reactions involving pure condensed phases and a...

Reactions Involving Pure Condensed Phases and a Gas Phase (Chap. 12)

In the last section we looked at reactions between gases– Control of the atmosphere is important for materials

engineering» Oxidation» Sulphidation» Reduction

In this section we will now look at reactions with pure liquids and solids with the gases

Ellingham diagrams are the basis for understanding the behaviour of:– Metals and oxides– Metals and sulphides– Other systems

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Equilibria between Pure Condensed Phases and Gas (Chap. 12.2)

Consider the equilibrium between a metal M and its oxide, MO, both solid in this case

The metal and oxide will be in equilibrium with their vapors, and again we are using this as an indicator of activity in the solids

Therefore, a gas phase equilibrium (just like in the last section of the course) is established:

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ΔG˚ means that this is the free energy change for reactants and products in their standard state (1 atm at the temperature of interest)

The gaseous reactants and products are not generally in their standard state at equilibrium– The metals and oxides are in equilibrium with their saturated

vapour pressures– The vapour pressures of metals and oxides are low– PO2 is low as we will see later when we look at data

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The next step is to relate the free energy of the vapours to the free energy of the solid phases, here just the metal

Metal and oxide molar volumes are insensitive to pressure in most situations, so the integral term is negligible

Now we make a subtle, but important, change in the standard state of the condensed phase, it is now just the pure species at the temperature of interest, and we do not have to specify 1 atm pressure (insensitive to pressure)

Standard state of O2 is still 1 atm at temperature of interest4


Returning to the free energy equilibrium condition for the gases:

We can substitute in our expressions for the difference in free energy of the vapour and the solid

This is the standard free energy change of the reaction

Metal and oxide are in their standard states (pure at temperature) (their activities are unity) and O2 at 1 atm at temperature

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Example of reduction of cuprous oxide; how low does PO2 need to be to reduce it

PO2 = 2.38 x 10-50 at 25˚C PO2 = 2.06 x 10-15 at 500˚C PO2 = 8.35 x 10-7 at 1000˚C PO2 = 0.60 at 2000˚C

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Decomposition of carbonates and hydroxides– Limestone CaCO3 decomposed or “calcined” to lime, CaO for

cement and slag forming components– Aluminum hydroxide Al(OH)3 calcined to Al2O3 in the

production of aluminum

Example of MgCO3

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Rotary kiln for calcination of aluminum hydroxide, similar to those used for limestone calcination in the cement industry

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Ellingham Diagrams (Ch. 12.4)

In most cases the free energy of formation of the oxides is a linear function of temperature

A is ΔH˚ and B is -ΔS˚ Convention is that the reactions are all based on 1 mole of

O2 on the left hand side, so the reactions are written as oxidation

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Example of silver oxidation

PO2 = 1 atm at 462 K, above this, oxide unstable

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Slope is positive – because ΔS˚ is negative– because there is a large decrease in volume – 1 mole of O2 on the left, none on the right– Change in the volume of metal and oxide are relatively small

Consider the oxidation of A to AO

The entropy of the gas is much greater than the condensed phases, so

The slopes of the curves in the Ellingham diagrams are all about the same, due to “loss” of 1 mole of gas

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Ellingham Diagrams (Ch. 12.4) ΔH˚ does not change very much with temperature (unless

there is a phase change (Chap. 12.5)) Consider Co and Mn: MnO much more stable than CoO due

to ΔH˚ for MnO is more negative

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In some cases lines on the Ellingham diagram can intersect because the slopes are slightly different, e.g. NiO & CoO

AO more stable than BO2 at high temperature

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Nomograph for PO2 is very convenient– Lines on ΔG˚ vs. T for various PO2

– Change in ΔG˚ going from 1 atm at T to PO2 at T

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Very convenient superimpose on Ellingham diagram to read off PO2 at the intersection

Just the same as calculating from:

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Nomographic scale is placed around the outside of the diagram

Convenient for using a ruler

Note you must start at 0 K and 0 kJ for these lines

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Effect of Phase Transformations (Chap. 12.5)

Phase changes result in elbows, consider melting of A

New ΔH˚ is more negative because melting endothermic

New ΔS˚ is also more negative, slope more positive

Note should be AO2(s) in figure

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Consider case of oxide melting:

Add reactions New ΔH˚ is more positive

(smaller)

New ΔS˚ is also smaller, so slope smaller

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Shape depends on which melts at lower temperature

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“Elbows” are larger for vaporization because of much larger changes in enthalpy and entropy

Add the 2 reactions

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KJSJH

FeClFeCl

ob

ob

gl

/46.97

500,126)(2)(2

+=∆

=∆

=

TTG

FeClClFeo

gs

78.33000,160)46.9768.63(600,126400,286

)(22)(

−−=−++−=∆

=+

Sulphides

In kcal

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Fluorides

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Chlorides (I)

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Chlorides (II)

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Bromides &Iodides

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Nitrides

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reactions involving pure condensed phases and a...

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