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Thermodynamics differentiation?

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RoseChm05E

Chemical
Jul 21, 2006
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If we have a correction term to be added to the excess Gibbs energy of mixing for UNIQUAC as follows, how do we get to the correction of the activity coefficient?

GE,HB/RT = sum(xi[-2ln(1+ ai*F)+(aiF/(1+aiF)) - Ci,pure])

where F = sum(xj*aj/(1+aj*F))

Yes, F looks like it is circularly defined...You have to implicitly differentiate it, i.e. use the chain rule. First you have to rearrange everything in terms of y:

GE,HB/RT = (1/n)*sum(ni[-2ln(1+ ai*F)+(aiF/(1+aiF)) - Ci,pure])

F = (1/n)*sum(nj*aj/(1+aj*F)) = (n1*a1/(1+a1*F)+n2*a2/(1+a2*F))

We know that the derivative of a sum = the sum of the derivatives, and we seem to have a form similar to the Wilson equation's derivation in my text. If we multiply everything by n and take the partial derivatives,

G_E,HB/RT = n1a1/(1+a1F)-2ln(1+a1F)-C1pure+n2a2/(1+a2F)-2ln(1+a2F)-C2pure)

When we derive this, Ci goes away as a constant; a's are constants too, but go with F's and n's so stay in place.

We have two "sections" to derive: sum(ni(aiF/(1+aiF)) and -2ni*ln(1+ai*F). Let's call these quantities D and E.

(dD/dni) with constant T, P, ni =/= j = aiF/(1+aiF) etc...

At this point I've tried messing with derivates some more, but am not sure about how to eliminate F (assuming I have to.) What should I do?
 
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