Entropic Mechanism of Indirect Coupling

in the frame of classical chemical thermodynamics and kinetics, there is no formal restrictions on performing chemical work using an indirect mechanism of coupling. However, quantitative analysis demonstrates that in a system functioning cyclically under equilibrium (or quasi-equilibrium) conditions the efficiency of this mechanism cannot be high. High stoichiometric ratios are possible only in the case of open system functioning under nonequilibrium conditions. 

Let us first consider an isolated reactor the mixture of A, B, and C molecules that are undergoing monomolecular reversible transformations 

Let us consider the relaxation of a system (with initial concentrations of the reagents a, h, and c) to chemical equilibrium. For certainty, we assume that the energy-accepting reaction (BoC) was initially at equilibrium, i.e., c/b = K2, while the energy-donating one (AoB) was out of equilibrium. After reaching the equilibrium state the concentrations of the reagents will obey the equations (b n — n )/(a — n ) = Ki and c -h n )/(b n — nj = K2. From these relations we finally obtain 

This means that the coupling efficiency could be high only if X2 1, i.e., for a thermodynamically favorable reaction. Meanwhile, for any energy-accepting reaction K2 1) high efficiency cannot be reached. 

We can see that the relative yield of C molecules depends only on the thermodynamics of coupled reactions. It follows from these formulas that it is practically impossible to reach high conversion factor values if an energyconsuming reaction (2.39) is characterized by a negative value of the standard chemical affinity, = RTln K2 0. For example, if s/2 = — 30 kJ/mole (standard affinity for the reaction of ATP synthesis from ADP and Pi), then 2 6 10 and consequently rj 1. This result illustrates why the mechanism of indirect coupling of the two chemical reactions having a common intermediate cannot be efficient. Actually, according to the above-mentioned example, in order to form one molecule C we need to add more than 10 molecules B. 

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