An Ellingham diagram is a plot of change in standard free energy with respect to temperature for various reactions like the formation of oxides oxides,, sulfides etc. of various elements. They were first constructed by Harold Ellingham in 1944.[1] 1944. [1] In metallurgy metallurgy,, the Ellingham diagram is used to predict the equilibrium temperature between a metal metal,, its oxide and oxygen oxygen,, and by extension, reactions of a metal with sulphur sulphur,, nitrogen and other non-metals non-metals.. The diagrams are useful in attempting to predict the conditions under which a metal ore will be reduced to the metal.
Salient features 1. Curves in the Ellingham diagrams diagrams for the formation of metallic metallic oxides are straight lines with a positive slope. 2. The lower the position of a metal in the Ellingham diagram diagram more is the stability of its oxide. For example, the Ellingham diagram for Al is found to be below Fe 2O3. 3. A metal found in the Ellingham diagram can act as a reducing agent for a metallic oxide found above it. Thus, Al can act as a reducing agent to Cr which is above it in the diagram. 4. The greater the gap between any two lines, the greater the efficiency of the reducing agent. 5. Stability of metallic oxides oxides decrease with increase in temperature. Highly unstable unstable oxides like Ag2O and HgO easily undergo thermal decomposition. 6. The formation formation enthalpy enthalpy of carbon dioxide (CO2) is almost a temperature-independent constant, while that of carbon monoxide (CO) has negative slope. According to Boudouard reaction, reaction , carbon monoxide is the dominant compound in higher temperatures, and the higher the temperature, the more efficient reductant carbon monoxide also is. 7. A substance whose formation formation enthalpy 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. (This reaction is employed in thermite thermite.) .) 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.
Ellingham diagrams follow from the Second Law of Thermodynamics [ΔG = ΔH - TΔS] and are a particular graphical form of it. ΔG is the Gibbs Free Energy Change,ΔH is the Enthalpy Change and ΔS is the Entropy Change] The Ellingham diagram plots the Gibbs free energy change (ΔG) for the oxidation reaction versus the temperature. In the temperature ranges commonly used, the metal and the oxide are in a condensed state (liquid or solid) with the oxygen gaseous, the reactions may be exothermic or endothermic, but the ΔG of the oxidation always becomes more negative with lower temperature, and thus the reaction becomes more probable statistically. At a sufficiently high temperature, the sign of ΔG may invert (becoming negative) and the oxide can spontaneously reduce to the metal. As with any chemical reaction prediction based on purely energetic grounds the reaction may or may not take place spontaneously on kinetic grounds if one or more stages in the reaction pathway have very high Activation Energies E A.
If two metals are present, two equilibriums have to be considered, so that the metal with the more negative ΔG reduces, the other oxidizes. Ellingham diagrams follow from the Second Law of Thermodynamics [ΔG = ΔH - TΔS] and are a particular graphical form of it. ΔG is the Gibbs Free Energy Change,ΔH is the Enthalpy Change and ΔS is the Entropy Change] The Ellingham diagram plots the Gibbs free energy change (ΔG) for the oxidation reaction versus the temperature. In the temperature ranges commonly used, the metal and the oxide are in a condensed state (liquid or solid) with the oxygen gaseous, the reactions may be exothermic or endothermic, but the ΔG of the oxidation always becomes more negative with lower temperature, and thus the reaction becomes more probable statistically. At a sufficiently high temperature, the sign of ΔG may invert (becoming negative) and the oxide can spontaneously reduce to the metal. As with any chemical reaction prediction based on purely energetic grounds the reaction may or may not take place spontaneously on kinetic grounds if one or more stages in the reaction pathway have very high Activation Energies E A. If two metals are present, two equilibriums have to be considered, so that the metal with the more negative ΔG reduces, the other oxidizes. An Ellingham diagram is a plot of ∆G versus temperature. Since ∆H and ∆S are essentially constant with temperature unless a phase change occurs, the free energy versus temperature plot can be drawn as a series of straight lines, where ∆S is the slope and ∆H is the y-intercept. The slope of the line changes when any of the materials involved melt or vaporize. Free energy of formation is negative for most metal oxides, and so the diagram is drawn with ∆G=0 at the top of the diagram, and the values of ∆G shown are all negative numbers. Temperatures where either the metal or oxide melt or vaporize are marked on the diagram. The Ellingham diagram shown is for metals reacting to form oxides (similar diagrams can also be drawn for metals reacting with sulfur, chlorine, etc., but the oxide form of the diagram is most common). The oxygen partial pressure is taken as 1 atmosphere, and all of the reactions are normalized to consume one mole of O2. The majority of the lines slope upwards, because both the metal and the oxide are present as condensed phases (solid or liquid). The reactions are therefore reacting a gas with a condensed phase to make another condensed phase, which reduces the entropy. A notable exception to this is the oxidation of solid carbon. The line for the reaction C+O2 ==> CO2 is a solid reacting with a mole of gas to produce a mole of gas, and so there is little change in entropy and the line is nearly horizontal. For the reaction 2C+O2 ==> 2CO we have a solid reacting with a gas to produce two moles of gas, and so there is a substantial increase in entropy and the line slopes rather sharply downward. Similar behavior can be seen in
parts of the lines for lead and lithium, both of which have oxides that boil at slightly lower temperatures than the metal does. There are three main uses of the Ellingham diagram: 1. Determine the relative ease of reducing a given metallic oxide to metal; 2. Determine the partial pressure of oxygen that is in equilibrium with a metal oxide at a given temperature; and 3. Determine the ratio of carbon monoxide to carbon dioxide that will be able to reduce the oxide to metal at a given temperature. The main application of Ellingham diagrams is indeed in the metallurgy industry, where it helps to select the best reducing agent for various ores in the extraction process. [edit] Reducing Agent for haematite During smelting of haematite ore in the blast furnace reduction takes place at the top of the furnace, where temperature is in the range of 600 oC to 700oC. From Ellingham diagram we notice that in this range carbon monoxide acts as a reducing agent since the process 2 CO + O 2 --> 2CO2 has a lower negative value of free energy change than the process 2C + O 2 --> 2CO. Thus in the blast furnace haematite is reduced by CO even though carbon is mixed with it. Fe 2O3 + 3CO --> 2Fe + 3CO 2 [edit] Reducing Agent for chromic oxide-carbon cannot be used At high temperature the Ellingham curve for the reaction 2C(s) + O 2(g) --> 2CO(g) slopes down and falls below the curves for all the metals. Hence, carbon can normally act as a reducing agent for all metal oxides at very high temperatures. But chromium formed at these temperatures react with carbon to form its carbide which gives undesirable properties to chromium metal obtained. Hence, for high temperature reduction of chromic oxide, carbon cannot be used. [edit] Alumino thermic process The Ellingham curve for Aluminum always lies below the curves of metals like Cr, Fe, etc. So Aluminum can be used as the reducing agent for oxides of all these metals. This fact can be illustrated as below: The free energies of formation of chromium oxide and aluminum oxide per mole of oxygen consumed are -540kJ and -827kJ respectively. The processes are:
•
•
(1)
(2)
So Aluminum oxide is more stable than chromium oxide. Equation (2) - Equation (1) gives
ΔG0 = − 287kJ Since Gibb's free energy is negative, aluminium acts as the reducing agent. In pyrometallurgy, Al is used as a reducing agent in alumino-thermic process to apps We shall consider the following reactions on the Ellingham diagram below: the oxidation of silver to form Ag2O (s); and the oxidation of cobalt to form CoO (s) and Co3O4: 4Ag (s) + O2 = 2Ag2O (s); 2Co + O2 (g) = 2CoO (s) 3Co + 2O2 (g) = Co3O4 (s).
The graph consists of lines of
Although ΔS and ΔH vary with temperature (see here to see why), the changes are so small as to be negligible, and so the lines are approximately straight. This means that they are of the form . We can immediately see that the standard free energy change is greater (more negative) for the cobalt reaction relative to that of silver at all temperatures. This means that at all temperatures the equilibrium constant is larger for the cobalt reaction-the composition is further weighted towards the products of the reaction. This is the reason that metals that appear higher up on the diagram are more stable than those metals
that appear lower down, and are more likely to be found in their pure solid form. The gradient of the two lines is approximately the same. We can see that the gradient of the lines is simply the standard entropy change for the reactions; . This is evident from the reactions, which both involve the elimination of one mole of gas - a large decrease in entropy. This is the reason for the positive slope of the lines. The reason for the change in slope is the change in phase of a component of the system, which alters the entropy change. As the standard free energy change for both reactions is still negative (below 460K) the large decrease in entropy must be counteracted by a large enthalpy of reaction. This is indeed the case. The intercept of the lines with 0K gives the enthalpy of the reaction: . We can hence see that the relative stability of the oxide of cobalt compared with the oxide of silver is due to the much larger standard enthalpy of reaction.
