Writing k1 a k1 b k2 b k2 c k3 c

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Unformatted text preview: exp((µC0  ­ µB0)/RT) If k ­1/k1<<1 then µB0  ­ µA0 < 0 and similarly µC0  ­ µB0 < 0 so that µC0 < µA0. But this contradicts the detailed balance equation for the pair AC: k ­3/k3 = exp((µA0  ­ µC0)/RT) << 1 which would imply µC0 < µA0. One way to think about this is that the detailed balance requires that the path A →B→C →A must be downhill in the Gibbs energy. However going downhill all the time one cannot return to the original state A. Writing: k1[...
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This note was uploaded on 04/24/2011 for the course CH 52635 taught by Professor Makarov during the Spring '11 term at University of Texas at Austin.

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