AT EQUILIBRIUM,  there is no net change in the ratio of Y to X, and the ˜G for both forward and backward reactions is zero.
The equilibrium Constant and ∆G° are Readily Derived from  each other
Inspection of the above equation reveals that the ∆G equals the value of ∆G° when the concentrations of Y and X are equal. But as any favorable reaction proceeds, the concentrations of the products will increase as the concentration of the substrates decreases. This change in relative concentrations will cause [X]/[Y] to become increasingly large, making the initially favorable ∆G less and less negative (the logarithm of a number x is positive for x > 1, negative for x < 1, and zero for x =1). Eventually, when ∆G = 0, a chemical equilibrium will be attained; here there is no net change in free energy to drive the reaction in either direction, inasmuch as the concentration effect just balances the push given to the reaction by ∆G°. As a result, the ratio of product to substrate reaches a constant value at chemical equilibrium (Figure 2–30).
We can define the equilibrium constant, K, for the reaction Y →X as where [X] is the concentration of the product and [Y] is the concentration of the reactant at equilibrium. Remembering that ∆G = ∆G° + RT ln [X]/[Y], and that ∆G = 0 at equilibrium, we see that
At 37°C, where RT = 2.58, the equilibrium equation is therefore: ∆G° = –2.58 ln K
Converting this equation from the natural logarithm (ln) to the more commonly used base 10 logarithm (log), we get ∆G° = –5.94 log K The above equation reveals how the equilibrium ratio of X to Y (expressed as the equilibrium constant, K) depends on the intrinsic character of the molecules, (as expressed in the value of ∆G° in kilojoules per mole). Note that for every 5.94 kJ/mole difference in free energy at 37°C, the equilibrium constant changes by a factor of 10 (Table 2–2). Thus, the more energetically favorable a reaction, the more product will accumulate if the reaction proceeds to equilibrium. More generally, for a reaction that has multiple reactants and products, such as A + B → C + D,
The concentrations of the two reactants and the two products are multiplied because the rate of the forward reaction depends on the collision of A and B and the rate of the backward reaction depends on the collision of C and D. Thus, at 37°C, ∆G° = –5.94 log where ∆G° is in kilojoules per mole, and [A], [B], [C], and [D] denote the concentrations of the reactants and products in moles/liter.
The free-energy Changes of Coupled Reactions are additive
We have pointed out that unfavorable reactions can be coupled to favorable ones to drive the unfavorable ones forward (see Figure 2–29). In thermodynamic terms, this is possible because the overall free-energy change for a set of coupled reactions is the sum of the free-energy changes in each of its component steps. Consider, as a simple example, two sequential reactions
X →Y and Y →Z whose ∆G° values are +5 and –13 kJ/mole, respectively. If these two reactions occur sequentially, the ∆G° for the coupled reaction will be –8 kJ/mole. This means that, with appropriate conditions, the unfavorable reaction X → Y can be driven by the favorable reaction Y → Z, provided that this second reaction follows the first. For example, several of the reactions in the long pathway that converts sugars into CO2 and H2O have positive ∆G° values. But the pathway nevertheless proceeds because the total ∆G° for the series of sequential reactions has a large negative value. Forming a sequential pathway is not adequate for many purposes. Often the desired pathway is simply X → Y, without further conversion of Y to some other product. Fortunately, there are other more general ways of using enzymes to couple reactions together. These often involve the activated carrier molecules that we discuss next.