Chemical and Conformational Equilibrium

A chemical reaction equation can be expressed as

$$\sum \nu_i C_i = 0$$ (1.84)

where $\nu_i$ is the stoichiometric coefficient for chemical species $C_i$. The coefficients $\nu_i$ are positive on the product side of the chemical equation and negative on the reactant side. At constant temperature and pressure, the Gibbs free energy change involves only chemical transformations:

$$dG = \sum \mu_i d N_i$$ (1.85)

If we differentiate Eq. 1.85 with respect to $N_1$, we obtain

$$\left(\frac{\partial G}{\partial N_1}\right) + \left(\frac{\... ...) \left(\frac{\partial N_2}{\partial N_1}\right) + \cdots = 0$$ (1.86)

at equilibrium. The changes in $N_i$ are related by

$$dN_i = \frac{\nu_i}{\nu_1} dN_1$$ (1.87)

which leads to

$$\sum \left(\frac{\partial G}{\partial N_i}\right) \frac{\nu_i}{\nu_1} = 0$$ (1.88)

Therefore

$$\sum \nu_i \mu_i = 0$$ (1.89)

Let's assume that the reacting molecules in the solution are at low to modest concentration. Thus we use the Henry's Law limit chemical potential Eq. 1.54. We also label $\gamma_i x_i$ the activity $a_i$. Then

$$\sum \nu_i (\mu_i^{\circ} + k_B T \ln a_i ) = 0$$ (1.90)

Rearranging this equation, we find

$$\prod a_i^{\nu_i} = \exp{\left(-\sum\nu_i\mu_i^{\circ}/k_BT\right)}$$ (1.91)

This formula relates the equilibrium activities of the chemical components to the infinite dilution chemical potentials. As mentioned above, we will provide microscopic expressions for the infinite dilution chemical potentials once we have the potential distribution theorem in hand.