Elingham diagrame and its application ,isolation chapter

ELINGHAM     DIAGRAME    AND    ITS    APPLICATIONS;-

Ellingham Diagram -Thermodynamics of Metallurgy

1.Gibbs equation helps us to predict the spontaneity of a reaction on the basis of enthalpy and entropy values directly.

2. Ellingham proposed the Ellingham diagram to predict the spontaneity of reduction of various metal oxides. Ellingham diagram was basically a curve which related the Gibbs energy value with the temperature. Gibbs energy is given as:    ΔG = ΔH – TΔS

Where ΔH is the change in enthalpy and ΔS is the change in entropy.

With respect to a reaction, Gibbs energy can be related to the equilibrium constant as:

ΔG^o= – RTlnK

Where K is the equilibrium constant.

3. when the reaction is exothermic, enthalpy of the system is negative, thus making Gibbs free energy negative. Hence, we can say that the reaction will proceed in the forward direction due to a positive value of the equilibrium constant. This law can be scaled for two different reactions taking place in a system too.  The overall reaction (combination of two reactions) will occur if and only if net ΔG (sum of ΔG’s of both the reactions) of the two possible reactions is negative.

Ellingham diagram for Reduction of Oxides:

1.   Ellingham diagram is a plot between Δ_fG^o and T for the formation of oxides of metals. A general reaction expressing oxidation is given by:

2xM(s) + O₂(g) → 2M_xO(s)

As is evident from the reaction, the gaseous amount of reactant is decreasing from left to right as the product formed is solid metal oxide on the right side. Hence, we can say that molecular randomness is also decreasing form left to right. Thus, ΔS is negative and ΔG shifts towards higher side despite rising T. Hence, for most of the reactions shown above for the formation of M_xO (s), the curve is positive.

2.   Except for the processes in which change of phase takes place, each plot is a straight line. This temperature at which change of phase takes place is indicated by a positive increase in the slope. For example, the melting is indicated by an abrupt change in the curve in Zn, ZnO plot.

3.   The metal oxide (M_xO) is stable at the point in a curve below which ΔG is negative. Above this point, the metal oxide is unstable and decomposes on its own.

4.   Feasibility of reductions of the oxide of the upper line by the element represented by the lower line is determined by the difference in the two Δ_rG⁰values after the point of intersection in the Ellingham diagram.

5.   In this reaction, the entropy (or) randomness decreases from left to right due to the consumption of gases. Hence, ∆S becomes negative. If the temperature is raised, then T∆S becomes more negative. So, ∆G becomes less negative.

6.   (ii) The Gibbs energy changes follow a straight line, unless the materials melt (or) vaporise. The temperature at which such a change occurs is indicated by an increase in the slope on the positive side.

7.   (iii) When the temperature is raised, a point will be reached where the graph crosses the line "∆G is zero." Below this temperature, the free energy of formation of the oxide is negative, so the oxide is stable. Above this temperature, the free energy of formation of the oxide is positive the oxide becomes unstable and will decompose into the metal and dioxygen.

8.   Any metal will reduce an oxide of another metal that lies above it in an Ellingham diagram. Ex: Al reduces FeO, CrO and NiO in termite reaction but Al will not reduce MgO at a temperature below 1500 ⁰C.

Limitations:

1.The reactants and products are in equilibrium, which is not often true.

2.It does not explained about the rate of the reaction.

Ease of Reduction

The position of the line for a given reaction on the Ellingham diagram shows the stability of the oxide as a function of temperature. Reactions closer to the top of the diagram are the most “noble” metals (for example, gold and platinum), and their oxides are unstable and easily reduced. As we move down toward the bottom of the diagram, the metals become progressively more reactive and their oxides become harder to reduce. A given metal can reduce the oxides of all other metals whose lines lie above theirs on the diagram. For example,

The 2Mg + O₂ ----à 2MgO line lies below the

Ti + O₂ ------à TiO₂   line, and so magnesium can reduce titanium oxide to metallic titanium.

Since the 2C + O₂-------à 2CO line is downward-sloping, it cuts across the lines for many of the other metals. This makes carbon unusually useful as a reducing agent, because as soon as the carbon oxidation line goes below a metal oxidation line, the carbon can then reduce the metal oxide to metal. So, for example, solid carbon can reduce chromium oxide once the temperature exceeds approximately 1225°C, and can even reduce highly-stable compounds like silicon dioxide and titanium dioxide at temperatures above about 1620°C and 1650°C, respectively. For less stable oxides, carbon monoxide is often an adequate reducing agent.

Equilibrium Partial Pressure of Oxygen

 The scale on the right side of the diagram labelled “Po₂” is used to determine what partial pressure of oxygen will be in equilibrium with the metal and metal oxide at a given temperature. The significance of this is that, if the oxygen partial pressure is higher than the equilibrium value, the metal will be oxidized, and if it is lower than the equilibrium value then the oxide will be reduced. To use this scale, you will need a straightedge. First, find the temperature you are interested in, and find the point where the oxidation line of interest crosses that temperature. Then, line up the straightedge with both that point, and with the point labelled “0” that is marked with short radiating lines (upper left corner of the diagram). Now, with the straightedge running through these two points, read off the oxygen partial pressure (in atmospheres) where the straightedge crosses the “Po₂” scale, and this is the equilibrium partial pressure. It is possible to reach the equilibrium oxygen partial pressure by use of a hard vacuum, purging with an inert gas to displace the oxygen, or using a scavenger chemical to consume the oxygen.

Ratio of CO/CO2 Needed for Reduction

When using carbon as a reducing agent, there will be a minimum ratio of CO to CO₂ that will be able to reduce a given oxide. The harder the oxide is to reduce, the greater the proportion of CO needed in the gases. To determine the CO/CO₂ ratio to reduce a metal oxide at a particular temperature, use the same procedure as for determining the equilibrium pressure of oxygen, except line up the straightedge with the point marked “C” (center of the left side of the diagram), and read the ratio off of the scale marked “CO/CO₂”.