# Proof of the chemical potential formula

My textbook as well as Wikipedia both state the formula $\mu_A = \mu_A^0 + RT\ln\{A\}$ which gives the chemical potential $\mu_A$ of a chemical species $A$ as function of the standard chemical potential $\mu_A^0$ and the activity (relative to standard conditions) $\{A\}$.

This formula can also be used to derive the Gibbs free energy formula as $\Delta_rG_{T, p} = \Delta_rG_{T, p}^0 + RT \ln Q_r$ where $Q_r$ is the reaction coefficient.

Both formulas appear to be widely used in chemistry, but I have not found a decent explanation for why they are correct. The texts I have seen did not provide sources for the claim and omitted explanation.

Is this an empirical formula, or is it a consequence of statistical mechanics? I would like to find a source with a clear proof or explanation for the origin of this equation. I know that some will define the activity precisely using this equation, but then the question becomes why the activity so-defined corresponds to concentration for ideal solutions. That is, why in an ideal solution, we have $\mu_A = \mu_A^0 + RT\ln c_A$ where $c_A$ is concentration of $A$ relative to standard concentration.

• My calculus and thermodynamics are too rusty to help. An answer explaining simplifications/assuming to move between the formulas would be nice besides the math itself. – MaxW Jun 17 '18 at 4:41
• In this Wiki article on the Nernst equation the initial section relates it to Gibbs free energy and then to your first equation in the the "Bolzman factors" section. en.wikipedia.org/wiki/Nernst_equation. So the connection between the two is reasonably apparent. – daniel Jun 17 '18 at 13:55
• Many thermodynamics textbooks, including Atkins PhysChem derives this for ideal solutions. – Greg Jun 17 '18 at 14:53
• The equations that relate chemical potential, standard chemical potential, and activity (or fugacity) are indeed direct definitions, or an algebraic manipulation away from the definition. If concentration is used by itself instead of activity, the equation may be known from experiment, is plausible from statistical mechanics, or again simply the definition of an ideal system. The standard chemical potential and activity are so defined that for a pure substance (or infinitely dilute mixture) activity reduces to some type of concentration depending on a chosen basis. – Linear Christmas Jun 17 '18 at 16:15
• @Greg I had not seen Atkins' book. Based on a glance at the table of contents, it looks quite thorough! – Tob Ernack Jun 17 '18 at 18:26

(a) The idea of potentials is familiar from mechanical and electrical systems. In an electric field the work required to move a charge $q$ from one location with potential $\theta_1$ to one with $\theta_2$ is $q(\theta_2-\theta_1)$. This expression has the form of a constant factor ($q$) times the change in potential. In a gravitational field the potential at height $h$ is $gh$ and so the work needed to lift a mass $m$ (the constant factor) from $h_1 \to h_2$ is $mg(h_2-h_1)= mg\Delta h$. We also know that these systems naturally move to position of lowest potential energy to reach equilibrium: stones do drop to the ground.

In a chemical system we can choose the mole number $n$ as the constant factor and then the molar free energy is the chemical potential which is usually given the symbol $\mu$.

The chemical potential controls mass equilibrium just as temperature controls thermal equilibrium. If there is a gradient of chemical potential then mass flows (or molecules rearrange). At equilibrium the chemical potential of all parts are equal. Similarly energy flows in a temperature gradient until equilibrium is achieved and the temperature is uniform throughout. In this sense the chemical potential is considered as providing the 'force' to drive chemical systems to equilibrium.

(b) If we wish to transfer $n$ moles of a gas from a state with molar free energy $G_1$ to one of $G_2$ then the work done (other than expansion) is the change in free energy $n(G_2-G_1)$. Suppose now that we isothermally transfer $n$ moles of an ideal gas from a vessel with pressure $p_1$ to one with pressure $p_2$ the work involved is $nRT\ln(p_2/p_1)$ which is also $n(G_2-G_1)$ so that $(G_2-G_1)=RT\ln(p_2/p_1)$.

Usually we relate the pressures to one at a standard state, say 1 atm. then letting $G_1 \to G^\mathrm{o}$ and $G_2 \to G$ then $G= G^\mathrm{o}+RT\ln(p)$ where $p$ is $p/(1\text{atm})$ which is a dimensionless ratio. As the molar free energy is the chemical potential,

$$\mu=\mu^{\mathrm{o}}+RT\ln(p)$$

(The equation $(G_2-G_1)=RT\ln(p_2/p_1)$ can also be derived by starting with $dG=Vdp-SdT$, at constant temperature ($dT=0$) and for an ideal gas by using the gas law substitute for $V$, and integrating from $p_1\to p_2$.)

(c) More formally the chemical potential can be defined via the work done in reversibly compressing a gas composed of $i$ species with mole numbers $n_i$. If the force $F$ is applied to a piston that moves a distance $dx$ then

$$Fdx= -pdV +\sum_i \mu_i dn_i$$

where $\mu_i$ is the chemical potential of species $i$ and is defined by this equation. This means that the chemical potential is the (reversible) rate of change of internal energy with mole number while keeping other variables ($S,V$) constant, thus since $dU=TdS-pdV+\sum_i \mu_i dn_i$, where $U$ is the internal energy, then $\displaystyle \mu_i= \left(\frac{\partial U}{\partial n_i}\right)_{T,S,n_j}$. (The subscript $n_j$ means that other mole number are held constant.)

It is more common, however, do define the chemical potential in terms of the Gibbs free energy and where $T,p$ are held constant.

For a pure substance your textbook will derive the equation

$$dG=Vdp-SdT$$

but if there is a mixture the number of moles can vary then it is necessary to account for energy changes due to this by adding a term to the energy and doing this produces

$$dG=Vdp-SdT+\sum_i \mu_i dn_i$$

which is sometimes referred to as the fundamental equation of chemical thermodynamics, and then the chemical potential is defined as

$$\displaystyle \mu_i= \left(\frac{\partial G}{\partial n_i}\right)_{p,T,n_j}$$

[discussion in (a) and (b) based on arguments in Lewis & Randall, 'Thermodynamics' publ McGraw-Hill.]

The Sackur-Tetrode equation can be written as $$S^{Trans}=R\ln \left( \frac{e}{C\Lambda^3} \right)$$ where $C$ is the concentration and $\Lambda$ the thermal wavelength. You can rewrite this as $$S^{Trans}=R\ln \left( \frac{e}{\Lambda^3 C^\circ\frac{C}{C^\circ}} \right) = R\ln \left( \frac{e}{C^\circ\Lambda^3} \right)-R\ln\left( \frac{C}{C^\circ} \right) = S^{\circ,Trans}-R\ln\left( \frac{C}{C^\circ} \right)$$ where $C^\circ$ is an arbitrary standard concentration. When you combine it with enthalpy and the other entropy terms you get $$G = G^{\circ}+RT\ln\left( \frac{C}{C^\circ} \right)$$