D22.3 ΔG° and K°
We know that there is a qualitative relationship between ΔrG° and equilibrium constant for a given reaction. The standard Gibbs free energy change for a reaction indicates whether a reaction is product-favored at equilibrium (ΔrG° < 0) or reactant-favored at equilibrium (ΔrG° > 0). A strongly product-favored reaction (large negative ΔrG°) has a large equilibrium constant (K>> 1) and a strongly reactant-favored reaction (large positive ΔrG°) has a very small equilibrium constant (K<<1, a very small fraction because K cannot be negative).
Quantitatively, this relationship between the equilibrium constant and ΔrG° is expressed by the equation:
Note that in these equations the equilibrium constant is represented by K°. The standard equilibrium constant, Kº, is formulated like Kc or Kp, but with all solution phase substance concentrations divided by the standard state concentration of 1 M and all gas phase substance pressures divided by the standard state pressure of 1 bar. Hence, Kº is truly unitless. Dividing by the standard-state concentration or pressure means that if concentrations in Kc are expressed in M (mol/L) the numerical values of Kº and Kc are the same. Similarly, if partial pressures in Kp are expressed in bar, the numerical values of Kº and Kp are the same.
|> 1||< 0||Product-favored at equilibrium.|
|< 1||> 0||Reactant-favored at equilibrium.|
|= 1||= 0||Reactants and products are equally abundant at equilibrium.|
Exercise: Gibbs Free Energy and Equilibrium
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