Ag-Si are immiscible alloys. I want to calculate partial molar enthalpies of dissolving 1 mole of Ag in Si at infinite dissolution and 1 mole of Si in Ag at infinite dissolution at different Temperatures and 1 atm pressure using Thermo Calc software .
Dear Surbhi Yadav ,
I am not sure if you want (A) the partial enthalpies of mixing at infinite dilution in the liquid, or (B) the heat from dissolving e.g. 1 mol of solid Si in a much larger amount of liquid Ag at 1200 K.
In both cases, the first step is to find a thermodynamic database. The only assessment for the Ag-Si binary I found is from 1988 (doi:10.1016/0040-6031(88)87048-5). This description was accepted in 10.1016/j.jct.2020.106172 and the database is provided as a supplementary material (although the Cu33Si7 phase causes spurious extrapolations in the Ag-Si binary so it's better to suspend it). Note that no miscibility gap is calculated in the liquid phase in the Ag-Si binary.
Please find below an answer for (A), which actually does not require Thermo-Calc:
The enthalpy of mixing is defined as:
(1) Hmix = H – (xAgHAg + xSiHSi)
A subregular solution model is used for the Ag-Si liquid phase. It follows from the CALPHAD formalism that:
(2) Hmix = xAgxSi(0a + 1a(xAg-xSi)) with 0a = -1963.3 J/mol and 1a = +11052 J/mol
By definition, the partial molar enthalpy of mixing of Ag is :
(3) partialHmix Ag = Hmix + (1 - xAg) dHmix / dxAg
Starting from (2) and using xSi = 1 - xAg will lead to:
(4) partialHmix Ag = 0a(1 - 2xAg + xAg2) + 1a(-1 + 6xAg - 9xAg2 + 4xAg3)
At infinite dillution, xAg = 0, and you finally have:
(5) partialHmix Ag inf dill = 0a - 1a
The same approach for Si leads to :
(6) partialHmix Si inf dill = 0a + 1a
So, the partial enthalpy of mixing at infinite dillution is temperature independent; for Ag in liquid Si it is -13.0 kJ/mol, and for Si in liquid Ag it's +9.1 kJ/mol.
Please note that I don't know about the reliability of the database that I used...
Kind regards,
Guillaume
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