Physical Interpretation of the Rate Equation

The preceding expression is acceptable in a formal sense, since it contains only concentrations of reactants and products, plus the initial concentration of the enzyme. However, it 

k 2/k2KiK — 1 /KiK2K and K K2K = where is the equilibrium constant (based on concentration) for the overall reaction, S P. Using these relationships, the rate equation becomes 

The term 1 — ([P]/[S] )/K is a measure of how far the reaction is from equilibrium, i. e., the extent to which the rate of the reverse reaction influences the net rate. This term has a value between 0 and 1. If [P]/[S] 1, the bracketed term is very close to 1, and the 

It is useful to put the rate equation for every reversible reaction into a form similar to Eqn. (5-13). The kinetic influence of the reverse reaction is quite easy to evaluate when the rate expression is in this form. 

Note that the equilibria for Reactions (5-H) and (5-J), leading to Eqns. (5-9) and (5-10), were written in opposite senses. The constant Ki is the equilibrium constant forformation of the enzyme-substrate complex E-S, from the substrate S, and the flee enzyme E. The constant 3 is the equilibrium constant for the decomposition of the enzyme -product complex into E and the product P. A common convention in catalysis is to use equilibrium constants based on the formation of the complex. The constant K is consistent with this convention, but K is not. If the equilibrium constant for thejbwwtibn of E-P from E and P is ATj, then ATj = 1/ 3.Using this relationship in Eqn. (5-13) leads to 

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