1. Consider a reaction A + B ßàC + D. The equilibrium constant at standard conditions is K’eq, equal to[C][D]/[B][A]. Thermodynamics gives us an expression to relate the standard free energy change at pH=7.0 (DG°’) and K’eq. It is: DG°’ = -RT ln Keq. We can characterize DG°’ under the following conditions (see Table 6.3 in the textbook):K’eq= 1, DG°’ = 0K’eq> 1, DG°’ = negativeK’eq< 1 (e.g., 0.1), DG°' = positive Suppose the K'eqthe reaction catalyzed by IDE is 1 × 103. Is the DG°' for this reaction zero, negative, or positive? What does that mean about the direction of the reaction under standard biochemical conditions?
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