In the Ellingham Diagram, the slope of the Metals/MetalOxides become more positive after Boiling Point, and thus dG increases,

But we know after boiling point, the metal(oxide) becomes liquid/gas and thus its dS values becomes more positive, as dS and T are also  increasing as a result T.dS should also increase. As a result from the equation-

dG = dH - TdS 

dG should become more negative, it means the value of dG should decrease and the slope of the line should go downwards, but why is it going upward in the Ellingham diagram?


I tried finding this on net, in many books, but didn't find the answer...can anyone/any teacher help me please!?


Ellingham diagram compares the ability of reducing agents to extract a particular metal from its oxide. The lines in Ellingham's diagram signifies about the Gibb's free energy change for the reaction of the formation of oxide. A reaction which has more positive Gibb's energy, will not be spontaneous.

As we have considered the reaction

M + O2 MO

if the reaction is non spontaneous, it means that MO will decompose to M and O. Actually we have considered the reverse reaction. we are concerned with the reduction of metal fro metal oxide, we assume the reverse reaction to find the same. Mainly we want to know that MO + X M + XO 

Considering the reverse reaction M + O2 MO

we see that Oxygen gas has more entropy than solid metal oxide, Thus entropy decreases and hence Gibb's free energy change is becoming more positive , with increasing temperature. We are not concerned with the boiling or melting of any oxide or metal. Increasing temperature does not always mean that the substance is boiling or melting.

The lines in diagram considers the free energy formation of the various metal oxides. The line or metal having , most positive Gibb's energy change will be unstable and hence it will prefer to stay as M and O2. Thus the other metal (X)can easily reduce that metal oxide(MO) to metal(M) and oxygen.

For e.g. by looking into the diagram , we can see that aluminium can reduce the zinc oxide to zinc metal as the free energy of formation of Al2O3 is more negative(it wants to stay as aluminium oxide than aluminium) and that of ZnO is less and thus it will prefer to stay as Zn and oxygen.

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I'm sorry! I had typed it with space but this website isn't allowing me to post it the same way
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Hi! Your question is very good one and even I had it a while ago. I read your question and noticed that you missed a point dG = dH - TdS Now T. dS as you said will increase but it will become more negative. This is because dS is more negative as one side you have liquid metal and gaseous oxide and the other side you have solid metal oxide dS = S2 - S1 Here S1 > S2 as entropy is more in reactant side Therefore dS will be negative. dG = dH + TdS So now dG will be more positive and hence slope is positive. It may seem confusing but think about it and I'm sure you'll get it. 😊
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