# 10.2: Chemical Potential

Equilibrium can be understood as accruing at the composition of a reaction mixture at which the aggregate chemical potential of the products is equal to that of the reactants. Consider the simple reaction

$A(g) \rightleftharpoons B(g)$

The criterion for equilibrium will be

$\mu_A=\mu_B$

If the gases behave ideally, the chemical potentials can be described in terms of the mole fractions of $$A$$ and $$B$$

$\mu_A^o + RT \ln\left( \dfrac{p_A}{p_{tot}} \right) = \mu_B^o + RT \ln\left( \dfrac{p_B}{p_{tot}} \right) \label{eq2}$

where Dalton’s Law has been used to express the mole fractions.

$\chi_i = \dfrac{p_i}{p_{tot}}$

Equation \ref{eq2} can be simplified by collecting all chemical potentials terms on the left

$\mu_A^o - \mu_B^o = RT \ln \left( \dfrac{p_B}{p_{tot}} \right) - RT \ln\left( \dfrac{p_A}{p_{tot}} \right) \label{eq3}$

Combining the logarithms terms and recognizing that

$\mu_A^o - \mu_B^o –\Delta G^o$

for the reaction, one obtains

$–\Delta G^o = RT \ln \left( \dfrac{p_B}{p_{A}} \right)$

And since $$p_A/p_B = K_p$$ for this reaction (assuming perfectly ideal behavior), one can write

$\Delta G^o = RT \ln K_p$

Another way to achieve this result is to consider the Gibbs function change for a reaction mixture in terms of the reaction quotient. The reaction quotient can be expressed as

$Q_p = \dfrac{\prod_i p_i^{\nu_i}}{\prod_j p_j^{\nu_j}}$

where $$\nu_i$$ are the stoichiometric coefficients for the products, and $$\nu_j$$ are those for the reactants. Or if the stoichiometric coefficients are defined by expressing the reaction as a sum

$0 =\sum_i \nu_i X_i$

where $$X_i$$ refers to one of the species in the reaction, and $$\nu_i$$ is then the stoichiometric coefficient for that species, it is clear that $$\nu_i$$ will be negative for a reactant (since its concentration or partial pressure will reduce as the reaction moves forward) and positive for a product (since the concentration or partial pressure will be increasing.) If the stoichiometric coefficients are expressed in this way, the expression for the reaction quotient becomes

$Q_p = \prod_i p_i^{\nu_i}$

Using this expression, the Gibbs function change for the system can be calculated from

$\Delta G =\Delta G^o + RT \ln Q_p$

And since at equilibrium

$\Delta G = 0$

and

$Q_p=K_p$

It is evident that

$\Delta G_{rxn}^o = -RT \ln K_p \label{triangle}$

It is in this simple way that $$K_p$$ and $$\Delta G^o$$ are related.

It is also of value to note that the criterion for a spontaneous chemical process is that $$\Delta G_{rxn}\ < 0$$, rather than $$\Delta G_{rxn}^o$$, as is stated in many texts! Recall that $$\Delta G_{rxn}^o$$ is a function of all of the reactants and products being in their standard states of unit fugacity or activity. However, the direction of spontaneous change for a chemical reaction is dependent on the composition of the reaction mixture. Similarly, the magnitude of the equilibrium constant is insufficient to determine whether a reaction will spontaneously form reactants or products, as the direction the reaction will shift is also a function of not just the equilibrium constant, but also the composition of the reaction mixture!

Example $$\PageIndex{1}$$:

Based on the data below at 298 K, calculate the value of the equilibrium constant ($$K_p$$) for the reaction

$2 NO(g) + O_2(g) \rightleftharpoons 2 NO_2(g)$

$$NO(g)$$ $$NO_2(g)$$
$$G_f^o$$ (kJ/mol) 86.55 51.53

Solution:

First calculate the value of $$\Delta G_{rxn}^o$$ from the $$\Delta G_{f}^o$$ data.

$\Delta G_{rxn}^o = 2 \times (51.53 \,kJ/mol) - 2 \times (86.55 \,kJ/mol) = -70.04 \,kJ/mol$

And now use the value to calculate $$K_p$$ using Equation \ref{triangle}.

$-70040\, J/mol = -(8.314 J/(mol\, K) (298 \, K) \ln K_p$

$K_p = 1.89 \times 10^{12}$

Note: as expected for a reaction with a very large negative $$\Delta G_{rxn}^o$$, the equilibrium constant is very large, favoring the formation of the products.