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Topic: Troubles with dG = dH - TdS (Ellingham diagrams)  (Read 3384 times)

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Offline steve627

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Troubles with dG = dH - TdS (Ellingham diagrams)
« on: July 04, 2012, 09:29:01 AM »
For those who don't know what an Ellingham diagram is, it's essentially a graph that plots dG as a function of T. The intercept is dH and the slope is dS. Anyway, as I understand it, the lines on the diagram are straight if there's no state change and that dH and dS are assumed to be independent of temperature. Whenever a kink occurs, a state change would've occured.

For example, consider the reaction 4 Cu + O2 --> 2 Cu2O. The line has 3 portions, where the two kinks are the melting points of Cu and Cu2O. The way I understand it is that the first portion is for Cu(s) and Cu2O(s), the second portion is Cu(l) and Cu2O(s), and the third portion is Cu(l) and Cu2O(l).
My question is, what is the appropriate way to construct a curve if I need an equation for non-standard states (ie. not solid)? Given that the available enthalpy and entropy data is for standard states only, I can only draw the dG function only for a solid to solid reaction (ie. the first portion of the line). I'm mostly interested in the liquid states for the metals and oxides.

For dH, I can simply take the enthalpies of formation of the species involved. In the above example, dH = 2 Hf [Cu2O(l)] - Hf [O2(g)] - 4 Hf [Cu(l)]
Then Hf [Cu2O(l)] = Hf [Cu2O(s)] + Hfus(298 K) = Hf [Cu2O(s)] + Cp,solid * (Tmp - 298 K) + Hfus (Tmp) + Cp,solid * (298 K - Tmp) , same for the other two.

I'm not sure what to do with dS as the numbers are a bit trickier to find.

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