Ellingham Diagram

ELLINGHAM DIAGRAM

NOTES ON THERMODYNAMIC PRINCIPLES OF METALLURGY:

Thermodynamic principles of Metallurgy:

Gibbs energy change, G for any process at a specified temperature helps us to understand the theory of metallurgical transformations and is expressed by the equation:

STHG where, H is the enthalpy change and S is the entropy change for the process at temperature T.

(When G = -ve, reaction is spontaneous, when G =+ve, reaction is non-spontaneous)

A reaction having +ve value of G can be made spontaneous by combining it with a reaction having a highly –ve value of G . This is called ‘coupling of reactions’.

Such coupling is easily understood through G o vs T plots for formation of oxides.

i.e: M(s) + O2 (g) MO(s)

This is known as Ellingham diagram.

The Ellingham diagram helps you to choose the right reducing agent to convert an oxide (ore) to metal. In other words it helps us to predict the feasibility of thermal reduction of an ore.

Understanding Ellingham diagram….

Consider the following reaction: M(s) + O2 (g) MO(s) H = -ve

1. In the above equation S is –ve since gas is consumed and solid is formed. Hence ST becomes +ve. But since H generally is a large –ve value, G is –ve and hence reaction is spontaneous. 2. But if we keep on increasing the temperature, ST term tends to a +ve value and thus the G value. 3. Thus the plot of G o vs T results in to a positive slope in case of most of the reactions of formation of oxides as below.

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4. If you observe the plot carefully, you will find that each plot is a straight line except a few like Mg-MgO , Zn-ZnO. There is an abrupt increase in G in these plots at a particular temperature. This point (temperature) in the plot refers to the melting point of the oxide. Since at m.p., solid changes to liquid, S increases abruptly leading to sudden increase in G value. 5. With the increase in temperature, G value moves towards +ve side (upwards). A point will arrive when the G value crosses zero and it becomes +ve. This point onwards the forward reaction (formation of MO) becomes non- spontaneous which means, the reverse reaction (formation of metal) becomes spontaneous above that temperature. 6. This means that if we heat the MO, beyond a particular temperature, M will be formed. But some highly reactive metals require very high temperature (impractical) to be formed and hence in such cases coupling with suitable reducing agents is carried out. 7. Converting the ore to metal simply by heating is only possible in the cases of Ag/ Ag 2O, Au/ Au2O, Hg/ HgO. In these cases MO converts to M by just heating (since these are less reactive). Heat 2Ag2O 4Ag + O2

Heat 2HgO 2Hg + O2

Limitations of Ellingham diagram (ED)….

1. ED talks about only feasibility of the reaction (since it is based on thermodynamics) and not rate of reaction (since it ignores kinetics) 2. In all reactions that are considered, the reactants and products are presumed to be in a state on equilibrium ( G o RT ln K ). Thus ED has only limited scope or application.

Choosing a suitable reducing agent (RA)….

1. Metal oxides as RA: Highly reactive metals (having high –ve G value for M MO) can act as RAs for those having

less –ve G value. In general, “the M involved in the formation of oxide placed lower in the ED can

reduce the MO of the M placed higher in the diagram”.

Eg: G (Cr-Cr2O3) is less –ve than G (Al-Al2O3)

Cr + O2 Cr2O3 G -820 …………………(1)

Al + O2 Al2O3 G -980 ……………….(2)

Since, Al-Al2O3 is place below Cr-Cr2O3, Al2O3 is more stable than Cr2O3 at a particular temperature.

Hence Al can reduce Cr2O3,. This can be done by reversing the equation (1) and adding to equation (2) as

follows:

Cr2O3 Cr + O2 G +820 kJ/mol

Al + O2 Al2O3 G -980 kJ/mol

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Cr2O3 + Al Cr + Al2O3 G -160 kJ/mol ……….spontaneous reaction.

2. Coke as RA: When C acts as RA, three reactions are possible

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C(s) + O2(g) CO2(g) ……………(i)

S 0, G = no change formation of CO2(g) not favored

2C(s) + O2(g) 2CO(g) ……………(ii)

S =+ve, G = becomes increasingly –ve formation of CO(g) is strongly favored

2CO(g) + O2(g) 2CO2(g) ………….(iii)

S = -ve , G = +ve formation of CO2(g) is not favored

Thus we infer that the reaction (ii) is the most favorable for conversion of MO to M.

Eg: ZnO(s) + C(s) Heat Zn(s) + CO(g)

Heat Fe2O3(s) + C(s) Fe(s) + CO(g)

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