures helps in predicting the temperature and reducing agent suitable for particular oxide.
  4. A metal found in the Ellingham diagram can act as a reducing agent for a metallic oxide found above it i.e., A given metal can reduce the oxides of other metals whose lines lie above theirs on the diagram. Thus, Al can act as a reducing agent to Cr which is above it in the diagram.
  5. A substance whose formation enthalpy is lower (ΔG line lower on diagram) at given temperature, will reduce one whose formation enthalpy is higher on the diagram. Hence metallic aluminum can reduce iron from iron oxide into metallic iron, aluminum itself oxidizing into aluminum oxide.
  6. The greater the gap between any two lines, the greater the efficiency of the reducing agent.
  7. Stability of metallic oxides decrease with increase in temperature. Highly unstable oxides like Ag2O and HgO easily undergo thermal decomposition.
  8. The intersection of two lines imply the equilibrium of oxidation and reduction reaction between two substances. Reduction with using a certain reductant is possible at the intersection point and higher temperatures where the ΔG line of the reductant is lower on diagram than the metallic oxide to be reduced. At the point of intersection the Gibbs energy is 0(zero), below this point the Gibbs energy is <0 and the oxides are stable,while above the point of intersection the Gibbs energy is >0 and so, the oxides are unstable.
  9. The formation enthalpy of carbon dioxide (CO2) is almost a temperature-independent constant, while that of carbon monoxide (CO) has negative slope. The line for CO intersects the lines of most of the metals. . Carbon monoxide is the dominant compound in higher temperatures, and the higher the temperature, the more efficient reductant carbon monoxide also is. This explains why carbon is considered as an energetic reducing agent at high temperature.


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by taking enery along Y-axis and temperature along X-axis,when a graph is plotted,we find curves of oxidation of metals......the line which is below a a better reducing agent.
as the line crosses the zero-line...its marks the unstable condition of the element.Eg.Mercury,Silver.

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