Unformatted text preview: pressions above are
NOT molar quantities, i.e., we need to keep in mind that GA = nAGm,A, G°A =
nAG°m,A, etc.. Now, the free energy change for the reaction may be written as
(P / P ) n C (P / P ) n D
, r G = , r G + RT ln é C n A D n B
ë P A /P ) (P B /P ) ù,
∆rG = (GC + GD) – (GA + GB)
= (nCGm,C + nDGm,D) – (nAGm,A + nBGm,B),
and ∆rG° is similarly defined. Now, at equilibrium, ∆rG = 0 and so, we get
é (P C / P ) C (P D / P ) D ù .
, r G = −RT ln ê
ë (P A /P ) A (P B /P ) B û eq
The partial pressures that enter into this expression are the values measured at
equilibrium and, therefore, is a constant at a given temperature. Note that, because
of the division of each...
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