E we need to keep in mind that ga nagma ga nagma etc

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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 ê( ë P A /P ) (P B /P ) ù, ú û where ∆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 n n é (P C / P  ) C (P D / P  ) D ù . , r G = −RT ln ê n nú 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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