This time, we note that the Delta is not actually referring to a difference on a measureable level, but a derivative (slope) of the change in Gibbs energy as the reaction proceeds.

We can also write

Which tells us that the Gibbs energy change is simply the difference between the chemical potential of the reactants and products, at the particular composition of the reaction mixture.  Obviously, at equilibrium, when all chemical potentials are equal,

Thus, a reaction system that has a negative Gibbs energy associated with it is called exergonic and will occur spontaneously.  A reaction system with a positive Gibbs energy associated is endergonic and will not proceed (in the forward direction) without being forced by coupling with an exergonic reaction to drive it.

Ideal Gas reactions

We can extend equation 8.2 for ideal gas situations easily, using equation 6.14 as follows

Thus, we can simplify this to be

Q is called the reaction Quotient and varies from zero (no products) to infinity (no reactants). The reaction quotient gives us the molar ratios of the system as it is.  A special case occurs at equilibrium: Q is replaced by K, the equilibrium constant when the system is at equilibrium and Δ_rG = 0.  Thus, we can rewrite equation 8.5 as

The Gibbs energy of reaction is the result of the difference in the Gibbs energies of reactants and products and also the Gibbs energy of mixing.  The resulting total Gibbs energy function has a minimum in it, corresponding to the equilibrium position for the system.  If there is negligible Gibbs energy of mixing then the reaction will proceed to completion.

We can rewrite these equations to allow for a more general reaction stoichiometry.  Also, we can replace the ratio of pressures p/p^o with the activities so that we can write

Where the values ν_j are the stoichiometric number of the balanced equation.  Stoichiometric numbers differ slightly from stoichiometric coefficients (which are only positive) in that they are negative for reactants and positive for products.  For example, in the Haber reaction,  3 H₂(g) + N₂(g) 2 NH₃(g), the reaction coefficients are -2, +3 and +1 for NH₃, H₂ and N₂, respectively.  Under the special conditions at equilibrium, we can rewrite equation 8.7 to be

This is called  the thermodynamic equilibrium constant and is always a unitless value, expressed using activities or fugacities.  In practice, we often use concentration values in units of mol/L and simply drop the units (if standard conc is 1 mol/L) or in units of molals and drop the units (if standard conc is defined as 1 molal) for ideal solutions.  Similarly, we use pressure units in bars and drop the units (standard pressure = 1 bar) for ideal gases.  We already know that these approximations are only good for ideal systems.  For example, electrolyte solutions would not be accurately modeled using this formulation.

Example:  what is the equilibrium constant expression for the reaction of CaCO₃ (s)CaO(s) + CO₂(g)?

Example: what is the equilibrium constant for the reaction for the Haber reaction?

We can use equation 8.6 to calculate K, as long as we first use equation 4.30 and the table of standard Gibbs energies of formation.

Equation 4.30, modified to use the more formalized stoichiometric number notation as we did for Q and K

Thus, K = 6.1×10⁵.  This is the exact thermodynamic equilibrium constant (within the limits of experimental measurement).  If we had calculated K using the pressure ratios p/pº to approximate activities, the accuracy of the calculation would have depended on how well our approximations a = p/p^o would work for the conditions.  This approximation would work best when the gas behaves ideally (low p, high T).

Example: Calculate the degree of dissociation of water at 2300K and 1.00 bar.  The standard Gibbs energy for the decomposition of water is Δ_rGº = +118.08 kJ/mol.  The reaction is

H₂O(g) H₂(g) + 1/2 O₂(g).

We calculate K from equation 8.6, in exponential form

We can now set up an ice table to figure out the fractional amount of reaction ξ for a total amount n of water.

now, after a bit of algebra, we can set up the equilibrium constant expression using the approximation to the activity of p/pº.

if we assume that the amount of reaction is very small (ξ << 1) we can simplify this expression.

which allows us to easily calculate χ = .0205 (Check assumption:  good within 5%)  so about 2% of the water is decomposed.

Generally, we want to use measurable values to calculate our mole fractions, such as mole fractions or molalities or molarities, rather than activities.  Thus, we can write approximations to the equilibrium constant expressions using activity coefficients.  Let's use molalities, for example, where we define standard molality as 1m.  we can write the activity for constituent j as a_j = γ_jb_j/b_jº, or more simply as a_j = γ_jb_j where the denominator disappears if we simply drop the units of molality.

consider a generalized equations n_A A + ν_B B ν_C C + n_D D

For conditions near ideal behavior, where the activity coefficients are all 1, we often just assume that K_g = 1 so that we get back to the simpler expression K ~ K_b. 

A more succinct way of writing the equilibrium constant expression, using the stoichiometric numbers is